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ForcesForces
By Neil Bronks
Force causes a body to change velocityhelliphellip accelerate
The unit is called the Newton (N)
Distance Speed and Time
Speed = distance (in metres)
time (in seconds)
D
TS
1) Dave walks 200 metres in 40 seconds What is his speed
2) Laura covers 2km in 1000 seconds What is her speed
3) How long would it take to run 100 metres if you run at 10ms
4) Steve travels at 50ms for 20s How far does he go
5) Susan drives her car at 85mph (about 40ms) How long does it take her to drive 20km
6) Convert 450ms into kmhr
bull A scalar quantity is a quantity that has magnitude only and has no direction in space
Scalars
Examples of Scalar Quantities
Length Area Volume Time Mass
bull A vector quantity is a quantity that has both magnitude and a direction in space
Vectors
Examples of Vector Quantities
Displacement Velocity Acceleration Force
Speed vs VelocitySpeed is simply how fast you are travellinghellip
Velocity is ldquospeed in a given directionrdquohellip
This car is travelling at a speed of 20ms
This car is travelling at a velocity of 20ms east
Scalar vs VectorScalar has only magnitudehellip mass
Vector has magnitude and direction helliphellip Weight
This car has a mass of 2000kg
This car has a Weight of 20000N
Distance and Displacement
Scalar- Distance travelled 200m
Vector- Displacement 120m
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Force causes a body to change velocityhelliphellip accelerate
The unit is called the Newton (N)
Distance Speed and Time
Speed = distance (in metres)
time (in seconds)
D
TS
1) Dave walks 200 metres in 40 seconds What is his speed
2) Laura covers 2km in 1000 seconds What is her speed
3) How long would it take to run 100 metres if you run at 10ms
4) Steve travels at 50ms for 20s How far does he go
5) Susan drives her car at 85mph (about 40ms) How long does it take her to drive 20km
6) Convert 450ms into kmhr
bull A scalar quantity is a quantity that has magnitude only and has no direction in space
Scalars
Examples of Scalar Quantities
Length Area Volume Time Mass
bull A vector quantity is a quantity that has both magnitude and a direction in space
Vectors
Examples of Vector Quantities
Displacement Velocity Acceleration Force
Speed vs VelocitySpeed is simply how fast you are travellinghellip
Velocity is ldquospeed in a given directionrdquohellip
This car is travelling at a speed of 20ms
This car is travelling at a velocity of 20ms east
Scalar vs VectorScalar has only magnitudehellip mass
Vector has magnitude and direction helliphellip Weight
This car has a mass of 2000kg
This car has a Weight of 20000N
Distance and Displacement
Scalar- Distance travelled 200m
Vector- Displacement 120m
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Distance Speed and Time
Speed = distance (in metres)
time (in seconds)
D
TS
1) Dave walks 200 metres in 40 seconds What is his speed
2) Laura covers 2km in 1000 seconds What is her speed
3) How long would it take to run 100 metres if you run at 10ms
4) Steve travels at 50ms for 20s How far does he go
5) Susan drives her car at 85mph (about 40ms) How long does it take her to drive 20km
6) Convert 450ms into kmhr
bull A scalar quantity is a quantity that has magnitude only and has no direction in space
Scalars
Examples of Scalar Quantities
Length Area Volume Time Mass
bull A vector quantity is a quantity that has both magnitude and a direction in space
Vectors
Examples of Vector Quantities
Displacement Velocity Acceleration Force
Speed vs VelocitySpeed is simply how fast you are travellinghellip
Velocity is ldquospeed in a given directionrdquohellip
This car is travelling at a speed of 20ms
This car is travelling at a velocity of 20ms east
Scalar vs VectorScalar has only magnitudehellip mass
Vector has magnitude and direction helliphellip Weight
This car has a mass of 2000kg
This car has a Weight of 20000N
Distance and Displacement
Scalar- Distance travelled 200m
Vector- Displacement 120m
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
bull A scalar quantity is a quantity that has magnitude only and has no direction in space
Scalars
Examples of Scalar Quantities
Length Area Volume Time Mass
bull A vector quantity is a quantity that has both magnitude and a direction in space
Vectors
Examples of Vector Quantities
Displacement Velocity Acceleration Force
Speed vs VelocitySpeed is simply how fast you are travellinghellip
Velocity is ldquospeed in a given directionrdquohellip
This car is travelling at a speed of 20ms
This car is travelling at a velocity of 20ms east
Scalar vs VectorScalar has only magnitudehellip mass
Vector has magnitude and direction helliphellip Weight
This car has a mass of 2000kg
This car has a Weight of 20000N
Distance and Displacement
Scalar- Distance travelled 200m
Vector- Displacement 120m
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
bull A vector quantity is a quantity that has both magnitude and a direction in space
Vectors
Examples of Vector Quantities
Displacement Velocity Acceleration Force
Speed vs VelocitySpeed is simply how fast you are travellinghellip
Velocity is ldquospeed in a given directionrdquohellip
This car is travelling at a speed of 20ms
This car is travelling at a velocity of 20ms east
Scalar vs VectorScalar has only magnitudehellip mass
Vector has magnitude and direction helliphellip Weight
This car has a mass of 2000kg
This car has a Weight of 20000N
Distance and Displacement
Scalar- Distance travelled 200m
Vector- Displacement 120m
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Speed vs VelocitySpeed is simply how fast you are travellinghellip
Velocity is ldquospeed in a given directionrdquohellip
This car is travelling at a speed of 20ms
This car is travelling at a velocity of 20ms east
Scalar vs VectorScalar has only magnitudehellip mass
Vector has magnitude and direction helliphellip Weight
This car has a mass of 2000kg
This car has a Weight of 20000N
Distance and Displacement
Scalar- Distance travelled 200m
Vector- Displacement 120m
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Scalar vs VectorScalar has only magnitudehellip mass
Vector has magnitude and direction helliphellip Weight
This car has a mass of 2000kg
This car has a Weight of 20000N
Distance and Displacement
Scalar- Distance travelled 200m
Vector- Displacement 120m
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Distance and Displacement
Scalar- Distance travelled 200m
Vector- Displacement 120m
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
bull Vector diagrams are shown using an arrow
bull The length of the arrow represents its magnitude
bull The direction of the arrow shows its direction
Vector Diagrams
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Vectors in opposite directions
6 m s-1 10 m s-1 = 4 m s-1
6 N 9 N = 3 N
Resultant of Two Vectors
Vectors in the same direction
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
The Parallelogram Law When two vectors are joined
tail to tail Complete the parallelogram The resultant is found by
drawing the diagonal
The Triangle Law When two vectors are joined
head to tail
Draw the resultant vector by completing the triangle
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Vector Addition
Speed in still air 120ms
Wind 50msResultant
R2 = 1202 + 502 = 14400 + 2500
= 16900
R = 130ms
Tan = 50120
= 2260
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Solution
Problem Resultant of 2 Vectors
Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)
Two forces are applied to a body as shown What is the magnitude and direction of the resultant force acting on the body
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagorasrsquo Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan ofDirection
1
ac
Resultant displacement is 13 N 67ordm with the 5 N force
13 N
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
bull When resolving a vector into components we are doing the opposite to finding the resultant
bull We usually resolve a vector into components that are perpendicular to each other
Resolving a Vector Into Perpendicular Components
y v
x
Here a vector v is resolved into an x component and a y component
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
bull Here we see a table being pulled by a force of 50 N at a 30ordm angle to the horizontal
Practical Applications
50 Ny=25 N
x=433 N30ordm
bull When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 433 N
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
bull If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are
bull x = v Cos θbull y = v Sin θ
Calculating the Magnitude of the Perpendicular Components
vy=v Sin θ
x=v Cos θ
θ
y
Proof
v
xCos
vCosx v
ySin
vSiny
x
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
60ordm
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown What is the horizontal
component of the force
Problem Calculating the magnitude of perpendicular components
Vert
ical
Com
ponent
Horizontal Component
Solution
N 576015Component Horizontal Cosx
N 99126015Component Vertical Siny
15 N
75 N129
9 N
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
bull A person in a wheelchair is moving up a ramp at constant speed Their total weight is 900 N The ramp makes an angle of 10ordm with the horizontal Calculate the force required to keep the wheelchair moving at constant speed up the ramp (You may ignore the effects of friction) (Stop here and freeze)
Solution
If the wheelchair is moving at constant speed (no acceleration) then the force that moves it up the ramp must be the same as the component of itrsquos weight parallel to the ramp
10ordm
10ordm80ordm
900 N
Complete the parallelogramComponent of weight
parallel to ramp N 2815610900 Sin
Component of weight perpendicular to ramp
N 3388610900 Cos
15628 N
88633 N
HW - 2003 HL Section B Q6
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
bull If a vector of magnitude v has two perpendicular components x and y and v makes and angle θ with the x component then the magnitude of the components are
bull x= v Cos θbull y= v Sin θ
Summary
vy=v Sin θ
x=v Cosθ
θ
y
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
AccelerationV-U
TA
Acceleration = change in velocity (in ms)
(in ms2) time taken (in s)
1) A cyclist accelerates from 0 to 10ms in 5 seconds What is her acceleration
2) A ball is dropped and accelerates downwards at a rate of 10ms2 for 12 seconds How much will the ballrsquos velocity increase by
3) A car accelerates from 10 to 20ms with an acceleration of 2ms2 How long did this take
4) A rocket accelerates from 1000ms to 5000ms in 2 seconds What is its acceleration
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Velocity-Time GraphsV
t1Constant Acceleration
V
t
2Constant Velocity
V
t
3Deceleration
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Velocity-time graphs
80
60
40
20
010 20 30 40 50
Velocity
ms
Ts
1) Upwards line =
Constant Acceleration
2) Horizontal line =
Constant Velocity
3) Shallow line =
Less Acceleration
4) Downward line =
Deceleration
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
80
60
40
20
0
1) How fast was the object going after 10 seconds
2) What is the acceleration from 20 to 30 seconds
3) What was the deceleration from 30 to 50s
4) How far did the object travel altogether
10 20 30 40 50
Velocity
ms
Ts
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
80
60
40
20
0
The area under the graph is the distance travelled by the object
10 20 30 40 50
Velocity
ms
Ts
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
8
0
6
0
4
0
2
0
0
10 20 30 40 50
Velocity
ms
Ts
40x20=800
Total Distance Traveled
=200+100+800+600=1700m
05x10x40=200
05x10x20=100
05x20x60=600
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Motion Formula
v = u + at A car starts from rest and accelerates for 12s at 2ms-2 Find the final velocity
U=30 s=200 and v = 0 find a=
v2 = u2 + 2as A car traveling at 30ms takes 200m to stop what is itrsquos deceleration
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900400=-225ms-2
Using V = U + at = 0 + 2x12 = 24ms
U=0 a=2 and t = 12 find v=
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Motion Formula
S = ut + 05at2
A train accelerates from rest at 10ms-2 for 12s find the distance it has traveled
Using S = ut + 05at2 = 0x12 +05x10x144 =720m
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Velocity and Acceleration
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
1t
lu
2t
lv
s
uva
2
22
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
HW
bull LC Ord 2008bull Q 1
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Friction is the force that opposes motion
The unit is called the
Newton (N)
Friction is the force between two bodies in
contact
Lubrication
reduces friction
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Lubrication reduces friction and separates the two
bodies
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Advantages and disadvantages of Friction
bull We can walk across a surface because of friction
bull Without friction walking is tough Ice is a prime example
bull It can also be a pain causing unwanted heat and reducing efficiency
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Friction1) What is friction
2) Give 3 examples where it is annoying
3) Give 3 examples where it is useful
4) What effect does friction have on the surfaces
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Recoil
Momentum of the ShootMomentum of Recoil =
Mass of Ball x Velocity of BallMass Canon x Velocity Canon =
V= 800150 =
2 x 400150 x Uc =
ub=400ms
m=2kg
Mass of canon=150kg
53ms
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Momentum
2kg 3kg
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
3kg x 10ms = 3kg x (-2ms) + 6kg x v
6v = 30 + 6
V = 6ms
10ms
V= ms
2 ms
3kg 6kg
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Internet Calculations
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
VERIFICATION OF THE PRINCIPLE OF CONSERVATION OF MOMENTUM
Velcro pad
Dual timer Photogate
Air track
t1 t2
Light beamCardl
Vehicle 1 Vehicle 2
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
1Set up apparatus as in the diagram2 Level the air-track To see if the track is level carry
out these testsa) A vehicle placed on a level track should not drift
toward either end Measure the mass of each vehicle m1 and m2
respectively including attachments using a balance
4 Measure the length l of the black card in metres5 With vehicle 2 stationary give vehicle 1 a gentle
push After collision the two vehicles coalesce and move off together
6 Read the transit times t1and t2 for the card through the two beams
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Calculate the velocity before the collision and after the collision
momentum before the collision=momentum after the collision
m1u = (m1 + m2) v
Repeat several times with different velocities and different masses
1t
lu
2t
lv
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
HW
bull LC Ordbull 2007 Q1
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Newtonrsquos Lawsbull 1 Every body stays in itrsquos state of rest
or constant motion until an outside force acts on it
bull 2 The rate of change of momentum is proportional to the applied force and in the direction of the applied force
bull F=mabull 3 To every action there is an equal and
opposite reaction
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Newton 2
t
mumvforce
t
uvmforce
)(
force Rate of change of Momentum
t
uvabut
)(
Forcema
Or Force=kma where k=constant
As this is the basic constant so we say k=1 and Force=ma
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
TO SHOW THAT a micro F
Card
l
t1 t2
s
Dual timer
Light beam
Photogate
Pulley
Air track
Slotted weights
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t1 time for card to pass first photo-gate
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
TO SHOW THAT a micro F
t1Dual timer
Light beam
Photogate
t2 time for card to pass second photo-gate
t2
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Procedure
Set up the apparatus as in the diagram Make sure the card cuts both light beams as it passes along the track
Level the air track Set the weight F at 1 N Release the vehicle Note the times t1 and t2
Remove one 01 N disc from the slotted weight store this on the vehicle and repeat
Continue for values of F from 10 N to 01 N Use a metre-stick to measure the length of the card l and
the separation of the photo gate beams s
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
1 Remember to include the following table to get full marks All tables are worth 3 marks when the Data has to be changed Draw a graph of am s-2 against FN Straight line though origin proves Newtons second law
1t
lu
2t
lv
s
uva
2
22
FN t1s t2s Vms Ums Ams2
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Newtonrsquos Laws on the Internet
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Balanced and unbalanced forces
Consider a camel standing on a road What forces are acting on it
Weight
Reaction
These two forces would be equal ndash we say that they are BALANCED The camel doesnrsquot move anywhere
Balanced and unbalanced forces
What would happen if we took the road away
Weight
Reaction
Balanced and unbalanced forces
What would happen if we took the road away
The camelrsquos weight is no longer balanced by anything so the camel falls downwardshellip
Weight
Balanced and unbalanced forces
1) This animal is either ________ or moving with _____ _____hellip
4) This animal ishellip
2) This animal is getting _________hellip
3) This animal is getting _______hellip
Let Go or Hang On A painter is high up on a ladder painting a house when unfortunately the ladder
starts to fall over from the vertical Determine which is the less harmful action for the painter to let go of the
ladder right away and fall to the ground or to hang on to the ladder all the way to
the ground
Engine forceFriction
Gra
vity
Reaction
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force (in N) = Mass (in kg) x Acceleration (in ms2)
F
AM
1) A force of 1000N is applied to push a mass of 500kg How quickly does it accelerate
2) A force of 3000N acts on a car to make it accelerate by 15ms2 How heavy is the car
3) A car accelerates at a rate of 5ms2 If it lsquos mass is 500kg how much driving force is the engine applying
4) A force of 10N is applied by a boy while lifting a 20kg mass How much does it accelerate by
Using F=ma
1000=500xa
a=2ms2
Using F=ma
3000=mx15
m=2000kg
Using F=ma
F=5x500
F=2500N
Using F=ma
10=20xa
a=05ms2
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Balanced and unbalanced forces
What would happen if we took the road away
The camelrsquos weight is no longer balanced by anything so the camel falls downwardshellip
Weight
Balanced and unbalanced forces
1) This animal is either ________ or moving with _____ _____hellip
4) This animal ishellip
2) This animal is getting _________hellip
3) This animal is getting _______hellip
Let Go or Hang On A painter is high up on a ladder painting a house when unfortunately the ladder
starts to fall over from the vertical Determine which is the less harmful action for the painter to let go of the
ladder right away and fall to the ground or to hang on to the ladder all the way to
the ground
Engine forceFriction
Gra
vity
Reaction
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force (in N) = Mass (in kg) x Acceleration (in ms2)
F
AM
1) A force of 1000N is applied to push a mass of 500kg How quickly does it accelerate
2) A force of 3000N acts on a car to make it accelerate by 15ms2 How heavy is the car
3) A car accelerates at a rate of 5ms2 If it lsquos mass is 500kg how much driving force is the engine applying
4) A force of 10N is applied by a boy while lifting a 20kg mass How much does it accelerate by
Using F=ma
1000=500xa
a=2ms2
Using F=ma
3000=mx15
m=2000kg
Using F=ma
F=5x500
F=2500N
Using F=ma
10=20xa
a=05ms2
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Balanced and unbalanced forces
1) This animal is either ________ or moving with _____ _____hellip
4) This animal ishellip
2) This animal is getting _________hellip
3) This animal is getting _______hellip
Let Go or Hang On A painter is high up on a ladder painting a house when unfortunately the ladder
starts to fall over from the vertical Determine which is the less harmful action for the painter to let go of the
ladder right away and fall to the ground or to hang on to the ladder all the way to
the ground
Engine forceFriction
Gra
vity
Reaction
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force (in N) = Mass (in kg) x Acceleration (in ms2)
F
AM
1) A force of 1000N is applied to push a mass of 500kg How quickly does it accelerate
2) A force of 3000N acts on a car to make it accelerate by 15ms2 How heavy is the car
3) A car accelerates at a rate of 5ms2 If it lsquos mass is 500kg how much driving force is the engine applying
4) A force of 10N is applied by a boy while lifting a 20kg mass How much does it accelerate by
Using F=ma
1000=500xa
a=2ms2
Using F=ma
3000=mx15
m=2000kg
Using F=ma
F=5x500
F=2500N
Using F=ma
10=20xa
a=05ms2
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Let Go or Hang On A painter is high up on a ladder painting a house when unfortunately the ladder
starts to fall over from the vertical Determine which is the less harmful action for the painter to let go of the
ladder right away and fall to the ground or to hang on to the ladder all the way to
the ground
Engine forceFriction
Gra
vity
Reaction
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force (in N) = Mass (in kg) x Acceleration (in ms2)
F
AM
1) A force of 1000N is applied to push a mass of 500kg How quickly does it accelerate
2) A force of 3000N acts on a car to make it accelerate by 15ms2 How heavy is the car
3) A car accelerates at a rate of 5ms2 If it lsquos mass is 500kg how much driving force is the engine applying
4) A force of 10N is applied by a boy while lifting a 20kg mass How much does it accelerate by
Using F=ma
1000=500xa
a=2ms2
Using F=ma
3000=mx15
m=2000kg
Using F=ma
F=5x500
F=2500N
Using F=ma
10=20xa
a=05ms2
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Engine forceFriction
Gra
vity
Reaction
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force (in N) = Mass (in kg) x Acceleration (in ms2)
F
AM
1) A force of 1000N is applied to push a mass of 500kg How quickly does it accelerate
2) A force of 3000N acts on a car to make it accelerate by 15ms2 How heavy is the car
3) A car accelerates at a rate of 5ms2 If it lsquos mass is 500kg how much driving force is the engine applying
4) A force of 10N is applied by a boy while lifting a 20kg mass How much does it accelerate by
Using F=ma
1000=500xa
a=2ms2
Using F=ma
3000=mx15
m=2000kg
Using F=ma
F=5x500
F=2500N
Using F=ma
10=20xa
a=05ms2
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force (in N) = Mass (in kg) x Acceleration (in ms2)
F
AM
1) A force of 1000N is applied to push a mass of 500kg How quickly does it accelerate
2) A force of 3000N acts on a car to make it accelerate by 15ms2 How heavy is the car
3) A car accelerates at a rate of 5ms2 If it lsquos mass is 500kg how much driving force is the engine applying
4) A force of 10N is applied by a boy while lifting a 20kg mass How much does it accelerate by
Using F=ma
1000=500xa
a=2ms2
Using F=ma
3000=mx15
m=2000kg
Using F=ma
F=5x500
F=2500N
Using F=ma
10=20xa
a=05ms2
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Force and accelerationIf the forces acting on an object are unbalanced then the object will accelerate like these wrestlers
Force (in N) = Mass (in kg) x Acceleration (in ms2)
F
AM
1) A force of 1000N is applied to push a mass of 500kg How quickly does it accelerate
2) A force of 3000N acts on a car to make it accelerate by 15ms2 How heavy is the car
3) A car accelerates at a rate of 5ms2 If it lsquos mass is 500kg how much driving force is the engine applying
4) A force of 10N is applied by a boy while lifting a 20kg mass How much does it accelerate by
Using F=ma
1000=500xa
a=2ms2
Using F=ma
3000=mx15
m=2000kg
Using F=ma
F=5x500
F=2500N
Using F=ma
10=20xa
a=05ms2
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
1) A force of 1000N is applied to push a mass of 500kg How quickly does it accelerate
2) A force of 3000N acts on a car to make it accelerate by 15ms2 How heavy is the car
3) A car accelerates at a rate of 5ms2 If it lsquos mass is 500kg how much driving force is the engine applying
4) A force of 10N is applied by a boy while lifting a 20kg mass How much does it accelerate by
Using F=ma
1000=500xa
a=2ms2
Using F=ma
3000=mx15
m=2000kg
Using F=ma
F=5x500
F=2500N
Using F=ma
10=20xa
a=05ms2
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Net Force creates Acceleration
Fnet=200N
F=200NFnet=100N
Fnet=0N
Fnet=-200N
F=-200N
F=-100N
F=-200N F=200N
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
HW
bull LC Ordbull 2004 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Net Force creates Acceleration
F=200NFnet=100N
F=-100N800kg
As net force causes acceleration F=ma
100N = 800kga
a=100800 = 0125ms2
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Acceleration gives Net Force
Feng=5000Na=3ms2
Friction=900kg
As net force causes acceleration F=ma
Fnet = 900kg 3ms2
Fnet= 2700N
So Friction = Feng ndash 2700N
Friction=2300N
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
A car of mass 500kg has an engine that produces 3kN of force what is the friction if the car is accelerating at 11 ms2
If the engine stops how long before the car stops if it is travelling at 20ms when the engine cuts out
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Archimedes Principle
bull A body in a fluid experiences an up-thrust equal to the weight of liquid displaced
20N
12N
8N
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Internet Diagram
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Floatation
bull A floating body displaces its own weight in water
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Floatation
bull A floating body displaces its own weight in water
=10000t 10000t
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Measuring Liquid Density
bull A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids that is the ratio of the density of the liquid to the density of water
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Terminal VelocityConsider a skydiver
1) At the start of his jump the air resistance is _______ so he _______ downwards
2) As his speed increases his air resistance will _______
3) Eventually the air resistance will be big enough to _______ the skydiverrsquos weight At this point the forces are balanced so his speed becomes ________ - this is called TERMINAL VELOCITY
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Terminal Velocity
4) When he opens his parachute the air resistance suddenly ________ causing him to start _____ ____
5) Because he is slowing down his air resistance will _______ again until it balances his _________ The skydiver has now reached a new lower ________ _______
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Velocity-time graph for terminal velocityhellipVelocit
y
Time
Speed increaseshellip
Terminal velocity reachedhellip
Parachute opens ndash diver slows down
New lower terminal velocity reached
Diver hits the ground
On the Moon
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Weight vs MassEarthrsquos Gravitational Field Strength is 98ms2 In other words a 1kg mass is pulled downwards by a force of
98N
W
gM
Weight = Mass x acceleration due to gravity
(in N) (in kg) (in ms2)
1) What is the weight on Earth of a book with mass 2kg
2) What is the weight on Earth of an apple with mass 100g
3) Dave weighs 700N What is his mass
4) On the moon the gravitational field strength is 16Nkg What will Dave weigh if he stands on the moon
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Weight vs Mass
bull Mass is the amount of matter in usbull Same on Earth and Space
bull Weight is the pull of gravity on usbull Different on Earth and Space
900kg
900kg
9000 N 0 N
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Homework
bull LC Ordinary Levelbull 2002 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Galileorsquos Falling Balls
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Gravity all bodies have gravity we feel it only
from planet sized objects
bull Acceleration due to gravity is 981ms2
bull That means every falling body gets 981ms faster every second
T=0 v=0ms
T=1s v=981ms
T=2s v=1962ms
T=3s v=2943ms
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Internet
bull Even proved it in real life
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
All bodies fall at the same rate
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
MEASUREMENT OF g
h
SwitchElectromagnet
Ball bearing
Trapdoor
Electronic timer
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
When the switch opens the ball falls
The timer records the time from when the switch opens until trap door opens
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Set up the apparatus The millisecond timer starts when the ball is released and stops when the ball hits the trapdoor
Measure the height h as shown using a metre stick
Release the ball and record the time t from the millisecond timer
Repeat three times for this height h and take the smallest time as the correct value for t
Repeat for different values of h
Calculate the values for g using the equation Obtain an average value for g
Place a piece of paper between the ball bearing and the electromagnet to ensure a quick release
221 gth
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Finding Drag
The sky diver accelerates at 2ms2 what is his drag
Force due to gravity=80g
=80(98)=784 N
Net Force=ma=802=160N
Drag=784-160=624N
80kg
Drag
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Newtonrsquos Cannon
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Launching a satellite
The cannon ball is constantly falling towards the earth but earth curve is same as itrsquos path
The Moon orbits the Earth It is also in free fall
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Newtons Law of Gravitationbull This force is always positivebull Called an inverse square law
F m1m2
d2
WhereF = Gravitational Forcem1m2 = Product of massesd = Distance between their center of gravity
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Gravity Calculations
bull To make an equation we add a constant
bull G The UNIVERSAL GRAVITATIONAL CONSTANT
Example What is the force on a man of mass 100kg standing on the surface of Mars
Mars mass=66x1023 kg and radius=34x106mG=667x10-11 Nm2kg-2
F = G m1 m2
d2
= 667x10-11 x 66x1023 x100
(34x106)2
= 380N
F = G m1 m2
d2
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
bull 2010 Question 6 [Higher Level]bull (Radius of the earth = 636 times 106 m acceleration due
to gravity at the earthrsquos surface = 981 m sminus2
bull Distance from the centre of the earth to the centre of the moon = 384 times 108 m
bull Assume the mass of the earth is 81 times the mass of the moon)
bull bull State Newtonrsquos law of universal gravitationbull Use this law to calculate the acceleration due to
gravity at a height above the surface of the earth which is twice the radius of the earth
bull Note that 2d above surface is 3d from earthrsquos centre
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
bull A spacecraft carrying astronauts is on a straight line flight from the earth to the moon and after a while its engines are turned off
bull Explain why the spacecraft continues on its journey to the moon even though the engines are turned off
bull Describe the variation in the weight of the astronauts as they travel to the moon
bull At what height above the earthrsquos surface will the astronauts experience weightlessness
bull The moon orbits the earth every 273 days What is its velocity expressed in metres per second
bull Why is there no atmosphere on the moon
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
HW
bull LC Ord 2008bull Q 6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Hookes Law
123456789
1011121314151617181920
Force
Extension
More force means more Extension - they are
proportional
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Hookes Law Calculation123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
123456789
1011121314151617181920
Force=0N
Length=5cm
Ext=0cm
Force=6N
Length=8cm
Ext=3cm
Force=12N
Length=11cm
Ext=6cm
Force
=24N
Length=17cm
Ext
=12cm
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Hookes Law ExampleForce =Constant (k) x Extension
Example a A mass of 3kg causes an extension of 03m what is the spring constant
3x98 = k x 03K=98Nm
B What is the extension if 40N is put on the same springForce = Spring Constant x Extension
40 = 98 x sS = 4098 = 041 m
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Homework
bull LC Ordbull 2003 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Work doneWhen any object is moved around work will need to be done on it to get it to move (obviously)
We can work out the amount of work done in moving an object using the formula
Work done = Force x Distance Movedin J in N in m
W
DF
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Kinetic energyAny object that moves will have kinetic energy
The amount of kinetic energy an object has can be found using the formula
Kinetic energy = frac12 x mass x velocity squared
in J in kg in ms
KE = frac12 mv2
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Some example questionshellipA 70kg boy is running at about 10ms What is his kinetic energy
Using KE=frac12mv2=05x70x10x10=3500J
A braking force of 1000N is applied by a driver to stop his car The car covered 50m before it stopped How much work did the brakes do
Work Done = force x distance = 1000x50 = 50000J
What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5ms
Using KE=frac12mv2=05x01x5x5=125J
A crane is lifting a 50kg load up into the air with a constant speed If the load is raised by 200m how much work has the crane
done
Work Done = force x distance = 50x981x200 = 98100J
KE = frac12 mv2
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Potential energyAn object has potential energy because of itrsquos position or condition
That means it is high or wound up
The formula is for high objects
Potential energy = mass x g x height
PE = mgh
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Work Done = Energy Converted
Work Done raising an object = PE Stored
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Consider an oscillating pendulum
Consider an oscillating pendulum
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
PE = mgh
KE = frac12 mv2
At the top of the oscillation the pendulum bob stops All itrsquos energy is PE
At the bottom the bob
has no PE only KE
PE at top=KE at bottom
h
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
mgh = frac12 mv2
PE at top=KE at bottom
H=10cm
mgh = frac12 mv2
gh = frac12 v2
v2 = 2gh v2 = 2(98)01
v = 14ms
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Powerbull The rate at which work
is donebull POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m If the jet has mass 200tonnes find the work done and the powerWork Done = Force x Distance = 200x1000x981x4000
=7 x 109 Joules
Power = Work Done Time = 7 x 109 Joules 120
= 583 x 107 Watts
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
HW
bull LC Ord 2007bull Q 6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
PressurePressure depends on two things
1) How much force is applied and
2) How big (or small) the area on which this force is applied is
Pressure can be calculated using the equation
Pressure (in Nm2) = Force (in N)
Area (in m2)
F
AP
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Some example questionshellip1) A circus elephant weighs 10000N and can stand on one
foot This foot has an area of 50cm2 How much pressure does he exert on the floor (in Pa)
2) A 50kg woman copies the elephant by standing on the heel of one of her high-heeled shoes This heel has an area of 1cm2 How much pressure does she exert on the floor
Pressure=Forcearea = 500N 00001m2 = 5000000 Pa
Extension task
Atmospheric pressure is roughly equivalent to 1kg pressing on every square centimetre on our body What does this equate to in units called Pascals (1 Pascal = 1Nm2)
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Pressure ndash in Fluids
Pressure increases with depth
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Pressure and DepthAs the frog
goes deeper there is a
greater weight of water above
it
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Atmospheric Pressurebull The earth is covered with
layer of Gasbull We are at the bottom of a
gas ocean 200km deepbull The effect of this huge
column of gas is 1 Tonne of weight on our shoulders
bull This is calledbull ATMOSPHERIC
PRESSURE
Heavy
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Proving Atmospheric Pressure
Very full glass of water
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Proving Atmospheric Pressure
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Proving Atmospheric Pressure
Now the atmospheric Pressure holds the card in place
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
The Barometer
bull The weight of the air holds up the mercury
bull If we use water the column is 104m high
bull 1 Atmosphere is 760mm of Hg
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
The Altimeterbull As we go higher there is
less air above usbull There is less
Atmospheric pressurebull We can measure the
altitude using a barometer with a different scale
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Aneroid Barometerbull Works on
changes in size of small can(Get it)
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Pressure and Volume in gasesThis can be expressed using the equation
Initial Pressure x Initial Volume = Final Press x Final Vol
PIVI = PFVF
1) A gas has a volume of 3m3 at a pressure of 20Nm2 What will the pressure be if the volume is reduced to 15m3
2) A gas increases in volume from 10m3 to 50m3 If the initial pressure was 10000Nm2 what is the new pressure
3) A gas decreases in pressure from 100000 Pascals to 50000 Pascals The final volume was 3m3 What was the initial volume
4) The pressure of a gas changes from 100Nm2 to 20Nm2 What is the ratio for volume change
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Pressure and Volume in gases
Pressure Volume Pressure x volume
20 10 200
200 1 200
4 50 200
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Internet Demo
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Boyles Law
Pressure is inversely proportional to volume
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
VERIFICATION OF BOYLErsquoS LAW 1
Bicycle pump
Reservoir of oil
Pressure gauge
Tube with volume of air trapped by oil
Volumescale
Valve
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Using the pump increase the pressure on the air in the tube Close the valve and wait 20 s to allow the temperature of the enclosed air to reach equilibrium
Read the volume V of the air column from the scale
Take the corresponding pressure reading from the gauge and record the pressure P of the trapped air
Reduce the pressure by opening the valve slightly ndash this causes an increase the volume of the trapped air column Again let the temperature of the enclosed air reach equilibrium
Record the corresponding values for the volume V and pressure P
Repeat steps two to five to get at least six pairs of readings
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
P
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Hydraulic systems
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Hydraulic systems
Pressure is constant throughout this liquid
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Hydraulic systems
Basically a smaller force on piston A will produce a larger force on piston B because the pressure of the liquid is constant
1) If the area of the slave piston is ten times bigger than the master piston what force will be needed to lift an object weighing 1000N
Pressure in Slave = 100010=100Pa
Pressure in Master = Force1 = 100Pa
Force in the master only 100N amazing
Magic
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
2006 Question 12 (a) [Higher Level]Define pressure Is pressure a vector quantity or a scalar quantity Justify your answerState Boylersquos law A small bubble of gas rises from the bottom of a lake The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 101 times 105 Pa The temperature of the lake is 4 oC Calculate the pressure at the bottom of the lakeCalculate the depth of the lake (acceleration due to gravity = 98 m sndash2 density of water = 10 times 103 kg mndash3)
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
HW
bull LC Ordbull 2005 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Center of Gravity
bull Things stay standing (STABLE) because their Center of Gravity acts through their base
bull The perpendicular from the COG passes inside the support
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Unstable Equilibrium
bull Things fall over because the center of gravity is outside the base
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Moments (Also called TORQUE)
=Force x Perpendicular distance
FORCE
Fulcrum
Perpendicular distance
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Moments=Force x Perpendicular distance
FORCE =10N
Perpendicular
distance=5m
= 10N x 5m = 50Nm
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
More than two forces
bull First prove all coplanar forces on a body in equilibrium add up to zero
(Forces Up = Forces Down)bull Then take moments about one end (Clockwise moments=Anti-clockwise moments)
10 60 907050
15N
15N
5N 10N 5N
N
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
bull First law coplanar forces bull Forces Up = Forces Down 15 + x = 15 + 5 +10 + 5 x = 20 N
10 60 907050
15N
15N
5N 10N 5N
N
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
bull Second law coplanar forces bull Take moments about AClockwise Moments = Anticlockwise Moments 10x15 + 50x5 + 70x10 + 90x5 = 60x15 + dx20150 + 250 + 700 + 450 = 900 + dx201550-900 = dx20 so d=325cm
10 60 907050
15N
15N
5N 10N 5N
20N
A325
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
INVESTIGATION OF THE LAWS OF EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Newton balance
Newton balance
Support
Paperclips
w1
w2 w3
w4
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
1 Use a balance to find the centre of gravity of the metre stick and its weight
2 The apparatus was set up as shown and a equilibrium point found
3 Record the reading on each Newton balance
4 Record the positions on the metre stick of each weight each Newton balance and the centre of gravity of the metre stick
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
For each situation (1) Forces up = Forces downie the sum of the readings on the balances should be equal to the sum of the weights plus the weight of the metre stick
(2)The sum of the clockwise moments about an axis through any of the chosen points should be equal to the sum of the anticlockwise moments about the same axis
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Internet
bull Ok its not the most exciting thing doing all the calculations so here Walter has done them for us and we just play and see how they are laid out
bull Notice the units of torque are included as we should
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
bull 2011 Question 6 (b) [Higher Level]bull State the conditions necessary for the equilibrium of
a body under a set of co-planar forcesbull Three children position themselves on a uniform
see-saw so that it is horizontal and in equilibrium bull The fulcrum of the see-saw is at its centre of gravity bull A child of mass 30 kg sits 18 m to the left of the
fulcrum and another child of mass 40 kg sits 08 m to the right of the fulcrum
bull Where should the third child of mass 45 kg sit in order to balance the see-saw
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
HW
bull LC Ordbull 2003 Q12(a)bull Last hw before xmas (honest)
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Couples of Forces
bull Two equal forces that cause a solid to rotate around an axis
bull Moment = Force x Distance
bull Moment = 5Nx006m bull Moment = 03 Nm
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Motion in a circle
Velocity always at 90o to the force or acceleration
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Circular Motion
bull Angular Velocitybull =θtbull Units of Radians
per secondbull Angle time
A particle goes round a circle in 4s what is itrsquos angular velocity
t
second24
2rads
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Circular Motion
second2rads
bull Linear Velocity(V)bull msbull V= rbull r=radius of motionbull Always changing as
direction is always changing this creates acceleration
bull If the radius is 6m
smrv 4292
6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous example a= 6 (2)2
=148ms2
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Centripetal ForceIf we have an acceleration we must have
a forceCentripetal force f = ma = m r 2
Tension in string of weight spun around head
Force on tyres (Or camel) as we go around corner
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Centripetal Acceleration
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Satellites balance forcesbull Balance of Gravity and Centripetalbull ((GMm)d2)=mv2d
GravityF=-GmMr2
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Period of Orbit
((GMm)d2)=mv2d
(GM)d=v2
(GM)d=(2dT)2
T2=42 d3 GM
Equate The Forces
V=Distance
time
T=Period
(Time for
Orbit)
Cancel
Mass of
satellite
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
d
mv
d
GMm 2
2
1
2mv
d
GMm
d
mv
d
GMm 2
2
22 2
T
dv
d
GM
322 4
GM
dT
In a test we do it like this
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Example of Orbits
Using T2=42 d3 GM(4x24x60x60)2=42 d3 (2x1031)(67x10-11)
d3 = 1x1030 d = 1x1010m
Height =h= d - r =1x1010m - 2x108m= 98x109m
What is the parking orbit height above Saturn if it is 200000km in radius It rotates every 4
days and has mass 2x1031Kg The Universal gravitational Constant is 67x10-11
d
r
h
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Geostationary or Clarke Orbit
Same period and angular velocity as the planet surface so stays above same spot What is itrsquos height above the earth
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Simple Harmonic Motionbull Is a vibration where the
acceleration is proportional to the displacement
a -sbull Further from centre =more acceleration
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Hookersquos Law as SHMForce Extension
F -s
ma -s
If mass is constant
a -s
So motion under Hookes law is SHM
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
HW
bull LC Ordbull 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Pendulum
bull If we displace the bob by a small angle it vibrates with SHM
Split cork
lBob
2030
Timer
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
(slope)
4 g
slope 4
4
2
22
22
gl
Tg
lT
l
T2
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
Time to go over hwbull LC Ord 2008 Q1bull LC Ord 2007 Q1bull LC Ord 2004 Q6bull LC Ord 2002 Q6bull LC Ord 2008 Q6bull LC Ord 2003 Q6bull LC Ord 2007 Q6bull LC Ord 2005 Q6bull LC Ord 2003 Q12(a)bull LC Ord 2006 Q6
