AP PHYSICS 1 EXAM REVIEW - Mrs. Libretto's Physics ...mrslibretto.weebly.com/uploads/1/3/3/1/13312103/...AP PHYSICS 1 EXAM REVIEW By Karyn Libretto (Northport High School) Format of

AP PHYSICS 1 EXAM REVIEW By Karyn Libretto (Northport High School)

Format of Exam

¨  3 hours longs ¤  Part 1 (90 minutes)

n 50 multiple- choice questions n  Calculator & entire Reference Table allowed

¤  10 minute break ¤  Part 2 (90 minutes)

n 5 free-response questions n 1 experimental design question n 1 qualitative / quantitative translation n 3 short-answer question

n  Calculator & entire Reference Table allowed n  The two sections are weighted equally

Content of the Exam

¤ Newtonian Mechanics n Kinematics n Newton’s Laws of Motion n Torque n Rotational Motion & Angular Momentum n Gravitation & Circular Motion n Work, Energy & Power n Linear Momentum n Oscillations (SHM), Mechanical Waves & Sound n Introduction to Electric Circuits

¨  We have A LOT to cover so please help me set the pace! ¤ If you understand and want me to move faster say

so! ¤ If you don’t understand tell me to slow down or

repeat. n I don’t mind!

Topic #1: Methods & Tools

!

Metric & Prefix Conversions

A student wishing to determine experimentally the acceleration g due to gravity has an apparatus that holds a small sphere above a recording plate, as shown at right. When the sphere is released, a timer automatically begins recording the time of fall. The timer automatically stops when the sphere strikes the recording plate. The student measures the time of fall for different values of the distance D shown at right and records the data in the table. These data points are also plotted on the graph.

Graphing: Best Fit Line

¨  Line or curve with equal amount of points located above and below.

a) On the grid above, sketch the smooth curve that best represents the student’s data.

Graphing: Linearization


¨  The student can use these data for distance D and time t to produce a second graph from which the acceleration g due to gravity can be determined.

b) If only the variables D and t are used, what quantities should the student graph in order to produce a linear relationship between the two quantities?

Since distance & time are related by

d = ½ gt2, you should graph

distance vs (time)2

(You may also graph √d vs t)


Time2 (s2) 0.196 .1024 .2116 .3481 .3969

Dist

ance

(m)

2.5

2.0

1.5

1.0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 Time2 (s2)

c) On the grid below, plot the data points for the quantities you have identified in part (b), and sketch the best straight-line fit to the points. Label your axes and show the scale that you have chosen for the graph.

Graphing: Physical Significance of Slope & Area under Curve

Velo

city

(m/s

)

Time (s)

slope =

ΔyΔx

= ΔvΔt

= ?


Velo

city

(m/s

)

Time (s)

slope =

ΔyΔx

= ΔvΔt

= a = acceleration


Velo

city

(m/s

)

Time (s)

AreaΔ = 1/2 b i h = ?


Velo

city

(m/s

)

Time (s)

AreaΔ = 1/2 b i h

AreaΔ = 1/2 t i v = displacement

(d = v i t)

v = vi + vf

2 =

vf2

= 1/2 vf


Dist

ance

(m)

Time (s)

Pote

ntia

l Ene

rgy

(J)

Height (m)

Pres

sure

(Pa)

Volume (m3)

Forc

e (N

)

Time (s)

Forc

e (N

)

Spring Stretch (m) En

ergy

per

Pho

ton

(J)

Frequency (Hz)

?

? ?


Dist

ance

(m)

Time (s)

Pote

ntia

l Ene

rgy

(J)

Height (m)

Pres

sure

(Pa)

Volume (m3)

Forc

e (N

)

Time (s)

Forc

e (N

)


ergy

per

Pho

ton

(J)

Frequency (Hz)

? ?

?


Dist

ance

(m)

Time (s)

Pote

ntia

l Ene

rgy

(J)

Height (m)

Pres

sure

(Pa)

Volume (m3)

Forc

e (N

)

Time (s)

Forc

e (N

)


ergy

per

Pho

ton

(J)

Frequency (Hz)

? ?

Work on Gas


Dist

ance

(m)

Time (s)

Pote

ntia

l Ene

rgy

(J)

Height (m)

Pres

sure

(Pa)

Volume (m3)

Forc

e (N

)

Time (s)

Forc

e (N

)


ergy

per

Pho

ton

(J)

Frequency (Hz)

?

Work on Gas

Impulse


Dist

ance

(m)

Time (s)

Pote

ntia

l Ene

rgy

(J)

Height (m)

Pres

sure

(Pa)

Volume (m3)

Forc

e (N

)

Time (s)

Forc

e (N

)


ergy

per

Pho

ton

(J)

Frequency (Hz)

Work on Gas

Impulse Work or PEs


Dist

ance

(m)

Time (s)

Pote

ntia

l Ene

rgy

(J)

Height (m)

Pres

sure

(Pa)

Volume (m3)

Forc

e (N

)

Time (s)

Forc

e (N

)


ergy

per

Pho

ton

(J)

Frequency (Hz)

Work on Gas

Impulse Work or PEs fo

Φ


Dist

ance

(m)

2.5

2.0

1.5

1.0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 Time2 (s2)

. . . and back to our graph

What is the physical significance of our slope?

Graphing: Physical Significance of Slope d) Using the slope of your graph in part (c), calculate the acceleration g due to gravity in this experiment.

¨  Linear general equation: y = mx + b ¨  Specific equation: d = (4.94 m/s2) t2 + 0m ¨  Mathematical model: d = ½ gt2 (d = ½ at2)

¨  So. . . g = 2 (slope) = 2 (4.94 m/s2) = 9.88 m/s2 ( 9 – 11 m/s2 was acceptable on exam)

slope =

ΔyΔx

= 1.0 m - 0.2 m

.212 s2 - .05 s2 = 4.94 m/s2

slope =

ΔyΔx

= dt2

= 1/2 g

Error

¨  Systematic Error - an error associated with a particular instrument or experimental technique that causes the measured value to be off by the same amount each time.

¤  affects the accuracy of results ¤  can be eliminated by fixing source of error ¤  shows up as non-zero y-intercept on a graph

Error

¨  Random Uncertainty - an uncertainty produced by unknown and unpredictable variations in the experimental situation.

¤  affects the precision of results ¤  can be reduced by taking repeated trials but not eliminated ¤  shows up as error bars on a graph

Error

e) State one way in which the student could improve the accuracy of the results if the experiment were to be performed again. Explain why this would improve the accuracy.

¨  The student could increase the number of trials for each distance value and then averages the results. ¤  This reduces personal and random errors

¨  The student may perform the experiment in a vacuum to reduce air resistance. ¤  This reduces systematical errors.

Kinematics

¨  Average Speed (v)– the rate of change of distance

¨  Average Velocity ( )– the rate of change of displacement

v = d

t v =

v + v02

v = x - x0t - t0

= Δx

Δt

Kinematics

¨  When is the distance traveled by an object equal to its displacement?

¨  When is the average speed of an object equal to

the magnitude of its average velocity?

¨  When can an object be traveling at a constant speed but not at a constant velocity?

Kinematics


When it travels in a straight line ¨  When is the average speed of an object equal to

the magnitude of its average velocity?


Kinematics



the magnitude of its average velocity? When it travels in a straight line


Topic 1: Newtonian Mechanics



the magnitude of its average velocity? When it travels in a straight line


When it is turning

Kinematics


¨  Two friends bicycle 3.0 kilometers north and then turn to bike 4.0 kilometers west in 30. Minutes. ¤ What is their average speed?

¤ What is their average velocity?

Kinematics




Kinematics

v = d

t =

7.0 km0.50 hr

= 14 km/hr




Kinematics

v = d

t =

7.0 km0.50 hr

= 14 km/hr

v = Δx

t =

x2 + y2

t =

(4.0 km)2 + (3.0 km)2

0.50 hr =

5.0 km0.50 hr

= 10. km/hr at 53° west of north

θ = tan−1 O

A = tan−1

4.0 km3.0 km

= 53°


Kinematics ¨  Acceleration (a) – the rate of change of velocity

a = v - v0t - t0

= Δv

Δt


¨  What are the three different ways an object can accelerate?

¨  If an object is accelerating, is its. . . ¤ speed changing? ¤ velocity changing?

¨  Can an object have a velocity but no acceleration?

¨  Can an object have acceleration but no velocity?

Kinematics


¨  What are the three different ways an object can accelerate? Speed up, slow down, & turn

¨  If an object is accelerating, is its. . . ¤ speed changing? ¤ velocity changing?



Kinematics



¨  If an object is accelerating, is its. . . ¤ speed changing? maybe ¤ velocity changing?



Kinematics



¨  If an object is accelerating, is its. . . ¤ speed changing? maybe ¤ velocity changing? Yes



Kinematics



¨  If an object is accelerating, is its. . . ¤ speed changing? maybe ¤ velocity changing? Yes

¨  Can an object have a velocity but no acceleration? Yes, if it’s traveling at constant velocity (cruise control)


Kinematics



¨  If an object is accelerating, is its. . . ¤  speed changing? maybe ¤ velocity changing? Yes

¨  Can an object have a velocity but no acceleration? Yes, if it’s traveling at constant velocity (cruise control)

¨  Can an object have acceleration but no velocity? Yes, at the instant its motion starts or a vertical projection at

is apex (changing direction)

Kinematics


¨  Is a negative acceleration the same thing as deceleration?

Kinematics


¨  Is a negative acceleration the same thing as deceleration? No / Sometimes – deceleration means slowing down but negative

acceleration means accelerating in the negative direction. (If an object is moving to right and slowing down it is decelerating & has negative

acceleration)

General Rule ¨  If velocity and acceleration are in the same direction, than the object

is speeding up. ¨  If velocity and acceleration are in opposite directions, than the

object is slowing down.

Kinematics


Kinematics

Velocity vs. Time Graphs

Forwards Backwards

Slope is above the x-axis

Slope is below the x-axis


Kinematics


Kinematics AB BC

CD DE

AB DE

BD

AB CD

BC DE

C

A E


Kinematics


Kinematics AB DE

BD

NONE

NONE

NONE

NONE

B D

A C E What would a position time graph of an accelerating object look like?


Kinematics

!


Kinematics

!

!


Kinematics

The graph above shows the velocity versus time for an object moving in a straight line. At what time after t = 0 does the object again pass through its initial position?


Kinematics

Negative displacement

The graph above shows the velocity versus time for an object moving in a straight line. At what time after t = 0 does the object again pass through its initial position?

Positive displacement of equal area

Answer: Between 1 and 2 seconds


¨  The Law of Falling Bodies - in the absence of air resistance, all bodies near the surface of the Earth fall with the same constant acceleration (g). ¤ g = 9.8 m/s2 near the surface of the Earth ¤ direction for vector g = - depends on frame of

reference ¤ Approximate g as10 m/s2 on MC section of exam

Kinematics


Kinematics


¨  A stone is dropped from rest from the top of a tall building. After 3.00 s of free-fall, what is the displacement y of the stone? What is its velocity?

Kinematics


¨  A stone is dropped from rest from the top of a tall building. After 3.00 s of free-fall, what is the displacement y of the stone? What is its velocity?

Kinematics

y = y0 + v0t + ½ at2

y = ½ (-9.8 m/s2)(3.0 s)2

y = - 44.1 m = 44.1 m down v = v0 + at

v = (-9.8 m/s2)(3.0 s)

v = -29.4 m/s


Kinematics


¨  Examining half of flight ¤  ttop = ½ total time ¤ vf = 0 m/s

¨  Examining entire flight ¤ Use ttotal ¤ vf = - vi

Kinematics

Always treat g as being

negative for these

problems!!!


y = y0 + v0t + 1/2at2

Kinematics

Final displacement is negative (downward)

Initial position is zero

g is negative

t = - (v0) ± (v0)

2 - 4 12

a⎛⎝⎜

⎞⎠⎟

(y)

2 12

a⎛⎝⎜

⎞⎠⎟

= - (v0) ± (v0)

2 - 2(a)(y)(a)


Kinematics Analyze the horizontal and vertical components of the projectile.


Kinematics Analyze the horizontal and vertical components of the projectile.

v =

dt

y = y0 +v0t + 1/2 at2

&v = v0 +at


Kinematics What is the initial vertical velocity of a projectile launched horizontally?


Kinematics


Kinematics

vix = v cos θ

viy = v sin θ


¨  Maximum height occurs at ¤ _____ total time. ¤ _____ degrees.

¨  Maximum range occurs at ¤ _____ total time. ¤ _____ degrees.

¨  Maximum hang time occurs at ¤ _____ degrees.

Kinematics


¨  Maximum height occurs at ¤ 1/2 total time. ¤ 90 degrees.

¨  Maximum range occurs at ¤  total time. ¤ 45 degrees.

¨  Maximum hang time occurs at ¤ 90 degrees.

Kinematics


Statics & Dynamics


Newton’s Laws of Motion ¨  An object continues in uniform motion in a straight line or remains at

rest unless a resultant (net) external force acts on it. ¨  When unbalanced forces act on an object, the object will accelerate

in the direction of the resultant (net) force. The acceleration will be directly proportional to the net force and inversely proportional to the object’s mass.

¨  When two bodies A and B interact, the force that A exerts on B is

equal and opposite to the force that B exerts on A.

Statics & Dynamics


Statics & Dynamics


¨  Mass (m) – the amount of matter that comprises an object and determines its resistance to a change in its motion (inertia) ¤ Units – kg

¨  Weight (W or Fg) – the magnitude of the force of gravity acting upon an object. ¤ Units - N

Statics & Dynamics

g =

!Fgm

¨  Gravitational mass – property of mass that produces a gravitational field. ¤ measured with the use of a double-pan or triple-beam

balance

¨  Inertial mass – an object’s resistance to acceleration ¤ measured with the use of an inertial balance, or spring-

loaded pan

a = ∑

!F

m = !Fnetm


¨  Density (ρ) - a measurement of how tightly matter is compacted.

¤ Units – kg/m3 ρ = m

V

V = ℓwh V = πr2ℓ V = 4

3πr3

Rectangular Solid Cylinder Sphere


¨  Normal Force (FN)– an electromagnetic repulsive contact force between two objects, acting perpendicular to their surface of contact.

n Tensional Force (FT) – the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object.

Statics & Dynamics

What’s the normal force acting on these block?


Statics & Dynamics

Newton’s Laws of Motion

¨  Frictional Force (Ff) – an oppositional force due to an electromagnetic force of attraction between surfaces.

!Ff ≤ u

!Fn

Newtonian Mechanics

¨  Net Force (ΣF, Fnet) – the sum of all forces acting on an object (resultant force)

a = ∑

!F

m = !Fnetm

General Strategy: ¤ One body problem or two body

problem? ¤ Sketch and label a free-body

diagram. ¤ Decide if the object is in

equilibrium or not. ¤ Break forces into perpendicular

components, if appropriate. ¤ Write a net force equation for

each dimension and solve.

¨  Equilibrium:

ΣF = 0

¨  Non-equilibrium:

ΣF = m a

Newton’s Laws of Motion

How?


¨  2. A 12.0-kg lantern is suspended from the ceiling by two wires, each at an angle of 300 from the vertical. What is the tension in each wire?

Statics & Dynamics


¨  2. A 12.0-kg lantern is suspended from the ceiling by two wires, each at an angle of 300 from the vertical. What is the tension in each wire?

Statics & Dynamics

Σ F = 0 N(FT cos θ + FT cos θ ) + -Fg = 0 N2FT cos θ = mg

FT = (12 kg)(9.8 m/s2)

2 i cos 30°FT = 68 N


¨  A sled is pulled with a force at an angle on a frictionless floor. Write an expression for the normal force

Statics & Dynamics


¨  A sled is pulled with a force at an angle on a frictionless floor. Write an expression for the normal force

Statics & Dynamics

FTy

Fg

FN

ΣFy = 0 N FTy + FN + -Fg = 0 N

FN = Fg - FT sinθ FN = mg - FT sinθ


Elevators

¨  In each case, the scale will read the normal or reaction force, not the weight ¨  Apparent weightlessness: sensation due to lack of

normal force

Statics & Dynamics


¨  Calculate the acceleration of the man. Statics & Dynamics


¨  Calculate the acceleration of the man. Statics & Dynamics

Σ F = ma FN + -Fg = m a 1000N + (–700 N) = (70 kg) a a = +4.3 m/s2


Two Body Problems ¨  Two masses are hung as shown over a frictionless, massless

pulley. Determine their acceleration and the tension in the rope.

Statics & Dynamics


Two Body Problem

Statics & Dynamics

Both blocks accelerate together, so treat as one large block.

– +


Two Body Problem

Statics & Dynamics

Both blocks accelerate together, so treat as one large block.

– +

Σ F = (m1 + m2) aFg1 + -Fg2 = (m1+ m2) a

a = m1g + -(m2g)

(m1+ m2)

a = 7 kg(9.8 m/s2)⎡⎣ ⎤⎦ + - 9 kg(9.8 m/s

2)⎡⎣ ⎤⎦(7 kg + 9 kg)

a = -1.23 m/s2

Negative = left


Two Body Problem

Statics & Dynamics

Σ F = (m1 + m2) aFg1 + -Fg2 = (m1+ m2) a

a = m1g + -(m2g)

(m1+ m2)

a = 7 kg(9.8 m/s2)⎡⎣ ⎤⎦ + - 9 kg(9.8 m/s

2)⎡⎣ ⎤⎦(7 kg + 9 kg)

a = -1.23 m/s2

Negative = counterclockwise


Two Body Problem

Statics & Dynamics

To find tension pick ONE block!

Σ F = m1 aFT1 + -Fg = m1 a

FT1 = (m1 a) + Fg = (m1a) + (m1g)

FT1 = 7 kg (+1.23 m/s2)⎡⎣ ⎤⎦ + 7 kg (9.8 m/s

2)⎡⎣ ⎤⎦FT1 = 77 N = 77 N up

–

+

Block 1 accelerates up so a is +

Block 2 accelerates down so a is -

Σ F = m2 aFT2 + -Fg = m2 a

FT2 = (m2 a) + Fg = (m2a) + (m2g)

FT2 = 9 kg (-1.23 m/s2)⎡⎣ ⎤⎦ + 9 kg (9.8 m/s

2)⎡⎣ ⎤⎦FT2 = 77 N = 77 N up

FT = m (a + g)


Two Body Problem with Friction

¨  Find the acceleration of this system and the tension in the rope if the coefficient of kinetic friction between the box and the tabletop is 0.20.

Statics & Dynamics



Statics & Dynamics

Σ F = (mA + mB)a-FgB + FfA = (mA + mB)a

-FgB + µFNA = (mA + mB)a

-FgB + µ -FgA = (mA + mB)a

a = -FgB + µ(-mA i g)

(mA + mB)

a = -mBg + µ(-mA i g)

(mA + mB)

a = -19.6 N + 0.20(5.0 kg i 9.8 m/s2)

(5.0 kg + 2.0 kg)a = -1.4 m/s2



Statics & Dynamics

Σ F = maFT - Fg= m a

FT= m g + m a

FT= m (g + a)

FT= 2.0 kg (9.8 m/s2 + -1.4 m/s2)

FT= 16.8 N

Block B accelerates down so a is -


Inclined Planes Statics & Dynamics

Fg⊥

Fg II

Fg

Fg - weight

Fg = mg

Fg II - pulls object downhill

Fg II = Fg sin θ

Fg⊥ - pulls object into hill

Fg⊥ = Fg cos θ


¨  A girl accelerates down a frictionless slide inclined at 38°. Determine the acceleration of the girl.

Statics & Dynamics


¨  A girl accelerates down a frictionless slide inclined at 38°. Determine the acceleration of the girl.

Statics & Dynamics

Σ F = ma

a = Σ Fm

a = -FgIIm

= m i g i sinθ

ma = -(9.8 m/s2)(sin 38°)a = 6.0 m/s2


¨  Write an expression for the steepest angle at which a box can rest on a hill before sliding (the angle of repose).

Statics & Dynamics


¨  Write an expression for the steepest angle at which a box can rest on a hill before sliding (the angle of repose).

Statics & Dynamics

Ff = µ FNmg sinθ = µ mg cosθ

µ = sinθcosθ

µ = tanθθ = tan-1µs

¨  Period ( T ) – time it takes to complete one cycle around a circle

Gravitation & Circular Motion

T =

time# revolutions

v =

dt

= circumference

period =

2rT

¨  The direction of the object’s instantaneous velocity is always tangent to the circle in the direction of motion.

¨  Its instantaneous velocity is always perpendicular to a radius drawn to the point of tangency.

¨  Since the direction of the object’s motion is always changing, its velocity is always changing therefore the object is always accelerating.

¨  The force causing the change in direction is an inwardly directed (centripetal) force.

¨  According to Newton’s second law of motion, an object always accelerates in the direction of net force so the acceleration must also be inwardly directed (centripetal).

¨  If the object is constantly subject to a net force and always accelerating, then it can never be in equilibrium.


Fgrav

Fnorm Ffric

Ftens


¨  Determine the maximum speed at which a car can safely negotiate an unbanked turn.


¨  Determine the maximum speed at which a car can safely negotiate an unbanked turn.


FF = Fc

µ iFN = mv2

r

v = µ img i r

m

v = µ i g i r

¨  Write an expression for the speed of the chair in terms of g, L, and θ .




FTy = mg AND FTx = mv2

r

FT cos θ = mg AND FT sinθ = mv2

rmg

cosθ = FT =

mv2

r i sinθ

m i g i r i sinθm(cosθ )

= v

v = L i g i tanθ

v = (Lsinθ ) i g i tanθ

r = L sinθ




r = L sinθ This formula shows that the angle and the radius depend on the velocity alone and not the mass of the chairs. A good job too if you think about an actual ride - if this wasn't true the wires would all get tangled up if the people in the chairs had different masses.

v = (Lsinθ ) i g i tanθ

¨  What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away?


¨  What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away?


FFY = Fg AND Fcx = FN

FFY = mg AND FN = mv2

rµ iFN = mg

µ i mv2

r⎛⎝⎜

⎞⎠⎟

= mg

µ = mgrmv2

µ = r i gv2


Treat inward as + (positive)

∑F = mv2

r

Fn - Fg = mv2

r

Fn = mv2

r + Fg

∑F = mv2

r

Fn + Fg = mv2

r

Fn = mv2

r - Fg

Bottom: Top:

¨  Write an expression for the minimum speed the cyclist needs in order to complete the loop.


¨  Write an expression for the minimum speed the cyclist needs in order to complete the loop.

Calculate at top, which is the critical point for making the loop.


∑F = FN + FgFc = FN + Fgmv2

r = FN + mg

FN = mv2

r - mg

At minimum or critical speed FN = 0

0 = mv2

r - mg

mv2

r = mg

v = r i g

Roller coasters need to EXCEED the critical velocity, whereas dryers need a lower velocity

to allow the cloths to fall and >luff.

You can use the above equation in reverse to see how fast you should go to loose contact with the road when driving over a hill to feel weightless.


¨  Newton’s Law of Universal Gravitation - Every point mass attracts every single other point mass by a force pointing along the line intersecting both points.

!Fg =

Gm1m2r2

G = 6.67 x 10-11 m3/kg s2


¨  A force of 400 N is measured between two objects. Determine the new force of attraction when:

¤  both masses double.

¤  the separation distance doubles.

¤  the separation is decreased to 1/3rd of its original value






Fgrav=

Gm1m2r2

= 1 ⋅2 ⋅212

= 4 = 1600 N






Fgrav=

Gm1m2r2

= 1 ⋅2 ⋅212

= 4 = 1600 N

Fgrav=

Gm1m2r2

= 1 ⋅1 ⋅122

= 14

= 100 N






Fgrav=

Gm1m2r2

= 1 ⋅2 ⋅212

= 4 = 1600 N

Fgrav=

Gm1m2r2

= 1 ⋅1 ⋅122

= 14

= 100 N

Fgrav=

Gm1m2r2

= 1 ⋅1 ⋅1(1 / 3)2

= 9 = 3600 N

g - acceleration due to gravity or gravitational field strength



The gravitational potential energy at an infinite distance away is ZERO. The gravitational potential energy near a planet is then

negative, since gravity does positive work as the mass approaches. This negative potential is indicative of a "bound state"; once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape.

Ug = -

Gm1m2r

ΔUg = mgΔy


¨  Orbital Speed of a Satellite ¤ Write an expression for this “orbital speed.”



¨  Orbital Speed of a Satellite ¤ Write an expression for this “orbital speed.”


FC = FgmSv

2

r =

GmEmSr2

v = GmE

r


¨  How fast would this 20. meter diameter space station have to spin to simulate earth’s gravity?



¨  How fast would this 20. meter diameter space station have to spin to simulate earth’s gravity?


ac = v2

rv = r i a

v = (10. m)(9.8 m/s2) v = 9.9 m/s


Torque Gravitation & Circular Motion

Torque

Torque

Sign convention: CW -, CCW +

τ = r⊥F = rF sinθ


¨  A 20. kg person and a 5.0 kg dog want to balance on a 8.0 m uniform 12. kg board that is centered on a pivot. If the dog sits at one end of the see-saw, how far away does the person have to sit to balance it?



¨  A 20. kg person and a 5.0 kg dog want to balance on a 8.0 m uniform 12. kg board that is centered on a pivot. If the dog sits at one end of the see-saw, how far away does the person have to sit to balance it?


Forces ΣF = 0

Torques


Torque 1.  forces through the axis of rotation are not torques

2.  don’t neglect the weight of the board (pole) if it is not supported at its center of mass – force and torque

3.  use the see-saw law if the only torques are due to weight and the object is in rotational equilibrium



¨  How much force is the person applying to hold up the end of the 12 kg board that is 8.0 meters long? How much normal force is the fulcrum applying?





1.0 m

FN FA = τA

Fg = τg



Forces Torques


ΣF = 0 N FA + FN + -‐FG = 0

Solve torques >irst



Forces Torques


ΣF = 0 N FA + FN + -‐FG = 0

Solve torques >irst

Στ = 0 Nm -‐FGr1 + FAr2 = 0 Nm

FGr1 = FAr2 m g r1 = FAr2

(12 kg)(9.8 m/s2)(1.0 m) = FA(5.0 m) FA = 24 N



Forces Torques


ΣF = 0 N FA + FN + -‐FG = 0

Solve torques >irst

24 N + FN – 120 N = 0

FN = 96 N

Στ = 0 Nm -‐FGr1 + FAr2 = 0 Nm

FGr1 = FAr2 m g r1 = FAr2

(12 kg)(9.8 m/s2)(1.0 m) = FA(5.0 m) FA = 24 N


¨  A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of plank hanging over the right support (see the drawing). To what distance x can a person who weights 450 N walk on the overhanging part of the plank before it just begins to tip?



¨  A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of plank hanging over the right support (see the drawing). To what distance x can a person who weights 450 N walk on the overhanging part of the plank before it just begins to tip?


As the plank just begins to tip, a normal force will cease to exist at the far left end. Thus, we can treat this like a see-saw

Στ = 0 Nm Σmr = Σmr m1r1 = m2r2

(22.5 kg)(1.4 m)= (45 kg)(x) r = 0.70 m

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