Physics Malaysian Matriculation Semester 1 Notes Complete

PHYSICS CHAPTER 2

CHAPTER 2:CHAPTER 2:Kinematics of linear motionKinematics of linear motion

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PHYSICS CHAPTER 2

Kinematics of linear motionKinematics of linear motion

2.1 Linear Motion2.2 Uniformly Accelerated Motion2.3 Free Falling Body2.4 Projectile Motion

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine and distinguish between and distinguish between

i.i. distance and displacement distance and displacement ii.ii. speed and velocityspeed and velocityiii.iii. instantaneous velocity, average velocity and uniform instantaneous velocity, average velocity and uniform

velocity.velocity.iv.iv. instantaneous acceleration, average acceleration and instantaneous acceleration, average acceleration and

uniform acceleration. uniform acceleration. SketchSketch graphs of displacement-time, velocity-time and graphs of displacement-time, velocity-time and

acceleration-time.acceleration-time. DetermineDetermine the distance travelled, displacement, velocity the distance travelled, displacement, velocity

and acceleration from appropriate graphs.and acceleration from appropriate graphs.

Learning Outcome:2.1 Linear Motion

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2.1. Linear motion (1-D)

2.1.1. Distance, d scalar quantity. is defined as the length of actual path between two pointslength of actual path between two points. For example :

The length of the path from P to Q is 25 cm.

P

Q

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vector quantity is defined as the distance between initial point and final the distance between initial point and final

point in a straight linepoint in a straight line. The S.I. unit of displacement is metre (m).

Example 1:An object P moves 20 m to the east after that 10 m to the south and finally moves 30 m to west. Determine the displacement of P relative to the original position.Solution :Solution :

2.1.2 Displacement,s

N

EW

S

O

P

θ

θ

20 m

10 m

10 m 20 m

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The magnitude of the displacement is given by

and its direction is

2.1.3 Speed, v is defined the rate of change of distancerate of change of distance. scalar quantity. Equation:

interval timedistance of changespeed =

ΔtΔdv =

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is a vector quantity. The S.I. unit for velocity is m s-1.

Average velocity, Average velocity, vvavav

is defined as the rate of change of displacementthe rate of change of displacement. Equation:

Its direction is in the same direction of the change in same direction of the change in displacementdisplacement.

2.1.4 Velocity,v

interval timentdisplaceme of change=avv

ΔtΔsvav =

12

12av tt

ssv−−=

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Instantaneous velocity, Instantaneous velocity, vv is defined as the instantaneous rate of change of the instantaneous rate of change of

displacementdisplacement. Equation:

An object is moving in uniform velocitymoving in uniform velocity if

ts

0tv

∆∆

→∆=

limit

constant=dtds

dtdsv =

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Therefore

Q

s

t0

s1

t1

The gradient of the tangent to the curve at point Q

= the instantaneous velocity at time, t = t1

Gradient of s-t graph = velocity

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vector quantity The S.I. unit for acceleration is m s-2.

Average acceleration, Average acceleration, aaavav

is defined as the rate of change of velocitythe rate of change of velocity. Equation:

Its direction is in the same direction of motionsame direction of motion. The accelerationacceleration of an object is uniformuniform when the magnitude magnitude

of velocity changes at a constant rate and along fixed of velocity changes at a constant rate and along fixed direction.direction.

2.1.5 Acceleration, a

interval time velocityof change=ava

12

12av tt

vva−−=

ΔtΔvaav =

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Instantaneous acceleration, Instantaneous acceleration, aa is defined as the instantaneous rate of change of velocityinstantaneous rate of change of velocity. Equation:

An object is moving in uniform acceleration moving in uniform acceleration if

tv

0ta

∆∆

→∆=

limit

constant=dtdv

2

2

dtsd

dtdva ==

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Deceleration,Deceleration, aa is a negative accelerationnegative acceleration. The object is slowing downslowing down meaning the speed of the object speed of the object

decreases with timedecreases with time.

Therefore

v

t

Q

0

v1

t1

The gradient of the tangent to the curve at point Q

= the instantaneous acceleration at time, t = t1

Gradient of v-t graph = acceleration

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Displacement against time graph (Displacement against time graph (s-ts-t))2.1.6 Graphical methods

s

t0

s

t0(a) Uniform velocity (b) The velocity increases with time

Gradient = constant

Gradient increases with time

(c)s

t0

Q

RP

The direction of velocity is changing.

Gradient at point R is negative.

Gradient at point Q is zero.

The velocity is zero.

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Velocity versus time graph (Velocity versus time graph (v-tv-t))

The gradient at point A is positive – a > 0(speeding up) The gradient at point B is zero – a= 0 The gradient at point C is negative – a < 0(slowing down)

t1 t2

v

t0 (a) t2t1

v

t0 (b) t1 t2

v

t0 (c)

Uniform velocityUniform acceleration

Area under the v-t graph = displacement

BC

A

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From the equation of instantaneous velocity,

Therefore

dtdsv =

∫∫ = vdtds

∫= 2

1

t

tvdts

graph under the area dedsha tvs −=

Simulation 2.1 Simulation 2.2 Simulation 2.3

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A toy train moves slowly along a straight track according to the displacement, s against time, t graph in figure 2.1.

a. Explain qualitatively the motion of the toy train.b. Sketch a velocity (cm s-1) against time (s) graph.c. Determine the average velocity for the whole journey.d. Calculate the instantaneous velocity at t = 12 s.

Example 2 :

0 2 4 6 8 10 12 14 t (s)

2

4

68

10

s (cm)

Figure 2.1Figure 2.1

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Solution :Solution :0 to 6 s :

6 to 10 s : 10 to 14 s :

b.

0 2 4 6 8 10 12 14 t (s)

0.68

1.50

v (cm s−1)

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Solution :Solution :c.

d.

12

12

ttssvav −

−=

s 14 tos 10 from velocity average=v

12

12

ttssv

−−=

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A velocity-time (v-t) graph in figure 2.2 shows the motion of a lift.

a. Describe qualitatively the motion of the lift.b. Sketch a graph of acceleration (m s-1) against time (s).c. Determine the total distance travelled by the lift and its displacement.d. Calculate the average acceleration between 20 s to 40 s.

Example 3 :

05 10 15 20 25 30 35 t (s)

-4-2

2

4

v (m s −1)


40 45 50

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Solution :Solution :a. 0 to 5 s : Lift moves upward from rest with

acceleration of 0.4 m s−2. 5 to 15 s : The velocity of the lift from 2 m s−1 to

4 m s−1 but the acceleration to 0.2 m s−2. 15 to 20 s : Lift 20 to 25 s : Lift 25 to 30 s : Lift 30 to 35 s : Lift moves

35 to 40 s : Lift moving

40 to 50 s :

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Solution :Solution :b.

t (s)5 10 15 20 25 30 35 40 45 500

-0.4-0.2

0.2

0.6

a (m s−2)

-0.6

-0.8

0.8

0.4

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Solution :Solution :c. i.

05 10 15 20 25 30 35 t (s)

-4-2

2

4

v (m s −1)

40 45 50A1

A2 A3

A4 A5

v-t ofgraph under the area distance Total =54321 AAAAA ++++=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )45152145

214105

211042

2152

21distance Total +++++++=

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Solution :Solution :c. ii.

d.

v-t ofgraph under the areant Displaceme =

54321 AAAAA ++++=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )45152145

214105

211042

2152

21ntDisplaceme −++−+++++=

12

12

ttvvaav −

−=

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1. Figure 2.3 shows a velocity versus time graph for an object constrained to move along a line. The positive direction is to the right.

a. Describe the motion of the object in 10 s.b. Sketch a graph of acceleration (m s-2) against time (s) for the whole journey.c. Calculate the displacement of the object in 10 s.

ANS. : 6 mANS. : 6 m

Exercise 2.1 :

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2. A train pulls out of a station and accelerates steadily for 20 s until its velocity reaches 8 m s−1. It then travels at a constant velocity for 100 s, then it decelerates steadily to rest in a further time of 30 s.a. Sketch a velocity-time graph for the journey.b. Calculate the acceleration and the distance travelled in each part of the journey.c. Calculate the average velocity for the journey.Physics For Advanced Level, 4th edition, Jim Breithaupt, Nelson Thornes, pg.15, no. 1.11

ANS. : 0.4 m sANS. : 0.4 m s−−22,0 m s,0 m s−−22,-0.267 m s,-0.267 m s−−22, 80 m, 800 m, 120 m; , 80 m, 800 m, 120 m; 6.67 m s6.67 m s−−11..

Exercise 2.1 :

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Derive and applyDerive and apply equations of motion with uniform equations of motion with uniform

acceleration:acceleration:

Learning Outcome:2.2 Uniformly accelerated motion

atuv +=2

21 atuts +=

asuv 222 +=

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2.2. Uniformly accelerated motion From the definition of average acceleration, uniform (constantconstant)

acceleration is given by

where v : final velocityu : initial velocity

a : uniform (constant) accelerationt : time

atuv += (1)

tuva −=

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From equation (1), the velocity-time graph is shown in figure 2.4:

From the graph, The displacement after time, s = shaded area under the

graph = the area of trapezium

Hence,

velocity

0

v

utimetFigure 2.4Figure 2.4

( )tvu21s += (2)

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By substituting eq. (1) into eq. (2) thus

From eq. (1),

From eq. (2),

( )[ ]tatuus ++=21

(3)2

21 atuts +=

( ) atuv =−

( )tsuv 2=+

multiply

( ) ( ) ( )attsuvuv

=−+ 2

asuv 222 += (4)

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Notes: equations (1) – (4) can be used if the motion in a straight motion in a straight

line with constant acceleration.line with constant acceleration. For a body moving at constant velocity, ( ( aa = 0) = 0) the

equations (1) and (4) become

Therefore the equations (2) and (3) can be written asuv =

vts = constant velocityconstant velocity

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A plane on a runway takes 16.2 s over a distance of 1200 m to take off from rest. Assuming constant acceleration during take off, calculatea. the speed on leaving the ground,b. the acceleration during take off.Solution :Solution :

a. Use

Example 4 :

s 2.16=t

?=v

( )tvus +=21

0=u

m 1200=s

?=a

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Solution :Solution :b. By using the equation of linear motion,

asuv 222 +=

OROR2

21 atuts +=

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A bus travelling steadily at 30 m s−1 along a straight road passes a stationary car which, 5 s later, begins to move with a uniform acceleration of 2 m s−2 in the same direction as the bus. Determinea. the time taken for the car to acquire the same velocity as the bus,b. the distance travelled by the car when it is level with the bus.Solution :Solution :

a. Given Use

Example 5 :

21 ms 2 0; ;constant s m 30 −− ==== ccb auv

cccc tauv +=1s m 30 −== bc vv

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b.

From the diagram,

c

b

1s m 30 −=bv

0=cus 0=bt s 5=bt

2s m 2 −=cab

bvb

c

bv

ttb =bc ss =

bc ss =

bbcccc tvtatu =+ 2

21

Thereforetvs bc =

;ttb = 5−= ttc

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A particle moves along horizontal line according to the equation

Where s is displacement in meters and t is time in seconds.

At time, t =2.00 s, determinea. the displacement of the particle,b. Its velocity, andc. Its acceleration.Solution :Solution :a. t =2.00 s ;

Example 6 :

ttts 23 243 +−=

ttts 23 243 +−=

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Solution :Solution :b. Instantaneous velocity at t = 2.00 s,

Use

Thus

dtdsv =

( )tttdtdv 243 23 +−=

( ) ( ) 22.0082.009 2 +−=v

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Solution :Solution :c. Instantaneous acceleration at t = 2.00 s,

Use

Hence

dtdva =

( ) 82.0018 −=a

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1. A speedboat moving at 30.0 m s-1 approaches stationary buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m s-2 by reducing the throttle.a. How long does it take the boat to reach the buoy?b. What is the velocity of the boat when it reaches the buoy?No. 23,pg. 51,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.

ANS. : 4.53 s; 14.1 m sANS. : 4.53 s; 14.1 m s−−11

2. An unmarked police car travelling a constant 95 km h-1 is passed by a speeder traveling 140 km h-1. Precisely 1.00 s after the speeder passes, the policemen steps on the accelerator; if the police car’s acceleration is 2.00 m s-2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?No. 44, pg. 41,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition.

ANS. : 14.4 sANS. : 14.4 s

Exercise 2.2 :

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3. A car traveling 90 km h-1 is 100 m behind a truck traveling 75 km h-1. Assuming both vehicles moving at constant velocity, calculate the time taken for the car to reach the truck.No. 15, pg. 39,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition.

ANS. : 24 sANS. : 24 s4. A car driver, travelling in his car at a constant velocity of

8 m s-1, sees a dog walking across the road 30 m ahead. The driver’s reaction time is 0.2 s, and the brakes are capable of producing a deceleration of 1.2 m s-2. Calculate the distance from where the car stops to where the dog is crossing, assuming the driver reacts and brakes as quickly as possible.

ANS. : 1.73 mANS. : 1.73 m

Exercise 2.2 :

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Describe and useDescribe and use equations for free falling body. equations for free falling body.

For For upward and downwardupward and downward motion, use motion, useaa = = −−gg = = −−9.81 m s9.81 m s−−22

Learning Outcome:2.3 Free falling body

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2.3. Free falling body is defined as the vertical motion of a body at constant the vertical motion of a body at constant

acceleration, acceleration, gg under gravitational field under gravitational field without air without air resistanceresistance..

In the earth’s gravitational field, the constant acceleration known as acceleration due to gravityacceleration due to gravity or free-fall free-fall

accelerationacceleration or gravitational accelerationgravitational acceleration. the value is gg = 9.81 m s= 9.81 m s−−22

the direction is towards the centre of the earth towards the centre of the earth (downward).(downward).

Note: In solving any problem involves freely falling bodies or free

fall motion, the assumption made is ignore the air assumption made is ignore the air resistanceresistance.

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Sign convention:

Table 2.1 shows the equations of linear motion and freely falling bodies.

Table 2.1Table 2.1

Linear motion Freely falling bodiesatuv += gtuv −=

as2uv 22 += gs2uv 22 −=2at

21uts += 2gt

21uts −=

+

- +

-

From the sign convention thus,

ga −=

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An example of freely falling body is the motion of a ball thrown vertically upwards with initial velocity, u as shown in figure 2.5.

Assuming air resistance is negligible, the acceleration of the ball, a = −g when the ball moves upward and its velocity velocity decreases to zerodecreases to zero when the ball reaches the maximum maximum height, height, HH.

H

uv

velocity = 0


uv =

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The graphs in figure 2.6 show the motion of the ball moves up and down.

Derivation of equationsDerivation of equations At the maximum height or

displacement, H where t = t1, its velocity,

hence

therefore the time taken for the ball reaches H,


t0

vu

−u

t1 2t1

t0

a

−g

t1 2t1

t

s

0

H

t1 2t1

v =0

gtuv −=1gtu −=0

0=v

gut1 =

Simulation 2.4

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To calculate the maximum height or displacement, H:use either

maximum height,

Another form of freely falling bodies expressions are

211 gtuts

21−=

gsuv 22 2−=Where s = H

gHu 20 2 −=

OROR

guH2

2

=

gtuv −=gsuv 222 −=

2

21 gtuts −=

gtuv yy −=yyy gsuv 222 −=2

21 gttus yy −=

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A ball is thrown from the top of a building is given an initial velocity of 10.0 m s−1 straight upward. The building is 30.0 m high and the ball just misses the edge of the roof on its way down, as shown in figure 2.7. Calculatea. the maximum height of the stone from point A.b. the time taken from point A to C.c. the time taken from point A to D.d. the velocity of the stone when it reaches point D.(Given g = 9.81 m s−2)

Example 7 :

A

B

C

D

u =10.0 m s−1

30.0 m


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Solution :Solution :a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s−1 thus

b. From point A to C, the vertical displacement, sy= 0 m thus

y2y

2y gsuv 2−=

2yy gttus

21−=

A

B

C

D

u

30.0 m

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Solution :Solution :c. From point A to D, the vertical displacement, sy= −30.0 m thus

By using

2yy gttus

21−=

A

B

C

D

u

30.0 m2a

4acbb 2 −±−=t

ORTime don’t Time don’t have have negative negative value.value.

a b c

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Solution :Solution :d. Time taken from A to D is t = 3.69 s thus

From A to D, sy = −30.0 m

Therefore the ball’s velocity at D is

A

B

C

D

u

30.0 m

gtuv yy −=( ) ( ) ( )3.699.8110.0 −=yv

OR

y2

y2

y gsuv 2−=( ) ( )( )30.09.81210.0 −−= 22

yv

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A book is dropped 150 m from the ground. Determinea. the time taken for the book reaches the ground.b. the velocity of the book when it reaches the ground.(given g = 9.81 m s-2)Solution :Solution :

a. The vertical displacement issy = −150 m

Hence

Example 8 :

uy = 0 m s−1

150 mm 150−=ys

2yy gttus

21−=

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Solution :Solution :b. The book’s velocity is given by

Therefore the book’s velocity is

gtuv yy −=

OR

y2

y2

y gsuv 2−=m 150−=ys

0=yu

?=yv

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1. A ball is thrown directly downward, with an initial speed of 8.00 m s−1, from a height of 30.0 m. Calculate a. the time taken for the ball to strike the ground,b. the ball’s speed when it reaches the ground.

ANS. : 1.79 s; 25.6 m sANS. : 1.79 s; 25.6 m s−−11

2. A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in figure 2.8.

From what height above the top of the windows did the stone fall?

ANS. : 1.75 mANS. : 1.75 m

Exercise 2.3 :

m 2.2


to travel this distance took 0.30 s

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Describe and useDescribe and use equations for projectile, equations for projectile,

CalculateCalculate time of flight, maximum height, range, time of flight, maximum height, range, maximum range, instantaneous position and velocity.maximum range, instantaneous position and velocity.

Learning Outcome:2.4 Projectile motion

θuux cos=θuu y sin=

0=xagay −=

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2.4. Projectile motion A projectile motion consists of two components:

vertical component (y-comp.) motion under constant acceleration, ay= −g

horizontal component (x-comp.) motion with constant velocity thus ax= 0

The path followed by a projectile is called trajectory is shown in figure 2.9.

v

u

θsx= R

sy=H

ux

v2yuy

v1x

v1y

v2x

v1

θ1

v2

θ2

t1 t2

B

A

P Q

C

y

xFigure 2.9Figure 2.9

Simulation 2.5

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From figure 2.9, The x-component of velocityx-component of velocity along AC (horizontal) at any

point is constant,constant,

The y-component (vertical) of velocity variesy-component (vertical) of velocity varies from one point to another point along AC.but the y-component of the initial velocity is given by

θuux cos=

θuu y sin=

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Table 2.2 shows the x and y-components, magnitude and direction of velocities at points P and Q.

Velocity Point P Point Q

x-comp.

y-comp.

magnitude

direction

11 gtuv yy −=θuuv xx1 cos==

22 gtuv yy −=θuuv xx2 cos==

( ) ( )2y1

2x11 vvv +=

= −

x1

y111 v

vθ tan

( ) ( )2y2

2x22 vvv +=

= −

x2

y212 v

vθ tan

Table 2.2Table 2.2

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The ball reaches the highest point at point B at velocity, v where x-component of the velocity, y-component of the velocity, y-component of the displacement,

Use

2.4.1 Maximum height, H

θuuvv xx cos===0=yv

yyy gsuv 222 −=

( ) gHu 2sin0 2 −= θ

guH

2sin 22 θ=

Hsy =

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At maximum height, H Time, t = ∆t’ and vy= 0

Use

2.4.2 Time taken to reach maximum height, ∆t’

gtuv yy −=( ) 'sin0 tgu ∆−= θ

gut θsin' =∆

2.4.3 Flight time, ∆t (from point A to point C)

'2 tt ∆=∆

gθut sin2=∆

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Since the x-component for velocity along AC is constant hence

From the displacement formula with uniform velocity, thus the x-component of displacement along AC is

2.4.4 Horizontal range, R and value of R maximum

tus xx =

θcosuvu xx ==

( ) ( )tuR ∆= θcos

( )

=

guuR θθ sin2cos

( )θθ cossin22

guR =

and Rsx =

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From the trigonometry identity,

thus

The value of R maximum when θθ = = 4545°° and sin 2sin 2θθ = = 11 therefore

θθθ cossin22sin =

θ2sin2

guR =

guR

2

max =

Simulation 2.6

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Figure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction.

Horizontal component along path AB.

Vertical component along path AB.

2.4.5 Horizontal projectile

h

xA B

u u

vxv

yv


constant velocity, === xx vuuxsx = nt,displaceme

0u y = velocity,initialhsy −= nt,displaceme

Simulation 2.7

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Time taken for the ball to reach the floor (point B), Time taken for the ball to reach the floor (point B), tt By using the equation of freely falling bodies,

Horizontal displacement, Horizontal displacement, xx Use condition below :

2yy gttus

21−=

2gt0h21−=−

ght 2=

The time taken for the ball free fall to point A

The time taken for the ball to reach point B=

(Refer to figure 2.11)


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Since the x-component of velocity along AB is constant, thus the horizontal displacement, x

Note : In solving any calculation problem about projectile motion,

the air resistance is negligibleair resistance is negligible.

tus xx =

=

ghux 2

and xsx =

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Figure 2.12 shows a ball thrown by superman with an initial speed, u = 200 m s-1 and makes an angle, θ = 60.0° to the horizontal. Determinea. the position of the ball, and the magnitude and direction of its velocity, when t = 2.0 s.

Example 9 :

Figure 2.12Figure 2.12 xO

u

θ = 60.0°

y

R

H

v2y

v1x

v1y v2xQv1

P

v2

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b. the time taken for the ball reaches the maximum height, H and

calculate the value of H.

c. the horizontal range, Rd. the magnitude and direction of its velocity when the ball

reaches the ground (point P).e. the position of the ball, and the magnitude and direction of its

velocity at point Q if the ball was hit from a flat-topped hill with the time at point Q is 45.0 s.

(given g = 9.81 m s-2)Solution :Solution :The component of Initial velocity :

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Solution :Solution :a. i. position of the ball when t = 2.0 s ,

Horizontal component :

Vertical component :

therefore the position of the ball is

2yy gttus

21−=

tus xx =

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Solution :Solution :a. ii. magnitude and direction of ball’s velocity at t = 2.0 s ,



Magnitude,

Direction,

gtuv yy −=

1xx uv −== s m 100

( ) ( ) 2222 153100 +=+= yx vvv

=

= −−

100153tantan 11

x

y

vv

θ

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Solution :Solution :b. i. At the maximum height, H :

Thus the time taken to reach maximum height is given by

ii. Apply

gtuv yy −=

0=yv

gttus yy 21−=

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Solution :Solution :c. Flight time = 2×(the time taken to reach the maximum height)

Hence the horizontal range, R is

d. When the ball reaches point P thusThe velocity of the ball at point P,Horizontal component:Vertical component:

( )17.62=t

tus xx =

11 s m 100 −== xx uv

0=ys

gtuv yy −=1

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Solution :Solution :Magnitude,

Direction,

therefore the direction of ball’s velocity is

e. The time taken from point O to Q is 45.0 s. i. position of the ball when t = 45.0 s,


( ) ( ) 2221

211 172100 −+=+= yx vvv

−=

= −−

100172tantan 1

1

11

x

y

vv

θ

300=θ from positive x-axis anticlockwisefrom positive x-axis anticlockwise

tus xx =

PHYSICS CHAPTER 2

71

Solution :Solution :Vertical component :

therefore the position of the ball is (4500 m, (4500 m, −−2148 m)2148 m)e. ii. magnitude and direction of ball’s velocity at t = 45.0 s ,



2yy gttus

21−=

gtuv yy −=2

12 s m 100 −== xx uv

PHYSICS CHAPTER 2

72

Solution :Solution :Magnitude,

Direction,

therefore the direction of ball’s velocity is

( ) ( ) 222 269100 −+=v

22

222 yx vvv +=

= −

x

y

vv

θ2

21tan

PHYSICS CHAPTER 2

73

A transport plane travelling at a constant velocity of 50 m s−1 at an altitude of 300 m releases a parcel when directly above a point X on level ground. Calculatea. the flight time of the parcel,b. the velocity of impact of the parcel,c. the distance from X to the point of impact.(given g = 9.81 m s-2)Solution :Solution :

Example 10 :

300 m

d

1s m 50 −=u

X

PHYSICS CHAPTER 2

74

Solution :Solution :The parcel’s velocity = plane’s velocity

thus

a. The vertical displacement is given by

Thus the flight time of the parcel is

1s m 50 −== uux

1s m 50 −=uand 1s m 0 −=yu

2

21 gttus yy −=

PHYSICS CHAPTER 2

75

Solution :Solution :b. The components of velocity of impact of the parcel:

Horizontal component:Vertical component:

Magnitude,

Direction,

therefore the direction of parcel’s velocity is

1s m 50 −== xx uv

( ) ( )7.829.810 −=yvgtuv yy −=

−=

= −−

506.77tantan 11

x

y

vv

θ

( ) ( ) 2222 6.7750 −+=+= yx vvv

PHYSICS CHAPTER 2

76

Solution :Solution :c. Let the distance from X to the point of impact is d.

Thus the distance, d is given by

tus xx =

PHYSICS CHAPTER 2

77


Use gravitational acceleration, g = 9.81 m s−2

1. A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in figure 2.13. If he shoots the ball at a 40.0° angle above the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.

ANS. : 10.7 m sANS. : 10.7 m s−−11

Exercise 2.4 :

PHYSICS CHAPTER 2

78

2. An apple is thrown at an angle of 30° above the horizontal from the top of a building 20 m high. Its initial speed is 40 m s−1. Calculatea. the time taken for the apple to strikes the ground,b. the distance from the foot of the building will it strikes

the ground,c. the maximum height reached by the apple from the

ground.ANS. : 4.90 s; 170 m; 40.4 mANS. : 4.90 s; 170 m; 40.4 m

3. A stone is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s−1 at 40° above the horizontal. How far above or below its original level will the stone strike the opposite wall?

ANS. : 10.3 m below the original level.ANS. : 10.3 m below the original level.

Exercise 2.4 :

PHYSICS CHAPTER 2

79

THE END…Next Chapter…

CHAPTER 3 :Force, Momentum and Impulse

PHYSICS CHAPTER 3

1

MOMENTUM AND IMPULSE

CHAPTER 3

PHYSICS CHAPTER 3

3.0 MOMENTUM AND IMPULSE

3.1 Momentum and impulse

3.2 Conservation of linear momentum

2

PHYSICS CHAPTER 3

3

At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:

DefineDefine momentum.

DefineDefine impulse and use F-t graph to determine impulse

Use Use

Learning Outcome:3.1 Momentum and impulse

PHYSICS CHAPTER 3

4

3.1.1 Linear momentum,

is defined as the product between mass and velocitythe product between mass and velocity. is a vector quantity. Equation :

The S.I. unit of linear momentum is kg m skg m s-1-1. The direction of the momentumdirection of the momentum is the samesame as the direction direction

of the velocityof the velocity. It can be resolve into vertical (y) component and horizontal (x)

component.

p

vmp =

xp

pyp

θ

θmvθppx coscos ==θmvθppy sinsin ==

PHYSICS CHAPTER 3

5

3.1.2 Impulse, Let a single constant force, constant force, FF acts on an object in a short time

interval (collision), thus the Newton’s 2nd law can be written as

is defined as the product of a force, the product of a force, FF and the time, and the time, tt OR the change of momentumthe change of momentum.

is a vector quantityvector quantity whose directiondirection is the samesame as the constant forceconstant force on the object.

J

constant ===∑ dtpdFF

12 pppddtFJ −===

momentum final: 2pwheremomentum initial: 1p

force impulsive :F

PHYSICS CHAPTER 3

6

The S.I. unit of impulse is N sN s or kg m skg m s−−11. If the forceforce acts on the object is not constantnot constant then

Since impulse and momentum are both vector quantities, then it is often easiest to use them in component form :

dtFdtFJ av

t

t

== ∫ 2

1

where force impulsive average :avF

( ) ( )xxx1x2xavx uvmppdtFJ −=−==

( ) ( )yyy1y2yavy uvmppdtFJ −=−==

( ) ( )zzz1z2zavz uvmppdtFJ −=−==

consider 2-D consider 2-D collision onlycollision only

PHYSICS CHAPTER 3

7

When two objects in collision, the impulsive force, F against time, t graph is given by the Figure 3.20.

1t 2tFigure 3.20Figure 3.20 t0

F

Shaded area under the F−t graph = impulse

Picture 3.1

Picture 3.2

Picture 3.3

PHYSICS CHAPTER 3

8

A 0.20 kg tennis ball strikes the wall horizontally with a speed of 100 m s−1 and it bounces off with a speed of 70 m s−1 in the opposite direction.a. Calculate the magnitude of impulse delivered to the ball by the wall,b. If the ball is in contact with the wall for 10 ms, determine the magnitude of average force exerted by the wall on the ball.Solution :Solution :

Example 3.1 :

Wall (2)11

1s m 100 −=1u

111s m 70 −=1v

022 == uv

kg 0.201 =m

PHYSICS CHAPTER 3

9

Solution :Solution :a. From the equation of impulse that the force is constant,

Therefore the magnitude of the impulse is 34 N s34 N s.

b. Given the contact time,

12 ppdpJ −==( )111 uvmJ −=

dtFJ av=

PHYSICS CHAPTER 3

10

An estimated force-time curve for a tennis ball of mass 60.0 g struck by a racket is shown in Figure 3.21. Determinea. the impulse delivered to the ball,b. the speed of the ball after being struck, assuming the ball is being served so it is nearly at rest initially.

Example 3.2 :

0.2 1.8 ( )mst0

( )kNF

1.0

18


PHYSICS CHAPTER 3

11

Solution :Solution :a. From the force-time graph,

b. Given the ball’s initial speed,

graph under the area tFJ −=

0=u( )uvmdpJ −==

kg 1060.0 3−×=m

PHYSICS CHAPTER 3

12

1. A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m.a. Calculate the impulse delivered to the ball during impact.b. If the ball is in contact with the slab for 2.00 ms, determine the average force on the ball during impact.

ANS. : 0.47 N s; 237. 1 NANS. : 0.47 N s; 237. 1 N2. A golf ball (m = 46.0 g) is struck with a force that makes an

angle of 45° with the horizontal. The ball lands 200 m away on a flat fairway. If the golf club and ball are in contact for 7.00 ms, calculate the average force of impact. (neglect the air resistance.)

ANS. : 293 NANS. : 293 N

Exercise 3.1 :

PHYSICS CHAPTER 3

13


3.

A tennis ball of mass, m = 0.060 kg and a speed, v = 28 m s−1 strikes a wall at a 45° angle and rebounds with the same speed at 45° as shown in Figure 3.22. Calculate the impulse given by the wall.

ANS. : 2.4 N s to the left or ANS. : 2.4 N s to the left or −−2.4 N s2.4 N s

Exercise 3.1 :

PHYSICS CHAPTER 3

14

3.2 Conservation of linear momentum

14

3.2.1 Principle of conservation of linear momentum states “In an isolated (closed) system, the total momentum In an isolated (closed) system, the total momentum

of that system is constantof that system is constant.” OR“When the net external force on a system is zero, the total When the net external force on a system is zero, the total momentum of that system is constantmomentum of that system is constant.”

In a Closed system,

From the Newton’s second law, thus

0 ==∑ dtpdF

0=∑ F

0=pd

PHYSICS CHAPTER 3

15

According to the principle of conservation of linear momentum, we obtain

OR

The total of initial momentum = the total of final momentumThe total of initial momentum = the total of final momentum

∑∑ = fi pp

constant=p

constant=∑ xpconstant=∑ yp

Therefore then

PHYSICS CHAPTER 3

16

Linear momentum in one dimension collisionLinear momentum in one dimension collisionExample 3.3 :

Figure 3.14 shows an object A of mass 200 g collides head-on with object B of mass 100 g. After the collision, B moves at a speed of 2 m s-1 to the left. Determine the velocity of A after Collision.SolutionSolution : :

1s m 6 −=Au

AB

1s m 3 −=Bu


∑∑ = fi pp

1s m 6 ;kg 0.100 ;kg 0.200 −−=== ABA umm11 s m 2 ;s m 3 −− −== BB vu

PHYSICS CHAPTER 3

17

Linear momentum in two dimension collisionLinear momentum in two dimension collisionExample 3.4 :

A tennis ball of mass m1 moving with initial velocity u1 collides with a soccer ball of mass m2 initially at rest. After the collision, the tennis ball is deflected 50° from its initial direction with a velocity v1 as shown in figure 3.15. Suppose that m1 = 250 g, m2 = 900 g, u1 = 20 m s−1 and v1 = 4 m s−1. Calculate the magnitude and direction of soccer ball after the collision.


1u

Before collision After collision

m1 m2

m1 1v

50

Simulation 3.3

PHYSICS CHAPTER 3

18

Solution :Solution :

From the principle of conservation of linear momentum,

The x-component of linear momentum,

∑∑ = fi pp

x22x11x22x11 vmvmumum +=+

;s m 20 ;kg 0.900 ;kg 0.250 1−=== 121 umm0 ;s m 4 ;0 1 5θvu 112 === −

∑∑ = fxix pp

PHYSICS CHAPTER 3

19

Solution :Solution :The y-component of linear momentum,

Magnitude of the soccer ball,

Direction of the soccer ball,

yy vmvm 22110 +=∑∑ = fyiy pp

( ) ( )2y2

2x22 vvv +=

=

= −−

4.840.851tantan 1

x2

y212 v

vθ

PHYSICS CHAPTER 3

20

1. An object P of mass 4 kg moving with a velocity 4 m s−1 collides elastically with another object Q of mass 2 kg moving with a velocity 3 m s−1 towards it.a. Determine the total momentum before collision.b. If P immediately stop after the collision, calculate the final velocity of Q.c. If the two objects stick together after the collision, calculate

the final velocity of both objects.ANS. : 10 kg m sANS. : 10 kg m s−−11; 5 m s; 5 m s−−11 to the right; 1.7 m s to the right; 1.7 m s−−11 to the right to the right2. A marksman holds a rifle of mass mr = 3.00 kg loosely in his

hands, so as to let it recoil freely when fired. He fires a bullet of mass mb = 5.00 g horizontally with a velocity 300 m s-1. Determinea. the recoil velocity of the rifle,b. the final momentum of the system.

ANS. : ANS. : −−0.5 m s0.5 m s−−11; U think.; U think.

Exercise 3.2 :

PHYSICS CHAPTER 3

21

3.

In Figure 3.16 show a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second block, with mass 1.80 kg. Speeds of 0.630 m s−1 and 1.40 m s-1, respectively, are thereby given to the blocks. Neglecting the mass removed from the first block by the bullet, determinea. the speed of the bullet immediately after it emerges from the first block andb. the initial speed of the bullet.ANS. : 721 m sANS. : 721 m s−−11; 937.4 m s; 937.4 m s−−11


1.20 kg 1.80 kg

0.630 m s-1 1.40 m s-1

Before

After

PHYSICS CHAPTER 3

22


4. A ball moving with a speed of 17 m s−1 strikes an identical ball that is initially at rest. After the collision, the incoming ball has been deviated by 45° from its original direction, and the struck ball moves off at 30° from the original direction as shown in Figure 3.17. Calculate the speed of each ball after the collision.

ANS. : 8.80 m sANS. : 8.80 m s−− 1 1; 12.4 m s; 12.4 m s−−11

Exercise 3.2 :

PHYSICS CHAPTER 3

23

3.2.2 Collision

is defined as an isolated event in which two or more bodies an isolated event in which two or more bodies (the colliding bodies) exert relatively strong forces on each (the colliding bodies) exert relatively strong forces on each other for a relatively short timeother for a relatively short time.

Two types of collisions :

Elastic collisionElastic collision

Inelastic (non-elastic) collisionInelastic (non-elastic) collision

PHYSICS CHAPTER 3

24

Elastic collisionElastic collision is defined as one in which the total kinetic energy (as well as one in which the total kinetic energy (as well as

total momentum) of the system is the same before and after total momentum) of the system is the same before and after the collisionthe collision.

Figure 3.18 shows the head-on collision of two billiard balls.

11 22

Before collision

At collision

After collision

11 2222um11um

11 2222vm11vm


Simulation 3.4

PHYSICS CHAPTER 3

25

The properties of elastic collisionproperties of elastic collision are

a. The total momentum is conservedtotal momentum is conserved.

b. The total kinetic energy is conservedtotal kinetic energy is conserved.

OR

∑∑ = fi pp

∑∑ = fi KK

222

211

222

211 vmvmumum

21

21

21

21 +=+

PHYSICS CHAPTER 3

26

Inelastic (non-elastic) collisionInelastic (non-elastic) collision is defined as one in which the total kinetic energy of the one in which the total kinetic energy of the

system is not the same before and after the collision (even system is not the same before and after the collision (even though the total momentum of the system is conserved)though the total momentum of the system is conserved).

Figure 3.19 shows the model of a completely inelastic completely inelastic collisioncollision of two billiard balls.

11 22At collision

After collision (stick together)

11 22v


Before collision 11 2211um 0=2u

2m

Simulation 3.5

PHYSICS CHAPTER 3

27

Caution: Not allNot all the inelastic collision is stick togetherstick together. In fact, inelastic collisions include many situationsmany situations in which

the bodies do not stickbodies do not stick

The properties of inelastic collisionproperties of inelastic collision area. The total momentum is conservedtotal momentum is conserved.

b. The total kinetic energy is not conservedtotal kinetic energy is not conserved because some of the energy is converted to internal energyinternal energy and some of it is transferred away by means of sound or heatsound or heat. But the total total energy is conservedenergy is conserved.

OR

∑∑ = fi pp

∑∑ = fi EE energy losses += ∑∑ fi KK

PHYSICS CHAPTER 3

28

Ball A of mass 400 g and velocity 4 m s-1 collides with ball B of mass 600 g and velocity 10 m s-1. After collision, A and B will move together. Determine the final velocity of both balls if A and B moves in the opposite direction initiallySolution : Solution : mmAA = 0.4 kg, u = 0.4 kg, uAA = 4 m s = 4 m s-1 -1 , m, mB B = 0.6 kg, u= 0.6 kg, uB B = -10 m s= -10 m s-1-1, , inelastic collisioninelastic collision

By using the principle of conservation of linear momentum, thus

Example 3.5 :

Before collision AA BBBuAu

After collision AA BB?=v

∑∑ = fi pp

vmmumum BABBAA )( +=+

PHYSICS CHAPTER 3

29


Final velocity of both balls is - 4.4 m s -1

BA

BBAA

mmumumv

++=

PHYSICS CHAPTER 3

30

A ball A of mass 1 kg moving at a velocity of 4 m s-1 collides with ball B of mass 2 kg which at rest. Calculate the velocity of both balls after collision if the collision is an elastic collision. Solution :Solution :

Given mA = 1 kg, uA = 4 m s-1, mB = 2 kg, uB = 0 m s-1, elastic collision

Example 3.6 :

Before collision AA BB

10 −= msuBAu

After collision AA BB?=Bv?=Av

PHYSICS CHAPTER 3

31

Solution :Solution :Apply principle of conservation of momentum,

Apply principle of conservation of kinetic energy,

BBAABBAA vmvmumum +=+)(2)(1)0(2)4(1 BA vv +=+

222 )(21)(

21)(

21

BBAAAA vmvmum +=

222 )()()( BBAAAA vmvmum +=

1B s m v2 -4 −=Av ……..(1)

PHYSICS CHAPTER 3

32

222 )(2)(1)4(1 BA vv +=Solution :Solution :

2216 BA vv += ………..(2)

Substitute equation (1) into equation (2)22 2)24(16 BB vv +−=

22 2)2(416 BB vv +−=22)2(28 BB vv +−=

PHYSICS CHAPTER 3

33

Substitute into equation (1),

238 BB vv =

167.2 −= msvB

(2.67) 2 -4=Av


PHYSICS CHAPTER 3

34


CHAPTER 4 :Forces

PHYSICS CHAPTER 4

1

CHAPTER 4:

FORCES

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PHYSICS CHAPTER 4

4. FORCES

4.1 Basic of Forces and Free Body Diagram

4.2 Newton’s Laws of Motion

2

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PHYSICS CHAPTER 4

3

At the end of this chapter, students should be able to:

Identify the forces acting on a body in different situations.

Weight

Tension

Normal force

Friction

Determine weight, static friction and kinetic friction

Draw free body diagram

Determine the resultant force

Learning Outcome:

4.1 Basic of Forces and Free Body Diagram

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PHYSICS CHAPTER 4

4.1 Basic of Forces and Free Body DiagramWeight,

is defined as the force exerted on a body under gravitational

field.

It is a vector quantity.

It is dependant on where it is measured, because the value of gvaries at different localities on the earth’s surface.

It always directed toward the centre of the earth or in the same

direction of acceleration due to gravity, g.

The S.I. unit is kg m s-2 or Newton (N).

Equation:

gmW

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PHYSICS CHAPTER 4

5

Tension, T

The tension force is the force that is transmitted through a

string, rope, cable or wire when it is pulled tight by forces

acting from opposite ends. The tension force is directed

along the length of the wire and pulls equally on the objects

on the opposite ends of the wire.

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Figure 4.1

PHYSICS CHAPTER 4

66

Normal (reaction) force,

is defined as a reaction force that exerted by the surface to

an object interact with it and the direction always

perpendicular to the surface.

An object lies at rest on a flat horizontal surface as shown in

Figure 4.2.

Ror N

N

gmW

0mgNFy

mgN

ThereforeFigure 4.2

Action: weight of an object is exerted on the

horizontal surface

Reaction: surface is exerted a force, N on the

object

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A free body diagram is defined as a diagram showing the chosen body by itself, with vectors

drawn to show the magnitude and directions of all the forces applied to the body by the other bodies

that interact with it.

PHYSICS CHAPTER 4

7

Friction

is defined as a force that resists the motion of one surface relative to another with which it is in contact.

is independent of the area of contact between the two surfaces..

is directly proportional to the reaction force.

OR

Coefficient of friction, is defined as the ratio between frictional force to reaction

force.

OR

is dimensionless and depends on the nature of the surfaces.

Nf

Nf force frictional:f

friction oft coefficien : μ

forcereaction : N

where

N

f

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PHYSICS CHAPTER 4

8

There are three types of frictional force :

Static, fs (frictional force act on the object before its move)

Kinetic, fk (frictional force act on the object when its move)

Rolling, fr (frictional force act on the object when its rolling)

Caution:

The direction of the frictional force exerted by a surface

on an object is always in the opposite direction of the

motion.

The frictional and the reaction forces are always

perpendicular.

Nf kk

Nf ss

Nf rr

skr fff where

thus skr

Can be ignored

Simulation 4.1

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AF_0516.html

AF_0516.html

PHYSICS CHAPTER 4

Example 4.1:

A mass is resting on a flat surface which has a normal force of

98N, with a coefficient of static friction of 0.35. What force

would it take to move the object?

9

Solution: N = 98N, μs = 0.35

Nf ss

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PHYSICS CHAPTER 4

10

Example 4.2:

A 15 kg piece of wood is placed on top of another piece of

wood. There is 35N of static friction measured between them.

Determine the coefficient of static friction between the two

pieces of wood.

Solution: N = mg = 15(9.81) = 147.15 N, Fs = 35 N

N

fss

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PHYSICS CHAPTER 4

Example 4.3

A dock worker loading crates on a ship finds that a 15 kg crate,

initially at rest on a horizontal surface, requires a 50 N

horizontal force to set it in motion. However, after the crate is in

motion, a horizontal force of 30 N is required to keep it moving

with a constant speed. The acceleration of gravity is 9.8 ms-2.

Find the coefficient of kinetic friction.

11

Solution:

Mass of crate = m = 15 kg

Force required to set the crate in motion = F1 = 50 N

Force required to keep the crate in moving at constant speed =

fk = 30 N

Acceleration of gravity = g = 9.81 ms-2

Normal force, N = mg = =

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PHYSICS CHAPTER 4

12

Solution:

From

N

fkk

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PHYSICS CHAPTER 4

Resultant force Is defined as a single force that represents the combined

effect of two or more forces

1313

The figure above shows three forces F1, F2 and F3 acted on a

particle O. Calculate the magnitude and direction of the

resultant force on particle O.

Example 4.4:y

30o

O

)N30(2F

)N10(1F

30o

x

)N40(3F

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PHYSICS CHAPTER 4

14

30o

Solution :

O

y

x

3F

30o

y3F

321 FFFFFr

yxr FFF

xxxx FFFF 321

yyyy FFFF 321

xF2

1F

2F

60o

yF2

x3F

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PHYSICS CHAPTER 4

15

Vector x-component y-component

1F

3F

2F

N 01 xF 11 FF y

N 011 yF

60cos302 xF

N 152 xF

60sin302 yF

N 622 yF

30cos403 xF

N 34.63 xF

30sin403 yF

N 203 yF

Vector

sum

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PHYSICS CHAPTER 4

16

y

xO

Solution :

The magnitude of the resultant force is

and

Its direction is 162 from positive x-axis OR 18 above negative x-

axis.

22

yxr FFF

x

y

F

Fθ 1tan

yF

xF

162

rF

18

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PHYSICS CHAPTER 4

17

1. Given three vectors P, Q and R as shown in Figure 4.3.

Calculate the resultant vector of P, Q and R.

ANS. : 49.4 m s2; 70.1 above + x-axis

Exercise 4.1:

Figure 4.3

y

x0

50 2s m 10 R

2s m 35 P

2s m 24 Q

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PHYSICS CHAPTER 4

18


State Newton’s First Law

Define mass as a measure of inertia.

Define the equilibrium of a particle.

Apply Newton’s First Law in equilibrium of forces

State and apply Newton’s Second Law

State and apply Newton’s Third Law.

Learning Outcome:

4.2 Newton’s laws of motion

td

vdm

td

mdvmv

td

d

td

dpF

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PHYSICS CHAPTER 4

19

4.2 Newton’s laws of motion

4.2.1 Newton’s first law of motion states “an object at rest will remain at rest, or continues to

move with uniform velocity in a straight line unless it is acted upon by a external forces”

OR

The first law gives the idea of inertia.

0FFnett

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PHYSICS CHAPTER 4

20

4.2.2 InertiaInertia

is defined as the tendency of an object to resist any change in its state of rest or motion.

is a scalar quantity.

Mass, m

is defined as a measure of a body’s inertia.


The S.I. unit of mass is kilogram (kg).

The value of mass is independent of location.

If the mass of a body increases then its inertia will increase.

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PHYSICS CHAPTER 4

21

Figures 4.4a and 4.4b show the examples of real experience of inertia.

Figure 4.4

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PHYSICS CHAPTER 4

22

4.2.3 Equilibrium of a particle

is defined as the vector sum of all forces acting on a particle (point) must be zero.

The equilibrium of a particle ensures the body in translational equilibrium and its condition is given by

This is equivalent to the three independent scalar equations along the direction of the coordinate axes,

There are two types of equilibrium of a particle. It is

Static equilibrium (v=0) body remains at rest (stationary).

Dynamic equilibrium (a=0) body moving at a uniform (constant) velocity.

0nettFF

Newton’s first

law of motion

0 , 0 , 0 zyx FFF

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PHYSICS CHAPTER 4

23

Problem solving strategies for equilibrium of a

particle

The following procedure is recommended when dealing with

problems involving the equilibrium of a particle:

Sketch a simple diagram of the system to help

conceptualize the problem.

Sketch a separate free body diagram for each body.

Choose a convenient coordinate axes for each body and

construct a table to resolve the forces into their

components.

Apply the condition for equilibrium of a particle in

component form :

Solve the component equations for the unknowns.

0xF 0yFand

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PHYSICS CHAPTER 4

2424

A load of 250 kg is hung by a crane’s cable. The load is pulled by a

horizontal force such that the cable makes a 30 angle to the

vertical plane. If the load is in the equilibrium, calculate

a. the magnitude of the tension in the cable,

b. the magnitude of the horizontal force. (Given g =9.81 m s2)

Solution :

Example 4.5:

30

F F

Free body diagram of the load :

gm

T

yT3060

xT

kg 250m

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PHYSICS CHAPTER 4

25

Solution :

1st method :

a.

Since the load is in the equilibrium, then

Thus

b. By substituting eq. (2) into eq. (1), therefore

0xF 060cos TF

kg 250m

Force x-component (N) y-component (N)

gm

0 9.81250mg

F

F 0

T 60cosT

60sinT

2453

0F

(1)

(2) 0yF 0245360sin T

060cos2833 F

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PHYSICS CHAPTER 4

26

30

Solution :

2nd method :

a. Since the load is in the equilibrium, then a closed triangle of

forces can be sketched as shown below.

b. 30sinT

F

kg 250m

30cosT

mg

30sin2833

F

F

gm T

From the closed triangle of forces, hence

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PHYSICS CHAPTER 4

27

Calculate the magnitude and direction of a force that balance the

three forces acted at point A as shown in Figure 4.5.

Example 4.6:

N 121FN 202F

N 303F

30.055.0

45.0A

Figure 4.5

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PHYSICS CHAPTER 4

28

Solution :

To find a force to balance the three forces means the system must

be in equilibrium hence

N 30 N; 20 N; 12 321 FFF


1F 55.0cos12

F

xF yF

6.88

55.0sin129.83

2F 30.0cos20

17.3

30.0sin2010.0

3F 45.0cos30

21.2

45.0sin3021.2

0 xF

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PHYSICS CHAPTER 4

29

Solution :

The magnitude of the force,

and its direction,

0 yF

021.210.09.83 yF

222

y

2

x FFF 1.3731.6

x

y1

F

Fθ tan

31.6

1.37tan 1θ

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PHYSICS CHAPTER 4

30

30

A window washer pushes his scrub brush up a vertical window at

constant speed by applying a force F as shown in Figure 4.6.

The brush weighs 10.0 N and the coefficient of kinetic friction is

k= 0.125. Calculate

a. the magnitude of the force F ,

b. the normal force exerted by the window on the brush.

Example 4.7:

F

50.0

Figure 4.6

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PHYSICS CHAPTER 4

31

Solution :

a. The free body diagram of the brush :

The brush moves up at constant speed (a=0) so that

Thus

0.125 ;N 10.0 kμW

W

F

N

kf

constant

speed


F

50.0cosF

kf

0Nμk

50.0sinF

W

0 10.0

N

N 0

N0.125

0amF

50.0cosFN 0 xF (1)

(2)10.00.12550.0sin NF 0 yF

50.0

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PHYSICS CHAPTER 4

32

Solution :

a. By substituting eq. (1) into eq. (2), thus

b. Therefore the normal force exerted by the window on the brush

is given by

10.050.0cos0.12550.0sin FF

50.0cosFN

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PHYSICS CHAPTER 4

33

Exercise 4.2:

Use gravitational acceleration, g = 9.81 m s2

1.

The system in Figure 5.8 is in equilibrium, with the string at the

centre exactly horizontal. Calculate

a. the tensions T1, T2 and T3.

b. the angle .

ANS. : 49 N, 28 N, 57 N; 29

Figure 4.7

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PHYSICS CHAPTER 4

34

Exercise 4.2:

2.

A 20 kg ball is supported from the ceiling by a rope A. Rope B

pulls downward and to the side on the ball. If the angle of A to

the vertical is 20 and if B makes an angle of 50 to the vertical

as shown in Figure 4.8, Determine the tension in ropes A and B.

ANS. : 134 N; 300 N

Figure 4.8

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PHYSICS CHAPTER 4

35

Exercise 4.2:

3.

A block of mass 3.00 kg is pushed up against a wall by a force

P that makes a 50.0 angle with the horizontal as show in

Figure 4.9. The coefficient of static friction between the block

and the wall is 0.250. Determine the possible values for the

magnitude of P that allow the block to remain stationary.

ANS. : 31.8 N; 48.6 N

Figure 4.9

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PHYSICS CHAPTER 4

36

Newton’s second law of motion

states “the rate of change of linear momentum of a moving

body is proportional to the resultant force and is in the

same direction as the force acting on it”

OR

its can be represented by

dt

pdF

where

momentumlinear in change : pd

interval time:dt

forceresultant : F

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PHYSICS CHAPTER 4

37

From the Newton’s 2nd law of motion, it also can be written as

Case 1:

Object at rest or in motion with constant velocity but with

changing mass. For example : Rocket

dt

vdm

dt

dmvF

dt

pdF

dt

vmdF

dt

vdm

dt

dmvF

mvp an

d

0dt

vd

dt

dmvF

and

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PHYSICS CHAPTER 4

38

Case 2:

Object at rest or in motion with constant velocity and constant

mass.

Thus

dt

vdm

dt

dmvF

Newton’s 1st law of motion

0 dt

pdF

constantp

0dt

dm 0

dt

vd

0F

where and

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PHYSICS CHAPTER 4

39

Case 3:

Object with constant mass but changing velocity.

The direction of the resultant force always in the same

direction of the motion or acceleration.

dt

vdm

dt

dmvF

0dt

dman

d

amF

dt

vdmF

dt

vda

and

where

objectan of mass : m

onaccelerati :a

forceresultant : F

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PHYSICS CHAPTER 4

40

Newton’s 2nd law of motion restates that “The acceleration of

an object is directly proportional to the nett force acting on

it and inversely proportional to its mass”.

OR

One newton(1 N) is defined as the amount of nett force that

gives an acceleration of one metre per second squared to a

body with a mass of one kilogramme.

OR 1 N = 1 kg m s-2

Notes:

is a nett force or effective force or resultant force.

The force which causes the motion of an object.

If the forces act on an object and the object moving at

uniform acceleration (not at rest or not in the

equilibrium) hence

amFFnett

m

Fa

F

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PHYSICS CHAPTER 4

41

Newton’s third law of motion

states “every action force has a reaction force that is equal in magnitude but opposite in direction”.

For example :

When the student push on the wall it will push back with the same force. (refer to Figure 4.10)

BAAB FF

A (hand)

B (wall)

BAF

ABF

Figure 4.10

is a force by the hand on the wall (action)Where

is a force by the wall on the hand (reaction)BAFABF

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PHYSICS CHAPTER 4

42

When a book is placed on the table. (refer to Figure 4.11)

If a car is accelerating forward, it is because its tyres are pushing backward on the road and the road is pushing forward on the tyres.

A rocket moves forward as a result of the push exerted on it by the exhaust gases which the rocket has pushed out.

In all cases when two bodies interact, the action and reaction forces act on different bodies.

Figure 4.11

Force by the book on the table (action)

Force by the table on the book (reaction)

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PHYSICS CHAPTER 4

43

Applications of Newton’s 2nd law of motion

From the Newton’s second law of motion, we arrived at equation

There are five steps in applying the equation above to solve problems in mechanics:

Identify the object whose motion is considered.

Determine the forces exerted on the object.

Draw a free body diagram for each object.

is defined as a diagram showing the chosen body by itself, with vectors drawn to show the magnitude and directions of all the forces applied to the body by the other bodies that interact with it.

Choose a system of coordinates so that calculations may be simplified.

Apply the equation above,

Along x-axis:

Along y-axis:

maFF nett

xx maF

yy maF

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PHYSICS CHAPTER 4

44

Three wooden blocks connected by a rope of negligible mass are

being dragged by a horizontal force, F in Figure 4.12.

Suppose that F = 1000 N, m1 = 3 kg, m2 = 15 kg and m3 = 30 kg.

Determine

a. the acceleration of blocks system.

b. the tension of the rope, T1 and T2.

Neglect the friction between the floor and the wooden blocks.

Example 4.8:

Figure 4.12

1T

m1 m2m3

2T

F

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PHYSICS CHAPTER 4

45

Solution :

a. For the block, m1 = 3 kg

For the block, m2 = 15 kg

For the block, m3 = 30 kg

a

amTFF 11x

(1)

amTTF 221x

(2)

1T

m1

m2

m3

2T

F

aTF 1x 31000

1T a

aTTF 21x 15

2T

a

amTF 32x

(3)

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PHYSICS CHAPTER 4

46

Solution :

a. By substituting eq. (3) into eq. (2) thus

Eq. (1)(4) :

b. By substituting the value of acceleration into equations (4) and

(3), therefore

045 aT1(4)

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PHYSICS CHAPTER 4

47

Two objects of masses m1 = 10 kg and m2 = 15 kg are connected

by a light string which passes over a smooth pulley as shown in

Figure 4.13. Calculate

a. the acceleration of the object of mass 10 kg.

b. the tension in the each string.

(Given g = 9.81 m s2)

Solution :

a. For the object m1= 10 kg,

Example 4.9:

Figure 4.13

m1

m2

1T

gmW 11

amgmTF 111y

(1)agT 1010 a where TTT 21

Simulation 4.2

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PHYSICS CHAPTER 4

48

Solution :

a. For the object m2= 15 kg,

Eq. (1) + (2) :

b. Substitute the value of acceleration into equation (1) thus

Therefore

2T

gmW 22

amTgmF 222y

(2)agT 1515 a

aTgFy 1515

1.96109.8110 T

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PHYSICS CHAPTER 4

49

Two blocks, A of mass 10 kg and B of mass 30 kg, are side by side

and in contact with each another. They are pushed along a smooth

floor under the action of a constant force F of magnitude 200 N

applied to A as shown in Figure 4.14. Determine

a. the acceleration of the blocks,

b. the force exerted by A on B.

Solution :

a. Let the acceleration of the blocks is a. Therefore

Example 4.10:

ammF BAx

N 200 kg; 30 kg; 10 Fmm BA

Figure 4.14

A BF

ammF BA

Simulation 4.3

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PHYSICS CHAPTER 4

50

Solution :

b. For the object A,

From the Newton’s 3rd law, thus

OR

For the object B,

amFFF ABAx

5.010200 BAFF

a

BAF

A

BABF

a

amFF BABx

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PHYSICS CHAPTER 4

51

1. A block is dragged by forces, F1 and F2 of the magnitude

20 N and 30 N respectively as shown in Figure 4.15. The

frictional force f exerted on the block is 5 N. If the weight of

the block is 200 N and it is move horizontally, determine the

acceleration of the block.

(Given g = 9.81 m s2)

ANS. : 1.77 m s2

Exercise 4.3:

50

a

1F

2F

f

20

Figure 4.15

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PHYSICS CHAPTER 4

52

2. One 3.5 kg paint bucket is hanging by a massless cord from

another 3.5 kg paint bucket, also hanging by a massless cord

as shown in Figure 4.16. If the two buckets are pulled upward

with an acceleration of 1.60 m s2 by the upper cord, calculate

the tension in each cord.

(Given g = 9.81 m s2)

ANS. : 39.9 N; 79.8 N

Exercise 4.3:

Figure 4.16

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PHYSICS CHAPTER 4

53

THE END…

Next Chapter…CHAPTER 5 :

Work, Energy and Power

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CHAPTER 5 WORK, ENERGY AND POWER

1

CHAPTER 5:

Work, Energy and Power

(3 Hours)


2


(a)Define and use work done by a force.

(b) Determine work done from the force-

displacement graph.

Learning Outcome:

5.1 Work (1 hour)

sFW


3

5.1 Work, W

Work done by a constant force

is defined as the product of the component of the

force parallel to the displacement times the

displacement of a body.

OR

is defined as the scalar (dot) product between

force and displacement of a body.


4

sFW

θFssθFW coscos

force of magnitude:F

body theofnt displaceme : s

sFθ

and between angle the:

Where,

Mathematically :


5

It is a scalar quantity.

Dimension :

The S.I. unit of work is kg m2 s2 or joule (J).

The joule (1 J) is defined as the work done by a force of 1 N

which results in a displacement of 1 m in the direction of

the force.

sFW

22TML W

22 s m kg 1m N 1J 1


6

Work done by a variable force

Figure 5.1 shows a force, F whose magnitude

changes with the displacement, s.

For a small displacement, s1 the force remains

almost constant at F1 and work done therefore

becomes W1=F1 s1 .


7

To find the total work done by a variable force, W when the

displacement changes from s=s1 to s=s2, we can divide the displacement into N small successive displacements :

s1 , s2 , s3 , …, sN

Thus

FN

F4

s4 sN

s1 s2

F/N

s0

F1

s1

W1

NN2211 sFsFsFW ...

Figure 5.1


8

When N , s 0, therefore

2

1

s

sFdsW

graphnt displaceme-force under the area theW

F/N

s/ms1 s20

Work = Area


9

Applications of work’s equation

Case 1 :

Work done by a horizontal force, F on an object (Figure 4.2).

Case 2 :

Work done by a vertical force, F on an object (Figure 4.3).

0θF

s

Figure 5.2

θFsW cosFsW

and

90θθFsW cosJ 0W

andF

s

Figure 5.3


10

Case 3 :

Work done by a horizontal forces, F1 and F2 on an object

(Figure 5.4).

Case 4 :

Work done by a force, F and frictional force, f on an object

(Figure 5.5).

0cos sFW 11

0cos sFW 22

sFWW nettnett

1F

2F

s

Figure 5.4 sFsFWWW 2121

21nett FFF sFFW 21 and

cos mafθFFnett sFW nettnett

sfFWnett cos masWnett

f F

Figure 5.5 s

and

OR


11

Caution :

Work done on an object is zero when F = 0 or s = 0 and

= 90.


12

Sign for work.

If 0< <90 (acute angle) then cos > 0 (positive value)

therefore

W > 0 (positive) work done on the system ( by the external force) where energy is transferred to the system.

If 90< <180 (obtuse angle) then cos <0 (negative value) therefore

W < 0 (negative) work done by the system where energy is transferred from the system.

cosFsW


13

You push your physics reference book 1.50 m along a horizontal

table with a horizontal force of 5.00 N. The frictional force is 1.60 N.

Calculate

a. the work done by the 5.00 N force,

b. the work done by the frictional force,

c. the total work done on the book.

Solution :

a. Use work’s equation of constant force,

Example 5.1 :

m 1.50s

N 5.00F

N 1.60f

cosθFsWF 0θand

Example 5.1 :


14

Solution :

b.

c.

θfsW f cos

fF WWW

OR

sFW nett

sfFW

180θand

180cos1.501.60fW

2.407.50W

1.501.605.00W


15

A box of mass 20 kg moves up a rough plane which is inclined to

the horizontal at 25.0. It is pulled by a horizontal force F of

magnitude 250 N. The coefficient of kinetic friction between the box

and the plane is 0.300.

a. If the box travels 3.80 m along the plane, determine

i. the work done on the box by the force F,

ii. the work done on the box by the gravitational force,

iii. the work done on the box by the reaction force,

iv. the work done on the box by the frictional force,

v. the total work done on the box.

b. If the speed of the box is zero at the bottom of the plane,

calculate its speed when it is travelled 3.80 m.

(Given g = 9.81 m s2)

Example 5.2 :


16

Solution :

a. Consider the work done along inclined plane, thus

i.

m 3.800.300; ;N 250 ;kg 20 sμFm k

θsFW xF cos 0θand

0cos3.8025cos250FW

25

25

kf

N

F

s

gmW

yF

25cosmg

xF

25sinmg

a

25x

y


17

Solution :

a. ii.

iii.

iv.

θsmgWg cos25sin 180θand

180cos3.8025sin9.8120gW

θNsWN cos 90θand

θsfW kf cos 180θand

180cossNμW kf

smgFμW kf 25cos25sin

3.8025cos9.812025sin2500.300 fW


18

Solution :

a. v.

b. Given

By using equation of work for nett force,

Hence by using the equation of linear motion,

fNgF WWWWW

3230315861W

masW

3.8020223 a

asuv 22 2

0u

3.802.93202v


19

A horizontal force F is applied to a 2.0 kg radio-controlled car as it

moves along a straight track. The force varies with the

displacement of the car as shown in Figure 5.6. Calculate the work

done by the force F when the car moves from 0 to 7 m.

Solution :

Example 5.3 :

5

47

053 6

(N)F

5 (m)s

Figure 5.6

graph under the area sFW

4672

15356

2

1W


20

Exercise 5.1 :

1. A block of mass 2.50 kg is pushed 2.20 m along a frictionless

horizontal table by a constant 16.0 N force directed 25.0 below

the horizontal. Determine the work done on the block by

a. the applied force,

b. the normal force exerted by the table, and

c. the gravitational force.

d. Determine the total work on the block.

(Given g = 9.81 m s2)

ANS. : 31.9 J; (b) & (c) U think; 31.9 J

2. A trolley is rolling across a parking lot of a supermarket. You

apply a constant force to the trolley as it

undergoes a displacement . Calculate

a. the work done on the trolley by the force F,

b. the angle between the force and the displacement of the

trolley.

ANS. : 150 J; 108

N j40i30 F

m j3.0i9.0 s


21

Exercise 5.1 :

3.

Figure 5.7 shows an overhead view of three horizontal forces

acting on a cargo that was initially stationary but that now

moves across a frictionless floor. The force magnitudes are

F1 = 3.00 N, F2 = 4.00 N and F3 = 10.0 N. Determine the total

work done on the cargo by the three forces during the first

4.00 m of displacement.

ANS. : 15.3 J

3F

1F

2F

y

x

35

50

Figure 5.7


22


(a) Define and use kinetic energy,

(b) Define and use potential energy:

i. gravitational potential energy,

ii. elastic potential energy for spring,

(c) State and use the principle of conservation of energy.

(d) Explain the work-energy theorem and use the related

equation.

Learning Outcome:

2

2

1mvK

mghU

2

2

1kxU

5.2 Energy And Conservation Of Energy


23

Energy is defined as the system’s ability to do work.

The S.I. unit for energy is same to the unit of work (joule, J).

The dimension of energy


Table 5.1 summarises some common types of energy.

22 TMLWorknergyE

Forms of

EnergyDescription

ChemicalEnergy released when chemical bonds between atoms

and molecules are broken.

Electrical Energy that is associated with the flow of electrical charge.

HeatEnergy that flows from one place to another as a result of

a temperature difference.

InternalTotal of kinetic and potential energy of atoms or molecules

within a body.


24

Forms of

EnergyDescription

Table 5.1

Nuclear Energy released by the splitting of heavy nuclei.

Mass

Energy released when there is a loss of small amount

of mass in a nuclear process. The amount of energy

can be calculated from Einstein’s mass-energy

equation, E = mc2

Radiant Heat Energy associated with infra-red radiation.

SoundEnergy transmitted through the propagation of a series

of compression and rarefaction in solid, liquid or gas.

Mechanicala. Kinetic

b. Gravitational

potentialc. Elastic

potential

Energy associated with the motion of a body.

Energy associated with the position of a body in a

gravitational field.

Energy stored in a compressed or stretched spring.


25

Conservation of energy

5.2.1 Kinetic energy, K is defined as the energy of a body due to its motion.

Equation :

Work-kinetic energy theorem

Consider a block with mass, m moving along the horizontal

surface (frictionless) under the action of a constant nett force,

Fnett undergoes a displacement, s in Figure 4.8.

2

2

1mvK

body a ofenergy kinetic:K

body a of speed : vbody a of mass : m

where

s

nettF

m

Figure 5.8

maFF nett (1)


26

By using an equation of linear motion:

By substituting equation (2) into (1), we arrive

Therefore

states “the work done by the nett force on a body equals the change in the body’s kinetic energy”.

as uv 222

s

uva

2

22 (2)

2

22

s

uvmFnett

ifnett KKmumvsF 22

2

1

2

1

KWnett


27

A stationary object of mass 3.0 kg is pulled upwards by a constant

force of magnitude 50 N. Determine the speed of the object when it

is travelled upwards through 4.0 m.

(Given g = 9.81 m s2)

Solution :

The nett force acting on the object is given by

By applying the work-kinetic energy theorem, thus

Example 5.4 :

0 m; 4.0 ;N 50; kg 3.0 usFm

F

s

gm

F

gm

9.813.050 mgFFnett

ifnett KKW

02

1 2 mvsFnett

23.02

14.020.6 v


28

A block of mass 2.00 kg slides 0.750 m down an inclined plane that

slopes downward at an angle of 36.9 below the horizontal. If the

block starts from rest, calculate its final speed. You can ignore the

friction. (Given g = 9.81 m s2)

Solution :

Example 5.5 :

s

36.9

0 m; 0.750 ; kg 2.00 usm

N

gm36.9

36.9sinmg36.9cosmg

a

x

y


29

Solution :

Since the motion of the block along the incline surface thus nett

force is given by

By using the work-kinetic energy theorem, thus

36.9sinmgFnett

0 m; 0.750 ; kg 2.00 usm

36.9sin9.812.00nettF

ifnett KKW

02

1 2 mvsFnett

22.002

10.75011.8 v


30

An object of mass 2.0 kg moves along the x-axis and is acted on

by a force F. Figure 5.9 shows how F varies with distance

travelled, s. The speed of the object at s = 0 is 10 m s1.

Determine

a. the speed of the object at s = 10 m,

b. the kinetic energy of the object at s = 6.0 m.

Example 5.6 :

10

5

064 10

(N)F

7 (m)s

Figure 5.9


31

Solution :

a.


1s m 10 kg; 2.0 um

m 10 tom 0 fromgraph under the area sFW

57106102

11046

2

1W

if KKW

22

2

1

2

1mumvW

22 102.02

12.0

2

132.5 v


32

Solution :

b.


m 6 tom 0 fromgraph under the area sFW

10462

1W

if KKW

2

2

1muKW f

2102.02

150 fK


33

Exercise 5.2 :


1. A bullet of mass 15 g moves horizontally at velocity of

250 m s1.It strikes a wooden block of mass 400 g placed at rest

on a floor. After striking the block, the bullet is embedded in the

block. The block then moves through 15 m and stops. Calculate

the coefficient of kinetic friction between the block and the floor.

ANS. : 0.278

2. A parcel is launched at an initial speed of 3.0 m s1 up a rough

plane inclined at an angle of 35 above the horizontal. The

coefficient of kinetic friction between the parcel and the plane is

0.30. Determine

a. the maximum distance travelled by the parcel up the plane,

b. the speed of the parcel when it slides back to the starting

point.

ANS. : 0.560 m; 1.90 m s1


34

5.2.2 Potential Energy

is defined as the energy stored in a body or system because

of its position, shape and state.

Gravitational potential energy, U

is defined as the energy stored in a body or system because

of its position.

Equation :

The gravitational potential energy depends only on the height

of the object above the surface of the Earth.

mghU

energy potential nalgravitatio : U

position initial thefrombody a ofheight : h

where

body a of mass : mgravity todueon accelerati : g


35

Work-gravitational potential energy theorem

Consider a book with mass, m is dropped from height, h1 to

height, h2 as shown in the Figure 5.10.

states “ the change in gravitational potential energy as

the negative of the work done by the gravitational force”.

1h

gm

gm

2h

s

Figure 5.10

21g hhmgmgsW

The work done by the gravitational force

(weight) is

fig UUmghmghW 21

UUUW ifg

UW

Therefore in general,


36

Negative sign in the equation indicates that

When the body moves down, h decreases, the

gravitational force does positive work because U <0.

When the body moves up, h increases, the work done

by gravitational force is negative because U >0.

For calculation, use

if UUUW

energy potential nalgravitatio final : fUwhere

force nalgravitatio aby done work : W

energy potential nalgravitatio initial : iU


37

In a smooth pulley system, a force F is required to bring an

object of mass 5.00 kg to the height of 20.0 m at a constant

speed of 3.00 m s1 as shown in Figure 5.11. Determine

a. the force, F

b. the work done by the force, F.

(Given g = 9.81 m s-2)

Example 5.7 :

Figure 5.11

F

m 20.0


38

Solution :

a. Since the object moves at the constant

speed, thus

b. From the equation of work,

1s m 3.00constant m; 20.0 kg; 5.00 vhsm

0nettF

mgF

F

s

gm

F

gm

Constant

speedθFsW cos 0θand

OR

θFsW cosmghUW

0θand


39

Elastic potential energy, Us

is defined as the energy stored in in elastic materials as the

result of their stretching or compressing.

Springs are a special instance of device which can store

elastic potential energy due to its compression or

stretching.

Hooke’s Law states “the restoring force, Fs of spring is

directly proportional to the amount of stretch or

compression (extension or elongation), x if the limit of

proportionality is not exceeded”

OR xFs

kxFs

spring of force restoring the: sF

)(n compressioor stretch ofamount the: if -xxx

constant forceor constant spring the:k

where


40

Negative sign in the equation indicates that the direction of Fs

is always opposite to the direction of the amount of stretch or

compression (extension), x.

Case 1:

The spring is hung vertically and its is stretched by a suspended

object with mass, m as shown in Figure 5.12.

The spring is in equilibrium, thus

Initial position

Final position

sF

gmW

x

Figure 5.12

mgWFs


41Figure 5.13

(Equilibrium position)

Case 2:

The spring is attached to an object and it is stretched and

compre5sed by a force, F as shown in Figure 5.13.

sF

F

0x

0x

sF

F

x

x

negative is sFpositive is x

positive is sFnegative is x

0 sF0x

The spring is in equilibrium,

hence

FFs


42

Caution:

For calculation, use :

Dimension of spring constant, k :

The unit of k is kg s2 or N m1

From the Hooke’s law (without “” sign), a restoring force, Fs

against extension of the spring, x graph is shown in Figure 5.14.

FkxFs

2s MTx

Fk

force applied : Fwhere

F

sF

0 x1x

graph under the area xFW s

1FxW2

1 11 xkxW

2

1

s21 UkxW

2

1

Figure 5.14


43

The equation of elastic potential energy, Us for compressing or

stretching a spring is

The work-elastic potential energy theorem,

Notes :

Work-energy theorem states the work done by the nett

force on a body equals the change in the body’s total

energy”

OR

xF2

1kx

2

1U s

2s

ifnett EEEW

sUW 2

i

2

fsisf kx2

1kx

2

1UUW OR


44

A force of magnitude 800 N caused an extension of 20 cm on a

spring. Determine the elastic potential energy of the spring when

a. the extension of the spring is 30 cm.

b. a mass of 60 kg is suspended vertically from the spring.

(Given g = 9.81 m s-2)

Solution :

From the Hooke’s law,

a. Given x=0.300 m,

Example 5.8 :

m 0.200 N; 800 xF

kxFFs

0.20800 k

2

2

1kxU s

23 0.3001042

1sU


45

Solution :

b. Given m=60 kg. When the spring in

equilibrium, thus

Therefore

0nettF

mgFs

mgkx

9.8160104 3 x

2

2

1kxU s

23 0.1471042

1sU

sF

gmW

x


46

5.2.3 Principle of conservation of energy

states “in an isolated (closed) system, the total energy of

that system is constant”.

According to the principle of conservation of energy, we get

The initial of total energy = the final of total energy

Conservation of mechanical energy

In an isolated system, the mechanical energy of a system is the

sum of its potential energy, U and the kinetic energy, K of the

objects are constant.

OR

fi EE

constant UKE

OR

ffii UKUK


47

Before After

cm 30

x

Figure 5.15

A 1.5 kg sphere is dropped from a height of

30 cm onto a spring of spring constant,

k = 2000 N m1 . After the block hits the

spring, the spring experiences maximum

compression, x as shown in Figure 5.15.

a. Describe the energy conversion

occurred after the sphere is

dropped onto the spring until the

spring experiences maximum

compression, x.

b. Calculate the speed of the sphere just

before strikes the spring.

c. Determine the maximum compression, x.

(Given g = 9.81 m s-2)

Example 5.9 :


48

The spring is not stretched

hence Us = 0. The sphere is

at height h1 above ground

with speed, v just before

strikes the spring. Therefore

The sphere is at height h2

above the ground after

compressing the spring by x.

The speed of the sphere at

this moment is zero. Hence

The spring is not stretched

hence Us = 0. The sphere is

at height h0 above ground

therefore U = mgh0 and it is

stationary hence K = 0.

(2)

v

1h

(3)

x

2h

cm 30h

0h

(1)

01 mghE 212 mv

2

1mghE 2

23 kx2

1mghE

Solution :

a.


49

Solution :

b. Applying the principle of conservation of energy involving the

situation (1) and (2),

210 mvhhmg

2

1

21 EE

210 mvmghmgh

2

1

0.309.812 v

1m N 2000 m; 0.30 kg; 1.5 khm

and 10 hhh

ghv 2


50

Solution :

c. Applying the principle of conservation of energy involving the


2221 kxmvhhmg

2

1

2

1

32 EE

22

21 kxmghmvmgh

2

1

2

1

1m N 2000 m; 0.30 kg; 1.5 khm

and 21 hhx

222000

2

12.431.5

2

19.811.5 xx


51

A bullet of mass, m1=5.00 g is fired into a wooden block of mass,

m2=1.00 kg suspended from some light wires as shown in Figure

5.16. The block, initially at rest. The bullet embeds in the block, and

together swing through a height, h=5.50 cm. Calculate

a. the initial speed of the bullet.

b. the amount of energy lost to the surrounding.

(Given g = 9.81 m s2)

Example 5.10 :

Figure 5.16

1m2m

21 mm

h1u


52

(1)

1m2m

1u02u

(3)

h

21 mm

012v

(2)

21 mm 12u

m 105.50 kg; 1.00kg; 105.00 23 hmm 21

32 EE

ghmmumm 211221 2

2

1

2105.509.8122 ghu12

UK

Solution :

a.

Applying the principle of conservation of energy involving the



53

Solution :

Applying the principle of conservation of linear momentum

involving the situation (1) and (2),

b. The energy lost to the surrounding, Q is given by

m 105.50 kg; 1.00kg; 105.00 23 hmm 21

21 pp

122111 ummum

1.041.00105.00105.00 33 1u

21 EEQ

22

1

2

11221

211 ummumQ

2323 1.041.00105.002

1209105.00

2

1 Q


54

Objects P and Q of masses 2.0 kg and 4.0 kg respectively are

connected by a light string and suspended as shown in Figure

5.17. Object Q is released from rest. Calculate the speed of Q at

the instant just before it strikes the floor.

(Given g = 9.81 m s2)

Example 5.11 :

Figure 5.17

P

Q

m 2

Smooth

pulley


55

Solution :

Applying the principle of conservation of mechanical energy,

0 m; 2 kg; 4.0kg; 2.0 QP uhmm

fi EE2

Q2

PPQ2

1

2

1vmvmghmghm

QPPQ KKUU

Initial

P

Q

m 2

Smooth

pulley

P

Qm 2

Smooth

pulley

v

v

Final

22 4.02

12.0

2

129.812.029.814.0 vv


56

Exercise 5.3 :


1. If it takes 4.00 J of work to stretch a spring 10.0 cm from its

initial length, determine the extra work required to stretch it an

additional 10.0 cm.

ANS. : 12.0 J

2. A book of mass 0.250 kg is placed on top of a light vertical

spring of force constant 5000 N m1 that is compressed by 10.0

cm. If the spring is released, calculate the height of the book rise

from its initial position.

ANS. : 10.2 m

3. A 60 kg bungee jumper jumps from a bridge. She is tied to a

bungee cord that is 12 m long when unstretched and falls a total

distance of 31 m. Calculate

a. the spring constant of the bungee cord.

b. the maximum acceleration experienced by the jumper.

ANS. : 100 N m1; 22 m s2


57

Exercise 5.3 :

4.

A 2.00 kg block is pushed against a light spring of the force

constant, k = 400 N m-1, compressing it x =0.220 m. When the

block is released, it moves along a frictionless horizontal surface

and then up a frictionless incline plane with slope =37.0 as

shown in Figure 5.18. Calculate

a. the speed of the block as it slides along the horizontal

surface after leaves the spring.

b. the distance travelled by the block up the incline plane before

it slides back down.

ANS. : 3.11 m s1; 0.81 m

Figure 5.18


58

Exercise 5.3 :

5.

A ball of mass 0.50 kg is at point A with initial speed, u =4 m s1

at a height of 10 m as shown in Figure 5.19 (Ignore the frictional

force). Determine

a. the total energy at point A,

b. the speed of the ball at point B where the height is 3 m,

c. the speed of the ball at point D,

d. the maximum height of point C so that the ball can pass over

it.

ANS. : 53.1 J; 12.4 m s1; 14.6 m s1; 10.8 m

u

m 10

A

B

C

DFigure 5.19


59


(a) Define and use power:

Average power,

Instantaneous Power,

(b) Derive and apply the formulae

(c) Define and use mechanical efficiency,

and the consequences of heat dissipation.

Learning Outcome:

5.3 Power and mechanical efficiency (1 hour)

dt

dWP

vFP

t

WPav

100%input

output

P

Pη


60

5.3 Power and mechanical efficiency

5.3.1 Power, P is defined as the rate at which work is done.

OR the rate at which energy is transferred.

If an amount of work, W is done in an amount of time t by a

force, the average power, Pav due to force during that time

interval is

The instantaneous power, P is defined as the instantaneous

rate of doing work, which can be write as

t

E

t

WPav

dt

dW

t

WP

0tlimit


61


The dimension of the power is

The S.I. unit of the power is kg m2 s3 or J s1 or watt (W).

Unit conversion of watt (W), horsepower (hp) and foot pounds

per second (ft. lb s1)

Consider an object that is moving at a constant velocity v along

a frictionless horizontal surface and is acted by a constant force,

F directed at angle above the horizontal as shown in Figure

5.20. The object undergoes a displacement of ds.

3222

TMLT

TML

t

WP

1s lb ft. 550 W746hp 1

Figure 5.20

F

sd


62

Therefore the instantaneous power, P is given by

OR

dt

dWP

vFP

dsθFdW cos

θFvP cos

and

dt

dsθFP

cos

dt

dsv and

where

vFθ

and between angle the:

force of magnitude:F velocityof magnitude : v


63

An elevator has a mass of 1.5 Mg and is carrying 15 passengers

through a height of 20 m from the ground. If the time taken to lift

the elevator to that height is 55 s. Calculate the average power

required by the motor if no energy is lost. (Use g = 9.81 m s2 and

the average mass per passenger is 55 kg)

Solution :

M = mass of the elevator + mass of the 15 passengers

M = 1500 + (5515) = 2325 kg

According to the definition of average power,

Example 5.12 :

t

MghPav

t

EPav

s 55 m; 20 Δth


64

An object of mass 2.0 kg moves at a constant speed of 5.0 m s1

up a plane inclined at 30 to the horizontal. The constant frictional

force acting on the object is 4.0 N. Determine

a. the rate of work done against the gravitational force,

b. the rate of work done against the frictional force,

c. the power supplied to the object. (Given g = 9.81 m s2 )

Solution :

Example 5.13 :

N 4.0 constant;s m 5.0 kg; 2.0 1 fvm

30

f

N

gmW

30cosmg

30sinmg

v

30x

y

s


65

Solution :

a. the rate of work done against the gravitational force is given by

t

θsmg

t

Wg cos30sin


180θand

t

smg

t

Wg 30sin

t

sv and

vmgt

Wg 30sin

5.030sin9.812.0

t

Wg

OR θvFt

Wg

gcos

180cos30sin vmgt

Wg


66

Solution :

b. The rate of work done against the frictional force is

c. The power supplied to the object, Psupplied

= the power lost against gravitational and frictional forces, Plost


180θandθfvt

W fcos

t

W

t

WP

fg

supplied


67

5.3.2 Mechanical efficiency, Efficiency is a measure of the performance of a machines,

engine and etc...

The efficiency of a machine is defined as the ratio of the useful

(output) work done to the energy input.

is a dimensionless quantity (no unit).

Equations:

100% in

out

E

Wη

OR

100% in

out

P

Pη

where system by the producedpower :outP

system a tosuppliedpower : inP


68

Notes :

In practice, Pout< Pin hence < 100%.

The system loses energy to its surrounding because it may

have encountered resistances such as surface friction or

air resistance.

The energy which is dissipated to the surroundings, may

be in the form of heat or sound.

A 1.0 kW motor is used to lift an object of mass 10 kg vertically

upwards at a constant speed. The efficiency of the motor is 75 %.

Determine

a. the rate of heat dissipated to the surrounding.

b. the vertical distance travelled by the object in 5.0 s.

(Given g = 9.81 m s2 )

Example 5.14 :


69

Solution :

a. The output power of the motor is given by

Therefore the rate of heat dissipated to the surrounding is

b.

Since the speed is constant hence the vertical distance in 5.0 s

is

W1000 75%; kg; 10.0 inPηm

%100in

out

P

Pη

1001000

75 outP

7501000dissipatedheat of Rate outin PP

θFvPout cos 0θwhere and mgF 0cosmgvPout

t

hv


70

Exercise 5.4 :


1. A person of mass 50 kg runs 200 m up a straight road inclined

at an angle of 20 in 50 s. Neglect friction and air resistance.

Determine

a. the work done,

b. the average power of the person.

ANS. : 3.36104 J; 672 W

2. Electrical power of 2.0 kW is delivered to a motor, which has an

efficiency of 85 %. The motor is used to lift a block of mass

80 kg. Calculate

a. the power produced by the motor.

b. the constant speed at which the block being lifted vertically

upwards by the force produced by the motor.

(neglect air resistance)

ANS. : 1.7 kW; 2.17 m s1


71

Exercise 5.4 :

3.

A car of mass 1500 kg moves at a constant speed v up a road

with an inclination of 1 in 10 as shown in Figure 5.21. All

resistances against the motion of the car can be neglected. If

the engine car supplies a power of 12.5 kW, calculate the

speed v.

ANS. : 8.50 m s1

Figure 5.21

10 1

PHYSICS CHAPTER 6

1

CHAPTER 6:

Circular motion

(3 Hours)

PHYSICS CHAPTER 6

2


Describe graphically the uniform circular motion.

In terms of velocity with constant magnitude (only the

direction of the velocity changes).

Learning Outcome:

6.1 Uniform circular motion (1 hour)

PHYSICS CHAPTER 6

3

6.1 Uniform circular motion is defined as a motion in a circle (circular arc) at a constant

speed.

Consider an object which does move with uniform circular

motion as shown in Figure 6.1.

Figure 6.1

r

θ

O

s

The length of a circular arc, s is given

by

rθs

pathcircular theof radius : r

subtends arc which theangle :θ where

radianin circle theof centre theto

PHYSICS CHAPTER 6

4

It is directed tangentially to the circular path and always

perpendicular to the radius of the circular path as shown in

Figure 6.2.

In uniform circular motion, the magnitude of the linear velocity

(speed) of an object is constant but the direction is

continually changing.

The unit of the tangential (linear) velocity is m s1.

6.1.1 Linear (tangential) velocity ,

r

O

v

r

v

r

vFigure 6.2

v

PHYSICS CHAPTER 6

5

The linear velocity, v is difficult to measure but we can measure

the period, T of an object in circular motion.

Period, T

is defined as the time taken for one complete revolution (cycle/rotation).

The unit of the period is second (s).

Frequency, f

is defined as the number of revolutions (cycles/rotations) completed in one second.

The unit of the frequency is hertz (Hz) or s1.

Equation :

Let the object makes one complete revolution in circular motion, thus

the distance travelled is (circumference of the circle),

the time interval is one period, T.

Tf

1

r2

PHYSICS CHAPTER 6

6

From the definition of speed,

If therefore

Note:

The unit of angular velocity (angular frequency) is rad s1

(radian per second).

Unit conversion of angle, :

interval time

distance of changev

T

rv

2OR rfv 2

fT

ω

22

rωv

pathcircular theof radius : rfrequency)(angular locity angular ve :ω

where

360rad 2

180rad

PHYSICS CHAPTER 6

7


Define and use centripetal acceleration and use

centripetal acceleration,

Define and solve problem on centripetal force,

Learning Outcome:

6.2 Centripetal force (2 hours)

r

va

2

c

r

mvF

2

c

PHYSICS CHAPTER 6

8

Figure 6.3 shows a particle moving with constant speed in a

circular path of radius, r with centre at O. The particle moves

from A to B in a time, t.

6.2.1 Centripetal (radial) acceleration,rc aa

or

Figure 6.3

1v

2v

The arc length AB is given by

The velocities of the particle at A

and B are v1 and v2 respectively

where

rΔΔs

r

ΔsΔ

vvv 21

(1)

PHYSICS CHAPTER 6

9

Let PQ and PR represent the velocity vectors v1 and v2respectively, as shown in Figure 6.4.

Then QR represent the change in velocity vector v of the particle in time interval t. Since the angle between PQ and PR is small hence

By equating (1) and (2) then

12 vvvΔ

2v

1v

P Q

RFigure 6.4

PQQR vΔΔv

v

ΔvΔ (2)

v

Δv

r

Δs

PHYSICS CHAPTER 6

10

Dividing by time, t, thus

Δt

Δv

v

1

Δt

Δs

r

1

v

a

r

v

r

va

2

c

pathcircular of radius : r

onaccelerati lcentripeta : cawhere

velocity gential)linear(tan : v

OR vra 2

c

frequency)(angular locity angular ve : ω

PHYSICS CHAPTER 6

11

ca

ca

ca

caca

ca

Figure 6.5

The centripetal acceleration is defined as the acceleration of

an object moving in circular path whose direction is

towards the centre of the circular path and whose

magnitude is equal to the square of the speed divided by

the radius.

The direction of centripetal (radial) acceleration is always

directed toward the centre of the circle and perpendicular to

the linear (tangential) velocity as shown in Figure 6.5.

PHYSICS CHAPTER 6

12

For uniform circular motion, the magnitude of the centripetal

acceleration always constant but its direction continuously

changes as the object moves around the circular path.

Because of

therefore we can obtain the alternative expression of centripetal

acceleration is

2

2

cT

ra

4

T

rv

2

r

a

2

Tr

c

2

PHYSICS CHAPTER 6

13

A motorbike moving at a constant speed 20.0 m s1 in a circular

track of radius 25.0 m. Calculate

a. the centripetal acceleration of the motorbike,

b. the time taken for the motorbike to complete one revolution.

Solution :

a. From the definition of the centripetal acceleration, thus

b. From the alternate formula of the centripetal acceleration, hence

Example 6.1 :

m 25.0 ;s m 20.0 1 rv

r

va

2

c

2

2

cT

ra

4

25.0

20.02

ca

2

25.0416.0

T

2

T

rv

2OR

PHYSICS CHAPTER 6

14

A car initially travelling eastward turns north by travelling in a

circular path at uniform speed as shown in Figure 6.6. The length

of the arc ABC is 235 m and the car completes the turn in 36.0 s.

Determine

a. the acceleration when the car is at B located at an angle of

35.0,

b. the car’s speed,

c. its average acceleration during the 36.0 s interval.

Example 6.2 :

Figure 6.6

PHYSICS CHAPTER 6

15

Solution :

a. The period of the car is given by

The radius of the circular path is

Therefore the magnitude of the centripetal acceleration is

s 36.0 m, 235 tsABC

36.044 tT

s 144T

rθsABC

2

π 235 r

2

2

cT

ra

4 2

2

144

1504πca

PHYSICS CHAPTER 6

16

Solution :

b. From the definition of the speed, thus

c. 1st method :

By using the triangle method for vector addition, thus the change

in the velocity is given by

s 36.0 m, 235 tsABC

t

s

t

sv ABC

36.0

235v

Cv

Av

AC vvv

2A

2

C vvv

226.536.53 v

θ

θ

PHYSICS CHAPTER 6

17

Solution :

Therefore the magnitude of the average acceleration is

and its direction :

s 36.0 m, 235 tsABC

t

vaav

36.0

9.24ava

A

C1

v

vθ tan

6.53

6.53tan 1θ

PHYSICS CHAPTER 6

18

Solution :

c. 2nd method :

x-component :

y-component :

s 36.0 m, 235 tsABC

t

vv

t

va AxCxx

xav

36.0

6.530

xava

t

vv

t

va

AyCyy

yav

36.0

06.53

yava

PHYSICS CHAPTER 6

19

Solution :

Therefore the magnitude of the average acceleration is

and

s 36.0 m, 235 tsABC

2yav

2

xavav aaa

220.1810.181 ava

xav

yav1

a

aθ tan

0.181

0.181tan 1θ

PHYSICS CHAPTER 6

20

A boy whirls a marble in a horizontal circle of radius 2.00 m and at

height 1.65 m above the ground. The string breaks and the marble

flies off horizontally and strikes the ground after traveling a

horizontal distance of 13.0 m. Calculate

a. the speed of the marble in the circular path,

b. the centripetal acceleration of the marble while in the circular

motion.

(Given g = 9.81 m s-2)

Solution :

Example 6.3 :

1.65 m

Before

13.0 m

u

After

u

r =2.00 m

1.65 m

PHYSICS CHAPTER 6

21

Solution :

a. From the diagram :

The time taken for the marble to strike the ground is

The initial speed of the marble after the string breaks is equal to

the tangential speed of the marble in the horizontal circle.

Therefore

0 ; yx uuu

29.812

101.65 t

2

2

1gttus yy

m 1.65 ; m 13.0 yx ss

0.58013.0 u

tus xx

PHYSICS CHAPTER 6

22

Solution :

b. From the definition of the centripetal acceleration, thus

r

u

r

vac

22

2.00

22.42

ca

PHYSICS CHAPTER 6

25

6.3 Centripetal force6.3.1 Equation of centripetal force

From Newton’s second law of motion, a force must be

associated with the centripetal acceleration. This force is

known as the centripetal force and is given by

amFF nett

cc amF

mvmrr

mvF 2

2

c

caa

vrr

va 2

2

c

where cFF

and

and

force lcentripeta :cFwhere

PHYSICS CHAPTER 6

26

ca

cF

cF

cF

ca

ca

v

v

v

The centripetal force is defined as a force acting on a body

causing it to move in a circular path of magnitude

and its always directed towards the centre of the circular

path.

Its direction is in the same direction of the centripetal

acceleration as shown in Figure 6.8.

Figure 6.8

r

mvF

2

c

PHYSICS CHAPTER 6

27

PHYSICS CHAPTER 6

28

cF

ca v

cF

cF

ca

ca

v

v

v

v

0Fc

0Fc

0ac

0ac

If the centripetal force suddenly stops to act on a body in the

circular motion, the body flies off in a straight line with the

constant tangential (linear) speed as show in Figure 6.9.

Note :

In uniform circular motion, the nett force on the system is

centripetal force.

The work done by the centripetal force is zero but the

kinetic energy of the body is not zero and given by

Figure 6.9

222 mr2

1mv

2

1K

Simulation 6.1

AF_0602.html

PHYSICS CHAPTER 6

29

As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion.

As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion.

As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion

PHYSICS CHAPTER 6

30

Without a centripetal force, an

object in motion continues along a

straight-line path.

With a centripetal force, an object in

motion will be accelerated and change its

direction.

PHYSICS CHAPTER 6

31

Note that the centripetal force is proportional to the square of the velocity, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle. If the centripetal force must be provided by friction alone on a curve, an increase in speed could lead to an unexpected skid if friction is insufficient.

PHYSICS CHAPTER 6

32

PHYSICS CHAPTER 6

33

Conical Pendulum

Example 6.4 :

Figure 6.10 shows a conical pendulum

with a bob of mass 80.0 kg on a 10.0 m

long string making an angle of 5.00 to the

vertical.

a. Sketch a free body diagram of the bob.

b. Determine

i. the tension in the string,

ii. the speed and the period of the bob,

iii. the radial acceleration of the bob.

(Given g =9.81 m s2)

6.3.2 Examples of uniform circular motion

Figure 6.10

PHYSICS CHAPTER 6

34

Solution :

a. The free body diagram of the bob :

b. i. From the diagram,

5.00 ;m 10.0 ;kg 80.0 θlm

gm

θT

θT cos

θT sin

0 yF

mgθT cos

ca

PHYSICS CHAPTER 6

35

The centripetal force is contributed

by the horizontal component of the

tension.

Solution :

b. ii.

5.00 ;m 10.0 ;kg 80.0 θlm

cx FF

r

mvθT

2

sin

θl

mvθT

2

sinsin

r

ll

rθ sin

θlr sin

m

θTlv

2sin

80.0

5.00sin10.07882

v

PHYSICS CHAPTER 6

36

Solution :

b. ii. and the period of the bob is given by

iii. From the definition of the radial acceleration, hence

5.00 ;m 10.0 ;kg 80.0 θlm

T

rv

2

T

θlv

sin2

T

5.00sin10.020.865

θl

va

2

rsin

5.00sin10.0

0.8652

ra

r

va

2

r

PHYSICS CHAPTER 6

37

Centre of

circle

Motion rounds a curve on a flat (unbanked) track (for car,

motorcycle, bicycle, etc…)

Example 6.5 :

A car of mass 2000 kg rounds a circular turn of radius 20 m. The

road is flat and the coefficient of friction between tires and the road

is 0.70.

a. Sketch a free body diagram of the car.

b. Determine the maximum car’s speed without skidding.

(Given g = 9.81 m s-2)

Solution :

a. The free body diagram of the car :

gm

N

f

0.70 ;m 20 ;kg 2000 μrm

ca

Picture 6.1

curveflattrack.jpg

PHYSICS CHAPTER 6

38

Solution :

b. From the diagram in (a),

y-component :

x-component : The centripetal force is provided by the frictional

force between the wheel (4 tyres) and the road.

Therefore

0.70 ;m 20 ;kg 2000 μrm

0yF mgN

r

mvF

2

x

r

mvf

2

μrgv r

mvμmg

2

PHYSICS CHAPTER 6

39

T

gm

r

ca

Motion in a horizontal circle

Example 6.6 :

A ball of mass 150 g is attached to one end of a string 1.10 m long.

The ball makes 2.00 revolution per second in a horizontal circle.

a. Sketch the free body diagram for the ball.

b. Determine

i. the centripetal acceleration of the ball,

ii. the magnitude of the tension in the string.

Solution :

a. The free body diagram for the ball :

Hz 2.00 ;m 1.10 ;kg 0.150 frlm

PHYSICS CHAPTER 6

40

Solution :

b. i. The linear speed of the ball is given by

Therefore the centripetal acceleration is

ii. From the diagram in (a), the centripetal force enables the ball

to move in a circle is provided by the tension in the string.

Hence

Hz 2.00 ;m 1.10 ;kg 0.150 frlm

rfT

rv

2

2

2.001.102v

r

va

2

c 1.10

13.82

ca

ccx maFF cmaT

PHYSICS CHAPTER 6

41

Motion in a vertical circle

Example 6.7 :

A small remote control car with mass 1.20 kg moves at a constant

speed of v = 15.0 m s1 in a vertical circle track of radius 3.00 m as

shown in Figure 6.12. Determine the magnitude of the reaction

force exerted on the car by the track at

a. point A,

b. point B.

(Given g = 9.81 m s2)

m 3.00

v

v

A

B

Figure 6.12

PHYSICS CHAPTER 6

42

Solution :

a. The free body diagram of the car at point A :

1s m 15.0 ;m 3.00 ;kg 1.20 vrm

gm

AN

ca

r

mvF

2

r

mvmgN

2

A

3.00

15.01.209.811.20

2

AN

PHYSICS CHAPTER 6

43

Solution :

b. The free body diagram of the car at point B :

1s m 15.0 ;m 3.00 ;kg 1.20 vrm

BN

ca

r

mvF

2

r

mvmgN

2

B

3.00

15.01.209.811.20

2

BN

gm

PHYSICS CHAPTER 6

44

A rider on a Ferris wheel moves in a vertical circle of radius,

r = 8 m at constant speed, v as shown in Figure 6.13. If the time

taken to makes one rotation is 10 s and the mass of the rider is

60 kg, Calculate the normal force exerted on the rider

a. at the top of the circle,

b. at the bottom of the circle.

(Given g = 9.81 m s-2)

Example 6.8 :

v

v

Figure 6.13

PHYSICS CHAPTER 6

45

Solution :

a. The constant speed of the rider is

The free body diagram of the rider at the top of the circle :

s 10 ;m 8 ;kg 60 Trm

T

rv

2 10

82πv

ca

gm

tN

r

mvF

2

r

mvNmg

2

t

8

5.03609.8160

2

tN

PHYSICS CHAPTER 6

46

Solution :

b. The free body diagram of the rider at the bottom of the circle :

s 10 ;m 8 ;kg 60 Trm

ca

gm

r

mvF

2

r

mvmgN

2

b

8

5.03609.8160

2

bNbN

PHYSICS CHAPTER 6

47

A sphere of mass 5.0 kg is tied to an inelastic string. It moves in a

vertical circle of radius 55 cm at a constant speed of 3.0 m s1 as

shown in Figure 6.14. By the aid of the free body diagram,

determine the tension in the string at points A, D and E.

(Given g = 9.81 m s-2)

Example 6.9 :

Figure 6.14

A

D

E 3.0 m s1

3.0 m s1

3.0 m s1

PHYSICS CHAPTER 6

48

Solution :

The free body diagram of the sphere at :

Point A,

Point D,

1s m 0.3 ;m 55.0 ;kg 0.5 vrm

ca

r

mvF

2

r

mvmgTA

2

0.55

3.05.09.815.0

2

AT

A

gmAT

ca

D

gm

DT

r

mvTD

2

0.55

3.05.02

DT

PHYSICS CHAPTER 6

49

Solution :

The free body diagram of the sphere at :

Point E,

Caution :

For vertical uniform circular motion only,

the normal force or tension is maximum at the bottom of

the circle.

the normal force or tension is minimum at the top of the

circle.

1s m 0.3 ;m 55.0 ;kg 0.5 vrm

ca r

mvmgTE

2

0.55

3.05.09.815.0

2

ET

E

gm

ET

PHYSICS CHAPTER 6

50

Exercise 6.2 :


1. A cyclist goes around a curve of 50 m radius at a speed of

15 m s1. The road is banked at an angle to the horizontal and

the cyclist travels at the right angle with the surface of the road.

The mass of the bicycle and the cyclist together equals 95 kg.

Calculate

a. the magnitude of the centripetal acceleration of the cyclist,

b. the magnitude of the normal force which the road exerts on

the bicycle and the cyclist,

c. the angle .

ANS. : 4.5 m s2; 1.02 kN; 24.6

PHYSICS CHAPTER 6

51

Exercise 6.2 :

2. A ball of mass 0.35 kg is attached to the end of a horizontal

cord and is rotated in a circle of radius 1.0 m on a frictionless

horizontal surface. If the cord will break when the tension in it

exceeds 80 N, determine

a. the maximum speed of the ball,

b. the minimum period of the ball.

ANS. : 15.1 m s1; 0.416 s

Figure 6.14

3. A small mass, m is set on the surface

of a sphere as shown in Figure 6.14.

If the coefficient of static friction is s

= 0.60, calculate the angle would

the mass start sliding.

ANS. : 31

m

θ

O

PHYSICS CHAPTER 6

52

Exercise 6.2 :

4. A ball of mass 1.34 kg is connected

by means of two massless string to

a vertical rotating rod as shown in

Figure 6.15. The strings are tied to

the rod and are taut. The tension in

the upper string is 35 N.

a. Sketch a free body diagram for

the ball.

b. Calculate

i. the magnitude of the tension

in the lower string,

ii. the nett force on the ball,

iii. the speed of the ball.

ANS. : 8.74 N; 37.9 N (radially

inward); 6.45 m s1

Figure 6.15

PHYSICS CHAPTER 6

53

THE END…


Gravitation

PHYSICS CHAPTER 7

1

CHAPTER 7:

Gravitation

(2 Hours)

ww

w.k

ms.m

atr

ik.e

du

.my/p

hysic

s

PHYSICS CHAPTER 7

In this chapter, we learns about

7.1 Gravitational force and field strength

7.2 Gravitational potential

7.3 Satellite motion in a circular orbit

2

PHYSICS CHAPTER 7

7.1 Gravitational Force and Field

Strength

7.1.1 Newton’s law of gravitation

7.1.2 Gravitational Field

7.1.3 Gravitational force and field strength

3

PHYSICS CHAPTER 7

4


State and use the Newton’s law of gravitation,

Learning Outcome:

7.1 Newton’s law of gravitation (1 hour)

2

21

r

mmGFg

ww

w.k

ms.m

atr

ik.e

du

.my/p

hysic

s

PHYSICS CHAPTER 7

5

7.1.1 Newton’s law of gravitation

States that a magnitude of an attractive force between two

point masses is directly proportional to the product of their

masses and inversely proportional to the square of the

distance between them.

OR mathematically,

2

1

rFg 21mmFg and

2

21

r

mmFg 2

21

r

mmGFg

force nalGravitatio:gF2 and 1 particle of masses:, 21 mm

2 and 1 particlebetween distance: r2211 kg m N x106.67 Constant nalgravitatio Universal: G

where

PHYSICS CHAPTER 7

6

The statement can also be shown by using the Figure 7.1.

where

2

211221

r

mmGFFF g

1m 2m

r12F

Figure 7.1

21F

2 particleon 1 particleby force nalGravitatio:12F

1 particleon 2 particleby force nalGravitatio:21F

Simulation 7.1

AF_1301.html

PHYSICS CHAPTER 7

7

Figures 7.2a and 7.2b show the gravitational force, Fg varies

with the distance, r.

Notes:

Every spherical object with constant density can be reduced to a point mass at the centre of the sphere.

The gravitational forces always attractive in nature and the forces always act along the line joining the two point masses.

gF

r0

gF

2

1

r0

21mGmgradient

Figure 7.2a Figure 7.2b

PHYSICS CHAPTER 7

8

A spaceship of mass 9000 kg travels from the Earth to the Moon

along a line that passes through the Earth’s centre and the Moon’s

centre. The average distance separating Earth and the Moon is

384,000 km. Determine the distance of the spaceship from the

Earth at which the gravitational force due to the Earth twice the

magnitude of the gravitational force due to the Moon.

(Given the mass of the Earth, mE=6.001024 kg, the mass of the

Moon, mM=7.351022 kg and the universal gravitational constant,

G=6.671011 N m2 kg2)

Example 7.1 :

PHYSICS CHAPTER 7

9

Solution :

Given

kg; 107.35 kg; 106.00 22M

24E mm

m 103.84 kg; 0900 8EMs rm

EmMmsm

x

EMr

xr EM

EsF

MsF

MsEs F2F

2EM

sM

2

sE 2xr

mGm

x

mGm

M

E

2

EM

2

2m

m

xr

x

22

24

28

2

107.352

106.00

103.84

x

x

m 103.32 8x

PHYSICS CHAPTER 7

10

Two spheres of masses 3.2 kg and 2.5 kg respectively are fixed at

points A and B as shown in Figure 7.3. If a 50 g sphere is placed

at point C, determine

a. the resultant force acting on it.

b. the magnitude of the sphere’s acceleration.

(Given G = 6.671011 N m2 kg2)

Example 7.2 :

Figure 7.3

A B

C

cm 8kg 3.2 kg 2.5

g 50

cm 6

PHYSICS CHAPTER 7

11

Solution :

a.

The magnitude of the forces on mC,

22

311

2

AC

CAA

1010

10503.2106.67

r

mGmF

N 101.07 9A

F

kg 1050kg; .52 kg; 3.2 3CBA

mmmm 1010m; 106 2

AC2

BC rr

0.6sin θ0.8cos θ

A B

C

m 108 2-

m 10 6 2

θ

θ

m 10 10 2

AF

BF

PHYSICS CHAPTER 7

12

Solution :

22

311

2

BC

CBB

106

10502.5106.67

r

mGmF

N 10.322 9B

F


AF

θF cosA θF sinA

0.8101.07 910108.56

0.6101.07 910106.42

BF

BF0

9102.32

kg 1050kg; .52 kg; 3.2 3CBA

mmmm 1010m; 106 2

AC2

BC rr

PHYSICS CHAPTER 7

13

Solution :

The magnitude of the nett force is

and its direction is

N 108.56 10 xF

22

yx FFF

N 10.96210.322106.42 9910 yF

29210 10.962108.56

N 10.083 9F

10

911

10.568

10.962tantan

x

y

F

Fθ

.973θ (254 from +x axis anticlockwise)

PHYSICS CHAPTER 7

14

Solution :

b. By using the Newton’s second law of motion, thus

and the direction of the acceleration in the same direction of the

nett force on the mC i.e. 254 from +x axis anticlockwise.

amF C

a39 1050103.08

28 s m 10.166 a

3

9

1050

103.08

a

PHYSICS CHAPTER 7

15

is defined as a region of space surrounding a body that has

the property of mass where the attractive force is

experienced if a test mass placed in the region.

Field lines are used to show gravitational field around an object

with mass.

For spherical objects (such as the Earth) the field is radial as

shown in Figure 7.4.

7.1.2 Gravitational Field

M

Figure 7.4

PHYSICS CHAPTER 7

16

The gravitational field in small region near the Earth’s surfaceare uniform and can be drawn parallel to each other as shown in Figure 7.5.

The field lines indicate two things:

The arrows – the direction of the field

The spacing – the strength of the field

Figure 7.5

The gravitational field is a conservative field in which the work done in moving a body from one point to another is independent of the path taken.

Note:

New

PHYSICS CHAPTER 7

17

Exercise 7.1 :

Given G = 6.671011 N m2 kg2

1. Four identical masses of 800 kg each are placed at the corners

of a square whose side length is 10.0 cm. Determine the nett

gravitational force on one of the masses, due to the other three.

ANS. : 8.2103 N; 45

2. Three 5.0 kg spheres are located in the xy plane as shown in

Figure 7.6.Calculate the magnitude

of the nett gravitational force

on the sphere at the origin due to

the other two spheres.

ANS. : 2.1108 N

Figure 7.6

PHYSICS CHAPTER 7

18

Exercise 7.1 :

3.

In Figure 8.7, four spheres form the corners of a square

whose side is 2.0 cm long. Calculate the magnitude and

direction of the nett gravitational force on a central sphere with

mass of m5 = 250 kg.

ANS. : 1.68102 N; 45

Figure 7.7

PHYSICS CHAPTER 7

19


Define gravitational field strength as gravitational force per unit mass,

Derive and use the equation for gravitational field strength.

Sketch a graph of ag against r and explain the change in agwith altitude and depth from the surface of the earth.

Learning Outcome:

7.1.3 Gravitational force and field strength

m

Fag

g

2r

MGa g

ww

w.k

ms.m

atr

ik.e

du

.my/p

hysic

s

PHYSICS CHAPTER 7

20

7.1.3 Gravitational field strength, is defined as the gravitational force per unit mass

of a body (test mass) placed at a point.

OR

It is a vector quantity.

The S.I. unit of the gravitational field strength is N kg1 or m s2.

ga

m

Fa

g

g

where

strength field nalGravitatio:ga

force nalGravitatio:gF

mass)(test body a of mass:m

PHYSICS CHAPTER 7

21

It is also known as gravitational acceleration (the free-fall acceleration).

Its direction is in the same direction of the gravitational force.

Another formula for the gravitational field strength at a point is given by

m

Fa

g

g and2g

r

GMmF

2gr

GMa

masspoint and massst between te distance : r

2g

1

r

GMm

ma

where

masspoint theof mass :M

PHYSICS CHAPTER 7

22

Figure 7.8 shows the direction of the gravitational field strength

on a point S at distance r from the centre of the planet.

2r

GMag

r

M

Figure 7.8

PHYSICS CHAPTER 7

23

The gravitational field in the small region near the Earth’s

surface( r R) are uniform where its strength is 9.81 m s2 and its direction can be shown by using the Figure 7.9.

Figure 7.9

2R

GMgag

Earth theof radius :Rwhere2s m 9.81onaccelerati nalgravitatio : g

PHYSICS CHAPTER 7

24

Determine the Earth’s gravitational field strength

a. on the surface.

b. at an altitude of 350 km.

(Given G = 6.671011 N m2 kg2, mass of the Earth,

M = 6.00 1024 kg and radius of the Earth, R = 6.40 106 m)

Solution :

a.

Example 7.3 :

RM

gaRr g m; 1040.6 6

26

2411

21040.6

1000.61067.6

R

GMg

The gravitational field strength is

1kg N 77.9 g OR2s m 77.9

rg

(Towards the centre of the Earth)

PHYSICS CHAPTER 7

25

Solution :

b.

2gr

GMa

26

2411

1075.6

106.001067.6

2g s m 78.8 a


R M

hRr 36 103501040.6

m 1075.6 6r

ga

h

r

The gravitational field strength is given by

PHYSICS CHAPTER 7

26

The gravitational field strength on the Earth’s surface is 9.81 N kg1.

Calculate

a. the gravitational field strength at a point C at distance 1.5R from

the Earth’s surface where R is the radius of the Earth.

b. the weight of a rock of mass 2.5 kg at point C.

Solution :

a. The gravitational field strength on the Earth’s surface is

The distance of point C from the Earth’s centre is

Example 7.4 :

1kg N 81.9 g

1

2kg N 81.9

R

GMg

RRRr 5.25.1

PHYSICS CHAPTER 7

27

Solution :

a. Thus the gravitational field strength at point C is given by

b. Given

The weight of the rock is

N 93.3W

2

Cr

GMag 25.2 R

GMag

225.6

1

R

GM

gmaW kg 5.2m

57.15.2

1kg N 57.181.925.6

1 ga



PHYSICS CHAPTER 7

28

Figure 8.10 shows an object A at a distance of 5 km from the object

B. The mass A is four times of the mass B. Determine the location

of a point on the line joining both objects from B at which the nett

gravitational field strength is zero.

Example 7.5 :

A

B

km 5

Figure 7.10

PHYSICS CHAPTER 7

29

Solution :

At point C,

BA3 4 m; 105 MMr

0nett

ga

2

B

23

B

105

4

x

M

x

M

m 10.671 3x

r

ABC

xr x

2ga

1ga

21 gg aa

2

B

2

A

x

GM

xr

GM

PHYSICS CHAPTER 7

30

Outside the Earth ( r > R)

Figure 8.11 shows a test mass which is outside the Earth and at

a distance r from the centre.

The gravitational field strength outside the Earth is

7.1.4 Variation of gravitational field strength on the

distance from the centre of the Earth

R

rM

Figure 8.11

2gr

GMa

2g

1

ra

PHYSICS CHAPTER 7

31

On the Earth ( r = R)

Figure 7.12 shows a test mass on the Earth’s surface.

The gravitational field strength on the Earth’s surface is

R

rM

Figure 7.12

2

2g s m 81.9 gR

GMa

PHYSICS CHAPTER 7

32

R

r

M

'M

Inside the Earth ( r < R)

Figure 7.13 shows a test mass which is inside the Earth and at

distance r from the centre.

The gravitational field strength inside the Earth is given by

Figure 7.13

2g

'

r

GMa

where

portion spherical of mass the: 'M radius, ofEarth theof r

PHYSICS CHAPTER 7

33

By assuming the Earth is a solid sphere and constant

density, hence

Therefore the gravitational field strength inside the Earth is

V

V

M

M

''

3

3

3

34

3

34'

R

r

R

r

M

M

MR

rM

3

3

'

2

3

3

gr

MR

rG

a

rR

GMa

3g ra g

PHYSICS CHAPTER 7

34

The variation of gravitational field strength, ag as a function of

distance from the centre of the Earth, r is shown in Figure 7.14.

Figure 7.14

R

ga

r0 R

gR

GMa

2g

ra g

2g

1

ra

PHYSICS CHAPTER 7

35


Define gravitational potential in a gravitational field.

Derive and use the formulae,

Sketch the variation of gravitational potential, V with distance, r from the centre of the earth.

Learning Outcome:

7.2 Gravitational potential (½ hour)

r

GMV

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PHYSICS CHAPTER 7

36

7.2 Gravitational potential

7.2.1 Work done by the external force

Consider an external force, F

is required to bring a test

mass, m from r1 to r2 ,

as shown in Figure 7.18.

At the distance r2 from the

centre of the Earth,

The work done by the

external force through

the small displacement

dr is

m

M

1r2r

F

gF

dr

Figure 7.18

gFF

0cosFdrdW

drFdW g

PHYSICS CHAPTER 7

37

Therefore the work done by the external force to bring test

mass, m from r1 to r2 is

2

1

r

rgdrFdW

2

12

r

rdr

r

GMmW

2r

GMmFg and

2

1

1r

rrGMmW

2

12

1r

rdr

rGMmW

21

1121 rr

GMmW rr

where

distance final: 2rdistance initial : 1r

PHYSICS CHAPTER 7

38

at a point is defined as the work done by an external force in bringing a test mass from infinity to a point per unit the test mass.

OR mathematically, V is written as:

It is a scalar quantity.

Its dimension is given by

7.2.2 Gravitational potential, V

m

WV

where

mass test theof mass :m point aat potential nalgravitatio :V

mass test a bringingin done work :W point a oinfinity t from

m

WV

M

TML 22

V

22TL V

PHYSICS CHAPTER 7

39

The S.I unit for gravitational potential is m2 s2 or J kg1.

Another formula for the gravitational potential at a point is given by

21

11

rrm

GMmV

m

WV and

21

11

rrGMmW

where 1r and rr 2

rm

GMmV

11

r

GMV

where

point ebetween th distance : rM mass,point theand

PHYSICS CHAPTER 7

40

The gravitational potential difference between point A and B

(VAB) in the Earth’s gravitational field is defined as the work

done in bringing a test mass from point B to point A per

unit the test mass.

OR mathematically, VAB is written as:

BABA

AB -VVm

WV

where

A.point toBpoint from

mass test thebringingin done work :BAW

Apoint at potential nalgravitatio : AV

Bpoint at potential nalgravitatio : BV

PHYSICS CHAPTER 7

41

Figure 7.19 shows two points A and B at a distance rA and rB

from the centre of the Earth respectively in the Earth’s

gravitational field.

M

A

BrA

rB

Figure 7.19

The gravitational potential

difference between the points A

and B is given by

BAAB VVV

BA

ABr

GM

r

GMV

AB

AB

11

rrGMV

PHYSICS CHAPTER 7

42

The gravitational potential difference between point B and A in

the Earth’s gravitational field is given by

The variation of gravitational potential, V when the test mass, mmove away from the Earth’s surface is illustrated by the graph

in Figure 7.20.

m

WVVV AB

ABBA

R

R

GM

r0

V

rV

1

Note:

The Gravitational potential at infinity is zero. 0V

Figure 7.20

PHYSICS CHAPTER 7

43

When in orbit, a satellite attracts the Earth with a force of 19 kN

and the satellite’s gravitational potential due to the Earth is

5.45107 J kg1.

a. Calculate the satellite’s distance from the Earth’s surface.

b. Determine the satellite’s mass.


M = 5.981024 kg and radius of the Earth , R = 6.38106 m)

Solution :

Example 7.7 :

R

gF

rh

173 kg J 10455 N; 1019 .VFg

PHYSICS CHAPTER 7

44

Solution :

a. By using the formulae of gravitational potential, thus

Therefore the satellite’s distance from the Earth’s surface is

r

GMV

m 1032.7 6r

66 1038.61032.7 h

m 104.9 5h

r

.. 24117 1098510676

1045.5

Rhr

173 kg J 10455 N; 1019 .VFg

PHYSICS CHAPTER 7

45

Solution :

b. From the Newton’s law of gravitation, hence

2r

GMmFg

kg 2552m

26

24113

10327

10985106761019

.

m..

173 kg J 10455 N; 1019 .VFg

PHYSICS CHAPTER 7

46


Explain satellite motion with:

velocity,

period,

Learning Outcome:

7.3 Satellite motion in a circular orbit (½ hour)

r

GMv

GM

rT

3

2

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PHYSICS CHAPTER 7

47

7.3 Satellite motion in a circular orbit7.3.1 Tangential (linear/orbital) velocity, v

Consider a satellite of mass, m travelling around the Earth of

mass, M, radius, R, in a circular orbit of radius, r with constant

tangential (orbital) speed, v as shown in Figure 7.22.

Figure 7.22

PHYSICS CHAPTER 7

48

The centripetal force, Fc is contributed by the gravitational force

of attraction, Fg exerted on the satellite by the Earth.

Hence the tangential velocity, v is given by

ccg maFF

r

mv

r

GMm 2

2

r

GMv

where

Earth theof mass :M

from satellite theof distance :rEarth theof centre the

constant nalgravitatio universal : G

PHYSICS CHAPTER 7

49

For a satellite close to the Earth’s surface,

Therefore

The relationship between tangential velocity and angular

velocity is

Hence , the period, T of the satellite orbits around the Earth is

given by

Rr and2gRGM

gRv

T

rrv

2

r

GM

T

r2

GM

rT

3

2

PHYSICS CHAPTER 7

50

Figure 8.23 shows a synchronous (geostationary) satellite which stays above the same point on the equator of the Earth.

The satellite have the following characteristics:

It revolves in the same direction as the Earth.

It rotates with the same period of rotation as that of the Earth (24 hours).

It moves directly above the equator.

The centre of a synchronous satellite orbit is at the centre of the Earth.

It is used as a communication satellite.

7.3.2 Synchronous (Geostationary) Satellite

Figure 8.23

PHYSICS CHAPTER 7

51

The weight of a satellite in a circular orbit round the Earth is half of

its weight on the surface of the Earth. If the mass of the satellite is

800 kg, determine

a. the altitude of the satellite,

b. the speed of the satellite in the orbit,


M = 6.001024 kg, and radius of the Earth , R = 6.40106 m)

Example 7.12 :

PHYSICS CHAPTER 7

52

Solution :

a. The satellite orbits the Earth in the circular path, thus

b. The speed of the satellite is given by

PHYSICS CHAPTER 7

53

The radius of the Moon’s orbit around the Earth is 3.8 108 m and

the period of the orbit is 27.3 days. The masses of the Earth and

Moon are 6.0 1024 kg and 7.4 1022 kg respectively. Calculate

the total energy of the Moon in the orbit.

Solution :

The period of the satellite is

The tangential speed of the satellite is

Example 7.13 :

s m 50.9 kg; 120 m; 1050.8 26 gmr

T

rv

2

13 s m 1024.4 v

36005.3T

s 12600T

12600

1050.82 6

v

PHYSICS CHAPTER 7

54

Solution :

A satellite orbits the planet in the circular path, thus

cg FF

2

2 r

mv

r

GMm

s m 50.9 kg; 120 m; 1050.8 26 gmr

r

GMv 2

and2gRGM

r

gRv

22

6

223

1050.8

50.91024.4

R

m 1001.4 6R

PHYSICS CHAPTER 7

55

Exercise 7.2 :

Given G = 6.671011 N m2 kg2

1. A rocket is launched vertically from the surface of the Earth

at speed 25 km s-1. Determine its speed when it escapes from

the gravitational field of the Earth.

(Given g on the Earth = 9.81 m s2, radius of the Earth ,

R = 6.38 106 m)

ANS. : 2.24104 m s1

2. A satellite revolves round the Earth in a circular orbit whose

radius is five times that of the radius of the Earth. The

gravitational field strength at the surface of the Earth is

9.81 N kg1. Determine

a. the tangential speed of the satellite in the orbit,

b. the angular frequency of the satellite.

(Given radius of the Earth , R = 6.38 106 m)

ANS. : 3538 m s1 ; 1.11104 rad s1

PHYSICS CHAPTER 7

56

Exercise 7.2 :

3. A geostationary satellite of mass 2400 kg is placed

35.92 Mm from the Earth’s surface orbits the Earth along a

circular path.

Determine

a. the angular velocity of the satellite,

b. the tangential speed of the satellite,

c. the acceleration of the satellite,

d. the force of attraction between the Earth and the satellite,

e. the mass of the Earth.

(Given radius of the Earth , R = 6.38 106 m)

ANS. : 7.27105 rad s1; 3.08103 m s1; 0.224 m s2;

537 N ; 6.001024 kg

PHYSICS CHAPTER 7

57

THE END…


Simple Harmonic Motion

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PHYSICS CHAPTER 8

CHAPTER 8: CHAPTER 8: Rotational of rigid bodyRotational of rigid body

(8 Hours)(8 Hours)

http

//:km

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http

//:km

s.m

atrik

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PHYSICS CHAPTER 8

2

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: a) Define and describea) Define and describe::

angular displacement (angular displacement (θθ)) average angular velocity (average angular velocity (ωωavav)) instantaneous angular velocity (instantaneous angular velocity (ωω)) average angular acceleration (average angular acceleration (ααavav) ) instantaneous angular acceleration (instantaneous angular acceleration (αα). ).

b) Relate b) Relate parameters in rotational motion with their corresponding parameters in rotational motion with their corresponding quantities in linear motion.quantities in linear motion. Write and use Write and use ::

c) c) UseUse equations for rotational motion with constant angular equations for rotational motion with constant angular acceleration.acceleration.

Learning Outcome:

8.1 Rotational Kinematics (2 hour)

http

//:km

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http

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s=rθ ; v=rω ;a t=rα ;ac=rω2= v 2

r

ω=ω0αt θ=ω0 t12 αt2 ω2=ω

022 αθ

PHYSICS CHAPTER 8

3

8.1 Parameters in rotational motion8.1.1 Angular displacement,θ is defined as an angle through which a point or line has an angle through which a point or line has

been rotated in a specified direction about a specified axis.been rotated in a specified direction about a specified axis. The S.I. unit of the angular displacement is radian (rad)radian (rad). Figure 8.1 shows a point P on a rotating compact disc (CD)

moves through an arc length s on a circular path of radius r about a fixed axis through point O.


PHYSICS CHAPTER 8

4

From Figure 8.1, thus

Others unit for angular displacement is degree (degree (°°)) and revolution (rev)revolution (rev). Conversion factor :

Sign convention of angular displacement : PositivePositive – if the rotational motion is anticlockwiseanticlockwise. NegativeNegative– if the rotational motion is clockwiseclockwise.

1 rev=2π rad=360°

θ=sr OR s=rθ

where θ : angle angular displacement in radians : arc lengthr : radius of the circle

PHYSICS CHAPTER 8

5

Average angular velocity, Average angular velocity, ωωavav

is defined as the rate of change of angular displacementthe rate of change of angular displacement. Equation :

Instantaneous angular velocity, Instantaneous angular velocity, ωω is defined as the instantaneous rate of change of angular the instantaneous rate of change of angular

displacementdisplacement. Equation :

8.1.2 Angular velocity

ωav=θ2−θ1

t2−t1= Δθ

Δt

ω= limitΔt 0

ΔθΔt =

dθdt

where θ2 : final angular displacement in radian

Δt : time intervalθ1 : initial angular displacement in radian

PHYSICS CHAPTER 8

6

It is a vector quantityvector quantity. The unit of angular velocity is radian per second (rad sradian per second (rad s-1-1)) Others unit is revolution per minute (rev minrevolution per minute (rev min−−11 or rpm) or rpm)

Conversion factor:

Note : Every partEvery part of a rotating rigid body has the same angular same angular

velocityvelocity.Direction of the angular velocityDirection of the angular velocity Its direction can be determine by using right hand grip ruleright hand grip rule

where

1 rpm =2π60 rad s−1=

π30 rad s−1

ThumbThumb : direction of angular velocityangular velocityCurl fingersCurl fingers : direction of rotationrotation

PHYSICS CHAPTER 8

7

Figures 8.2 and 8.3 show the right hand grip rule for determining the direction of the angular velocity.



ω

ω

PHYSICS CHAPTER 8

8

The angular displacement,θ of the wheel is given by

where θ in radians and t in seconds. The diameter of the wheel is 0.56 m. Determine

a. the angle, θ in degree, at time 2.2 s and 4.8 s,b. the distance that a particle on the rim moves during that time interval,c. the average angular velocity, in rad s−1 and in rev min−1 (rpm), between 2.2 s and 4.8 s,d. the instantaneous angular velocity at time 3.0 s.

Example 8.1 :

θ=5t2− t

PHYSICS CHAPTER 8

9


a. At time, t1 =2.2 s :

At time, t2 =4.8 s :

r=d2 =

0 .562 =0 . 28 m

θ1=5 2. 22− 2.2θ1=22 rad

θ2=5 4 .82−4 .8 θ2=110 rad

PHYSICS CHAPTER 8

10


b. By applying the equation of arc length,

Therefore

c. The average angular velocity in rad s−1 is given by

r=d2 =

0 .562 =0 . 28 m

s=rθΔs=rΔθ=r θ2−θ1 Δs=0 . 28 110−22

ωav=ΔθΔt

=θ2−θ1t2−t1

ωav=110−22 4 . 8−2. 2

PHYSICS CHAPTER 8

11

Solution :Solution :c. and the average angular velocity in rev min−1 is

d. The instantaneous angular velocity as a function of time is

At time, t =3.0 s :

ωav=33 . 9 rad1 s 1 rev

2π rad 60 s1 min

ω=ddt

5t2−t ω=

dθdt

ω=10 t −1

ω=10 3 .0 −1

PHYSICS CHAPTER 8

12

A diver makes 2.5 revolutions on the way down from a 10 m high platform to the water. Assuming zero initial vertical velocity, calculate the diver’s average angular (rotational) velocity during a dive.(Given g = 9.81 m s−2)Solution :Solution :

Example 8.2 :

u y=0θ0=0

10 m

waterθ1=2. 5 rev

PHYSICS CHAPTER 8

13

Solution :Solution :From the diagram,Thus

Therefore the diver’s average angular velocity is

θ1=2.5×2π=5π rads y=−10 m

s y=u y t−12 gt2

−10=0−12

9 . 81 t 2

ωav=θ1−θ0

t

ωav=5π−01 . 43

PHYSICS CHAPTER 8

14

Average angular acceleration, Average angular acceleration, ααavav

is defined as the rate of change of angular velocitythe rate of change of angular velocity. Equation :

Instantaneous angular acceleration, Instantaneous angular acceleration, αα is defined as the instantaneous rate of change of angular the instantaneous rate of change of angular

velocityvelocity. Equation :

8.1.3 Angular acceleration

α av=ω2−ω1

t2−t1= Δω

Δt

α= limitΔt 0

ΔωΔt =

dωdt

where ω2 : final angular velocity

Δt : time intervalω1 : initial angular velocity

PHYSICS CHAPTER 8

15Figure 8.4Figure 8.4

ω

It is a vector quantityvector quantity. The unit of angular acceleration is rad srad s−−22. Note:

If the angular acceleration, αα is positivepositive, then the angular velocity, ωω is increasingincreasing.

If the angular acceleration, αα is negativenegative, then the angular velocity, ωω is decreasingdecreasing.

Direction of the angular accelerationDirection of the angular acceleration If the rotation is speeding upspeeding up, αα and ωω in the same directionsame direction

as shown in Figure 8.4.

α

PHYSICS CHAPTER 8

16


ω

If the rotation is slowing downslowing down, αα and ωω have the opposite opposite directiondirection as shown in Figure 8.5.

Example 8.3 :The instantaneous angular velocity, ω of the flywheel is given by

where ω in radian per second and t in seconds. Determinea. the average angular acceleration between 2.2 s and 4.8 s,b. the instantaneous angular acceleration at time, 3.0 s.

α

ω=8t3−t 2

PHYSICS CHAPTER 8

17

Solution :Solution :a. At time, t1 =2.2 s :

At time, t2 =4.8 s :

Therefore the average angular acceleration is

ω1=8 2 . 2 3−2 . 22

ω1=80 . 3 rad s−1

ω2=8 4 .8 3−4 . 8 2

α av=ω2−ω1

t 2−t1

α av=862−80 .34 . 8−2. 2

PHYSICS CHAPTER 8

18

Solution :Solution :b. The instantaneous angular acceleration as a function of time is

At time, t =3.0 s :

α=ddt

8t3− t2

α=dωdt

α=24 3 . 0 2−2 3 . 0

PHYSICS CHAPTER 8

19

Exercise 8.1 :1. If a disc 30 cm in diameter rolls 65 m along a straight line

without slipping, calculatea. the number of revolutions would it makes in the process,b. the angular displacement would be through by a speck of

gum on its rim.ANS. : 69 rev; 138ANS. : 69 rev; 138ππ rad rad2. During a certain period of time, the angular displacement of a

swinging door is described by

where θ is in radians and t is in seconds. Determine the angular displacement, angular speed and angular accelerationa. at time, t =0,

b. at time, t =3.00 s.ANS. : ANS. : 5.00 rad, 10.0 rad s5.00 rad, 10.0 rad s−−11, 4.00 rad s, 4.00 rad s−−22; 53.0 rad, 22.0 rad s; 53.0 rad, 22.0 rad s−−11, ,

4.00 rad s4.00 rad s−−22

θ=5 . 0010 . 0t2. 00 t2

PHYSICS CHAPTER 8

20

8.1.2 Relationship between linear and rotational motion 8.1.2 Relationship between linear velocity, v and

angular velocity, ω When a rigid body is rotates about rotation axis O , every

particle in the body moves in a circle as shown in the Figure 8.6.

v

sθ

y

x

rP

O


PHYSICS CHAPTER 8

21

Point P moves in a circle of radius r with the tangential velocity v where its magnitude is given by

The directiondirection of the linear (tangential) velocitylinear (tangential) velocity always tangent to the circular pathtangent to the circular path.

Every particle on the rigid body has the same angular speedsame angular speed (magnitude of angular velocity) but the tangential speedtangential speed is notnot the samesame because the radiusradius of the circle, rr is changing changing dependdepend on the position of the particleposition of the particle.

v=dsdt

v=r dθdt

s=rθ

v=rω

and

Simulation 7.1

PHYSICS CHAPTER 8

22

at

aca

x

y

P

O

If the rigid bodyrigid body is gaining the angular speedgaining the angular speed then the tangential velocitytangential velocity of a particle also increasingincreasing thus twotwo component of accelerationacceleration are occurredoccurred as shown in Figure 8.7.

8.1.2 Relationship between tangential acceleration, at and angular acceleration, α


PHYSICS CHAPTER 8

23

The components are tangential acceleration, tangential acceleration, aatt and

centripetal acceleration, centripetal acceleration, aacc given by

but

The vector sum of centripetal and tangential accelerationvector sum of centripetal and tangential acceleration of a particle in a rotating body is resultant (linear) acceleration, resultant (linear) acceleration, aa given by

and its magnitude,

a t=dvdt

a t=r dωdt

a t=rα

v=rωand

ac=v2

r=rω2=vω

a=a tac

∣a∣=a t2ac

2Vector formVector form

PHYSICS CHAPTER 8

24

8.1.3 Rotational motion with uniform angular acceleration Table 8.1 shows the symbols used in linear and rotational

kinematics.

Table 8.1Table 8.1

Linear motion Quantity Rotational

motion

s θDisplacementDisplacement

u ω0Initial velocityInitial velocity

v ωFinal velocityFinal velocity

a αAccelerationAcceleration

t tTimeTime

PHYSICS CHAPTER 8

25

Table 8.2 shows the comparison of linear and rotational motion with constant acceleration.

Linear motion Rotational motion

a=constant

v=uat

α=constant

ω=ω0αt

s=ut12 at2 θ=ω0 t1

2 αt 2

v2=u22 as ω2=ω022 αθ

s=12

vu t θ=12 ωω0 t

where θ in radian. Table 8.2Table 8.2

PHYSICS CHAPTER 8

26

A car is travelling with a velocity of 17.0 m s−1 on a straight horizontal highway. The wheels of the car has a radius of 48.0 cm. If the car then speeds up with an acceleration of 2.00 m s−2 for 5.00 s, calculatea. the number of revolutions of the wheels during this period,b. the angular speed of the wheels after 5.00 s.Solution :Solution :a. The initial angular velocity is

and the angular acceleration of the wheels is given by

Example 8.4 :

u=17 . 0 m s−1 , r=0 . 48 m , a=2 . 00 m s−2 , t=5 . 00 su=rω0

17 . 0=0 . 48 ω0

2 . 00=0 . 48 αa=rα

PHYSICS CHAPTER 8

27

Solution :Solution :a. By applying the equation of rotational motion with constant angular acceleration, thus

therefore

b. The angular speed of the wheels after 5.00 s is

θ=ω0 t12 αt 2

θ=229 rad

u=17 . 0 m s−1 , r=0 . 48 m , a=2 . 00 m s−2 , t=5 . 00 s

θ=35 . 4 5 . 0012

4 . 17 5 . 002

ω=ω0αtω=35 . 44 . 17 5 . 00

PHYSICS CHAPTER 8

28

The wheels of a bicycle make 30 revolutions as the bicycle reduces its speed uniformly from 50.0 km h-1 to 35.0 km h-1. The wheels have a diameter of 70 cm.a. Calculate the angular acceleration.b. If the bicycle continues to decelerate at this rate, determine the time taken for the bicycle to stop. Solution :Solution :

Example 8.5 :

θ=30×2π=60π rad , r=0 .702 =0 . 35 m ,

u=50 . 0 km1 h 103 m

1 km 1 h3600 s =13 . 9 m s−1 ,

v=35 .0 km1 h 103 m

1 km 1 h3600 s =9 . 72 m s−1

PHYSICS CHAPTER 8

29

Solution :Solution :a. The initial angular speed of the wheels is

and the final angular speed of the wheels is

therefore

b. The car stops thus Hence

u=rω013 .9=0 .35 ω0

v=rω9 . 72=0 . 35 ω

ω2=ω022 αθ

27 . 8 2=39 . 722α 60π

ω=0 ω0=27 .8 rad s−1and

ω=ω0αt0=27 .8 −2. 13 t

PHYSICS CHAPTER 8

30

A blade of a ceiling fan has a radius of 0.400 m is rotating about a fixed axis with an initial angular velocity of 0.150 rev s-1. The angular acceleration of the blade is 0.750 rev s-2. Determinea. the angular velocity after 4.00 s,b. the number of revolutions for the blade turns in this time interval,c. the tangential speed of a point on the tip of the blade at time, t =4.00 s,d. the magnitude of the resultant acceleration of a point on the tip of the blade at t =4.00 s.Solution :Solution :

a. Given t =4.00 s, thus

Example 8.6 :

r=0 . 400 m , ω0=0. 150×2π=0 .300π rad s−1 ,α=0 .750×2π=1. 50π rad s−2

ω=ω0αtω=19 . 8 rad s−1ω= 0 . 300π1 .50π 4 .00

PHYSICS CHAPTER 8

31

Solution :Solution :b. The number of revolutions of the blade is

c. The tangential speed of a point is given by

θ=ω0 t12 αt 2

θ=41. 5 radθ=0 . 300 π 4 . 001

21 .50 π 4 . 00 2

v=rωv=0 . 400 19 . 8

PHYSICS CHAPTER 8

32

Solution :Solution :d. The magnitude of the resultant acceleration is

a=ac2a

t2

a=v 2

r 2

rα 2

a=7 .922

0 . 400 2

0 . 400×1. 50π 2

PHYSICS CHAPTER 8

33

Calculate the angular velocity ofa. the second-hand,b. the minute-hand andc. the hour-hand,of a clock. State in rad s-1.d. What is the angular acceleration in each case?Solution :Solution :a. The period of second-hand of the clock is T = 60 s, hence

Example 8.7 :

ω=2πT ω=

2π60

PHYSICS CHAPTER 8

34

Solution :Solution :b. The period of minute-hand of the clock is T = 60 min = 3600 s, hence

c. The period of hour-hand of the clock is T = 12 h = 4.32 ×104 s, hence

d. The angular acceleration in each cases is

ω=2π3600

ω=2π4 . 32×104

PHYSICS CHAPTER 8

35

A coin with a diameter of 2.40 cm is dropped on edge on a horizontal surface. The coin starts out with an initial angular speed of 18 rad s−1 and rolls in a straight line without slipping. If the rotation slows down with an angular acceleration of magnitude 1.90 rad s−2, calculate the distance travelled by the coin before coming to rest.Solution :Solution :

The radius of the coin is

Example 8.8 :

d=2 . 40×10−2 m

ω0=18 rad s−1

s

α=−1 .90 rad s−2ω=0 rad s−1

r=d2 =1 .20×10−2 m

PHYSICS CHAPTER 8

36

Solution :Solution :The initial speed of the point at the edge the coin is

and the final speed isThe linear acceleration of the point at the edge the coin is given by

Therefore the distance travelled by the coin is

u=rω0

u=1 . 20×10−2 18

v=0 m s−1

a=rαa=1. 20×10−2 −1. 90

v2=u22as0=0 .216 22 −2 .28×10−2 s

PHYSICS CHAPTER 8

37

Exercise 8.2 :1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev

min-1 about its central axis. Determinea. its angular speed,b. the tangential speed at a point 3.00 cm from its centre,c. the radial acceleration of a point on the rim,d. the total distance a point on the rim moves in 2.00 s.

ANS. : ANS. : 126 rad s126 rad s−−11; 3.77 m s; 3.77 m s−−11; 1.26 ; 1.26 ×× 10 1033 m s m s−−22; 20.1 m; 20.1 m

2. A 0.35 m diameter grinding wheel rotates at 2500 rpm. Calculatea. its angular velocity in rad s−1,b. the linear speed and the radial acceleration of a point on the

edge of the grinding wheel.ANS. : ANS. : 262 rad s262 rad s−−11; 46 m s; 46 m s−−11, 1.2 , 1.2 ×× 10 1044 m s m s−−22

PHYSICS CHAPTER 8

38

Exercise 8.2 :3. A rotating wheel required 3.00 s to rotate through 37.0

revolution. Its angular speed at the end of the 3.00 s interval is 98.0 rad s-1. Calculate the constant angular acceleration of the wheel.

ANS. : ANS. : 13.6 rad s13.6 rad s−−22

4. A wheel rotates with a constant angular acceleration of 3.50 rad s−2.a. If the angular speed of the wheel is 2.00 rad s−1 at t =0, through what angular displacement does the wheel rotate in 2.00 s.b. Through how many revolutions has the wheel turned during this time interval?c. What is the angular speed of the wheel at t = 2.00 s?

ANS. : ANS. : 11.0 rad; 1.75 rev; 9.00 rad s11.0 rad; 1.75 rev; 9.00 rad s−−11

PHYSICS CHAPTER 8

39

Exercise 8.2 :5. A bicycle wheel is being tested at a repair shop. The angular

velocity of the wheel is 4.00 rad s-1 at time t = 0 , and its angular acceleration is constant and equal −1.20 rad s-2. A spoke OP on the wheel coincides with the +x-axis at time t = 0 as shown in Figure 8.8.

a. What is the wheel’s angular velocity at t = 3.00 s?b. What angle in degree does the spoke OP make with the positive x-axis at this time?

ANS. :ANS. : 0.40 rad s0.40 rad s−−11; 18; 18°°


x

y

PO

PHYSICS CHAPTER 8

40

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and use Define and use torque. torque.

State and use State and use conditions for equilibrium of rigid body:conditions for equilibrium of rigid body:

Learning Outcome:

8.2 Equilibrium of a uniform rigid body (2 hour)

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∑∑∑ === 0 , 0 , 0 τFF yx

PHYSICS CHAPTER 8

41

8.2.1 Torque (moment of a force), The magnitude of the torquemagnitude of the torque is defined as the product of a the product of a

force and its perpendicular distance from the line of action force and its perpendicular distance from the line of action of the force to the point (rotation axis)of the force to the point (rotation axis).OR

Because ofwhere r : distance between the pivot point (rotation

axis) and the point of application of force.Thus

τ

Fdτ =

force theof magnitude :Farm)(moment distancelar perpendicu : d

torque theof magnitude : τwhere

θsinrd =

sinθτ Fr=rF and between angle : θwhere

OR Fr ×=τ

PHYSICS CHAPTER 8

42

It is a vector quantityvector quantity. The dimension of torque is

The unit of torqueunit of torque is N mN m (newton metre), a vector productvector product unlike the joule (unit of work)joule (unit of work), also equal to a newton metre, which is scalar productscalar product.

Torque is occurred because of turning (twisting) effects of turning (twisting) effects of the forces the forces on a body.

Sign convention of torque: PositivePositive - turning tendency of the force is anticlockwiseanticlockwise. NegativeNegative - turning tendency of the force is clockwiseclockwise.

The value of torque dependstorque depends on the rotation axisrotation axis and the magnitude of applied forcemagnitude of applied force.

[ ] [ ][ ] 22TMLdF −==τ

PHYSICS CHAPTER 8

43

Case 1 :Case 1 : Consider a force is applied to a metre rule which is pivoted at

one end as shown in Figures 8.9a and 8.9b.

Figure 8.9aFigure 8.9a

F

F

θ

Figure 8.9bFigure 8.9b

Pivot point (rotation axis)

Fdτ =

θrd sin=

θFrFdτ sin==

(anticlockwise)

(anticlockwise)r

Point of action of a force

Line of action of a force

d

PHYSICS CHAPTER 8

44

O

Figure 8.10Figure 8.102θ

Case 2 :Case 2 : Consider three forces are applied to the metre rule which is

pivoted at one end (point O) as shown in Figures 8.10.

Caution : If the line of action of a force is through the rotation axisline of action of a force is through the rotation axis

then

1F

1θ

111 θrd sin=

321 ττττ ++=∑ O

Therefore the resultant (nett) torque is

3F

2F 1r

0sin === 333333 θrFdFτ

222 θrd sin=

111111 θrFdFτ sin==222222 θrFdFτ sin−=−=

2r

2211 dFdFτ −=∑ O

θFrτ sin=0=τ

and 0=θSimulation 5.1

PHYSICS CHAPTER 8

45

Determine a resultant torque of all the forces about rotation axis, O in the following problems.a.

Example 8.9 :

m 5

N 10=2Fm 5 N 30=1F

m 3

m 3

N 20=3F

m 10

m 6O

PHYSICS CHAPTER 8

46

b.Example 8.9 :

m 5

N 10=2F

m 5

N 30=1F

m 3

m 3

N 25=4F

N 20=3F

m 10

α

m 6O β

PHYSICS CHAPTER 8

47

m 5m 5

m 10

m 6O

Solution :Solution :a.

Force Torque (N m), τo=Fd=Frsinθ

1F

( ) ( ) 90330 −=−

2F ( )( ) 50510 +=+

N 10=2FN 30=1F

N 20=3F

m 3=1d

m 5=2d

3F

0The resultant torque:

PHYSICS CHAPTER 8

48

m 5

m 10

m 3

m 6

m 5

Solution :Solution :b.

Force Torque (N m), τo=Fd=Frsinθ

1F ( ) ( ) 90330 −=−

2F

( )( )( ) 51.50.515520sin ==βrF33F

0 The resultant torque:

N 10=2F

N 30=1F

0.51553

3sin22

=+

=βO

N 20=3F

N 25=4Fα

β

m 3=1dβ

m 5=r

4F

0

3d

PHYSICS CHAPTER 8

49

8.2 Equilibrium of a rigid body8.2.1.1 Non-concurrent forces is defined as the forces whose lines of action do not pass the forces whose lines of action do not pass

through a single common point. through a single common point. The forces cause the rotational motionrotational motion on the body. The combination of concurrent and non-concurrent forces cause

rolling motionrolling motion on the body. (translational and rotationaltranslational and rotational motion)

Figure 8.11 shows an example of non-concurrent forces.

2F

3F

1F


4F

PHYSICS CHAPTER 8

50

8.2.1.2 Equilibrium of a rigid body Rigid bodyRigid body is defined as a body with definite shape that a body with definite shape that

doesn’t change, so that the particles that compose it stay in doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is fixed position relative to one another even though a force is exerted on itexerted on it.

If the rigid body is in equilibriumrigid body is in equilibrium, means the body is translational and rotational equilibriumtranslational and rotational equilibrium.

There are two conditionstwo conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must The vector sum of all forces acting on a rigid body must

be zero.be zero.∑ == 0nettFF

OR

∑∑∑ === 0 , 0 , 0 zyx FFF

PHYSICS CHAPTER 8

51

The vector sum of all external torques acting on a rigid The vector sum of all external torques acting on a rigid body must be zero about any rotation axisbody must be zero about any rotation axis.

This ensures rotational equilibriumrotational equilibrium. This is equivalent to the three independent scalar

equations along the direction of the coordinate axes,

Centre of gravity, CG Centre of gravity, CG is defined as the point at which the whole weight of a body the point at which the whole weight of a body

may be considered to actmay be considered to act. A force that exerts on the centre of gravityexerts on the centre of gravity of an object will

cause a translational motiontranslational motion.

∑ ==τ 0nettτ

∑∑∑ === 0 , 0 , 0 zyx τττ

PHYSICS CHAPTER 8

52

Figures 8.14 and 8.15 show the centre of gravity for uniformcentre of gravity for uniform (symmetric) objectobject i.e. rod and sphere

rodrod – refer to the midway point between its endmidway point between its end.

spheresphere – refer to geometric centregeometric centre.

2l

2l

CG

CGl



PHYSICS CHAPTER 8

53

8.2.4 Problem solving strategies for equilibrium of a rigid body

The following procedure is recommended when dealing with problems involving the equilibrium of a rigid body: Sketch a simple diagramSketch a simple diagram of the system to help

conceptualize the problem. Sketch a separate free body diagramSketch a separate free body diagram for each body. Choose a convenient coordinate axesChoose a convenient coordinate axes for each body and

construct a tableconstruct a table to resolve the forces into their components and to determine the torque by each force.

Apply the condition for equilibrium of a rigid bodyApply the condition for equilibrium of a rigid body :

SolveSolve the equationsequations for the unknowns.

∑ = 0xF ∑ = 0yF; and ∑ = 0τ

PHYSICS CHAPTER 8

54

A hanging flower basket having weight, W2 =23 N is hung out over the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is counterbalanced by a body of weight, W1 as shown in Figure 8.14. If the mass of the beam is 3.0 kg, calculatea. the weight, W1 needed,

b. the force exerted on the beam at point O.(Given g =9.81 m s−2)

Example 8.10 :

1W2W

A BO35 cm 75 cm


PHYSICS CHAPTER 8

55

Solution :Solution :The free body diagram of the beam :

Let point O as the rotation axis.

N 23 ;kg 3 == 2Wm

0.75 mA B

OCG

1W

2W

N

gm

0.35 m

0.55 m 0.55 m

0.20 m0.20 m

Force y-comp. (N) Torque (N m), τo=Fd=Frsinθ

1W

1W−

gm ( )( )9.813− ( ) ( ) 5.880.2029.4 −=−

( ) 11 WW 0.750.75 −=−

2W

23− ( )( ) 8.050.3523 =+

N

N 029.4−=

PHYSICS CHAPTER 8

56

Solution :Solution :Since the beam remains at rest thus the system in equilibrium. a. Hence

b. 0=∑ yFand

∑ = 0Oτ

05.888.050.75 =−+− 1W

029.423 =+−−− NW1

( ) 029.4232.89 =+−−− N

PHYSICS CHAPTER 8

57

A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure 8.15. The height of the end A of the ladder is 8.0 m from the rough floor.a. Determine the horizontal and vertical forces the floor exerts on the end B of the ladder when a firefighter of mass 60 kg is 3.0 m from B.b. If the ladder is just on the verge of slipping when the firefighter is 7.0 m up the ladder , Calculate the coefficient of static friction between ladder and floor.(Given g =9.81 m s−2)

Example 8.11 :

A

B

smooth wall

rough floorFigure 8.15Figure 8.15

PHYSICS CHAPTER 8

58

Solution :Solution :a. The free body diagram of the ladder : Let point B as the rotation axis.

kg 60 ;kg 5.0 == fl mm

A

B

CG

gm f

1N

gml

2N

α

m 8.0m 10

m 3.0

m 5.0

Force x-comp. (N)

y-comp. (N)

Torque (N m), τB=Fd=Frsinθ

gml

1N1N

0.8108sin ==α

sf−

gm f

49.1− 0.6106sin ==β

2N

sf

0

589−0

2N

0

0

m 6.0

αβ( ) ( ) βsin5.049.1

β

β

147=

0

( ) ( ) βsin3.05891060=

( ) αN1 sin10−1N8−=0

0 sf

PHYSICS CHAPTER 8

59

Solution :Solution : Since the ladder in equilibrium thus

0=∑ Bτ

081060147 =−+ 1NN 151=1N

0=∑ xF

0=− s1 fNHorizontal force:Horizontal force:

0=∑ yF

058949.1 =+−− 2NVertical force:Vertical force:

PHYSICS CHAPTER 8

60

m 10

A

B

m 8.0

m 6.0

m 5.0 α

β

Solution :Solution :b. The free body diagram of the ladder : Let point B as the rotation axis.

0.6sin 0.8;sin == βα

gm f

gml

sf

m 7.0

Force x-comp. (N)

y-comp. (N)

Torque (N m), τB=Fd=Frsinθ

gml

1N1N

2s Nμ−

gm f

49.1−

2N

sf

0

589−0

2N

0

0

α

( ) ( ) βsin5.049.1

β

β147=

0

( ) ( ) βsin7.05892474=

( ) αN1 sin10−1N8−=0

0

2N

1N

PHYSICS CHAPTER 8

61

Solution :Solution : Consider the ladder stills in equilibrium thus

0=∑ Bτ

082474147 =−+ 1NN 328=1N

0=∑ xF

0=− 2s1 NμN

0=∑ yF

058949.1 =+−− 2NN 638=2N

( ) ( ) 0638328 =− sμ

PHYSICS CHAPTER 8

62


A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 8.16. A cable at an angle 30° with the beam helps to support the light.a. Sketch a free body diagram of the beam.b. Determine i. the tension in the cable, ii. the force exerted on the beam by the pole.(Given g =9.81 m s−2)

Example 8.12 :

PHYSICS CHAPTER 8

63

Solution :Solution :a. The free body diagram of the beam :

b. Let point O as the rotation axis.

kg 10.0 ;kg 20.0 == bf mm

Force x-comp. (N) y-comp. (N) Torque (N m), τo=Fd=Frsinθgm f ( )l196−0 196−

O CG

gm f

gmb

T

S

30

ll0.5

gmb ( )( ) ll 49.10.598.1 −=−0 98.1−

T

TlTl 0.530sin =30cosT− 30sinT

S

xS yS 0

PHYSICS CHAPTER 8

64

Solution :Solution :b. The floodlight and beam remain at rest thus i.

ii.

0=∑ Oτ

00.549.1196 =+−− Tlll

0=∑ xF

0cos =+− xS30T

N 424=xS0=∑ y

F030sin98.1196 =++−− yST

N 49.1=yS

PHYSICS CHAPTER 8

65

Solution :Solution :b. ii. Therefore the magnitude of the force is

and its direction is given by

2y

2x SSS +=

( ) ( ) 22S 49.1424 +=

= −

x

y

SS

θ 1tan

= −

42449.1tan 1θ

PHYSICS CHAPTER 8

66

Exercise 8.3 :Use gravitational acceleration, g = 9.81 m s−2

1.

Figure 8.17 shows the forces, F1 =10 N, F2= 50 N and F3= 60 N are applied to a rectangle with side lengths, a = 4.0 cm and b = 5.0 cm. The angle γ is 30°. Calculate the resultant torque about point D.

ANS. : -3.7 N mANS. : -3.7 N m

D

A B

C γ

1F

3F

2F


a

b

PHYSICS CHAPTER 8

67


Exercise 8.3 :2.

A see-saw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 8.18. The fulcrum is under the centre of gravity of the board. Determinea. the magnitude of the force exerted by the fulcrum on the

board,b. where the father should sit from the fulcrum to balance the system.

ANS. : 1128 N; 1.31 mANS. : 1128 N; 1.31 m

PHYSICS CHAPTER 8

68

3.

A traffic light hangs from a structure as show in Figure 8.19. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the traffic light is 12.0 kg. Determine a. the tension in the horizontal massless cable CD, b. the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole.

ANS. : 248 N; 197 N, 248 NANS. : 248 N; 197 N, 248 N


Exercise 8.3 :

PHYSICS CHAPTER 8

69

4.

A uniform 10.0 N picture frame is supported by two light string as shown in Figure 8.20. The horizontal force, F is applied for holding the frame in the position shown.a. Sketch the free body diagram of the picture frame.b. Calculate i. the tension in the ropes, ii. the magnitude of the horizontal force, F .

ANS. : 1.42 N, 11.2 N; 7.20 NANS. : 1.42 N, 11.2 N; 7.20 N

Exercise 8.3 :


F

50.0

cm 15.0

cm 30.0

PHYSICS CHAPTER 8

70

Learning Outcome:

8.3 Rotational dynamics (1 hour)

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define Define the moment of inertia of a rigid body about an axis,the moment of inertia of a rigid body about an axis,

State and useState and use torque, torque,

∑=

=n

1iiirmI 2

Iατ =

PHYSICS CHAPTER 8

71

8.3.1 Centre of mass, moment of inertia and torque8.3.1.1 Centre of mass (CM) is defined as the point at which the whole mass of a body the point at which the whole mass of a body

may be considered to be concentratedmay be considered to be concentrated. Its coordinate ((xxCMCM, , yyCMCM)) is given the expression below:

xCM=∑i=1

n

mi xi

∑i=1

n

mi ; y CM=∑i=1

n

mi y i

∑i=1

n

mi where mi : mass of the i th particle

x i : x coordinate of the i th particley i : y coordinate of the i th particle

PHYSICS CHAPTER 8

72

Two masses, 3 kg and 5 kg are located on the y-axis at y =1 m and y =5 m respectively. Determine the centre of mass of this system.Solution :Solution :

Example 8.13 :

0

1 m

y=

5 mm1=3 kg; m2=5 kg

m1

m2

yCM=∑i=1

2

mi y i

∑i=1

2

mi

=m1 y1m2 y2

m1m2

yCM=3 1 5 5

35

CM3 . 5 m

PHYSICS CHAPTER 8

73

A system consists of three spheres have the following masses and coordinates :

(1) 1 kg, (3,2) ; (2) 2 kg, (4,5) and (3) 3 kg, (3,0).Determine the coordinate of the centre of mass of the system.Solution :Solution :The x coordinate of the CM is

Example 8.14 :

m1=1 kg; m2=2 kg; m3=3 kg

xCM=∑i=1

3

mi x i

∑i=1

3

mi

=m1 x1m2 x2m3 x3

m1m2m3

xCM=1 3 2 4 3 3

123

PHYSICS CHAPTER 8

74

Solution :Solution :The y coordinate of the CM is

Therefore the coordinate of the CM is

yCM=∑i=1

3

mi y i

∑i=1

3

y i

=m1 y1m2 y2m3 y3

m1m2m3

yCM=1 2 2 5 3 0

123

PHYSICS CHAPTER 8

75

Figure 8.21 shows a rigid body about a fixed axis O with angular velocity ω.

is defined as the sum of the products of the mass of each the sum of the products of the mass of each particle and the square of its respective distance from the particle and the square of its respective distance from the rotation axisrotation axis.

8.3.1.2 Moment of inertia, I

m1

m2

mn

m3

r1r 2

r3

r nO

ω


PHYSICS CHAPTER 8

76

OR

It is a scalar quantityscalar quantity. Moment of inertia, Moment of inertia, II in the rotational kinematics is analogousanalogous

to the mass, mass, mm in linear kinematics. The dimensiondimension of the moment of inertia is M LM L22. The S.I. unitS.I. unit of moment of inertia is kg mkg m22. The factorsfactors which affect the moment of inertia, I of a rigid body:

a. the massmass of the body,b. the shapeshape of the body,c. the positionposition of the rotation axisrotation axis.

I=m1 r12m2r2

2m3 r32. .. mn rn

2=∑i=1

n

mi ri2

I : moment of inertia of a rigid body about rotation axism : mass of particler : distance from the particle to the rotation axis

where

PHYSICS CHAPTER 8

77

Moments of inertia of various bodiesMoments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects

about axes through the centre of mass.

Shape Diagram Equation

I CM=MR 2

I CM=12 MR2

Hoop or ring or thin cylindrical

shell

Solid cylinder or disk

CM

CM

PHYSICS CHAPTER 8

78

CM




I CM=1

12 ML2Uniform rod or

long thin rod with rotation axis through the

centre of mass.

CM

I CM=25 MR2Solid Sphere

PHYSICS CHAPTER 8

79




I CM=23 MR2Hollow Sphere or

thin spherical shell

CM

Table 8.3Table 8.3

PHYSICS CHAPTER 8

80

Four spheres are arranged in a rectangular shape of sides 250 cm and 120 cm as shown in Figure 8.22.

The spheres are connected by light rods . Determine the moment of inertia of the system about an axisa. through point O,b. along the line AB.

Example 8.15 :

250 cm

60 cm

60 cm

2 kg 3 kg

4 kg5 kg

OA B


PHYSICS CHAPTER 8

81

Solution :Solution :a. rotation axis about point O,

Since r1= r2= r3= r4= r thus

and the connecting rods are light therefore

m1=2 kg; m2=3 kg; m3=4 kg; m4=5 kg

r1 0 . 6 m

m1 m2

m3m4

O

r 2

r 4 r3

1. 25 m

r=0 . 6 21 .25 2=1 .39 mI O=m1 r1

2m2 r22m3 r3

2m4 r 42

I O=r2 m1m2m3m4 =1 .39 2 2345

PHYSICS CHAPTER 8

82

Solution :Solution :b. rotation axis along the line AB,

r1= r2= r3= r4= r=0.6 m therefore

m1=2 kg; m2=3 kg; m3=4 kg; m4=5 kg

I AB=m1 r12m2 r 2

2m3 r32m4 r4

2

I AB=0 . 6 2 2345

m1 m2

m3m4

A B

r1 r 2

r 4 r3

I AB=r2 m1m2m3m4

PHYSICS CHAPTER 8

83

Relationship between torque,Relationship between torque,ττ and angular acceleration, and angular acceleration, αα Consider a force, F acts on a rigid body freely pivoted on an

axis through point O as shown in Figure 8.23.

The body rotates in the anticlockwise direction and a nett torque is produced.

8.3.2 Torque,τ

m1

m2

mn r1

r 2

r nO

a1

ana2

F


PHYSICS CHAPTER 8

84

A particle of mass, m1 of distance r1 from the rotation axis O will experience a nett force F1 . The nett force on this particle is

The torque on the mass m1 is

The total (nett) torque on the rigid body is given by

F1=m1 a1

F1=m1 r1 αa1=r1 αand

τ 1=r1 F1 sin 90°

τ 1=m1 r12 α

∑ τ=∑i=1

n

mi ri2α

∑ τ=m1r12 αm2 r2

2α .. .mn r n2α

∑ τ=Iα

∑i=1

n

mi ri2=Iand

PHYSICS CHAPTER 8

85

From the equation, the nett torquenett torque acting on the rigid body is proportionalproportional to the body’s angular accelerationangular acceleration.

Note :

Nett torque ,∑ τ=Iα

Nett force,∑ F=ma

is analogous to the

PHYSICS CHAPTER 8

86

Forces, F1 = 5.60 N and F2 = 10.3 N are applied tangentially to a disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure 8.24.

Calculate,a. the nett torque on the disc.b. the magnitude of angular acceleration influence by the disc.( Use the moment of inertia, )

Example 8.16 :


I CM=12 MR2

F1

O30 .0 cm

F2

PHYSICS CHAPTER 8

87

Solution :Solution :a. The nett torque on the disc is

b. By applying the relationship between torque and angular acceleration,

R=0 . 30 m ; M=5 . 00 kg

∑ τ=τ1τ 2∑ τ=−RF1RF2=R −F1F2 ∑ τ= 0 . 30 −5 .6010 . 3

∑ τ=12

MR2α∑ τ=Iα

1. 41=12

5 .00 0 . 30 2α

PHYSICS CHAPTER 8

88

A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. The moment of inertia of the wheel about the axis is0.050 kg m2. A light string wrapped around the wheel is attached to a 2.0 kg block that slides on a horizontal frictionless surface. A horizontal force of magnitude P = 3.0 N is applied to the block as shown in Figure 8.25. Assume the string does not slip on the wheel.

a. Sketch a free body diagram of the wheel and the block.b. Calculate the magnitude of the angular acceleration of the wheel.

Example 8.17 :


PHYSICS CHAPTER 8

89

Solution :Solution :a. Free body diagram :

for wheel,

for block,

R=0 . 20 m ; I=0 .050 kg m 2 ; P=3 .0 N; m=2 .0 kg

W

TS

NT

W b

P

a

PHYSICS CHAPTER 8

90

Solution :Solution :b. For wheel,

For block,

By substituting eq. (1) into eq. (2), thus

∑ τ=IαRT=Iα T=

IαR

(1)

∑ F=ma P−T=ma (2)

R=0 . 20 m ; I=0 .050 kg m 2 ; P=3 .0 N; m=2 .0 kg

a=RαP− IαR =ma and

P− IαR =mRα

PHYSICS CHAPTER 8

91

An object of mass 1.50 kg is suspended from a rough pulley of radius 20.0 cm by light string as shown in Figure 8.26. The pulley has a moment of inertia 0.020 kg m2 about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley. After 0.3 s, determinea. the linear acceleration of the object,b. the angular acceleration of the pulley,c. the tension in the string,d. the liner velocity of the object,e. the distance travelled by the object. (Given g = 9.81 m s-2)

Example 8.18 :


R

1.50 kg

PHYSICS CHAPTER 8

92

Solution :Solution :a. Free body diagram :

for pulley,

for block,

W

a

T

S ∑ τ=IαRT=Iα α=

aRand

RT=I aR

T=IaR2 (1)

T

m g

∑ F=mamg−T=ma (2)

PHYSICS CHAPTER 8

93


a. By substituting eq. (1) into eq. (2), thus

b. By using the relationship between a and α, hence

mg− IaR2 =ma

R=0 .20 m ; I=0 . 020 kg m 2 ; m=1 . 50 kg;u=0; t =0 .3 s

a=Rα

1 .50 9 .81 −0 .020 a0 . 20 2 =1 .50 a

7 . 36=0 . 20 α

PHYSICS CHAPTER 8

94


c. From eq. (1), thus

d. By applying the equation of liner motion, thus

e. The distance travelled by the object in 0.3 s is

R=0 .20 m ; I=0 . 020 kg m 2 ; m=1 . 50 kg;u=0; t =0 .3 s

v=uatv=0 7 .36 0 . 3

T=IaR2 T=

0 . 020 7 . 36 0 .20 2

s=ut12 at 2

s=012

7 . 36 0 .3 2

PHYSICS CHAPTER 8

95


1. Three odd-shaped blocks of chocolate have following masses and centre of mass coordinates: 0.300 kg, (0.200 m,0.300 m); 0.400 kg, (0.100 m. -0.400 m); 0.200 kg, (-0.300 m, 0.600 m).Determine the coordinates of the centre of mass of the system of three chocolate blocks.

ANS. : ANS. : (0.044 m, 0.056 m)(0.044 m, 0.056 m)2. Figure 8.27 shows four masses that are held at

the corners of a square by a very light frame. Calculate the moment of inertia of the system about an axis perpendicular to the planea. through point A, andb. through point B.

ANS. : ANS. : 0.141 kg m0.141 kg m22; 0.211 kg m; 0.211 kg m22

80 cm

80 cm

150 g 150 g

70 g

70 g

40 cm

A

B


PHYSICS CHAPTER 8

96

2 . 00 m s−2

T 2

T 1

Exercise 8.4 :3. A 5.00 kg object placed on a

frictionless horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in Figure 8.28. The pulley has a radius of 0.250 m and moment of inertia I. The block on the table is moving with a constant acceleration of 2.00 m s−2.a. Sketch free body diagrams of both objects and pulley.b. Calculate T1 and T2 the tensions in the string.c. Determine I.

ANS. : 10.0 N, 70.3 N; 1.88 kg mANS. : 10.0 N, 70.3 N; 1.88 kg m22


PHYSICS CHAPTER 8

97

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Solve problem related to :Solve problem related to :

kinetic energy, kinetic energy,

work, work,

power, power,

Learning Outcome:8.4 Work and Energy of Rotational Motion (2 hours)

K r=12 Iω2

P=τω

W =τθ

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PHYSICS CHAPTER 8

98

8.4 Rotational kinetic energy and power8.4.1 Rotational kinetic energy, Kr Consider a rigid body rotating about the axis OZ as shown in

Figure 8.29.

Every particle in the body is in the circular motion about point O.

m1

m2

mn

m3

r1r 2

r3

r nO

v1

v2

v3

vn

Z


PHYSICS CHAPTER 8

99

The rigid body has a rotational kinetic energy which is the total total of kinetic energy of all the particles in the bodyof kinetic energy of all the particles in the body is given by

K r=12 m1 v1

212 m2 v2

212 m3 v 3

2.. .12 mn vn

2

K r=12 m1 r1

2ω212 m2 r 2

2 ω212 m3 r3

2 ω2. .. 12 mn rn

2 ω2

K r=12 ω2 m1r1

2m2r 22m3r3

2. . .mn rn2

K r=12 Iω2

K r=12

ω2∑i=1

n

mi r i2 ∑i=1

n

mi r i2=Iand

PHYSICS CHAPTER 8

100

From the formula for translational kinetic energy, Ktr

After comparing both equations thus

For rolling body without slippingrolling body without slipping, the total kinetic energy of total kinetic energy of the body, the body, KK is given by

K tr=12 mv2

ωω is analogous to vvII is analogous to mm

K=K trK r

K tr : translational kinetic energyK r : rotational kinetic energy

where

PHYSICS CHAPTER 8

101

A solid sphere of radius 15.0 cm and mass 10.0 kg rolls down an inclined plane make an angle 25° to the horizontal. If the sphere rolls without slipping from rest to the distance of 75.0 cm and the inclined surface is smooth, calculatea. the total kinetic energy of the sphere,b. the linear speed of the sphere,c. the angular speed about the centre of mass.(Given the moment of inertia of solid sphere is and the gravitational acceleration, g = 9.81 m s−2)

Example 8.19 :

I CM=25 mR2

PHYSICS CHAPTER 8

102


a. From the principle of conservation of energy,

R=0 .15 m ; m=10 . 0 kg

∑ E i=∑ E fmgh=K

K=mgs sin 25°

K=10 . 0 9 . 81 0 . 75 sin 25°

s=0 . 75 m

h=ssin 25°

v CM 25°

R

PHYSICS CHAPTER 8

103

Solution :Solution :b. The linear speed of the sphere is given by

c. By using the relationship between v and ω, thus

R=0 .15 m ; m=10 . 0 kg

K=K trK r K=12 mv2

12 Iω2 ω=

vRand

K=12

mv212 2

5mR2 v

R 2

K=7

10 mv2

31 .1=7

1010 . 0 v2

v=Rω 2 .11=0 .15 ω

PHYSICS CHAPTER 8

104

The pulley in the Figure 8.30 has a radius of 0.120 m and a moment of inertia 0.055 g cm2. The rope does not slip on the pulley rim. Calculate the speed of the 5.00 kg block just before it strikes the floor. (Given g = 9.81 m s−2)

Example 8.20 :

2 .00 kg

5 . 00 kg

7 . 00 m


PHYSICS CHAPTER 8

105

Solution :Solution :The moment of inertia of the pulley,

m1=5 .00 kg ;m2=2.00 kg; R=0 .120 m; h=7 . 00 m

I= 0 . 055 g 1 cm210−3 kg1 g 10−4m2

1 cm2 =5 . 5×10−9 kg m2

m2

m1

7 . 00 m

Initial

m2

m17 . 00 m

v

v

Final

∑ E i=U1 ∑ E f =K tr 1K tr 2K rU2

PHYSICS CHAPTER 8

106


By using the principle of conservation of energy, thus

∑ E i=∑ E fU1=K tr1K tr 2K rU2

m1 gh=12 m1 v2

12 m2 v2

12 Iω2m2 gh

m1−m2 gh=12

v2 m1m2 12

I vR

2

5 . 00−2 . 00 9 . 81 7 . 00=12

v 2 5 . 002. 00 12

5 . 5×10−9 v0 . 120

2

m1=5 . 00 kg ;m2=2. 00 kg; R=0 .120 m;h=7 . 00 m; I =5 .5×10−9 kg m2

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107

Consider a tangential force, F acts on the solid disc of radius R freely pivoted on an axis through O as shown in Figure 8.31.

The work done by the tangential force is given by

8.4.2 Work, W


F

ds

O

dθRR

dW =FdsdW =FRdθ

∫ dW=∫θ1

θ2 τdθ W =∫θ1

θ2 τdθ

ds=Rdθand

PHYSICS CHAPTER 8

108

If the torque is constant thus

Work-rotational kinetic energy theorem states

W =τ θ2−θ1W =τ∫θ1

θ2 dθ

W =τΔθ

τ : torqueΔθ : change in angular displacement

where

W : work done

W =ΔK r= K r f −K r iW =

12 Iω2−

12 Iω0

2

is analogous to the W =Fs

PHYSICS CHAPTER 8

109

From the definition of instantaneous power,

Caution : The unitunit of kinetic energy, work and powerkinetic energy, work and power in the

rotationalrotational kinematics is samesame as their unitunit in translational translational kinematics.

8.4.3 Power, P

P=dWdt dW =τdθand

P=τdθdt

P=τω

dθdt =ωand

is analogous to the P=Fv

PHYSICS CHAPTER 8

110

A horizontal merry-go-round has a radius of 2.40 m and a moment of inertia 2100 kg m2 about a vertical axle through its centre. A tangential force of magnitude 18.0 N is applied to the edge of the merry-go- round for 15.0 s. If the merry-go-round is initially at rest and ignore the frictional torque, determinea. the rotational kinetic energy of the merry-go-round,b. the work done by the force on the merry-go-round,c. the average power supplied by the force.(Given g = 9.81 m s−2)Solution :Solution :

Example 8.21 :

FR=2 . 40 m

PHYSICS CHAPTER 8

111


a. By applying the relationship between nett torque and angular acceleration, thus

Use the equation of rotational motion with uniform angular acceleration,

Therefore the rotational kinetic energy for 15.0 s is

∑ τ=IαRF=Iα 2 . 40 18 . 0 =2100α

ω=ω0αtω=0 2 .06×10−2 15 .0 ω=0 .309 rad s−1

K r=12 Iω2 K r=

12

2100 0 .309 2

R=2 . 40 m ; I=2100 kg m2 ; F=18 . 0 N;t=15 .0 s; ω0=0

PHYSICS CHAPTER 8

112


b. The angular displacement, θ for 15.0 s is given by

By applying the formulae of work done in rotational motion, thus

c. The average power supplied by the force is

W =τθ

θ=ω0 t12 αt 2

Pav=Wt

W =2 . 40 18 . 0 2 . 32

R=2 . 40 m ; I=2100 kg m2 ; F=18 . 0 N;t=15 .0 s; ω0=0

θ=012

2. 06×10−215 . 0 2

W =RFθ

Pav=10015 . 0

PHYSICS CHAPTER 8

113

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and useDefine and use angular momentum, angular momentum,

State and useState and use the principle of conservation of angular the principle of conservation of angular momentummomentum

Learning Outcome:

8.5 Conservation of angular momentum (1 hour)

L=Iω

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PHYSICS CHAPTER 8

114

8.5 Conservation of angular momentum8.5.1 Angular momentum, is defined as the product of the angular velocity of a body the product of the angular velocity of a body

and its moment of inertia about the rotation axisand its moment of inertia about the rotation axis. OR

It is a vectorIt is a vector quantity. Its dimension is M LM L22 T T−−11

The S.I. unit of the angular momentum is kg mkg m22 s s−−11.

L

where

L=IωL : angular momentumI : moment of inertia of a bodyω : angular velocity of a body

is analogous to the p=mv

PHYSICS CHAPTER 8

115

The relationship between angular momentum, L with linear momentum, p is given by vector notation :

magnitude form :

Newton’s second law of motion in term of linear momentum is

hence we can write the Newton’s second law in angular form as

and states that a vector sum of all the torques acting on a a vector sum of all the torques acting on a rigid body is proportional to the rate of change of angular rigid body is proportional to the rate of change of angular momentummomentum.

L=r×p=r ×m v

L=rpsin θ=mvr sin θwhere

θ :the angle between {r with {v ¿ ¿r : distance from the particle to the rotation axis

∑ F=Fnett=d pdt

∑ τ=τ nett=d Ldt

PHYSICS CHAPTER 8

116

states that a total angular momentum of a system about an a total angular momentum of a system about an rotation axis is constant if no external torque acts on the rotation axis is constant if no external torque acts on the systemsystem.OR

Therefore

8.5.2 Principle of conservation of angular momentum

Iω=constant

∑ τ=d Ldt

=0

d L=0

If the ∑ τ=0

dL=∑ L f−∑ Li

∑ L i=∑ L f

and

PHYSICS CHAPTER 8

117

A 200 kg wooden disc of radius 3.00 m is rotating with angular speed 4.0 rad s-1 about the rotation axis as shown in Figure 8.32 . A 50 kg bag of sand falls onto the disc at the edge of the wooden disc.

Calculate,a. the angular speed of the system after the bag of sand falling onto the disc. (treat the bag of sand as a particle)b. the initial and final rotational kinetic energy of the system. Why the rotational kinetic energy is not the same?(Use the moment of inertia of disc is )

Example 8.22 :

ω0

Before

Rω

After

R


12 MR2

PHYSICS CHAPTER 8

118

Solution :Solution :a. The moment of inertia of the disc,

The moment of inertia of the bag of sand,

By applying the principle of conservation of angular momentum,

R=3 . 00 m ;ω0=4 . 0 rad s−1 ; mw=200 kg; mb=50 kg

I w=12 mw R2=

12

200 3. 00 2

I w ω0=I wI bω

I b=mb R2= 50 3 . 002

∑ L i=∑ L f

I w=900 kg m2

I b=450 kg m2

900 4 .0 = 900450 ω

PHYSICS CHAPTER 8

119

Solution :Solution :b. The initial rotational kinetic energy,

The final rotational kinetic energy,

thus It is because the energy is lost in the form of heat from the the energy is lost in the form of heat from the friction between the surface of the disc with the bag of friction between the surface of the disc with the bag of sand.sand.

R=3 . 00 m ;ω0=4 . 0 rad s−1 ; mw=200 kg; mb=50 kg

K r i=12 I w ω0

2=12

900 4 .0 2

K r f=12 I wI b ω2=

12

900450 2 . 672K r i=7200 J

K r i≠K r f

PHYSICS CHAPTER 8

120

A raw egg and a hard-boiled egg are rotating about the same axis of rotation with the same initial angular velocity. Explain which egg will rotate longer.Solution :Solution :The answer is hard-boiled egghard-boiled egg.

Example 8.23 :

PHYSICS CHAPTER 8

121

Solution :Solution :ReasonReasonRaw egg : When the egg spins, its yolk being denser moves away from the axis of rotation and then the moment of inertia of the egg increases because ofFrom the principle of conservation of angular momentum,

If the I is increases hence its angular velocity, ω will decreases.

Hard-boiled egg :The position of the yolk of a hard-boiled egg is fixed. When the egg is rotated, its moment of inertia does not increase and then its angular velocity is constant. Therefore the egg continues to spin.

I=mr2

Iω=constant

PHYSICS CHAPTER 8

122

A student on a stool rotates freely with an angular speed of 2.95 rev s−1. The student holds a 1.25 kg mass in each outstretched arm that is 0.759 m from the rotation axis. The moment of inertia for the system of student-stool without the masses is 5.43 kg m2. When the student pulls his arms inward, the angular speed increases to 3.54 rev s−1. a. Determine the new distance of each mass from the rotation axis.b. Calculate the initial and the final rotational kinetic energy of the system.Solution :Solution :

Example 8.24 :

ω0=2 .95 rev1 s 2π rad

1 rev =18 . 5 rad s−1

ω=3 . 54 rev1 s 2π rad

1 rev =22. 2 rad s−1

PHYSICS CHAPTER 8

123Before

ω0 ω

After

r a r a


r br b

mm

m=1. 25 kg ; ω0=18. 5 rad s−1 ; I ss=5. 43 kg m2 ;r b=0 .759 m ; ω=22. 2 rad s−1 ;

PHYSICS CHAPTER 8

124


a. The moment of inertia of the system initially is

The moment of inertia of the system finally is

By using the principle of conservation of angular momentum, thus

I i=I ssI m I i=I ssmrb2mr

b2 =I ss2 mr

b2

I i=5 .43 2 1 .25 0 .759 2=6 . 87 kg m2

I f =I ss2 mra2

= 5 .43 2 1.25 ra2

I i ω0=I f ω∑ Li=∑ Lf

6 . 87 18 .5 =5 .432 .5ra222 .2

m=1. 25 kg ; ω0=18. 5 rad s−1 ; I ss=5. 43 kg m2 ;r b=0 .759 m ; ω=22.2 rad s−1 ;

I f =5 . 432 .5ra2

PHYSICS CHAPTER 8

125


b. The initial rotational kinetic energy is given by

and the final rotational kinetic energy is

K r i=12 I i ω02

=12

6 .87 18 . 52

K r i=1 .18×103 J

K r f=12 I f ω2

=12

5 . 432 .5 0 .344 222. 2 2

m=1. 25 kg ; ω0=18. 5 rad s−1 ; I ss=5. 43 kg m2 ;r b=0 .759 m ; ω=22.2 rad s−1 ;

PHYSICS CHAPTER 8

126


1. A woman of mass 60 kg stands at the rim of a horizontal turntable having a moment of inertia of 500 kg m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about the frictionless vertical axle through its centre. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m s−1 relative to the Earth.a. In the what direction and with what value of angular speed

does the turntable rotate?b. How much work does the woman do to set herself and the

turntable into motion?ANS. : ANS. : 0.360 rad s0.360 rad s−−11 ,U think; 99.9 J ,U think; 99.9 J

PHYSICS CHAPTER 8

127

Exercise 8.5 :2. Determine the angular momentum of the Earth

a. about its rotation axis (assume the Earth is a uniform solid sphere), andb. about its orbit around the Sun (treat the Earth as a particle

orbiting the Sun). Given the Earth’s mass = 6.0 x 1024 kg, radius = 6.4 x 106 m

and is 1.5 x 108 km from the Sun.ANS. : ANS. : 7.1 x 107.1 x 103333 kg m kg m22 s s−−11; 2.7 x 10; 2.7 x 104040 kg m kg m22 s s−−11

3. Calculate the magnitude of the angular momentum of the second hand on a clock about an axis through the centre of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a thin rod rotating with angular velocity about one end. (Given the moment of inertia of thin rod about the axis through the CM is )

ANS. : ANS. : 4.71 x 104.71 x 10−−66 kg m kg m22 s s−−11

112 ML2

PHYSICS CHAPTER 8

128

Linear Motion Relationship Rotational MotionSummary:

m

v=rω ω=dθdt

a=dvdt α=

dωdt

∑ F=ma ∑ τ=Iα

W =Fs W =τθ

v=dsdt

a=rα

τ=rF sin θ

P=Fv P=τω

p=mv L=IωL=rpsin θ

II=∑i=1

n

mi ri2

129

PHYSICS CHAPTER 8


CHAPTER 9 :Simple Harmonic Motion (SHM)

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Physics Malaysian Matriculation Semester 1 Notes Complete