AP Physics

Chapter 11 Review

Rotational Motion

2€

L = mvRL = 1000Kg 50 m

sec( ) 100m( )

L = 5.0 ×106 Kg⋅m 2

sec

1. A car of mass 1000 kg moves with a speed of 50 on a circular track of radius 100 m. What is the magnitude of its angular momentum relative to the center of the race track?

€

msec

€

L = Iω

L = 12 mR2( )ω

L = 12 10Kg( ) 1m( )2 10 rad

sec( )

L = 50 Kg⋅m 2

sec

3

2. A solid cylinder of radius R = 1.0 m and mass 10 kg rotates about its axis. When its angular velocity is 10 , its angular momentum is ?

€

radsec

4

From Conservation of Energy,

€

v = 2gh = 2( ) 9.8 msec2( ) 2m( ) = 6.26 m

sec

L = mvR = 1.5Kg( ) 6.26 msec( ) 2m( )

L =18.8 Kg⋅m 2

sec

3. In the figure, a 1.5-kg weight swings in a vertical circle at the end of a string having negligible weight. The string is 2 m long. If the weight is released with zero initial velocity from a horizontal position, its angular momentum at the lowest point of its path relative to the center of the circle is approximately

4. A massless rope is wrapped around a uniform cylinder that has radius R and mass M, as shown in the figure. Initially, the unwrapped portion of the rope is vertical and the cylinder is horizontal. The linear acceleration of the cylinder is

5

6

#4

€

FNet = ma

Mg− IaR2 = Ma

Mg = Ma +IaR2

Mg = M +I

R2

a

Mg

M +I

R2

= a

€

Mg

M +12 MR2

R2

= a

MgM + 1

2 M( )= a

Mg32 M

= a

23

g = a

#5 7

5. A merry-go-round of radius R = 2.0 m has a moment of inertia and is rotating at 10 A child whose mass is 25 kg jumps onto the edge of the merry-go-round. The new angular speed (in ) of the merry-go-round is approximately€

I = 250Kg ⋅m2

€

revmin .

€

revmin

€

Li = Lf

Iiωi = I fω f

IMGRω i = IMGR + IC( )ω f

IMGRωi

IMGR + IC( )=ω f

250Kg ⋅m2

250Kg ⋅m2 + 25Kg 2m( )2 10 revmin( ) =ω f

7.1 revmin =ω f

8

€

FNet = ma

m2g− IaR2 − m1g = m1 + m2( )a

m2 − m1( )

m1 + m2 +I

R2

g = a

1Kg( )

3Kg +5Kg ⋅m 2

1m( )2

g = a

g8

= a

6.Two blocks, m1 = 1kg and m2 = 2kg, are connected by a light string as shown in the figure. If the radius of the pulley is 1 meter and its moment of inertia is the acceleration of the system is ?

€

5 Kg ⋅m2,

€

m1

€

m2

#7 €

mg

€

T cosθ

€

T sinθ€

θ

€

L = mvR

7. A conical pendulum bob of mass m, is set into motion in a circular path in a horizontal plane as shown in the figure. The angular momentum of the bob about the vertical axis through the point P is€

P

9

#7 7

€

mg

€

T cosθ

€

T sinθ€

θ

€

l

€

R€

θ

€

R = l sinθ

€

T =mg

cosθ

€

T sinθ =mv 2

RRT sinθ

m= v

10

11

€

L = mvR

L = m RT sinθm

l sinθ( )

L = ml sinθ( ) mg

cosθsinθ

m

l sinθ( )

L = mgl sin2θ( )

cosθ

l sinθ( )

L = ml sin2θgl

cosθ

#8 12

8. A thin rod of mass 2m and length l is struck at one end by a ball of clay of mass m, moving with speed v as shown in the figure. The ball sticks to the rod. After the collision, the angular momentum of the clay-rod system about A, the midpoint of the rod, is

90°

A l

€

ω

€

Li = Lf

mv l2

= 512 ml2ω f

v = 56 lω f

6v5l

=ω f

Find the angular velocity of the system after the collision.

13

€

I = 112 2m( )l 2 + m l

2( )2

I = 16 ml 2 + 1

4 ml 2( )I = 5

12 ml2

€

L = Iω

L = 512 ml 2ω

L = 512 ml 2 6v

5lL = 1

2 mlv

14

€

Li = Lf

mvR = 12 MR2 + mR2[ ]ω

mvR12 MR2 + mR2[ ]

=ω

0.1Kg( ) 5 msec( ) 0.2m( )

12 1Kg( ) 0.2m( )2

+ 0.1Kg( ) 0.2m( )2[ ]=ω

9. A particle of mass m = 0.10kg and speed v = 5.0 collides and sticks to the end of a uniform solid cylinder of mass M = 1.0kg and radius R = 20 cm. If the cylinder is initially at rest and is pivoted about a frictionless axle through its center, what is the final angular velocity? €

msec

€

0.1 Kg⋅m 2

sec

0.024Kg ⋅m2 =ω

4.16 radsec =ω

15

€

Li = Lf

IS + 2 5Kg( ) 1m 2( )[ ]ωi = IS + 2 5Kg( ) 0.1m 2( )[ ]ω f

15 revsec

5.1=ω f

3 revsec =ω f

10. A skater extends her arms horizontally, holding a 5-kg mass in each hand. She is rotating about a vertical axis with an angular velocity of one revolution per second. If she drops her hands to her sides, what will the final angular velocity (in ) be if her moment of inertia remains approximately constant at 5 Kg˙m2, and the distance of the masses from the axis changes from 1 m to 0.1 meters?

€

revsec

#1116

€

UTop = K + Kθ

mgh = 12 mv2 + 1

2 Iω 2

mgh = 12 mv2 + 1

212 mR2( ) v

R

2

gh = 34 v 2

43 gh = v

11. A solid cylinder rolls down an incline as shown in the figure. The l i n e a r acceleration of its center of mass is

θ

€

43 gh = v d =

hsinθ

v 2 = vo2 + 2ad

43 gh = 2a h

sinθ

23 g = a 1

sinθ23 gsinθ = a

17

12. A solid sphere, spherical shell, solid cylinder and a cylindrical shell all have the same mass, m and radius R. If they are all released from rest at the same elevation and roll without stopping, which reaches the bottom of an inclined plane first?

θ

€

UTop = K + Kθ1

mgh = 12 mv 2 + 1

2 Iω 2

mgh = 12 mv 2 + 1

2 mR2( ) vR

2

gh = v 2

gh = vRing

18

15€

UTop = K + Kθ1

mgh = 12 mv2 + 1

2 Iω 2

mgh = 12 mv2 + 1

225 mR2( ) v

R

2

gh = 710 v 2

107 gh = v

#12

Sphereθ

€

UTop = K + Kθ1

mgh = 12 mv 2 + 1

2 Iω 2

mgh = 12 mv 2 + 1

212 mR2( ) v

R

2

gh = 34 v 2

43 gh = v

cylinder

1. Sphere

2. Cylinder

3. Ring

16

21

€

0.3Kg

€

0.2Kg

€

2m#13

€

xCofG =mx∑m∑

=0.3Kg 2m( )

0.5Kg= 1.2m from the 0.2Kg mass.

€

I = mR2 = 0.2Kg( ) 1.2m( )2+ 0.3Kg( ) 0.8m( )2∑

I = 0.288 + 0.192( )Kg ⋅m 2

I = 0.48Kg ⋅ m2

13. Two particles (m1 = 0.20 kg, m2 = 0.30 kg) are positioned at the ends of a 2.0-m long rod of negligible mass. What is the moment of inertia of this rigid body about an axis perpendicular to the rod and through the center of mass?

14. A uniform rod (m = 2.0 kg, l = 3 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60° below the horizontal?

€

τ = Iα

mgcos60 l2

= 13 ml 2α

3g2l

cosθ =α =3 9.8 m

sec2( )2 3m( )

cosθ

α = 2.45 radsec2

22

€

θ€

2m

€

mg

15. A uniform rod (m = 1.5 kg) is 2.0 meters long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the instant it is released? What is the angular speed the instant the rod reaches the vertical position?

€

τ = Iα

mg l2

cosθ = 13 ml2( )α

g 12

cosθ = 13 l( )α

3g2l

cos30° =α

6.37 radsec2 =α

#15a

23

€

l2

€

l4

€

h =3l4

€

UTop = KBottom

mg 3l4

= 12 Iω2

mg 3l4

= 12

13 ml 2( )ω 2

g 32

= 13 l( )ω2

9g2l

=ω

#15b

24

16. Stars originate as large bodies of slowly rotating gas. Because of gravity, these clumps of gas slowly decrease in size. The angular velocity of a star increases as it shrinks because of

a. conservation of angular momentumb. conservation of linear momentumc. conservation of energyd. the law of universal gravitatione. conservation of mass

25

17. The object shown below has mass m and velocity v. The direction of its angular momentum vector with respect to an axis perpendicular to the page through point O is:

v

O

Right Hand Rule!Into the page

26

27

m

2m

O

€

2r

€

r

18. Two objects move in parallel circles around a rotation axis O with equal tangential speeds. r2 = r. r1 = 2r. m2 = 2m; m1 = m. Forces of equal magnitude are applied opposite to their velocities to stop them. Which statement is correct?

a. m1 will stop first because it has the smaller mass.b. m2 will stop first because it has the smaller radius.c. m1 will stop first because the torque on it is greater.d. m2 will stop first because it has the smaller moment of inertia.e. both objects will stop at the same time because the angular accelerations are equal.

19. Which of the following are different physical quantities but have the same units?

I. Work II. Kinetic Energy III. Torque

a. I and II only d. I, II, and IIIb. I and III only e. None of thesec. II and III only

28

€

W = F ⋅ x = Nm = J

K = 12 mv2 =

Kg ⋅m2

sec2 = Nm = J

τ = F⊥x = Nm

29

60°

20. A uniform thin rod of mass m and length l is suspended form one end by a frictionless pivot so that it can swing freely in the plane of the paper, as shown below. When the rod is at rest it is struck by a clay ball of equal mass m with an initial velocity vo at an angle of 60° with the vertical. The clay ball strikes the rod at the free end and sticks to it.

€

vo

26

€

Li = Lf

mvrsinφ = Irod + Imass( )ω f

mvo sin60°l = 13 ml2 + ml2( )ω f

vo sin60° = 43 l

vl

34 vo sin60° = v

20.2 The linear vilocity v of the free end just after the collision would be most nearly?

We could also use the cos 30°!

60°

€

vo

31

21. The engine of a car generates a constant power P. If the car starts from rest, and after t seconds its velocity is v, the speed of the car at 2t would be:

€

P =dWdt

and W = ΔK

dW = Pt = 12 mv 2 − 0

P is constant, soooooo

€

t∝v 2

and v∝ tat 2t,

€

v2

v=

2tt

v2 = 2v

32

€

Li = Lf

Imgr + IP( )ω i = Imgrω f

Imgr + IP( )

Imgr

ω i =ω f

400Kg ⋅m2 + 75Kg( ) 2m( )2( )400Kg ⋅m2 ω i =ω f

74ω i =ω f

22. A merry-go-round

is rotating at ωo with a 75 Kg person standing on the edge of the platform. The person then walks to the center of the platform. The new angular velocity of the merry-go-round is most nearly?

€

I = 400Kg ⋅m 2;R = 2m( )

33

€

Li = Lf

12 I( )ωo = I + 1

2 I( )ω f

12 I( )32 I

ωo =ω f

13ωo =ω f

23. A disk with a moment of inertia relative to a vertical axis through its center is dropped onto a disk having half its moment of inertia which is rotating with an angular velocity ωo. What is the final angular velocity of the system?

24. Two blocks of masses m1 and m2 are connected by a light cord that passes over a pulley of mass M, as shown. Block m2 slides on a frictionless horizontal surface. The blocks and pulley are initially at rest. When m2 is released, the blocks accelerate and the pulley rotates. The total angular momentum of the system of the two blocks and the pulley relative to the axis of rotation of the pulley is:

a. proportional to the radius of the pulley.b. proportional to the velocity of the blocks.c. proportional to the length of the strings.d. proportional to L1, the length of string from the pulley to m1.e. proportional to L2, the length of string from the pulley to m2.f. conserved, because the Earth doesn't move.g. true for all of the above.h. true only to (a) and (b) above.

€

m1

€

m2

35

€

ω =dθdt

= 2 0.4( )e2t

α =dωdt

= 4 0.4( )e2t

e2 0( ) =1α = 4 0.4( ) =1.6 rad

sec2

€

a = Rαa =1.6 rad

sec2 0.04m( )a = 0.064 m

sec2

€

θ = 0.4e2t

What is the magnitude of the total linear acceleration at t = 0 on a point on the object that is 4 cm from the axis?