Study Questions/Problems Week 7

Chapters 10 introduces the motion of extended bodies,

necessitating a description of rotation---something a point mass can’t do. This chapter covers many aspects of rotation;

hence, a long list of problems.

Chapter 10: Conceptual Questions 1-10

Conceptual Exercises 4, 7, 8, 9, 11 Problems 1, 3, 6, 10, 14, 18, 25, 29, 33, 41, 44, 48, 51,

54, 55, 57, 58, 60, 63

Answers/solutions for all even numbered CQs and CEs, and

for each of the problems listed above are on the following pages—don’t peek until you have done your best to solve a problem.

Chapter 10: Rotational Kinematics and Energy

Answers to Even-Numbered Conceptual Questions 2. Yes. In fact, this is the situation whenever you drive in a circular path with constant speed.

4. Every point on Earth has the same angular speed. Therefore, the smallest linear speed occurs where the distance from the axis of rotation is smallest; namely, at the poles.

6. The moment of inertia of an object changes with the position of the axis of rotation because the distance from the axis to all the elements of mass have been changed. It is not just the shape of an object that matters, but the distribution of mass with respect to the axis of rotation.

8. Spin the two spheres with equal angular speeds. The one with the larger moment of inertia—the hollow sphere—has the greater kinetic energy, and hence will spin for a longer time before stopping.

10. (a) What determines the winner of the race is the ratio I mr

2 , as we see in the discussion

just before Conceptual Checkpoint 10-4. This ratio is MR2

/ MR2= 1 for the first hoop and

2M( )R2

/ 2M( )R2

= 1 for the second hoop. Therefore, the two hoops finish the race at the

same time. (b) As in part (a), we can see that the ratio I/mr2 is equal to 1 regardless of the radius. Thus, all hoops, regardless of their mass or radius, finish the race in the same time.

Answers to Even-Numbered Conceptual Exercises 2. The angular speed of an object with constant angular acceleration changes linearly with

time. Therefore, at the time t/2 the angular speed of this object is /2.

4. (a) Betsy and Jason have the same angular speed, . Therefore, the ratio is 1. (b) Jason’s

linear speed

vt= r( ) is one half of Betsy’s linear speed, giving a ratio of 1/2. (c) Jason’s

centripetal acceleration

acp

= v2

r = r2( ) is one half of Betsy’s centripetal acceleration. The

ratio, therefore, is 1/2.

6. At the top of a tall building, your distance from the axis of rotation of the Earth is greater than it was on the ground floor. Therefore, your linear speed at the top of the building is greater than it is on the ground floor.

8. The long, thin minute hand—with mass far from the axis of rotation—has the greater moment of inertia.

10. The center of the outer quarter moves in a circle that has twice the radius of a quarter. As a result, the linear distance covered by the center of the outer quarter is twice the circumference of a quarter. Therefore, if the outer quarter rolls without slipping, it must complete two revolutions.

12. The moment of inertia of the Earth was increased slightly, because mass (water in the dam) was moved from a lower elevation to a higher elevation as the dam was filled. Raising this mass to a higher elevation moves it farther from the axis of rotation, which increases the moment of inertia.

14. The ranking is as follows: case 1 < case 2 < case 3. To be specific, the moment of inertia for case 1 (x axis) is I = (9.0 kg)(1.0 m)2 = 9.0 kg·m2; for case 2 (y axis) it is I = (2.5 kg)(2.0 m)2 = 10.0 kg·m2; and for case 3 (z axis) it is I = (9.0 kg)(1.0 m)2 + (2.5 kg)(2.0 m)2 = 19.0 kg·m2.

16. The ranking is as follows: disk 1 = disk 2 = disk 3. All uniform disks finish the race in the same time, regardless of their mass and/or radius.

See following pages for Problem Solutions

1. Picture the Problem: This is a units conversion problem.

Strategy: Multiply the angle in degrees by

radians

180° to get radians.

Solution:

30° rad

180°=

6rad 45°

rad

180°=

4rad 90°

rad

180°=

2rad 180°

rad

180°= rad

Insight: The quantity is the circumference of a circle divided by its diameter. = 3.1415926536 ...

3. Picture the Problem: The minute and hour hands of the clock rotate at constant angular speed.

Strategy: Determine the time for each hand to complete a revolution. Multiply the revolutions per time by 2 radians per revolution and convert the time units to seconds to obtain an angular speed in radians/second.

Solution: 1. (a) The minute

hand completes a cycle once

per hour:

1 rev

hr

2 rad

rev

1 hr

3600 s= 1.745 10

3 rad/s

2. (b) The hour hand

completes

two revolutions per day:

2 rev

day

2 rad

rev

1 day

24 hr

1 hr

3600 s= 1.454 10 4 rad/s

Insight: The angular speed of the hour hand is 12 times less than the angular speed of the minute hand

because it completes a cycle once every 12 hours.

6. Picture the Problem: The tire rotates about its axis through a certain angle.

Strategy: Use equation 10-2 to find the angular displacement.

Solution: Solve equation 10-2 for :

=s

r=

1.75 m

0.33 m= 5.3 rad

Insight: This angular distance corresponds to 304° or 84% of a complete revolution.

10. Picture the Problem: The floppy disk rotates about its axis at a constant angular speed.

Strategy: Use equation 10-5 to relate the period of rotation to the angular speed. Then use equation 10-12 to find the linear speed of a point on the disk’s rim.

Solution: 1. (a) Solve equation

10-5 for :

=2

T=

2

0.200s= 31.4 rad/s

2. (b) Apply equation 10-12

directly: v

T= r = 1

23.5 in( ) 31.4 rad/s( ) = 55 in/s 1 m 39.4 in = 1.4 m/s

3. (c) A point near the center will have the same angular speed as a point on the rim because the

rotation periods are the same.

Insight: While the angular speed is the same everywhere on the disk, the linear speed is greatest at the

rim. The read/write circuitry must compensate for the different speeds at which the bits of data will

move past the head.

14. Picture the Problem: The propeller rotates about its axis, increasing its angular velocity at a constant

rate.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the

angular acceleration.

Solution: Solve equation 10-11 for

:

=

2

0

2

2=

26 rad/s( )2

12 rad/s( )2

2 2.5 rev 2 rad rev( )= 17 rad/s

2

Insight: A speed of 26 rad/s is equivalent to 250 rev/min, indicating the motor is turning pretty slowly.

A typical outboard motor is designed to operate at 5000 rev/min at full throttle.

18. Picture the Problem: The discus thrower rotates about a vertical axis through her center of mass,

increasing her angular velocity at a constant rate.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the number of revolutions through which the athlete rotates and the time elapsed during the specified interval.

Solution: 1. (a) Solve equation

10-11 for :

=0=

2

0

2

2=

6.3 rad/s( )2

02

2 2.2 rad/s2( )

= 9.0 rad 1 rev 2 rad

= 1.4 rev

2. (b) Solve equation 10-8 for t:

t =0=

6.3 0 rad/s

2.2 rad/s2

= 2.9 s

Insight: Notice the athlete turns nearly one and a half times around. Therefore, she should begin her

spin with her back turned toward the range if she plans to throw the discus after reaching 6.3 rad/s. If

she does let go at that point, the linear speed of the discus will be about 6.3 m/s (for a 1.0 m long arm)

and will travel about 4.0 m if launched at 45° above level ground. Not that great compared with a

championship throw of over 40 m (130 ft) for a college woman.

25. Picture the Problem: The hour hand rotates about its axis at a constant rate.

Strategy: Convert the angular speed of the tip of the hour hand into a linear speed by

multiplying by its radius (equation 10-12).

Solution: Apply equation

10-12 directly:

vt= r = 8.2 cm( )

1 rev

12 h

2 rad

rev

1 h

3600 s= 0.0012 cm/s = 12 μm/s

Insight: The tip of a minute hand travels much faster, not only because its angular speed is 12 times

faster than the hour hand, but also because the minute hand is longer than the hour hand.

29. Picture the Problem: Jeff clings to a vine and swings along a vertical

arc as depicted in the figure at right.

Strategy: Use equation 10-12 to find the angular speed from the

knowledge of the linear speed and the radius. Use equation 6-15 to

find the centripetal acceleration from the speed and the radius of

motion.

Solution: 1. (a) Solve

equation

10-12 for :

=v

t

r=

8.50 m/s

7.20 m= 1.18 rad/s

2. (b) Apply equation 6-15

directly:

a

cp=

vt

2

r=

8.50 m/s( )2

7.20 m= 10.0 m/s2

3. (c) The centripetal force required to keep Jeff moving in a circle is provided by the vine.

Insight: The vine must actually do two things, support Jeff’s weight and provide his centripetal force.

That is why it is possible that the vine is strong enough to support him when he is hanging vertically

but not strong enough to support him while he is swinging. There’s no easy way for him to find out

without trying! But he should wear a helmet…

33. Picture the Problem: The reel rotates about its axis at a constant rate.

Strategy: Find the linear speed of the string by using the radius of the reel and the angular speed

according to equation 10-12.

Solution: 1. (a) Apply equation 10-12

directly:

vt= r = 3.7 cm( )

3.0 rev

s

2 rad

rev= 0.70 m/s

2. (b) If the radius of the reel were doubled but the angular speed remained the same, the linear speed

would also double as can be seen by an examination of equation 10-12.

Insight: If the fish grabbed the lure when there was 10 m of string between it and the reel, it will take

14 seconds to reel in the big one.

41. Picture the Problem: The tires roll without slipping at constant speed.

Strategy: Because the tires roll without slipping, equation 10-15 describes the direct relationship

between the center of mass speed and the angular velocity of the tires. Solve the equation for the

angular velocity.

Solution: Solve equation 10-15 for :

=v

t

r=

12 m/s

0.31 m= 39 rad/s

Insight: If the tires spin freely, as when the car is stuck in snow, the center of mass speed can be zero

while the angular velocity is very high.

44. Picture the Problem: Your car’s tires roll without slipping, increasing their velocity at a constant rate.

Strategy: Use the fact that the tires roll without slipping to find the angular acceleration and angular

velocity from their linear counterparts. Then use the kinematic equations for rotation (equations 10-8

through 10-11) to determine the angle through which the tire rotated during the specified interval.

Solution: 1. Solve equation 10-12

for 0

0=

v0

r=

17 m/s

0.33 m= 52 rad/s

2. Solve equation 10-14 for :

=a

r=

1.12 m/s2

0.33 m= 3.4 rad/s

2

3. Apply equation 10-10 directly:

=0t + 1

2t

2= 52 rad/s( ) 0.65 s( ) + 1

23.4 rad/s

2( ) 0.65 s( )2

= 35 rad = 5.5 rev

Insight: Another way to solve this question is to find the final angular speed (54 rad/s) and then use

= 1

2+

0( ) t to find the answer.

48. Picture the Problem: The fan blade rotates about its axis with a constant angular speed.

Strategy: Use equation 10-17 for the kinetic energy of a rotating object to solve for the moment of

inertia.

Solution: Solve equation 10-17 for I:

I =2K

2=

2 4.1 J( )

12 rad/s( )2= 0.057 kg m2

Insight: If the angular speed of the fan blade were to double to 24 rad/s, the kinetic energy would

quadruple to 16 J.

51. Picture the Problem: The ball rotates about its center with a constant angular velocity.

Strategy: Use equation 7-6 to find the translational kinetic energy and equation 10-17 to find the

rotational kinetic energy of the curveball.

Solution: 1. Apply equation 7-6

directly: K

t= 1

2Mv

2= 1

20.15 kg( ) 46 m/s( )

2= 160 J

2. Use I =

2

5MR

2 for a uniform

sphere

in equation 10-17:

Kr= 1

2I

2= 1

2

2

5MR

2( ) 2= 1

5MR

2 2

= 1

50.15 kg( ) 0.037 m( )

241 rad/s( )

2= 0.069 J

Insight: Only a tiny fraction of the total kinetic energy is used to spin the ball, but it has a marked

effect on the trajectory of the pitch!

54. Picture the Problem: The lawn mower blade rotates at constant speed about its axis.

Strategy: Find the rotational kinetic energy of the blade using equation 10-17, then set that energy equal to its gravitational potential energy and solve for

y

max. Treat the blade as a uniform rod rotating about its center, so that its

moment of inertia is

1

12ML

2 as indicated in Table 10-1.

Solution: 1. (a) Use equation 10-17 to find K

r:

Kr= 1

2I

2= 1

2

1

12ML

2( ) 2

= 1

240.58 kg( ) 0.56 m( )

2 3500 rev

min

2 rad

rev

1 min

60 s

2

Kr= 1000 J = 1.0 kJ

2. (b) Set K

r= U and solve for

y

max:

Kr= U = mgy

max

ymax

=K

r

mg=

1000 J

(0.58 kg)(9.81 m/s2 )= 180 m = 0.18 km

Insight: 180 m is 590 ft straight up! The rotation rate is made to be so high in order that linear speed of the outside tip

of the mower blade (100 m/s = 230 mi/h) is high enough to cleanly slice the blades of grass.

55. Picture the Problem: The block rises upward, slows down, and

comes to rest as the kinetic energy of the wheel and the block is

converted into the block’s gravitational potential energy.

Strategy: Use the formula from step 4 of Example 10-6 that was

derived from conserving kinetic energy between the initial and

final states. Solve the expression for the moment of inertia of the

wheel.

Solution: 1. Write out the

equation from

step 4 of Example 10-6:

h =v2

2g1+

I

mR2

2. Solve the expression for I:

I = mR2 2gh

v21

3. Insert the numerical values:

I = 2.1 kg( ) 0.080 m( )2 2(9.81 m/s2 )(0.074 m)

(0.33 m/s)21 = 0.17 kg m2

Insight: If the wheel were massless the block would rise to a height of only 5.6 mm. This is because a

substantial amount of energy is stored in the rotation of the massive wheel, and that energy can be used

to lift the block much further than if it had only its own kinetic energy to use.

57. Picture the Problem: The disk and the hoop roll without slipping down

the incline plane, starting from rest at height h above the level surface.

Strategy: Conserve mechanical energy, including both translational and rotational kinetic energy, in order to find the speeds of the objects at the bottom of the incline. Use equation 10-15 to relate the rotational speed to the linear speed of each object.

Solution: 1. (a) Set

E

i= E

f and let

= v r :

Ui+ K

i= U

f+ K

f

mgh + 0 = 0 + 1

2mv2

+ 1

2I 2

= 1

2mv2

+ 1

2I v r( )

2

2. For the disk

I =

1

2mr

2:

mgh = 1

2mv2

+ 1

2

1

2mr2( ) v r( )

2

gh = 1

2v2

+ 1

4v2

= 3

4v2

v = 4

3gh = 4

39.81 m/s

2( ) 0.82 m( ) = 3.3 m/s

3. (b) For the hoop I = mr

2:

mgh = 1

2mv2

+ 1

2mr2( ) v r( )

2

gh = 1

2v2

+ 1

2v2

= v2

v = gh = 9.81 m/s2( ) 0.82 m( ) = 2.8 m/s

Insight: Note that the final speeds are independent of the masses and the radii of the objects. A disk will always beat a

hoop as long as their centers of mass begin at the same height h above the level surface.

58. Picture the Problem: The larger mass falls and the smaller mass

rises until the larger mass hits the floor.

Strategy: Use conservation of mechanical energy, including the rotational energy of the pulley, to determine the mass of the pulley. Because the rope does not slip on the pulley, there is a direct relationship

v = r

pbetween the rotation of the pulley and the

linear speed of the rope and masses.

Solution: 1. (a) Equate the initial and final mechanical energies,

then solve for the mass of the pulley.

2. (b) Set

E

i= E

f and let

= v r

p:

Ui+ K

i= U

f+ K

f

m1gh + 0 + 0 = m

2gh + 1

2m

1v2

+ 1

2m

2v2

+ 1

2I

p

2

m1

m2( ) gh = 1

2m

1+ m

2( )v2+ 1

2

1

2m

prp

2( ) v rp( )

2

3. Rearrange

the equation

and solve for

m

p:

1

4m

pv2

= m1

m2( ) gh 1

2m

1+ m

2( )v2

mp=

4 m1

m2( ) gh 1

2m

1+ m

2( )v2

v2

=

4 2.0 kg( ) 9.81 m/s2( ) 0.75 m( ) 1

28.0 kg( ) 1.8 m/s( )

2

1.8 m/s( )2

mp= 2.2 kg

Insight: By the time the masses reach 1.8 m/s, 1.8 J or 12% of the 15 J of total kinetic energy is stored

in the kinetic energy of the pulley, so the pulley plays a minor role in the energy balance of the system.

60. Picture the Problem: The ball rolls without

slipping up the incline and continues rolling at a

constant but lower speed.

Strategy: Set the mechanical energy of the ball at the start equal to its mechanical energy at the top of the rack. Use the fact that it rolls without slipping to write its energy in terms of linear speed and solve for the final speed. Use

I =

2

5mr

2 for a solid sphere.

Solution: 1. (a) Set E

i= E

f and

solve for v, letting

= v r since

the ball rolls without slipping.

Ui+ K

i= U

f+ K

f

0 + 1

2mv

i

2+ 1

2I

i

2= mgh + 1

2mv

f

2+ 1

2I

f

2

1

2mv

i

2+ 1

2

2

5mr2( ) v

ir( )

2

= mgh + 1

2mv

f

2+ 1

2

2

5mr2( ) v

fr( )

2

2. Simplify the expression and

solve for v

f:

1

2mv

i

2+ 1

5mv

i

2= mgh + 1

2mv

f

2+ 1

5mv

f

2

7

10v

i

2= gh + 7

10v

f

2

vf= v

i

2 10

7gh

= 2.85 m/s( )2

10

79.81 m/s

2( ) 0.53 m( )

vf= 0.83 m/s

3. (b) If the radius of the ball were increased the speed found in part (a) would stay the same because

the expression depends only on the initial speed and the height of the rack.

Insight: Although a larger radius would mean a higher linear speed for the same angular speed, it

would also mean a larger moment of inertia. The effect of the radius therefore cancels out.

63. Picture the Problem: The cylinder rolls

down the ramp without slipping, gaining

both translational and rotational kinetic

energy.

Strategy: Use conservation of energy to find

total kinetic energy at the bottom of the

ramp. Then set that energy equal to the sum

of the rotational and translational energies.

Since the cylinder rolls without slipping, the

equation

= v r can be used to write the

expression in terms of linear velocity alone.

Use the resulting equation to find

expressions for the fraction of the total

energy that is rotational and translational

kinetic energy.

Solution: 1. (a) Set

E

i= E

f and solve

for K

f:

Ui+ K

i= U

f+ K

f

mgh + 0 = 0 + Kf

Kf= mgh = 2.0 kg( ) 9.81 m/s2( ) 0.75 m( )

= 14.7 J = 15 J

2. (b) Set K

f equal to

K

t+ K

r:

Kf= 1

2mv

2+ 1

2I

2= 1

2mv

2+ 1

2

1

2mr

2( ) v r( )2

= 1

2mv

2+ 1

4mv

2= 3

4mv

2

3. Determine K

r from steps 1 and 2:

Kr= 1

4mv

2= 1

3

3

4mv

2( ) = 1

3K

f= 1

314.7 J( ) = 4.9 J

4. (c) Determine K

t from steps 1 and 2:

Kt= 1

2mv

2= 2

3

3

4mv

2( ) = 2

3K

f= 2

314.7 J( ) = 9.8 J

Insight: The fraction of the total kinetic energy that is rotational energy depends upon the moment of

inertia. If the object were a hoop, for instance, with I = mr2

, the final kinetic energy would be half

translational, half rotational.