physics chapter 10 answers

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10 – 1

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 2 ,I mr as we see in the discussion just before

Conceptual Checkpoint 10-4. This ratio is 2 2/ 1MR MR = for the first hoop and ( ) ( )2 22 / 2 1M R M R⎡ ⎤ =⎣ ⎦ 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.

Solutions to Problems and Conceptual Exercises

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

Strategy: Multiply the angle in degrees by radians

180π⎛ ⎞

⎜ ⎟°⎝ ⎠ to get radians.

Solution: rad rad rad rad30 rad 45 rad 90 rad 180 rad

180 6 180 4 180 2 180π π π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞° = ° = ° = ° =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟° ° ° °⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Insight: The quantity π is the circumference of a circle divided by its diameter. 3.1415926536 ...π = 2. Picture the Problem: This is a units conversion problem.

Strategy: Multiply the angle in radians by 180

radiansπ°⎛ ⎞

⎜ ⎟⎝ ⎠

to get degrees.

Solution: 180 180 180 18030 0.70 126 1.5 270 5 900

6π π π π

π π π π° ° ° °⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ° = ° = ° = °⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Insight: The quantity π is the circumference of a circle divided by its diameter. 3.1415926536 ...π = 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 2π 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:

31 rev 2 rad 1 hr 1.745 10 rad/s hr rev 3600 s

π −× × = ×

2. (b) The hour hand completes two revolutions per day:

42 rev 2 rad 1 day 1 hr 1.454 10 rad/s day rev 24 hr 3600 s

π −× × × = ×

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.

Chapter 10: Rotational Kinematics and Energy James S. Walker, Physics, 4th Edition


10 – 2

4. Picture the Problem: The second hand completes one revolution every 60 seconds. Strategy: Multiply 1 rev 60 s by conversion factors to obtain the appropriate units.

Solution: 1. (a) Convert the units:

1 rev 3600 s 60 rev/h60 s h

⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2. (b) Convert the units: 1 rev 60 s 360 deg 360 deg/min

60 s min rev⎛ ⎞⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

3. (c) Convert the units:

1 rev 2 rad rad/s 0.1047 rad/s60 s rev 30

π π⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Insight: Another common unit of angular speed is rev/min or rpm. The second hand rotates at 1.00 rev/min. 5. Picture the Problem: This is a units conversion problem. Strategy: Convert the given angular speeds into units of rad/s and compare their magnitudes.

Solution: 1. Find ω for the tire: 3

tire2.00 10 deg rad 34.9 rad/s

s 180 degπω

⎛ ⎞⎛ ⎞×= =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

2. Find ω for the drill: drill

400.0 rev 2 rad 1 min 41.89 rad/smin rev 60 s

πω ⎛ ⎞⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

3. Find ω for the propeller: prop 40.0 rad/sω =

4. We can now rank the angular speeds from least to greatest: tire, propeller, drill

Insight: In order to compare any quantities you must always ensure they have been measured in the same units. 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 θ: 1.95 m 5.9 rad

0.33 msr

θ = = =

Insight: This angular distance corresponds to 339° or 94% of a complete revolution. 7. Picture the Problem: The Earth travels in a nearly circular path around the Sun, completing one revolution per year. Strategy: Convert the known angular speed of 1 rev/yr into units of rev/min.

Solution: Convert the units: 61 rev 1 yr 1 day 1 h 1.90 10 rev/minyr 365 days 24 h 60 min

ω −⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= = ×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

Insight: This angular speed corresponds to about 0.986 deg/day or 1.99×10−7 rad/s. A good “rule of thumb” in astronomy is that the Sun appears to move 1°/day against the background of the “fixed” stars.

8. Picture the Problem: The Earth rotates once on its axis every 24 hours.

Strategy: Convert the known angular speed of 1 rev/day into units of radians per second.

Solution: Convert the units: 51rev 2 rad 1 day 1 h 7.27 10 rad/s

day rev 24h 3600sπω −⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= = ×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ Insight: This angular speed corresponds to about 15° / hour. A “rule of thumb” in astronomy is that the “fixed” stars

will move across the sky at this rate (1° every 4 minutes, or 15 arcsec/s) due to Earth’s rotation. 9. Picture the Problem: The pulsar rotates about its axis, completing 1 revolution in 0.33 s. Strategy: Divide one revolution or 2π radians by the period in seconds to find the angular speed. Solution: Calculate ω using equation 10-3: 2 rad 2 rad 190 rad/s

0.033st Tθ π πω Δ

= = = =Δ

Insight: The rotation rate of the pulsar can also be described as 1800 rev/min.



10 – 3

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 ω :

2 2 31.4 rad/s0.200sT

π πω = = =

2. (b) Apply equation 10-12 directly: ( )( )12 3.5 in 31.4 rad/s 55 in/s 1 m 39.4 in 1.4 m/sTv rω= = × = × =

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.

11. Picture the Problem: The propeller rotates about its axis with constant angular acceleration.

Strategy: Use the kinematic equations for rotating objects and the given formula to find the average angular speed and angular acceleration during the specified time intervals. By comparison of the formula given in the problem,

( ) ( )2 2125 rad/s 42.5 rad/st tθ = + ,with equation 10-10, 210 0 2t tθ θ ω α= + + , we can identify 0 125 rad/sω = and

212 42.5 rad/sα = .

Solution: 1. (a) Use equations 10-3 and 10-10 to find avω :

( ) ( ) ( )( )

210 020

av

22

2av

t125 rad/s 0.010 s 42.5 rad/s 0.010 s 0

0.010 s125 rad/s 1.3 10 rad/s

t t

t t

ω α θθ θθω

ω

⎡ ⎤+ −−Δ ⎣ ⎦= = =Δ⎡ ⎤+ −⎣ ⎦=

= = ×

2. (b) Use equations 10-3 and 10-10 to find avω :

( )( ) ( )( )( )( ) ( )( )

22 210 2

22 210 0 0 02

0av

0

125 rad/s 1.010 s 42.5 rad/s 1.010 s 169.60 rad

125 rad/s 1.000 s 42.5 rad/s 1.000 s 167.50 rad

169.60 167.50 rad 210 rad/s 2.11.010 1.000 s

t t

t t

t t t

θ ω α

θ ω α

θ θθω

⎡ ⎤⎡ ⎤= + = + =⎣ ⎦ ⎣ ⎦⎡ ⎤⎡ ⎤= + = + =⎣ ⎦ ⎣ ⎦

−Δ −= = = = = ×

Δ − −210 rad/s

3. (c) Use equations 10-3 and 10-10 to find avω :

( )( ) ( )( )( )( ) ( )( )

22 210 2

22 210 0 0 02

0av

0

125 rad/s 2.010 s 42.5 rad/s 2.010 s 422.95 rad

125 rad/s 2.000 s 42.5 rad/s 2.000 s 420.00 rad

422.95 420.0 rad 295 rad/s 3.0 12.010 2.000 s

t t

t t

t t t

θ ω α

θ ω α

θ θθω

⎡ ⎤⎡ ⎤= + = + =⎣ ⎦ ⎣ ⎦⎡ ⎤⎡ ⎤= + = + =⎣ ⎦ ⎣ ⎦

−Δ −= = = = = ×

Δ − −20 rad/s

4. (d) The angular acceleration is positive because the angular speed is positive and increasing with time.

5. (e) Apply equation 10-6 directly:

20av

210 125 rad/s 85 rad/s1.00 0.00 st

ω ωα

− −= = =

Δ −

6. Apply equation 10-6 directly: 20

av295 210 rad/s 85 rad/s

2.00 1.00 stω ω

α− −

= = =Δ −

Insight: We violated the rules of significant figures in order to report answers with two significant figures. Such problems arise whenever you try to subtract two large but similar numbers to get a small difference. The answers are only known to one significant figure, but we reported two in order to show clearly that the angular acceleration is constant. Of course, we could also have determined from the equation given in the problem that because

212 42.5 rad/sα = , it must be true that 285.0 rad/s .α =



10 – 4

12. Picture the Problem: An object at rest begins to rotate with a constant angular acceleration.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to answer the conceptual question.

Solution: Equation 10-10 indicates that for an object that starts from rest and then rotates with constant angular acceleration, the angular displacement 21

2 tθ α= is proportional to the square of the time. We conclude that if the

object rotates through angle θ in time t then it rotates through angle 4θ in time 2.t

Insight: Using similar reasoning we can determine the object will rotate through angle 9θ in time 3t. 13. Picture the Problem: An object at rest begins to rotate with a constant angular acceleration.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to answer the conceptual question.

Solution: Equation 10-8 indicates that for an object that starts from rest and then rotates with constant angular acceleration, the angular speed tω α= is linearly proportional to the time. We conclude that if the object has angular

speed ω after time t then its angular speed will be 2ω at time 2.t

Insight: Using similar reasoning we can determine the object will rotate through angle 9θ in time 3t. 14. Picture the Problem: The weight rises and slows down, coming to rest before falling downward once again.

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

Solution: Solve equation 10-8 for t: 0

02

5.0 5.40 rad/s 5.0 s2.10 rad/s

t

t

ω ω αω ω

α

= +− − −

= = =−

Insight: The magnitude of ω will continue to increase until the mass hits the floor or the string runs out. 15. Picture the Problem: The wheel is given an initial angular velocity, slows down with constant angular acceleration,

and eventually comes to rest.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angle through which the wheel has rotated. The angular acceleration is given in the solution to Example 10-2 as 20.736 rad/s .α = −

Solution: Solve equation 10-11 for θΔ :

( ) ( )( )

2 22 20

0 2

2.45 rad/s 3.40 rad/s3.78 rad

2 2 0.736 rad/sω ω

θ θ θα

−−Δ = − = = =

−

Insight: This angle is 216° or 0.601 revolutions. The trick in the game is to set the initial angular velocity just right so that the wheel comes to rest after rotating through the correct angle to win the prize.

16. 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 :α ( ) ( )

( )

22 220 26 rad/s 12 rad/s

17 rad/s2 2 2.5 rev 2 rad rev

ω ωα

θ π

2 −−= = =

Δ ×

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.

17. 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 angle through which the propeller rotated.

Solution: Solve equation 10-9 for θΔ : ( ) ( )( )1 102 2 11 28 rad/s 2.4 s 47 rad 7.4 revtθ ω ωΔ = + = + = =

Insight: A speed of 28 rad/s is equivalent to 270 rev/min, indicating the motor is turning pretty slowly. A typical outboard motor is designed to operate at 5000 rev/min at full throttle.



10 – 5

18. Picture the Problem: The bicycle wheel rotates about its axis, slowing down with constant angular acceleration before

coming to rest.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angular acceleration and the time elapsed.

Solution: 1. (a) Solve equation 10-11 for :α

( )( )

( )

222 220

0

0 6.35 rad/s0.226 rad/s

2 2 14.2 rev 2 rad revω ω

αθ θ π

−−= = = −

− ×

2. (b) Solve equation 10-8 for t: 0

2

0 6.35 rad/s 28.1 s0.226 rad/s

tω ω

α− −

= = =−

Insight: The greater the friction in the axle, the larger the magnitude of the angular acceleration and the sooner the wheel will come to rest.

19. Picture the Problem: The ceiling fan rotates about its axis, slowing down with constant angular acceleration before

coming to rest.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the number of revolutions through which the fan rotates during the specified intervals. Because the fan slows down at a constant rate of accelera-tion, it takes exactly half the time for it to slow from 0.96 rev/s to 0.48 rev/s as it does to come to a complete stop.

Solution: 1. (a) Apply equation 10-9 directly: ( ) ( )( )1 102 2 0 0.96 rev/s 2.4 min 60 s min 69 revtθ ω ωΔ = + = + × =

2. (b) Apply equation 10-9 directly: ( ) ( )( )1 102 2 0.48 0.96 rev/s 1.2 min 60 s min 52 revtθ ω ωΔ = + = + × =

Insight: Note that it takes half the time to slow to half speed, but the fan rotates through much more than half of its total of 69 revolutions during that time. An alternative way to solve the problem is to use 0 tα ω ω= − to find

20.0067 rev/sα = − and then use 210 2t tθ ω αΔ = + to find θΔ for each of the specified intervals. Note that you can

stick with units of rev/s2 to find θΔ in units of revolutions instead of converting to radians and back again. 20. 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 θΔ :

( )( )

2 22 20

0 2

6.3 rad/s 09.0 rad 1 rev 2 rad

2 2 2.2 rad/s

1.4 rev

ω ωθ θ θ π

α−−

Δ = − = = = ×

=

2. (b) Solve equation 10-8 for t: 0

2

6.3 0 rad/s 2.9 s2.2 rad/s

tω ω

α− −

= = =

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.



10 – 6

21. Picture the Problem: The minute hand rotates faster than the hour hand, closing the

angular gap between the two hands until the gap is 45.0°.

Strategy: The hour and minute hands rotate at different rates. Use equation 10-3 to find expressions for the angular positions of the hour and minute hands as a function of time with 3:00 being the starting positions ( h,0 90.0θ = ° and m,0 0.0θ = ° ). Set the angular positions difference equal to 45.0° and solve for t.

Solution: 1. Use equation 10-3 to find an expression for h mθ θ− : ( ) ( )

m m,0 m

h h,0 h h m h,0 m,0 h m

t

t t

θ θ ω

θ θ ω θ θ θ θ ω ω

= +

= + ⇒ − = − + −

2. Find hω and mω : 1

h

1m

1 rev 1 h 360 0.500 min12.0 h 60 min rev1 rev 360 6.00 min

60 min rev

ω

ω

−

−

°= × × = °

°= × = °

3. Solve the expression in step 1 for t:

( ) ( )( )

( ) ( )( )

h m h,0 m,0

1h m

45.0 90.08.18 min

0.500 6.00 mint

θ θ θ θω ω −

− − − ° − °= = =

− − °

4. The two hands will be 45.0° apart 8.18 min after 3:00, or at 3:08:11.

Insight: If you set the angular position difference to zero you find the two hands will be exactly aligned at 3:16:22. 22. Picture the Problem: The centrifuge rotates about its axis, slowing down with constant angular acceleration and

coming to rest.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angular acceleration and the number of revolutions through which the centrifuge rotates before coming to rest.

Solution: 1. (a) Solve equation 10-8 for :α ( ) 20 0 3850 rev/min 1 min 60 s

6.29 rev/s10.2 st

ω ωα

− ×−= = =

2. (b) Solve equation 10-9 for θΔ : ( ) ( )( )1 102 2 0 3850 rev/min 1 min 60 s 10.2 s 327 revtθ ω ωΔ = + = + × =

Insight: Another way of expressing the angular acceleration is to say that it slows down at a rate of 377 rev/min/s. 23. Picture the Problem: The Earth rotates on its axis, slowing down with constant angular acceleration.

Strategy: Equation 10-6 gives an expression for the angular acceleration as a function of rotation rate and time. Determine the difference in rotation rates between 1906 and 2006 by approximating T T T+ Δ ≅ because 0.840 s is tiny compared with the time (3.16×107 s) it takes to complete 365 revolutions. Then use equation 10-6 to find the average angular acceleration over the 100 year time interval.

Solution: 1. Find the difference in angular speeds:

( )( )

( ) ( )( )( )( )

0 2

2

120

0.840 s365 rev 2 rad rev

365 d 24 h/d 3600 s/h

1.94 10 rad/s

T T T TT T T T T T T

θ θω ω θ θ

π

ω ω −

⎡ ⎤− + Δ −Δ⎛ ⎞− = − = ≅⎢ ⎥ ⎜ ⎟+ Δ + Δ ⎝ ⎠⎢ ⎥⎣ ⎦⎧ ⎫−⎪ ⎪= × ⎨ ⎬

⎡ ⎤⎪ ⎪⎣ ⎦⎩ ⎭− = − ×

2. Apply equation 10-6 directly:

( )12

22 2av 7

1.94 10 rad/s 6.14 10 rad/s100 y 3.16 10 s/yt

ωα−

−Δ − ×= = = − ×

Δ × ×

Insight: Your first instinct might be to find the angular speed in 1906 assuming a period of 24.000 hrs (7.272205217×10−5 rad/s) and figure out the angular speed in 2006 (7.272205023×10−5 rad/s), but as you can see, attempting to subtract these numbers requires us to ignore the rules for significant figures. Using the approximation outlined above allows us to avoid the subtraction problem and keep three significant figures.



10 – 7

24. Picture the Problem: The compact disk rotates about its axis, increasing its angular speed at a constant rate.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the average angular speed during the time interval and then the angle through which the disk spins during this interval.

Solution: 1. (a) Find the average angular speed over the time interval and use equation 10-9 to find θΔ .

2. (b) Apply equation 10-9 directly: ( ) ( )( )1 102 2 310 0 rev/min 3.3 s 1 min 60 s 8.5 revtθ ω ωΔ = + = + × =

Insight: An alternative way to solve the problem is to use equation 10-8 to find 21.72 rev/sα = and use α in equation 10-11 to find 7.8 revθΔ = for the specified interval. Note that you can stick with units of rev/s2 to find θΔ in units of revolutions instead of converting to radians and back again.

25. Picture the Problem: The saw blade rotates about its axis, slowing its angular speed at a constant rate until it comes to

rest.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angular acceleration of the saw blade and the angle through which the blade spins during this interval. Then use equation 10-2 to convert the angular distance to a linear distance.

Solution: 1. (a) Solve equation 10-8 for :α ( ) 20 0 4440 rev/min 1 min 60 s

29.6 rev/s2.50 st

ω ωα

− ×−= = = −

2. (b) Use equation 10-9 to find :θΔ ( ) ( )( )1 102 2 0 4440 rev/min 2.50 s 1 min 60 s 92.5 revtθ ω ωΔ = + = + × =

3. Convert θΔ to s: ( )( )12 10.0 in 12 in/ft 92.5 rev 2 rad/rev 242 fts r θ π= Δ = × =

4. (c) The blade completes exactly 92.5 revolutions, so a point on the rim ends up exactly opposite of where it started. Its displacement is therefore one blade diameter or 10.0 in.

Insight: If the blade had completed an integer number of revolutions, a point on the rim would end up exactly where it began and the displacement would be zero even though the distance it travels is hundreds of feet.

26. Picture the Problem: The drill bit rotates about its axis, increasing its angular speed at a constant rate.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angular acceleration, average angular speed during the time interval, and the angle through which the drill bit spins during this interval.

Solution: 1. (a) Solve equation 10-8 for :α ( ) 3 20 350,000 rev/min 1 min 60 s 0

2.8 10 rev/s2.1st

ω ωα

× −−= = = ×

2. (b) Use equation 10-9 to find :θΔ ( ) ( )( )1 102 2

3

350,000 0 rev/min 2.1 s 1 min 60 s

6.1 10 rev

tθ ω ωΔ = + = + ×

= ×

Insight: The angular acceleration could also be expressed as 1.7×104 rad/s2. Note that the bit spins thousands of times during the 2.1 seconds it is coming up to speed. At full speed it spins over 12,000 times in 2.1 seconds!

27. Picture the Problem: Jason is a distance R from the axis of rotation of a merry-go-round and Betsy is a distance 2R

from the axis.

Strategy: Use an understanding of rotational motion to answer the conceptual question.

Solution: 1. (a) Although the linear speeds of Jason and Betsy are different, their angular speeds are the same because they both ride on the same merry-go-round. Because each completes one revolution in the same amount of time, the rotational period of Jason is equal to the rotational period of Betsy.

2. (b) The best explanation is III. It takes the same amount of time for the merry-go-round to complete a revolution for all points on the merry-go-round. Statements I and II are each false.

Insight: Jason has a smaller linear speed tv r ω= and a smaller centripetal acceleration 2cpa r ω= than does Betsy.



10 – 8

28. Picture the Problem: Jason is a distance R from the axis of rotation of a merry-go-round and Betsy is a distance 2R

from the axis. Strategy: Use an understanding of the connections between linear and rotational quantities to answer the conceptual

question.

Solution: 1. (a) Find the ratio of angular speeds: Jason

Betsy

1ω ωω ω

= =

2. (b) Find the ratio of linear speeds: t, Jason

t, Betsy

12 2

v Rv R

ωω

= =

3. (c) Find the ratio of centripetal accelerations:

2cp, Jason

2cp, Betsy

122

a Ra R

ωω

= =

Insight: Jason completes the same number of revolutions as does Betsy but travels half the linear distance.

29. Picture the Problem: You stand on the top floor of Taipei 101, which is anchored to a rotating Earth.

Strategy: Use an understanding of rotational motion to answer the conceptual question.

Solution: 1. (a) Although your linear speed is larger when standing on the top floor (because you are located at a larger distance from the Earth’s axis of rotation), your angular speed is the same as that of the Earth. We conclude that your angular speed due to the Earth’s rotation is equal to your angular speed when you stand on the ground floor.

2. (b) The best explanation is I. The angular speed is the same at all distances from the axis of rotation. Statements II and III are each false.

Insight: In reference to statement III, it is true that air spins faster near the center of a hurricane than it does at the outer edge of the storm, but such a storm is a complex arrangement of moving fluids and not a solid rotating object like the Earth or a merry-go-round. A new concept called angular momentum (chapter 11) can be used to explain the differential rotation rates in a hurricane.

30. 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 using equation 10-12.

Solution: Apply equation 10-12 directly: ( )t1 rev 2 rad 1 h8.2 cm 0.0012 cm/s 12 m/s12 h rev 3600 s

v r πω μ⎛ ⎞⎛ ⎞⎛ ⎞= = = =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

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.

31. Picture the Problem: The two children sit in different places on the same merry-go-round, which is rotating about its axis at a constant rate.

Strategy: Find the angular speed of the children by dividing 2π radians (for a complete circle) by the time it takes to complete a revolution. Then use equation 10-12 together with the angular speed to find the linear speed.

Solution: 1. (a) Apply equation 10-3 to find 1 2and ω ω : 1 2

2 rad 1.4 rad/s4.5 sπω ω= = =

2. (b) Apply equation 10-12 directly: ( ) ( )t1 1 1 2.0 m 1.4 rad/s 2.8 m/sv rω= = =

3. Apply equation 10-12 directly: ( )( )t2 2 2 1.5 m 1.4 rad/s 2.1 m/sv r ω= = =

Insight: Child 1 experiences a greater linear speed and greater centripetal acceleration because he is at a larger radius.

32. Picture the Problem: The Frisbee rotates at a constant rate about its central axis. Strategy: Use equation 10-12 to find the angular speed from the knowledge of the linear speed and the radius.

Solution: Solve equation 10-12 for ω : ( )

t12

3.7 m/s 26 rad/s0.29 m

vr

ω = = =

Insight: The rotation of a Frisbee produces its unique, stable flight characteristics.



10 – 9

33. Picture the Problem: The two horses are located at different places on the same carousel, which is rotating about its

axis at a constant rate. Strategy: Find the angular speed of the horses by dividing 2π radians (for a complete circle) by the time it takes to

complete a revolution. Then use equation 10-12 together with the angular speed to find the linear speed.

Solution: 1. (a) Apply equation 10-3 to find 1 2and ω ω : 1 2

2 rad 0.14 rad/s45 sπω ω= = =

2. Apply equation 10-12 directly: ( )( )t1 1 1 2.75 m 0.14 rad/s 0.38 m/sv rω= = =

3. (b) Apply equation 10-12 directly: ( )( )t2 2 2 1.75 m 1.4 rad/s 0.24 m/sv r ω= = =

Insight: The outer horse has a greater linear speed and greater centripetal acceleration because it is at a larger radius. 34. 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.


t 8.50 m/s 1.18 rad/s7.20 m

vr

ω = = =

2. (b) Apply equation 6-15 directly: ( )22

2tcp

8.50 m/s10.0 m/s

7.20 mv

ar

= = =

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!

35. Picture the Problem: Jeff clings to a vine and swings along a vertical arc as depicted in the

figure at right.

Strategy: Use equation 10-13 to find Jeff’s centripetal acceleration and equation 10-14 to find his tangential acceleration. Add these two perpendicular vectors to find the total acceleration.

Solution: 1. Apply equation 10-13 directly: ( )( )22cp

2

7.20 m 0.850 rad/s

5.20 m/s

a rω= =

=

2. Apply equation 10-14 directly: ( )( )2t

2

7.20 m 0.620 rad/s

4.46 m/s

a rα= =

=

3. Add the two perpendicular vectors: ( ) ( )2 22 2 2 2 2cp t 5.202 m/s 4.464 m/s 6.85 m/sa a a= + = + =

4. Find the angle φ :

2cp–1 –1

2t

5.20 m/stan tan 49.44.46 m/s

aa

φ⎛ ⎞ ⎛ ⎞

= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Insight: The angle φ will increase with Jeff’s speed if his angular acceleration remains constant because cpa depends on the square of the tangential speed.



10 – 10

36. Picture the Problem: The compact disk rotates about its axis, increasing its angular speed at a constant rate.

Strategy: Find the angular acceleration of the disk using equation 10-8, and then find the tangential acceleration from equation 10-14.

Solution: 1. (a) Solve equation 10-8 for :α ( ) 20 4.00 rev/s 2 rad rev 0

8.38 rad/s3.00 st

πω ωα

× −−= = =

2. Apply equation 10-14 directly: ( )( )2 21t 2 0.12 m 8.38 rad/s 0.503 m/sa rα= = =

3. (b) The tangential acceleration depends only on the rate of change of angular speed, not the particular value of the angular speed. Therefore, the tangential acceleration remains 0.503 m/s2 irregardless of the angular speed.

Insight: In actual practice the CD player requires the motor to continuously change the angular speed of the disk. See problem 91 for details.

37. Picture the Problem: The compact disk rotates about its central axis at a constant angular speed.

Strategy: Use equation 10-12 to find the linear speed of a point on the outer rim of the CD, and then use equation 10-13 to find the centripetal acceleration. Use ratios to determine the linear speed and centripetal acceleration for a point that is half the distance to the rotation axis.

Solution: 1. (a) Apply equation 10-12 directly: ( )( )1t 2 0.120 m 5.05 rad/s 0.303 m/sv r ω= = =

2. (b)Apply equation 10-13 directly: ( )( )22 21cp 2 0.120 m 5.05 rad/s 1.53 m/sa rω= = =

3. (c) Use a ratio to find the new linear speed: ( )

12 2 2 1 1

2 12 21 1

1 0.303 m/s 0.152 m/s2

rv rv v

v r rωω

= = = ⇒ = = =

4. Use a ratio to find the new cpa : ( )

2 12 22 2 2 1 1

2 12 221 1

1 1.53 m/s 0.765 m/s2

ra ra a

a rrωω

= = = ⇒ = = =

Insight: The angular velocity is the same for all points on the CD regardless of the distance to the rotation axis.

38. Picture the Problem: The reel rotates about its axis at a constant rate. Strategy: Calculate the linear speed of the string from the radius of the reel and the angular speed.

Solution: 1. (a) Apply equation 10-12 directly: ( )t3.0 rev 2 rad3.7 cm 0.70 m/s

s revv r πω ⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ 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.

39. Picture the Problem: The Ferris wheel rotates at a constant rate, with the centripetal acceleration of the passengers always pointing toward the axis of rotation. The acceleration of the passenger is thus upward when they are at the bottom of the wheel and downward when they are at the top of the wheel.

Strategy: Use equation 10-13 to find the centripetal acceleration. The centripetal acceleration remains constant (as long as the angular speed remains the same) and points toward the axis of rotation.

Solution: 1. (a) Apply equation 10-13 directly: ( )

22 2

cp2 rad9.5 m 0.29 m/s

36 sa r πω ⎛ ⎞= = =⎜ ⎟

⎝ ⎠ 2. When the passenger is at the top of the Ferris wheel, the centripetal acceleration points downward toward the axis of

rotation. 3. (b) The centripetal acceleration remains 0.29 m/s2 for a passenger at the bottom of the wheel because the radius and

angular speed remain the same, but here the acceleration points upward toward the axis of rotation. Insight: In order to double the centripetal acceleration you need to increase the angular speed by a factor of 2 or

decrease the period by a factor of 2 . In this case a period of 25 seconds will double the centripetal acceleration.



10 – 11

40. Picture the Problem: The Ferris wheel rotates clockwise but is slowing down at a

constant rate. The Ferris wheel has a radius of 9.5 m and rotates once every 36 s.

Strategy: Find the tangential acceleration of the passenger at the top of the Ferris wheel and combine it with the centripetal acceleration to find the total acceleration.

Solution: 1. Use equation 10-13 to find cpa : ( )

22 2

cp2 rad9.5 m 0.29 m/s

36 sa r πω ⎛ ⎞= = =⎜ ⎟

⎝ ⎠ (downward)

2. Use equation 10-14 to find ta : ( )( )2 2t 9.5 m 0.22 rad/s 2.1 m/sa rα= = − = − (to the left)

3. (c) Combine the components to find a : ( ) ( )2 22 2 2 2 2cp t 0.29 m/s 2.1 m/s 2.1 m/sa a a= + = + − =

4. (d) Find the angleφ and relate it to the direction of motion (which is to the right):

2cp1 1

2t

0.29 m/stan tan 7.92.1 m/s

aa

φ − −⎛ ⎞ ⎛ ⎞= = = − °⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠

or 172° below the

direction of motion.

Insight: In this case the tangential acceleration is 7.2 times greater than the centripetal acceleration. The passengers will notice the slowing down more than they noticed the centripetal acceleration when it was rotating at a constant rate.

41. Picture the Problem: The ball moves in a circle of constant radius at constant speed.

Strategy: The motion is approximately horizontal so we can neglect the fact that the rope would be inclined a little bit below horizontal in order to support the weight of the ball. Set the rope tension equal to the centripetal force required to keep the ball moving in a circle and solve for the angular speed.

Solution: 1. (a) Set the string force cpF ma= and solve for ω :

( )( )

2cp

11 N 2.2 rad/s0.52 kg 4.5 m

F ma mr

Fmr

ω

ω

= =

= = =

2. (b) Because ω is inversely proportional to r, the maximum angular velocity will increase if the rope is shortened.

Insight: This is a fairly weak rope, since 11 N is only 2.5 lb. Still, the problem illustrates well how the centripetal force increases linearly with the distance from the rotation axis. Decreasing r decreases the force, or allows a higher ω for the same amount of force.

42. Picture the Problem: The sanding disk rotates about its axis at a constant rate.

Strategy: Convert the angular speed of the disk into the linear speed of its rim by multiplying by its radius (equation 10-12). Use the same equation together with equation 10-5 to determine the period of rotation for the given rim speed.

Solution: 1. (a) Apply equation 10-12 directly: ( )( )4t 0.00320 m 2.15 10 rad/s 68.8 m/sv rω= = × =

2. (b) Substitute 2 Tω π= and solve for T:

( )t

5

t

22 0.00320 m2 7.31 10 s 73.1 s

275 m/s

v r r T

rTv

ω πππ μ−

= =

= = = × =

Insight: An angular speed of 2.15×104 rad/s is equivalent to 205,000 rev/min, or 3420 rev/s! Such high speeds are necessary to get the linear speed of the rim of such a small tool up to a value where it polishes well.

Top of Ferris Wheel

φ

acpa

ta

r

v



10 – 12

43. Picture the Problem: The sanding disk rotates about its axis, increasing its angular velocity at a constant rate.

Strategy: Use equation 10-14 to find the tangential acceleration and 10-13 to find the centripetal acceleration of a point on the rim of the disk.

Solution: 1. Apply equation 10-14 directly: ( )( )2 2t 0.00320 m 232 rad/s 0.742 m/sa rα= = =

2. Apply equation 10-13 directly: ( )( )22 2 2cp 0.00320 m 640 rad/s 1300 m/s 1.3 km/sa rω= = = =

Insight: The centripetal acceleration is very much greater than the tangential acceleration for a point on the rim, or any other point for that matter! The two are equal only when 2232 rad/s 15.2 rad/s.ω = = Note that there are only two significant figures in step 2 because of the ambiguity in the value 640 rad/s.

44. Picture the Problem: The electron moves along a circular path at constant speed.

Strategy: Use equation 10-12 to convert the linear speed into an angular speed. Convert that angular speed to units of revolutions per second to answer the question in part (b). Then use equation 10-13 to find the centripetal acceleration.


616t

–11

2.18 10 m/s 4.12 10 rad/s5.29 10 m

vr

ω ×= = = ×

×

2. (b) Convert ω to rev/s: ( )( )16 15 154.12 10 rad/s 1 rev 2 rad 6.56 10 rev/s 6.56 10 orbitsω π= × = × = ×

3. (c) Apply equation 10-13 directly: ( )( )22 –11 16 22 2cp 5.29 10 m 4.12 10 rad/s 8.98 10 m/sa rω= = × × = ×

Insight: The electron is traveling at 0.727% the speed of light in this classical model. In the quantum mechanical model it is not moving at all, even though it has momentum! More amazing facts about quantum mechanics are in Chapter 30.

45. Picture the Problem: The wheel rotates about its axis, increasing its angular velocity at a constant rate.

Strategy: Set the tangential (equation 10-14) and centripetal (equation 10-13) accelerations equal to each other for a single point on the rim. Find an expression for the angular speed as a function of time using equation 10-8 ( 0 0ω = because the wheel starts from rest), and substitute the expression into the result of the first step. Then solve the equation for t.

Solution: 1. Set t cpa a= : 2 2t cp a r r aα ω α ω= = = ⇒ =

2. Substitute 0 tω α= + and solve for t: ( )2 2 2 21 1 t t t tα α α

α α= = ⇒ = ⇒ =

Insight: The greater the angular acceleration, the shorter the elapsed time before the angular and centripetal accelerations equal each other. After that the centripetal acceleration dominates because it is proportional to 2.α

46. Picture the Problem: As you drive down the highway, the top of your tires are moving with a speed v.

Strategy: Note the discussion surrounding figures 10-10 and 10-11, which show that the top of a wheel (that is rolling without slipping) is moving twice as fast as the center of the wheel.

Solution: The reading on the speedometer gives the speed of the axles of your car. This is the same as the speed of the occupants inside the car. If the top of the tires have a speed v, the axles have a speed 2 .v

Insight: If the tires spin freely, as when the car is stuck in snow, the rim of the tire is moving at high speed while the axle is at rest.



10 – 13

47. 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 ω : t 15 m/s 48 rad/s

0.31 mvr

ω = = =

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.

48. Picture the Problem: The drive wheel of the tricycle rolls without slipping at constant speed.

Strategy: Because the wheel rolls without slipping, equation 10-15 describes the direct relationship between the center of mass speed and the angular velocity of the driving wheel.

Solution: Apply equation 10-15 directly: ( )( )t 0.260 m 0.373 rev/s 2 rad rev 0.609 m/sv rω π= = × =

Insight: This speed corresponds to about 1.4 mi/h, half the normal walking speed of an adult. The larger wheels on adult bicycles allow for higher linear speeds for the same angular speed of the driving wheel.

49. Picture the Problem: A soccer ball rolls without slipping at constant speed.

Strategy: Use the known circumference of the ball to determine its radius, and use the distance and time information to find its center of mass velocity. Combine the two quantities using equation 10-15 to find the angular speed of the ball.

Solution: 1. Find r from the circumference:

0.700 m 0.1114 m2 2Crπ π

= = =

2. Determine the velocity: ( )( )( )14.0 yd 3 ft yd 1 m 3.281 ft

3.82 m/s3.35 s

xvt

Δ= = =

Δ

3. Solve equation 10-15 for ω :

3.82 m/s 34.3 rad/s0.1114 m

vr

ω = = =

Insight: The ball travels about 8.55 mi/h. The reason we can determine its angular speed is because it rolls without slipping. If it is kicked very hard, it will fly through the air, spinning freely, and there would be no relationship between its angular and center-of-mass velocities. We bent the rules for significant figures in step 1 to avoid rounding error.

50. 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 :ω 0

017 m/s 52 rad/s0.33 m

vr

ω = = =

2. Solve equation 10-14 for :α

221.12 m/s 3.4 rad/s

0.33 mar

α = = =

3. Apply equation 10-10 directly: ( )( ) ( )( )22 21 10 2 252 rad/s 0.65 s 3.4 rad/s 0.65 s

35 rad 5.5 rev

t tθ ω α= + = +

= =

Insight: Another way to solve this question is to find the final angular speed (54 rad/s) and then use ( )102 tθ ω ω= + to

find the answer.



10 – 14

51. Picture the Problem: The bicycle tire rolls without slipping, increasing its angular speed at a constant rate. Strategy: Because the tire rolls without slipping there is a direct relationship between its linear and angular acceleration

(equation 10-14). Use that equation together with the definition of acceleration (equation 3-5) to determine the angular acceleration.

Solution: 1. (a) Solve equation 10-14 for :α : ( )

( )( )28.90 0 m/s1 2.03 rad/s

0.360 m 12.2 sa vr r t

α−Δ

= = = =Δ

2. (b) If the radius of the tires had been smaller, the angular acceleration would have been greater than 2.03 rad/s2.

Insight: The linear acceleration in this case is 0.730 m/s2. If the bike were sliding without friction down an incline, we could say its acceleration equals sing θ and the slope must be θ = 4.26°. However, because the tire rolls we must take torque and moment of inertia into account (see Chapter 11). If we do so we find θ would still be 4.26° for a hoop-shaped wheel but only 2.13° if the wheel were a solid disk.

52. Picture the Problem: The minute hand of a clock is long, thin, and uniform; and the hour hand is short, thick, and

uniform. Each hand has the same mass and the same axis of rotation. Strategy: Apply the concept of moment of inertia to answer the conceptual question. Solution: 1. (a) The moment of inertia of any object depends upon its total mass as well as the distribution of that mass

in relation to the axis of rotation. Because 2i i ,I m r= ∑ the farther the mass is located from the axis of rotation, the

larger the moment of inertia. Because the mass of the longer minute hand is located farther from its axis of rotation, we conclude that the moment of inertia of the minute hand is greater than the moment of inertia of the hour hand.

2. (b) The best explanation is II. Having mass farther from the axis of rotation results in a greater moment of inertia. Statements I and III are each false.

Insight: If the hour hand were much more massive than the minute hand, it could have a larger moment of inertia than the minute hand even though it is shorter. In general, the moment of inertia of a thin, uniform rod of mass M and length L that pivots about one of its ends is 21

3 .M L 53. Picture the Problem: Tons of dust and small particles rain down onto the Earth from space every day. Strategy: Apply the concept of moment of inertia to answer the conceptual question. Solution: 1. (a) The moment of inertia 2

i iI m r= ∑ of any object depends upon its total mass as well as the distribution of that mass in relation to the axis of rotation. Even though the radius of the Earth does not increase appreciably due to space dust, its total mass does increase a little bit. We conclude that the Earth’s moment of inertia does increase due to the accumulation of the space dust.

2. (b) The best explanation is I. The dust adds mass to the Earth and increases its radius slightly. Statement II would only be true if the dust were connected to the Earth somehow as it fell toward the planet. However, the falling dust is not connected to the Earth and does not contribute to the Earth’s moment of inertia until it lands. Statement III is false.

Insight: The mass gain from space dust is a tiny fraction of the Earth’s mass. The Earth gains about 20,000 tons of dust per year, or approximately a 3×10−16

% annual gain in mass. 54. Picture the Problem: A bicycle wheel is rotated about an axis through its rim and parallel to its axle. Strategy: Apply the concept of moment of inertia to answer the conceptual question. Solution: 1. (a) The moment of inertia 2

i iI m r= ∑ of any object depends upon its total mass as well as the distribution of that mass in relation to the axis of rotation. When the bicycle wheel is rotated about an axis that passes through its rim, an appreciable fraction of its mass is located at a distance farther than one radius from the axis. We conclude that its moment of inertia about this axis is greater than its moment of inertia about its axle.

2. (b) The best explanation is III. Mass is farther from the axis when the wheel is rotated about the rim. Statement I is false and statement II is true, but irrelevant.

Insight: You can observe the change in moment of inertia by first attempting to spin a bicycle wheel about its axle. Then place your finger just inside the rim and attempt to rotate the wheel about your finger, a little bit like a hoola hoop. You’ll notice that it is much more difficult to spin the wheel in the second case due to the larger moment of inertia.



10 – 15

55. Picture the Problem: The bicycle wheel is assumed to have the shape of an ideal hoop.

Strategy: Use the known moment of inertia of a hoop ( )2I MR= to find the radius of the wheel.

Solution: Solve 2I MR= for R:

20.13 kg m 0.36 m0.98 kg

IRM

⋅= = =

Insight: This wheel would have a 0.72 m diameter or 28 inches. This is a bit bigger than the standard 26 inch or 0.66 m bicycle tire.

56. Picture the Problem: The grindstone is a solid disk that rotates about its axis at constant angular speed. Example 10-4

indicates that its moment of inertia is 4.30 kg·m2.

Strategy: Find the angular speed from the rotation period and then use the known moment of inertia and equation 10-17 to find the kinetic energy.

Solution: Apply equation 10-17 directly: ( )

22 21 1

2 22 rad4.30 kg m 4.81 J4.20 s

K I πω ⎛ ⎞= = ⋅ =⎜ ⎟⎝ ⎠

Insight: The angular speed of the disk is a gentle 14.3 rev/min, a fairly easy speed to maintain even with a foot-powered grindstone. However, the 0.913 m/s linear speed of the rim of the grindstone isn’t very effective at grinding axes!

57. 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:

( )( )

22 2

2 4.6 J2 0.054 kg m13 rad/s

KIω

= = = ⋅

Insight: If the angular speed of the fan blade were to double to 24 rad/s, the kinetic energy would quadruple to 16 J. 58. Picture the Problem: The hoop rolls without slipping on a horizontal surface at constant speed. Using the data from

Example 10-5, the hoop has a mass of 1.20 kg, a radius of 10.0 cm, and a linear speed of 1.41 m/s.

Strategy: Find the translational kinetic energy using equation 7-6 and the rotational kinetic energy using equation 10-17. Then add the two energies to find the total kinetic energy.

Solution: 1. (a) Apply equation 7-6 directly: ( )( )221 1t 2 2 1.20 kg 1.41 m/s 1.19 JK mv= = =

2. (b) Apply equation 10-17 directly: ( ) ( )22 2 2 2 21 1 1 1r 2 2 2 2 1.19 JK I mr mr v r mvω ω= = = = =

3. (c) Add the two energies: ( )( )221t r 22 1.20 kg 1.41 m/s 2.39 JK K K mv= + = × = =

Insight: If we simply added 1.19 + 1.19 J = 2.38 J in part (c) we would’ve introduced some rounding error. Note that exactly half of the hoop’s kinetic energy is translational and half is rotational, whereas in the Example 10-5, two-thirds of the disk’s kinetic energy was translational and one-third rotational.



10 – 16

59. Picture the Problem: The L-shaped object in the figure at right can be

rotated in one of the following three ways: case 1, rotation about the x axis;case 2, rotation about the y axis; and case 3, rotation about the z axis (which passes through the origin perpendicular to the plane of the figure).

Strategy: The moment of inertia of any object depends upon its total mass as well as the distribution of that mass in relation to the axis of rotation. Because 2

i i ,I m r= ∑ the farther the mass is located from the axis of rotation, the larger the moment of inertia. Calculate the moment of inertia for each of the three cases in order to determine the ranking.

Solution: 1. Find I for case 1: ( )( )22 21 i i 9.0 kg 1.0 m 9.0 kg mI m r= = = ⋅∑

2. Find I for case 2: ( )( )22 22 i i 2.5 kg 2.0 m 10 kg mI m r= = = ⋅∑

3. Find I for case 3: ( )( ) ( )( )2 22 23 i i 9.0 kg 1.0 m 2.5 kg 2.0 m 19 kg mI m r= = + = ⋅∑

4. By comparing the above calculations we arrive at the ranking I1 < I2 < I3.

Insight: Whenever the axis of rotation passes through the center of mass of an object, that object makes no contribution to the moment of inertia about that axis because r = 0.

60. Picture the Problem: The compact disk rotates about its central axis at a constant angular speed.

Strategy: First find the moment of inertia of the CD, modeling it as a uniform disk. Then find the kinetic energy using equation 10-17. Finally, use a ratio to find the new angular speed that would double the rotational kinetic energy.

Solution: 1. (a) Apply equation 10-17 directly, using 21

2I MR= for the CD: ( )

( )( ) ( )

2 2 21 1 12 2 2

2 214 0.012 kg 0.060 m 34 rad/s 0.012 J = 12 mJ

K I MRω ω= =

= =

2. (b) Use a ratio to find the new ω :

( )

newnew new old

old old oldold

new old

2 22

2

2 2 34 rad/s 48 rad/s

K I K KK KK I

ωω

ω ω

= = = =

= = =

Insight: A speed of 48 rad/s corresponds to 460 rev/min, typical for a music CD player. Computer CD-ROM drives and DVD drives can spin the disk at a much faster rate.

61. 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: ( )( )221 1t 2 2 0.15 kg 48 m/s 170 JK Mv= = =

2. Use 225I MR= for a uniform sphere

in equation 10-17: ( )

( )( ) ( )

2 2 2 2 21 1 2 1r 2 2 5 5

2 215 0.15 kg 0.037 m 42 rad/s 0.072 J

K I MR MRω ω ω= = =

= =

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!



10 – 17

62. Picture the Problem: The basketball rolls without slipping at constant speed on the level floor.

Strategy: Use equation 10-17 and the moment of inertia of a hollow sphere to find the rotational kinetic energy of the ball. Then use equation 7-6 to find the translational kinetic energy, and sum the two to find the total kinetic energy. Use a ratio to find out what fraction of the ball’s energy is rotational kinetic energy.

Solution: 1. (a) Apply equations 10-17 and 7-6 to find r t, ,K K and the total K:

( )( )22 2 21 1 2 1r 2 2 3 3

21t 2

2 2 251 1r t 3 2 6

K I MR v R Mv

K MvK K K Mv Mv Mv

ω= = =

== + = + =

2. Calculate the ratio r :K K

213r

256

25

MvKK Mv

= =

3. (b) The answer to part (a) will stay the same if the linear speed of the ball is doubled to 2v, because the ratio is independent of speed, radius, and mass.

Insight: The ratio is a constant because when an object rolls without slipping there is a direct relationship between its translation and rotation: tv rω= .

63. Picture the Problem: The Earth rotates on its axis, slowing down with constant angular acceleration.

Strategy: Determine the difference in rotation rates over the span of a century by approximating T T T+ Δ ≅ because 0.0023 s is tiny compared with the time (86,400 s) it takes to complete one revolution. Then use equation 10-6 to find the average angular acceleration over the 100-year time interval.

Solution: 1. Find the difference in angular speeds:

( )( )

( ) ( )( )( )( )

0 2

2

120

0.840 s365 rev 2 rad rev

365 d 24 h/d 3600 s/h

1.94 10 rad/s


θ θω ω θ θ

π

ω ω −

⎡ ⎤− + Δ −Δ⎛ ⎞− = − = ≅⎢ ⎥ ⎜ ⎟+ Δ + Δ ⎝ ⎠⎢ ⎥⎣ ⎦⎧ ⎫−⎪ ⎪= × ⎨ ⎬

⎡ ⎤⎪ ⎪⎣ ⎦⎩ ⎭− = − ×

2. Find the sum of the angular speeds: ( )

( )0 2

2 2T T T T TT T T T T T TT

θ θ θω ω θ θ⎡ ⎤+ + Δ + Δ⎛ ⎞+ = + = ≅ ≅⎢ ⎥ ⎜ ⎟+ Δ + Δ ⎝ ⎠⎢ ⎥⎣ ⎦

3. Multiply the results of steps 1 and 2: ( ) ( )

22 2

0 0 0 2 3

2 2T TTT Tθ θω ω ω ω ω ω θ Δ Δ⎛ ⎞⎛ ⎞− + = − = − = −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

4. Find the difference rKΔ over a time interval of 100 years:

( )

( )( ) ( ) ( )( )

2 22 2 2 21 1 1 1

r 0 02 2 2 2 3 3

2 2E E

3

2 224 622

2

2

0.331

0.331 5.97 10 kg 6.38 10 m 2 0.0023 s1.1 10 J

86,400 s

T I TK I I I IT T

M R TT

θ θω ω ω ω

θ

π

⎛ ⎞Δ ΔΔ = − = − = − = −⎜ ⎟

⎝ ⎠Δ

= −

× ×= − = − ×

5. Use equation 7-10 to find the energy loss rate (power):

2212

7

1.1 10 J 3.5 10 W 3.5 TW100 y 3.16 10 s/y

WPt

− ×= = = − × = −

× ×

Insight: Your first instinct might be to find the angular speed a hundred years ago assuming a period of 24.000 hrs (7.272205217×10−5 rad/s) and figure out the angular speed in 2006 (7.272205023×10−5 rad/s), but as you can see, attempting to subtract these numbers requires us to ignore the rules for significant figures. Using the approximation outlined above allows us to avoid the subtraction problem and keep two significant figures. The huge energy loss is due primarily to tidal friction, as the ocean tides dissipate the kinetic energy of the Earth’s rotation into heat.



10 – 18

64. 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 maxy . Treat the blade as a uniform rod rotating about its center, so that its moment of inertia is 21

12 ML as indicated in Table 10-1. Solution: 1. (a) Use equation 10-17

to find rK : ( )

( )( )

2 2 21 1 1r 2 2 12

221

24

r

3500 rev 2 rad 1 min0.65 kg 0.55 mmin rev 60 s

1100 J 1.1 kJ

K I ML

K

ω ω

π

= =

⎡ ⎤⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎝ ⎠⎣ ⎦= =

2. (b) Set rK U= and solve for maxy :

( )( )

r max

rmax 2

1100 J 173 m 0.17 km0.65 kg 9.81 m/s

K U mgyK

ymg

= =

= = = =

Insight: 170 m is 570 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.

65. Picture the Problem: A ball is released from rest on the frictionless side of the

curved surface depicted at right and comes to rest on the no-slip surface.

Strategy: Apply the concepts discussed in Conceptual Checkpoint 10-5 to answer the question.

Solution: In this case, all of the original gravitational potential energy is converted to translational kinetic energy when the ball reaches the bottom of the track. It then begins to spin as it skids on the no-slip surface until it rolls without slipping. This process dissipates some of the mechanical energy. The remaining mechanical energy is converted to gravitational potential energy, but because mechanical energy has been lost the height attained will be less. We conclude that when the ball comes to rest on the no-slip surface, its height is less than the height from which it was released.

Insight: According to the discussion in Conceptual Checkpoint 10-5 the ball does not reach its initial height when it is released from the no-slip side, either, but for a different reason. In that case some of its mechanical energy is stored as rotational energy and in this case some of its mechanical energy is converted to thermal energy by friction.

66. 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:

2

212v Ihg mR

⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2. Solve the expression for I: 2

2

2 1ghI mRv

⎛ ⎞= −⎜ ⎟⎝ ⎠

3. Insert the numerical values: ( )( )

( )( )( )

22 2

2

2 9.81 m/s 0.074 m2.1 kg 0.080 m 1 0.17 kg m

0.33 m/sI

⎡ ⎤⎢ ⎥= − = ⋅⎢ ⎥⎣ ⎦

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 farther than if it had only its own kinetic energy to use.



10 – 19

67. Picture the Problem: The yo-yo falls straight down, increasing both its translational

and rotational kinetic energy. The Active Example 10-3 indicates that the mass of the yo-yo is 0.056 kg, its moment of inertia is 5 22.9 10 kg m−× ⋅ ,and the radius r of the axle the string wraps around is 0.0064 m.

Strategy: Use conservation of energy, including both translational and rotational kinetic energy, to determine the height h through which the yo-yo falls.

Solution: 1. Set i fE E= and simplify:

( ) ( )

i i f f2 21 1

2 222 2 21 1 1

2 2 2

0K U K U

mgh mv Imv I v r v m I r

ω+ = +

+ = +

= + = +

2. Solve for h:

( )( )( ) ( )

2

2

2 –5 2

22

20.65 m/s 2.9 10 kg m0.056 kg 0.29 m

2 0.056 kg 9.81 m/s 0.0064 m

v Ih mmg r

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎡ ⎤× ⋅= + =⎢ ⎥

⎢ ⎥⎣ ⎦

Insight: Since the yo-yo fell 0.50 m in order to attain a speed of 0.85 m/s in the Active Example 10-3, we expect it will need to fall a shorter distance in order to attain a speed of only 0.65 m/s. Note that the distance of the fall is proportional to the square of the speed, so that ( ) ( )( ) ( )2 22 2

new old new old 0.50 m 0.65 m/s 0.85 m/s 0.29 m.h h v v= = = 68. Picture the Problem: Several disks of different masses and radii roll

without slipping down the incline plane, starting from rest at height h above the level surface.

Strategy: Apply the concepts discussed in Conceptual Checkpoint 10-4 to determine the ranking of the three disks.

Solution: In the discussion surrounding Conceptual Checkpoint 10-4 it was determined that the speed of a rolling object

at the bottom of an incline of height h is 2

2 .1

ghvI m r

=+

However, for any uniform disk 2 12I m r = regardless of its

mass or radius. We conclude that the speeds of the disks will be equal and the ranking of the order in which the three disks finish the race is disk 1 = disk 2 = disk 3.

Insight: Using a similar argument we can see that any uniform sphere will beat any uniform disk because for a uniform sphere 2 2

5I m r = regardless of the mass or radius of the sphere.



10 – 20

69. 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 i fE E= and let v rω = : ( )

i i f f22 2 21 1 1 1

2 2 2 20 0U K U Kmgh mv I mv I v rω

+ = +

+ = + + = +

2. For the disk 212I mr= : ( )( )

( )( )

22 21 1 12 2 2

2 2 231 12 4 4

24 43 3 9.81 m/s 0.82 m 3.3 m/s

mgh mv mr v rgh v v v

v gh

= +

= + =

= = =

3. (b) For the hoop 2I mr= : ( )( )

( )( )

22 21 12 2

2 2 21 12 2

29.81 m/s 0.82 m 2.8 m/s

mgh mv mr v rgh v v v

v gh

= +

= + =

= = =

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.

70. 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 pv r ω= between 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 i fE E= and let

pv rω = :

( ) ( ) ( )( )

i i f f2 2 21 1 1

1 2 1 2 p2 2 222 21 1 1

1 2 1 2 p p p2 2 2

0 0U K U K

m gh m gh m v m v I

m m gh m m v m r v r

ω+ = +

+ + = + + +

− = + +

3. Rearrange the equation and solve for pm :

( ) ( )( ) ( )

( )( )( ) ( )( )( )

2 21 1p 1 2 1 24 2

211 2 1 22

p 2

22 12

2

p

4

4 2.0 kg 9.81 m/s 0.75 m 8.0 kg 1.8 m/s

1.8 m/s2.2 kg

m v m m gh m m vm m gh m m v

mv

m

= − − +

⎡ ⎤− − +⎣ ⎦=

⎡ ⎤−⎣ ⎦=

=

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.



10 – 21

71. Picture the Problem: The ball rolls without slipping, starting from a height of

0.78 m above the bottom of the track. When it gets to the bottom it encounters a frictionless surface, so it continues rotating at the same angular speed it had when it arrived at the bottom, and slides its way up to a lower height than 0.78 m.

Strategy: The initial gravitational potential energy of the ball gets converted into rotational and translational kinetic energy at the bottom of the track. The rotational kinetic energy remains constant on the frictionless side of the track, and the translational kinetic energy converts into gravitational potential energy. Use energy conservation and the fact that it rolls without slipping to find the center of mass speed and angular speed at the bottom of the track. Then convert the translational kinetic energy into gravitational potential energy to find the height to which the ball rises on the frictionless side. Use 22

5I mr= for a solid sphere.

Solution: 1. (a) Set i fE E= and solve for v, letting v rω = because the ball rolls without slipping. Here “i” refers to the release point on the no-slip side and “f” refers to the bottom of the track.

( )( )

i i f f2 21 1

i 2 222 21 1 2

i 2 2 527

i 1010

i7

0 0U K U K

mgh mv Imgh mv m r v rmgh mv

v g h

ω+ = +

+ = + +

= +

=

=

2. Now find v rω = : ( )( )2

i

2

10 9.81 m/s 0.78 m101 17 0.029 m 7

114 rad/s 1.1 10 rad/s

g hvr r

ω ⎛ ⎞= = = ⎜ ⎟⎝ ⎠

= = ×

3. (b) Set i fE E= and solve for h. Here “i” refers to the bottom of the track and “f” refers to the turning point on the frictionless side of the track. ( ) ( )

i i f f2 2 21 1 1

f2 2 221

f 2102

i7i i

0

5 5 0.78 m 0.56 m2 2 7 7

U K U Kmv I mgh I

mgh mvghvh h

g g

ω ω+ = +

+ + = +=

= = = = =

Insight: Trouble will occur when the marble arrives once again at the bottom of the track, because it will still be rotating clockwise from when it was rolling to the right, but now it is moving to the left when it encounters the no-slip surface. It will lose its rotational kinetic energy at that point. This is somewhat analogous to a bowling alley, where a low friction, waxed surface at the start of the lane allows the ball to rotate while slipping, but a higher friction surface near the pins causes the ball to roll without slipping.



10 – 22

72. 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 22

5I mr= for a solid sphere.

Solution: 1. (a) Set i fE E= and solve for v, letting v rω = because the ball rolls without slipping. ( )( ) ( )( )

i i f f2 2 2 21 1 1 1

i i f f2 2 2 22 22 2 2 21 1 2 1 1 2

i i f f2 2 5 2 2 5

0U K U K

mv I mgh mv Imv mr v r mgh mv mr v r

ω ω+ = +

+ + = + +

+ = + +

2. Simplify the expression and solve for fv :

( ) ( )( )

2 2 2 21 1 1 1i i f f2 5 2 5

2 27 7i f10 10

2 10f i 7

2 2107

f

2.85 m/s 9.81 m/s 0.53 m

0.83 m/s

mv mv mgh mv mvv gh v

v v gh

v

+ = + += +

= −

= −

=

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.

73. Picture the Problem:The mass falls straight down, its speed reduced due to the

rotation of the disk. The physical situation is depicted at right.

Strategy: Set the initial mechanical energy equal to the final mechanical energy. The initial gravitational potential energy of the block will be converted into translational kinetic energy of the block and rotational kinetic energy of the pulley.

Solution: 1. (a) Set i fE E= and let v rω = :

( )( )

i i f f2 21 1

b b p2 222 21 1 1

b b p2 2 22 21 1

b b p2 4

0 0U K U K

m gh m v I

m gh m v m r v rm gh m v m v

ω+ = +

+ = + +

= +

= +

2. Now solve for v: ( )( )( )( )

( )

2 1 1b b p2 4

2b

b p

4 1.3 kg 9.81 m/s 0.50 m42 2 1.3 kg 0.31 kg

3.0 m/s

m gh v m m

m ghv

m m

= +

= =+ +

=

3. (b) Adding a small weight to the rim of the pulley will increase its moment of inertia, allowing the pulley to store a larger fraction of the total mechanical energy as its rotational energy and leaving less energy for the translational kinetic energy of the block. The speed of the block will therefore decrease.

Insight: If the pulley were massless and there was no friction, the speed of the block would be 2 3.1 m/sgh = .



10 – 23

74. Picture the Problem: The leg pivots about the hip from a vertical to a

horizontal position, as indicated in the figure.

Strategy: The gravitational potential energy of the leg becomes rotational energy, and mechanical energy is conserved as long as the leg muscles do no work. Use conservation of mechanical energy to find the angular speed of the leg when it hits the floor, and use that angular speed to find the linear speed of the foot. The initial height of the center of mass of the leg is ( )1

2 0.95 m 0.475 m= (0.95 m)/2 = 0.475 m. The moment of inertia of a

uniform rod pivoted about one end is 213I mL= (Table 10.1).

Solution: 1. Set i fE E= and let v rω = :

( )( )( )

i i f f21

22 21 1

2 3

2

0 0

6 9.81 m/s 0.475 m65.6 rad/s

0.95 m

U K U Kmgh I

mgh mL

ghL

ωω

ω

+ = ++ = +

=

= = =

2. Use equation 10-12 to find tv of the foot: ( )( )t 0.95 m 5.6 rad/s 5.3 m/sv rω= = =

Insight: The center of mass of the leg has a translational speed when the foot hits the floor. So why is there no translational kinetic energy in the equation of step 1? It’s because the axis of rotation, the hip, is not moving. The rotational kinetic energy of the leg 21

2 Iω already contains the kinetic energies of all portions of the leg due to its rotation about the hip, so we would only need to add a translational term if the hip itself had a nonzero velocity.

75. 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. Because the cylinder rolls without slipping, the equation

v rω = 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 i fE E= and solve for fK :

( )( )( )

i i f f

f2

f

0 02.0 kg 9.81 m/s 0.75 m

14.7 J 15 J

U K U Kmgh K

K mgh

+ = ++ = +

= =

= =

2. (b) Set fK equal to t rK K+ : ( )( )22 2 2 21 1 1 1 1f 2 2 2 2 2

2 2 231 12 4 4

K mv I mv mr v rmv mv mv

ω= + = +

= + = 3. Determine rK from steps 1 and 2: ( ) ( )2 231 1 1 1

r f4 3 4 3 3 14.7 J 4.9 JK mv mv K= = = = =

4. (c) Determine tK from steps 1 and 2: ( ) ( )2 231 2 2 2t f2 3 4 3 3 14.7 J 9.8 JK mv mv K= = = = =

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 2I mr= , the final kinetic energy would be half translational, half rotational.



10 – 24

76. Picture the Problem: The sphere 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. Because the sphere rolls without slipping, the equation

v rω = 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 i fE E= and solve for fK :

( )( )( )

i i f f

f2

f

0 02.5 kg 9.81 m/s 0.75 m

18.4 J 18 J

U K U Kmgh K

K mgh

+ = ++ = +

= =

= =

2. (b) Set fK equal to t rK K+ : ( )( )22 2 2 21 1 1 1 2f 2 2 2 2 5

2 2 271 12 5 10


ω= + = +

= + = 3. Determine rK from steps 1 and 2: ( ) ( )2 271 2 2 2

r f5 7 10 7 7 18.4 J 5.3 JK mv mv K= = = = =

4. (c) Determine tK from steps 1 and 2: ( ) ( )2 25 7 5 51t f2 7 10 7 7 18.4 J 13 JK mv mv K= = = = =

Insight: The sphere has a higher speed at the bottom of the ramp (3.24 m/s) than does the cylinder from problem 75 (3.13 m/s) because less of the sphere’s kinetic energy is stored as rotational energy, so that more of it is devoted to translational kinetic energy. The sphere always wins a race with a cylinder. Try it!

77. Picture the Problem: You stand on the observation deck of the Empire State Building in New York.

Strategy: Use the relation tv r ω= between linear speed and angular speed to answer the conceptual question.

Solution: For a solid object that is rotating, a point that is farther from the rotation axis has a higher linear speed than a point that is closer to the rotation axis. We conclude that your linear speed due to the Earth’s rotation is greater than when you were waiting for the elevators on the ground floor.

Insight: The different linear speeds at the top and bottom of a building produce a phenomenon called the Coriolis force when a ball is dropped. Because the ball has more eastward linear velocity when it is released at the top of the building than do points near the ground, the ball will deflect toward the east as it falls.

78. Picture the Problem Two eggs are spun on a kitchen counter , one raw and the other hard boiled. One of the eggs

spins considerably longer than the other.

Strategy: Consider the principles involved with the rotation of a rigid body to answer the conceptual question.

Solution: When you give a hard-boiled egg a spin the entire egg rotates with the same angular speed. When you spin a raw egg, however, the outside rotates more rapidly than the liquid interior. The drag between the interior and the exterior results in the raw egg slowing down more rapidly than the solid hard-boiled egg. We conclude that the raw egg is the one that stops spinning in a short time.

Insight: In another interesting trick, try spinning a hard-boiled egg (or a peanut M&M®) really fast with its long axis horizontal. If conditions are right it will spontaneously “sit up” and spin with its long axis vertical.

79. Picture the Problem When the Hoover Dam was completed its reservoir behind it filled with water.

Strategy: Consider the principles involved with the calculation of the moment of inertia of a rigid body to answer the conceptual question.

Solution: The accumulation of water at higher elevation than it had before the dam was built effectively redistributed



10 – 25

some of the Earth’s mass to a location that is farther from the axis of rotation. We conclude that the moment of inertia of the Earth did increase as a result of building the dam.

Insight: Increasing the moment of inertia of the Earth has the effect of reducing the rotation speed of the Earth, as we’ll see in chapter 11 when we consider the conservation of angular momentum.

80. Picture the Problem: As you stand on the equator, you are traveling in a circle with a radius equal to the radius of the

Earth and at a speed determined by the rotational period of the Earth. Strategy: Use Newton’s Second Law to find the normal force exerted by the Earth on the person, and set it equal to

zero. Solve the resulting expression for the angular speed and then find the rotational period of the Earth. Solution: 1. Write out Newton’s Second Law

for when you stand on the equator and set N = 0. Write 2

cpa Rω= (equation 10-13) and solve for ω :

cp

20

N mg ma

mg mR g Rω ω

= − = −

− = − ⇒ =

∑F

2. Now apply equation 10-5 directly:

( ) ( )2 6

2 2 2

9.81 m/s 6.37 10 m

5060 s 1 hr 24 min

Tg R

π π πω

= = =×

= =

Insight: Although it would be fun at first, a weightless condition at the equator would make life difficult because everything would be weightless, regardless of its mass. Imagine if the ocean water were not held firmly down by gravity, and neither were the ships!

81. Picture the Problem: The diver rotates about her center of mass at constant angular speed.

Strategy: Use equation 10-3 together with the angle through which the diver rotated and the time elapsed in order to find the average angular speed.

Solution: Apply equation 10-3 directly: ( )2.50 rev 2 rad/rev

6.8 rad/s2.3 st

πθω×Δ

= = =Δ

Insight: The time of flight reveals that if the diver stepped off the platform with zero initial velocity, it must have been 22 m high (71 feet)! More likely it was a standard board 3.0 m above the water, and the diver first rose upward, reached a maximum altitude, and then returned to the water.

82. Picture the Problem: The hula hoop rolls without slipping at constant speed on level ground.

Strategy: Set the kinetic energy equal to the sum of translational and rotational energies, then use the fact that it rolls without slipping to find an expression in terms of velocity alone. Then solve for the velocity.

Solution: Substitute v rω = into equation 10-19 and solve for v:

( )( )22 2 2 2 21 1 1 12 2 2 2

0.12 J 1.4 m/s0.065 kg

K mv I mv mr v r mv

Kvm

ω= + = + =

= = =

Insight: For a rolling object the fraction of the total kinetic energy that is rotational energy depends upon the moment of inertia. If the object were a hollow sphere, for instance, with 22

3I mr= , the final kinetic energy would be 60% translational and 40% rotational.

83. Picture the Problem: The plane flies in a level, circular path at constant speed.

Strategy: Set the centripetal acceleration equal to 7.00g and solve equation 6-15, 2cp ,a v r= for the radius of the

circle.

Solution: Solve equation 6-15 for r:

( )( )

22 2

2cp

245 m/s874 m

7.00 7.00 9.81 m/sv vra g

= = = =

Insight: This must be a jet plane because 245 m/s corresponds to 548 mi/h. A Cessna flying at 50 m/s (110 mi/h) can take a turn with a radius of only 23 m without passing out. Whether the aircraft could tolerate such a turn is another matter!



10 – 26

84. Picture the Problem: A quarter rolls without slipping around the edge of an

identical quarter.

Strategy: Use the relationship between center-of-mass speed and rotational speed of an object that rolls without slipping to answer the conceptual question.

Solution: 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.

Insight: It is true that the circumferences of the two quarters are the same, so at first glance it would seem as if the rolling quarter will make only one rotation. But note that when the rolling quarter begins, George Washington’s neck is close to the stationary quarter. After rolling halfway around, the top of Washington’s head is near the stationary quarter. At this point, however, the rolling quarter is back in its original orientation relative to the laboratory, and has completed one full revolution.

85. Picture the Problem: The object shown at right can be rotated in three

different ways: case 1, rotation about the x axis; case 2, rotation about the y axis; and case 3, rotation about the z axis.

Strategy: Calculate the moment of inertia in each of the three cases and compare their values to determine the ranking.

Solution: 1. Calculate I1:

( )( ) ( )

21 i i

2 212

21

2 2 2 3

10

I m r

M R M R

I M R

=

= +

=

∑

2. Calculate I2:

( )

22 i i

2 2 22 2 3 8

I m r

M R M R M R

=

= + =∑

3. Calculate I3: ( )( )22 2 213 i i 22 2 2 6I m r M R M R M R= = + =∑

4. By comparing the magnitudes of the three moments of inertia we arrive at the ranking 3 2 1 .I I I< < Insight: For a given angular speed ω it takes the least amount of energy 21

2K Iω= to spin this object about the z axis. 86. Picture the Problem: The image for this problem is shown at right.

Strategy: The disk has rotated about ¼ of 90° or 8π radians or 1 16 of a revolution. The initial position (leading edge, assuming it is going from left to right) of the BB is about 42.0 cm and the final position is about 64.5 cm. Find the time elapsed from the rotation rate of the disk, and then find the speed of the BB from the measured distance and the time elapsed.

Solution: 1. Solve equation 10-3 for tΔ : 1 16 rev50.4 rev/s

0.0012 s 1.2 ms

t θωΔ

Δ = =

= =

2. Now find the speed from equation 2-3: 0.645 0.420 m 190 m/s

0.0012 sxvt

Δ −= = =

Δ

Insight: We can’t do much better than two significant figures because of the difficulty in measuring the angle accurately. The 190 m/s speed is about 420 mi/h or about half the speed of sound (343 m/s).



10 – 27

87. Picture the Problem: The image for this problem is shown at right.

Strategy: Lines dropped vertically downward from the edges of the disk meet the ruler at about 40.5 cm and 54.5 cm, meaning the diameter of the disk is 14.0 cm and its radius is 7.00 cm. Use the known radius and angular speed to find the linear speed of the edge of the disk. The arc length is estimated to be about s = 2.6 cm from the photograph. The initial position (leading edge, assuming it is going from left to right) of the BB is about 42.0 cm and the final position is about 64.5 cm, so the BB has traveled a distance 22.5 cmxΔ = . The speed of the BB is the ratio of the distance it travels to the distance the edge of the wheel travels, multiplied by the speed of the wheel’s edge. Equation 10-12 can be used to answer part (c) and equation 10-13 to answer part (d).

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

t 0.0700 m 50.4 rev/s 2 rad/rev

0.0700 m 317 rad/s 22.1 m/s

v rω π= = ×

= =

2. (b) Use a ratio to find BBv :

( )( ) ( )BB wheel

22.5 cm22.1 m/s 190 m/s

2.6 cmxv v

sΔ

= = =

3. (c) Solve equation 10-12 for r:

( )190 m/s 0.60 m 60 cm317 rad/s

tvr

ω= = = =

4. (d) Combine equations 6-16 and 10-13 to find the centripetal force on a 1.0 g lump of putty:

( )( )( )

2cp cp

2

cp

0.0010 kg 0.0700 m 317 rad/s

7.0 N 1.6 lb

F ma mr

F

ω= =

=

= =

Insight: The wheel either would need to be 8.7 times larger or be rotating 8.7 times faster in order for its edge to travel at the same speed as the BB. If the putty can’t stick with at least 7.0 N of force it will fly off the rotating disk.

88. Picture the Problem: The hour hand rotates at one revolution per 12 hours and the minute hand rotates at one

revolution per hour. Strategy: The hour and minute hands rotate at different rates. Use equation 10-3 to find expressions for the angular

positions of the hour and minute hands as a function of time with 2:00 being the starting positions [ ( )2

h,0 12 360 60.0θ = ° = ° and m,0 0.0θ = ° ]. Set the angular positions difference equal to zero and solve for t.

Solution: 1. (a) At 2:10 P.M. the minute hand points to the 2 but the hour hand has moved toward the 3, so they aren’t quite lined up yet. The meeting time must be after 2:10 P.M.

2. (b) At 2:15 P.M. the minute hand points to the 3 but the hour hand hasn’t reached it yet. That means the minute hand has already passed by the hour hand. The meeting time must be before 2:15 P.M.

3. (c) Use equation 10-3 to find an expression for h mθ θ− :

( ) ( )

m m,0 m

h h,0 h

h m h,0 m,0 h m

tt

t

θ θ ω

θ θ ω

θ θ θ θ ω ω

= +

= +

− = − + −

4. Find hω and mω : 1

h

1m

1 rev 1 h 360 0.500 min12.0 h 60 min rev1 rev 360 6.00 min

60 min rev

ω

ω

−

−

°= × × = °

°= × = °

5. Solve the expression in step 2 for t:

( ) ( )( )

( ) ( )( )

h m h,0 m,0

1h m

0.0 60.010.9 min

0.500 6.00 mint

θ θ θ θ

ω ω −

− − − ° − °= = =

− − °

6. The two hands will be aligned 10.9 min after 2:00, or at 2:10:54 P.M.

Insight: Use a similar approach to find that the next time the hands will align will be 3:16:22 P.M.



10 – 28

89. Picture the Problem: The diver flies the air in a parabolic trajectory, rotating about her center of mass until she hits the water.

Strategy: The time the diver is in the air is determined solely by the initial height and initial vertical speed (zero in this case). Use equation 4-6 to find the time of flight. The number of revolutions she makes is then equal to her angular speed multiplied by the time of flight (equation 10-3).

Solution: 1. (a) Solve equations 4-6 for t: 210 0 2

210 2

0

0 0

2

yy y v t gt

y gt

t y g

= + −

= + −

=

2. Now solve equation 10-3 for θΔ :

( )av av 0

2

2

2 3.0 m2.2 rad 1 rev 0.27 revs 2 rad 9.81 m/s

t y gθ ω ω

π

Δ = Δ =

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

3. (b) It does not depend on her initial speed. Her initial speed determines how far she travels horizontally before hitting the water, but has no effect on the time she falls.

Insight: The diver is in the air for 0.78 s. That means she needs an angular speed of 2 rad 0.78 s 8.1 rad/sπ = in order to complete a rotation and land feet first.

90. Picture the Problem: The lump of clay moves in a circle with a constant angular speed. Strategy: Use equation 10-5 to find the angular speed 2 .Tω π= Then use this angular speed in equation 10-12 to find

the linear speed and in equation 10-13 to find the centripetal acceleration. Solution: 1. (a) Apply equation 10-12

directly, letting 2 Tω π= : ( ) ( )t 2 2 0.068 m 0.52 s 0.82 m/sv r r Tω π π= = = × =

2. (b) Repeat step 1 for equation 10-13: ( )

2 22 2

cp2 20.068 m 9.9 m/s

0.52 sa r r

Tπ πω ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3. (c) If the period of rotation is doubled, the angular speed ω is cut in half. This will cut tv in half and cut cpa to a fourth.

Insight: The potter’s wheel is rotating at 115 rev/min, fast enough to make the centripetal force on the lump of clay to be greater than its weight.

91. Picture the Problem: The geometry of the CD reader is indicated in the

figure at right.

Strategy: Use equation 10-12 to find a relationship between the angular speed of the disk and the linear speed at the point the laser beam strikes the disk. Use the initial and final angular speeds and the time elapsed to determine the angular acceleration.

Solution: 1. (a) As the laser moves from the center outward, the increasing radius requires the angular speed of the disk to decrease in order to maintain the same linear speed.

2. (b) Solve equation 10-12 for ω : t 1.25 m/s 50.0 rad/s0.0250 m

vr

ω = = =

3. (c) Repeat step 2 for the new radius: t 1.25 m/s 20.8 rad/s

0.0600 mvr

ω = = =

4. (d) Apply equation 10-6 directly: 3 2f i

av20.8 50.0 rad/s 7.31 10 rad/s

66.5 min 60 s/mint tω ωωα −−Δ −

= = = = − ×Δ Δ ×

Insight: While CD players still use this variable angular speed method, most CD-ROM drives in computers use a constant angular velocity system.



10 – 29

92. Picture the Problem: The pigeon drops straight downward for 14 m, rotating about its center of mass. Strategy: The time it takes the pigeon to fall is determined solely by the vertical distance and initial vertical speed (zero

in this case). Use equation 4-6 to find the time of fall. The number of revolutions the bird makes is then equal to its angular speed multiplied by the time of fall (equation 10-3).

Solution: 1. Solve equation 4-6 for t: 210 0 2

210 2

0

0 0

2

yy y v t gt

y gt

t y g

= + −

= + −

=

2. Now solve equation 10-3 for θΔ :

( )( )

av av 0

2

2

2 14 m12 rad 1 rev 3.2 revs 2 rad 9.81 m/s

t y gθ ω ω

π

Δ = Δ =

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Insight: The bird falls for 1.7 s. That means she needs an angular speed of 8 rad 1.7 s 15 rad/sπ = in order to complete the fourth rotation and finish its fall in the upright position.

93. Picture the Problem: The marble rolls down the incline, increasing its linear and angular velocities at a constant rate. Strategy: Use the fact that the marble rolls without slipping to find its angular acceleration from its linear counterpart

using equation 10-14. Then use equation 10-8 to determine the final angular speed after accelerating from rest.

Solution: 1. (a) Solve equation 10-14 for :α ( )

22 2 2

12

3.3 m/s 412 rad/s 4.1 10 rad/s0.016 m

ar

α = = = = ×

2. (b) Apply equation 10-8 directly:

( ) ( )2

20 1

2

3.3 m/s0 1.5 s 6.2 10 rad/s0.016 m

at tr

ω ω α= + = + = = ×

Insight: We could also use conservation of energy in a manner similar to problem 71 to find that the marble has rolled down a vertical height of ( )27

10 1.7 mh v g= = and has traveled a distance of 212 3.7 md at= = so the angle of the

incline must be ( )1sin 1.7 3.7 28θ −= = ° above horizontal. 94. Picture the Problem: The ball rolls without slipping along the

horizontal table top, then falls off the edge, traveling along a parabolic arc until it hits the floor.

Strategy: Use the table height together with equation 4-6 to determine the time of fall. Use the number of rotations and the time elapsed to find the angular speed of the ball while it was in the air. Assuming that the angular speed was unchanged from when it was rolling on the table, equation 10-15 can be used to find the linear speed from the angular speed.

Solution: 1. Solve equation 4-6 for t: 210 0 2

210 020 0 2

yy y v t gt

y gt t y g

= + −

= + − ⇒ =

2. Determine the angular speed ω from equation 10-3:

02t y gθ θω Δ Δ

= =Δ

3. Apply equation 10-15 directly: ( ) ( )

( ) ( )20

0.37 rev 2 rad/rev0.032 m 0.20 m/s

2 2 0.66 m 9.81 m/sv r r

y gπθω

×Δ= = = =

Insight: If you solve the intermediate steps you’ll find that t = 0.37 s and 6.3 rad/s.ω = If the ball is to complete one full revolution while in the air, it needs an angular speed of 17 rad/s or a linear speed of 0.55 m/s while rolling.



10 – 30

95. Picture the Problem: The two students walk in different directions along

the same circular path and at different angular speeds until they meet each other.

Strategy: Use equation 10-9 to write expressions for the angular position of each student as a function of time. Determine the time at which the positions are equal and then determine the location at which they meet. Let the clockwise direction be positive, so that cw 0.045 rad/sω = and

ccw 0.023 rad/s.ω = − Let the student who walks counterclockwise begin at

0,ccw 2 radθ π= instead of zero. Otherwise, his angular position would become negative and would never be equal to the angular position of the other student.

Solution: 1. (a) Set cw ccwθ θ= and solve for t:

( ) ( )

0,cw cw 0,ccw ccw

0,ccw 0,cw

cw ccw

2 0 rad 92 s0.045 rad/s 0.023 rad/s

t t

t

θ ω θ ωθ θ πω ω

+ = +− −

= = =− − −

2. (b) Use the time found in step 1 to find the angular position:

( )( )( )

cw 0,cw cw 0 0.045 rad/s 92 s

4.2 rad 180 rad 240 clockwise from North

tθ θ ω

π

= + = +

= × ° = °

3. (c) Use equation 10-12 to find an expression for the radius r:

1 2 1 2

1 2

1 2

0.23 m/s 10 m0.045 0.023 rad/s

v v r rv v

r

ω ω

ω ω

− = −

−= = =

− −

Insight: If the counterclockwise student were to double her angular speed to − 0.046 rad/s, the two would meet in 69 s and at 180° clockwise from north.

96. Picture the Problem: The yo-yo spins about its rotation axis, decreasing its angular speed at a constant rate. Strategy: Use the angle through which the yo-yo rotated and the average angular speed during the time interval to

determine the time elapsed according to equation 10-9. Then use equation 10-8 to make a ratio with which to find the time elapsed for the new change in angular speed.

Solution: 1. (a) Solve equation 10-9 for t: ( )( ) ( )

10 02

0

0

2 120 rev 2 rad rev25 s

35 25 rad/s

t

t

θ θ ω ωθ θ π

ω ω

= + +− 2 ×

= = =+ +

2. (b) Use a ratio to find the new time: ( )

( )( )( )

0new newnew old

old 0 old

15 25 rad/s1 25 s

35 25 rad/st

t tt

ω ω αω ω α

− −= = = ⇒ = =

− −

Insight: As long as the angular acceleration is constant it will always take the same amount of time for the yo-yo to change its angular speed by 10 rad/s.

97. Picture the Problem: The tire rolls without slipping, increasing its linear and angular speeds at a constant rate. Strategy: Use the given speeds to find the linear acceleration of the tire, and then relate it to the rotational acceleration

of the tire using equation 10-14.

Solution: 1. (a) Use equation 3-5 to find a: ( ) 245 mi/h 0.447 m/s mi/h 0

2.2 m/s9.1 s

vat

× −Δ= = =

Δ

2. Use equation 10-14 to find :α

( )2

22.2m/s 6.9 rad/s0.32 m

ar

α = = =

3. (b) If the radius of the tire is cut in half then the angular acceleration doubles.

Insight: Verify for yourself that if the car has a mass of 1200 kg it would need a power of 27 kW or 36 hp to accomplish the acceleration in the given amount of time.

N

E θ

start



10 – 31

98. Picture the Problem: The cylinder and the sphere roll 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. Because the sphere and the cylinder roll without slipping, the equation v rω = can be used to write the expression in terms of linear velocity alone. Use the resulting equations to find the speeds of each object at the bottom of the 0.75 m high ramp.

Solution: 1. (a) The sphere has a smaller moment of inertia than the cylinder, so it will store less of its energy as rotational energy and more as translational. We therefore expect the sphere to have the higher speed.

2. (b) Set i fE E= and solve for fK :

( )( )( )

i i f f

f2

f

0 02.0 kg 9.81 m/s 0.75 m 14.7 J 15 J

U K U Kmgh K

K mgh

+ = ++ = +

= = = =

3. Set fK equal to t rK K+ for the cylinder: ( )( )22 2 2 21 1 1 1 1f 2 2 2 2 2

2 2 231 12 4 4


ω= + = +

= + =

4. Solve the equation in step 3 for cylinderv : ( )( )

fcylinder

4 14.7 J43.1 m/s

3 3 2.0 kgK

vm

= = =

5. Set fK equal to t rK K+ : ( )( )22 2 2 21 1 1 1 2f 2 2 2 2 5

2 2 271 12 5 10


ω= + = +

= + =

6. Solve the equation in step 3 for spherev : ( )

( )f

sphere

10 14.7 J103.2 m/s

7 7 2.0 kgK

vm

= = =

Insight: The sphere always wins a race with a cylinder, regardless of their mass or radii. Try it! 99. Picture the Problem: The centrifuge rotates with a constant angular speed. Strategy: Use equation 10-13 and the known angular speed and centripetal acceleration to determine the diameter of the

centrifuge. Then use Newton’s Second Law to find the force exerted by the centrifuge in order to accelerate the 15.0-gram sample under these conditions.

Solution: 1. (a) Solve equation 10-13 for r: ( )

( )( ) ( )

2cp2 2rev 21 min

min 60 s rev

6840 9.81 m/s2 2 2 0.334 m

6050

ad r

ω π= = = =

⎡ ⎤⎣ ⎦

2. (b) Use Newton’s Second Law to find the force: ( ) ( )( )2cp 0.0150 kg 6840 9.81 m/s 1.01 kNF ma= = =

Insight: The high rate of rotation produces an enormous acceleration, almost 7000 times stronger than gravity, so that the force exerted on a 15.0-gram sample (that weighs just 0.53 ounces) is 227 lb!

100. Picture the Problem: When the clutch engages, the yo-yo converts its rotational kinetic energy into gravitational

potential energy as it climbs up the string. Strategy: Use conservation of energy between the bottom of the string and the top to find the angular speed required for

the yo-yo to climb the string. Solution: Set i fE E=

and solve for ω :

( )( )( )

i i f f21

2

22

–5 2

0 0

2 0.11 kg 9.81 m/s 1.0 m2 170 rad/s 1.7 10 rad/s7.4 10 kg m

U K U KI mgh

mghI

ω

ω

+ = +

+ = +

= = = = ×× ⋅

Insight: The angular speed of 170 rad/s corresponds to 1600 rev/min, a substantial spin rate.



10 – 32

101. Picture the Problem: The centrifuge rotates about its axis, increasing its angular speed at a constant rate.

Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angular acceleration and the angle through which the centrifuge rotates during the specified interval.

Solution: 1. (a) Solve equation 10-8 for α : 20 550 430 rad/s 15 rad/s8.2 st

ω ωα

− −= = =

2. (b) Solve equation 10-9 for θΔ : ( )( ) 31 10 02 2( ) 430 550 rad/s 8.2 s 4.0 10 radtθ θ ω ω− = + = + = ×

Insight: The final angular speed of 550 rad/s corresponds to 5300 rev/min, and the angle of 4000 radians corresponds to 640 revolutions in 8.2 s, an average of 78 rev/s or 490 rad/s.

102. Picture the Problem: The honeybee’s wings oscillate, pivoting about the base of the wing in a roughly circular motion.

Strategy: Assume the angle through which a wing moves from top to bottom is 23120 rad.π° = Then the total angle

through which each wing moves in one second is ( )( )( ) 100023 3250 beats 2 cycle/beat rad/cycle rad.π π=

Assume the wing’s angular speed is zero at each end of its up-and-down motion, so that the maximum angular speed is twice the average angular speed. Use the total angle through which each wing moves in one second to find the average angular speed and double it to find the maximum angular speed. Then assume each wing is 1.0 cm long in order to find the maximum speed of the wing tip.

Solution: 1. (a) Determine maxω from the θΔ and time information discussed above:

( )max av

1000 3 rad2 2 2 2100 rad/s

1.0 stπθω ω Δ

= = = =Δ

2. (b) Apply equation 10-12 with 0.010 m:r = ( )( )0.010 m 2100 rad/s 21 m/sv rω= = =

Insight: The rapid oscillations of the honeybee’s wings allow it to fly at fairly rapid speeds as well as hover in midair.

103. Picture the Problem: The Sun moves in a circular path about the center of the Milky Way galaxy with a constant angular speed.

Strategy: Use equation 10-3 to determine the angular speed of the Sun from the orbit period. Then use equation 10-12 together with the speed and angular speed information to find the orbit distance of the Sun.

Solution: 1. (a) Find ω from the orbit period:

( )( )16

6 7

2 rad 8.3 10 rad/s240 10 yr 3.16 10 s/yrt

θ πω −Δ= = = ×

Δ × ×

2. (b) Solve equation 10-12 for r: 17

16

137 mi/s 1.7 10 mi8.3 10 rad/s

vrω −= = = ×

×

Insight: This orbit distance is the same as 2.7×1017 km, or 28,000 light years!

104. Picture the Problem: The ceiling fan rotates about its axis, speeding up with constant angular acceleration, then rotating with constant angular speed, then slowing down with constant angular acceleration.

Strategy: Use equation 10-9 to find the angle through which the fan rotates during the specified intervals. The average angular speed is ( ) ( )1 1

av 2 20 1.9 rev/s 0.95 rev/sω ω= + = = during the acceleration and deceleration intervals, and

av 1.9 rev/sω = during the constant angular speed interval.

Solution: Apply equation 10-9 to find θΔ : ( )( )( )( )( )( )

1 av,1 1

2 av,2 2

3 av,3 3

1 2 3

0.95 rev/s 15 s 14 rev

1.9 rev/s 5.5 min 60 s/min 630 rev

0.95 rev/s 2.4 min 60 s/min 140 rev

780 rev

t

t

t

θ ω

θ ω

θ ω

θ θ θ θ

Δ = Δ = =

Δ = Δ = × =

Δ = Δ = × =

Δ = Δ + Δ + Δ =

Insight: There are other ways to find the answer, including finding the angular acceleration and then using 21

0 2t tθ ω αΔ = + , but whenever you know the initial and final angular speeds it is easiest to use av tθ ωΔ = Δ .



10 – 33

105. Picture the Problem: The compartment moves in a circular path at constant angular speed.

Strategy: Use equation 10-13 to find the centripetal acceleration inside the compartment, and then equation 10-12 to find the linear speed.

Solution: 1. (a) Apply equation 10-13 directly:

( )( )22cp

2 2

28 ft 0.305 m/ft 10 rev/min 2 rad/rev 1 min/60 s

9.36 m/s 1.0 9.81 m/s 0.95

a r

g g

ω π= = × × ×

= × =

2. (b) Apply equation 10-12 directly: ( )( )t 28 ft 0.305 m/ft 10 rev/min 2 rad/rev 1 min/60 s

8.9 m/s

v rω π= = × × ×

=

Insight: In 22 hours the four people completed 13,200 revolutions and traveled 710 km or 440 miles! 106. Picture the Problem: The Crab pulsar rotates on its axis, slowing down with constant angular acceleration.

Strategy: Equation 10-6 gives an expression for the angular acceleration as a function of rotation rate and time. Determine the difference in rotation rates over the span of a year by approximating T T T+ Δ ≅ because 1.26×10−5 s is tiny compared with the time span of a year (3.16×107 s). Then use equation 10-6 to find the average angular acceleration over the 1-year time interval.

Solution: 1. (a) Find the difference in angular speeds:

( )( )

( )( )( )

0 2

5

2

0

1.26 10 s2 rad

0.0330 s

0.0727 rad/s


θ θω ω θ θ

π

ω ω

−

⎡ ⎤− + Δ −Δ⎛ ⎞− = − = ≅⎢ ⎥ ⎜ ⎟+ Δ + Δ ⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤− ×⎢ ⎥=⎢ ⎥⎣ ⎦

− = −

2. Apply equation 10-6 directly: ( )

9 2av 7

0.0727 rad/s 2.30 10 rad/s1.00 y 3.16 10 s/yt

ωα −Δ −= = = − ×

Δ × ×

3. (b) Use equation 10-8 to find the time it will take the pulsar to stop:

( ) ( )09 2

10 37

0 2 rad 0.0330 s2.30 10 rad/s

1 y8.28 10 s 2620 y 2.62 10 y3.16 10 s

tπω ω

α −

−−= =

− ×

= × × = = ××

4. (c) Solve equation 10-8 for 0ω assuming it is currently the year 2010: ( ) ( )

( )( )( )

0

9 2 7

2 rad 0.0330 s

2.30 10 rad/s 2010 1054 y 3.16 10 s/y

260 rad/s

tω ω απ

−

= −

= −

− × − ×

=

5. Use equation 10-5 to find 0T : 0

0

2 2 0.0242 s 24.2 ms260 rad/s

T π πω

= = = =

Insight: The pulsar is slowing down very gradually, because its period has only increased by 36% over a period of 952 years!



10 – 34

107. Picture the Problem: The rod swings downward, pivoting about one end.

Strategy: Use conservation of mechanical energy to determine the angular speed of the rod when it is vertical. Then apply equation 10-12 to find the linear speed tv of the free end of the rod. Let y = 0 correspond to the position of the center of mass of the rod when it is vertical, so that

0 2y L= at the start. The rod has only rotational kinetic energy at the end, not translational, because the pivot point is stationary. The moment of inertia of a rod pivoted about one end is 21

3I ML= .

Solution: 1. (a) Set i fE E= and solve for ω :

( )

i i f f

212 21

3

30 2 0

K U K U

MgL MgL gMg L II LML

ω ω

+ = +

+ = + ⇒ = = =

2. (b) Apply equation 10-12 directly: t

3 3gv r L gLL

ω= = =

Insight: The angular speed of the rod actually decreases with increasing length, but the tangential speed increases with increasing length.

108. Picture the Problem: The rod swings downward, pivoting about one end.

Strategy: Use conservation of mechanical energy to determine the angular speed of the rod and ball combination when it is vertical. Then apply equation 10-12 to find the linear speed tv of the free end of the rod. Let y = 0 correspond to the position of the tip of the rod when it is vertical, so that 0y L= at the start and 2y L= for the rod at the end. The rod and ball combination has only rotational kinetic energy at the end, not translational, because the pivot point is stationary. The moment of inertia of a rod and ball combination is ( )2 21

3 2I ML M d= + .

Solution: 1. (a) Set i fE E= and solve for ω :

( ) ( ) ( ) ( )i i f f

212

25 12 2

1 12 2 2

2 213

0 2 2 2

3 22 2

2 22

K U K U

MgL M gL I Mg L M g L d

MgL MgL Mgd IMgL Mgd MgL Mgd

I ML Md

ω

ω

ω

+ = +

+ + = + + −

− + =

+ += =

+

2. Divide top and bottom by M, and factor 12 gL

out of the numerator and 213 L out of the denominator:

( )( )

( )( )

122

22 2 213

1 4 1 4321 6 1 6

gL d L d LgLL d L d L

ω⎡ ⎤+ +

= = ⎢ ⎥+ +⎢ ⎥⎣ ⎦

3. Apply equation 10-12 directly:

( )( )

( )( )t 2 2

1 4 1 43 31 6 1 6

d L d Lgv r L gLL d L d L

ω⎡ ⎤ ⎡ ⎤+ +

= = =⎢ ⎥ ⎢ ⎥+ +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

4. (b) When 0d = the speed of the rod is t 3v gL= . It will have that same value when the fraction

( )( )2

1 41

1 6

d L

d L

+=

+.

5. Set the fraction equal to one: ( ) ( )( )

21 4 1 6

4 6 2 3

d L d L

d L d L

+ = +

= ⇒ =

Insight: The center of percussion is about 2/3 the length of the rod from the pivot point. A similar sweet spot exists on baseball bats (which are NOT uniform rods). When the ball is struck by that point on the bat, the ball’s mass does not affect the angular velocity of the bat very much and causes very little discomfort for the batter’s hands.



10 – 35

109. Picture the Problem: The plank translates 1.0 m while rolling on two soup

cans as indicated in the diagram at right.

Strategy: Imagine a can rolls one entire circumference, 2 .rπ As shown in Figure 10-9, its center of mass will also move a distance 2 .rπ Meanwhile the top edge of the can is “rolling” without slipping against the surface of the plank, propelling the plank forward another distance 2 .rπ The plank therefore moves 2 rπ due to the rolling and another 2 rπ because the top edge of the can has moved that far relative to the floor (along with the center of mass). Therefore, the plank moves twice as far as the center of mass of the can. Use that fact to find the distance the center of mass moves. The arc length through which the can rim rotates equals the distance through which the center of mass moves (see Figure 10-9).

Solution: 1. (a) The center of mass moves half as far as the plank: ( )1

2 1.0 m 0.50 mxΔ = =

2. (b) Solve equation 10-2 for θ:

( )12

0.50 m 1 rev15 rad 2.4 rotations0.065 m 2 rad

sr

θπ

= = = × =

Insight: The argument for why the center of mass moves only half as far as the plank (relative to the floor) is related to the argument of why the top edge of a rolling wheel moves twice as fast as the center (see Figure 10-11).



10 – 36

110. Picture the Problem: The Ferris wheel rotates at a constant rate,

with the centripetal acceleration of the passengers always pointing toward the axis of rotation. Free body diagrams of the passenger are indicated at each of the three positions indicated in the problem.

Strategy: Use equation 10-13 to find the centripetal acceleration. The centripetal acceleration remains constant (as long as the angular speed remains the same) and points toward the axis of rotation. Then use Newton’s Second Law to find the normal force exerted by the seat on the person at each of the indicated positions. The seat force is exactly equal to the normal force at the top and at the bottom of the Ferris wheel, but halfway up the wheel the seat force is the vector sum of the normal force N pointing straight upward and the friction force sf that points toward the center of the wheel and provides the centripetal force.

Solution: 1. (a) Apply equation 10-13 directly: ( )

22 21

cp 28.1 rev 2 rad 1 min12 m 4.32 m/s

min rev 60 sa r πω ⎛ ⎞= = × × =⎜ ⎟

⎝ ⎠

2. Write Newton’s Second Law in the vertical direction at the top of the wheel to find N . At this location there is no horizontal acceleration or force: ( ) ( )( )

cp

2cp 65 kg 9.81 4.32 m/s

360 N 0.36 kN upward

yF N mg ma

N m g a

= − = −

= − = −

= =

∑

3. (b) Write Newton’s Second Law in the vertical direction at the bottom of the wheel to find N . At this location there is no horizontal acceleration or force: ( ) ( )( )

cp

2cp 65 kg 9.81 4.32 m/s

920 N 0.92 kN upward

yF N mg ma

N m g a

= − =

= + = +

= =

∑

4. (c) The seat exerts two forces on the person, an upward force to balance the weight and in inward static friction force to provide the centripetal acceleration. Use Newton’s Second Law to find the components of the seat force:

s cp s cp

ˆ0

ˆ y

x

F N mg mg

F f ma ma

= +

= − = ⇒ =

= = ⇒ =

∑∑

sS N f

N y

f x

5. Use the components to find the magnitude and direction of the seat force:

( ) ( )

( ) ( ) ( )

2 22 2cp

2 22 2

21 1 1

2cp

65 kg 4.32 m/s 9.81 m/s 0.70 kN

9.81 m/stan tan tan4.32 m/s

66 above the horizontal

x y

y

x

S S S ma mg

S mgS ma

θ − − −

= + = +

= + =

⎛ ⎞⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= °

Insight: In order to double the centripetal acceleration you need to increase the angular speed by a factor of 2 or decrease the period by a factor of 2 ; in this case a rotation rate of 11 rpm will double the centripetal acceleration. We bent the rules for significant figures a bit in step 1 to avoid rounding errors in subsequent steps.

N

W

v

r

N

Wv

S

W

v



10 – 37

111. Picture the Problem: The ball rolls without slipping down the

incline, gaining speed, then is launched horizontally off the edge, traveling along a parabolic arc until it hits the floor.

Strategy: Use conservation of energy to determine the center of mass speed of the sphere at the bottom of the ramp. Then use equation 4-9 to determine the horizontal range of the ball as it travels through the air. Use the table height together with equation 4-6 to determine the time of fall. Use the angular speed of the ball and the time of fall to find the number of rotations it makes before landing.

Solution: 1. (a) Set i fE E= and simplify:

( )( )

i i f f2 21 1

2 222 21 1 2

2 2 52 2 271 1

2 5 10

0 0K U K U

mgh mv Imv mr v r

mgh mv mv mv

ω+ = +

+ = + +

= +

= + = 2. Solve the expression from step 1 for v: ( )( )210 10

7 7 9.81 m/s 0.61 m 2.92 m/sv gh= = =

3. Apply equation 4-9 to find d: ( ) ( )table

0 2

2 1.22 m22.92 m/s 1.5 m

g 9.81 m/sh

d v= = =

4. (b) Solve equation 4-6 for t:

( )

210 0 2

2 table1table 2 2

2 1.22 m20 0 0.50 s

9.81 m/s

yy y v t gt

hh gt t

g

= + −

= + − ⇒ = = =

5. Determine the angular speed ω from equation 10-12: ( )

t12

2.92 m/s 34 rad/s0.17 m

vr

ω = = =

6. Find the number of revolutions: ( )( )34 rad/s 0.50 s 17 rad 1 rev 2 rad 2.7 revtθ ω π= = = × =

7. (c) If the ramp were made frictionless, the sphere would slide, not roll. It would therefore store no energy in its rotation, and all of its gravitational potential energy would become translational kinetic energy. It would therefore launch from the table edge with a higher speed and the landing distance d would increase.

Insight: If the ramp were frictionless, the launch speed would be 2gh instead of 107 gh , an 18% increase in speed

and therefore an 18% increase in the landing distance d. 112. Picture the Problem: A human-powered centrifuge platform rotates

with constant angular speed in order to produce “artificial gravity” for astronauts. The figure at right shows the centripetal acceleration (in gs) produced by platforms of four different radii.

Strategy: The centripetal acceleration of a point on a rotating platform is given by 2

cp .a r ω= Therefore, for a given ω the platform with the largest radius will produce the greatest cp .a

Solution: If we examine the figure for a particular angular speed, we see that platform 4 produces the smallest centripetal acceleration

cpa and platform 1 the largest. The largest cpa corresponds to the

largest radius r. We arrive at the ranking 4 3 2 1 .r r r r< < <

Insight: Doubling the angular speed for a particular radius will quadruple the centripetal acceleration.



10 – 38

113. Picture the Problem: A human-powered centrifuge platform rotates with constant angular speed in order to produce

“artificial gravity” for astronauts. The figure at right shows the centripetal acceleration (in gs) produced by platforms offour different radii.

Strategy: The centripetal acceleration of a point on a rotating platform is given by 2cp .a r ω= Use this expression to

determine ω.

Solution: Solve 2cpa r ω= for ω:

2cp 5 9.81 m/s

6.25 ft 0.305 m/ft5.07 rad 1 rev 60 s 48.4 rev/min

s 2 rad min

ar

ω

π

×= =

×

= × × =

Insight: Cutting the angular speed in half for a 6.25 ft radius will cut the centripetal acceleration by a factor of four. 114. Picture the Problem: A human-powered centrifuge platform rotates


Strategy: The previous problem determined that a centripetal acceleration of 5.00g could be produced by the test model at an angular speed of 48.5 rev/min. Determine which of the four curves passes through this data point.

Solution: An examination of the figure shows that curve 3 passes through the point (48.5 rpm, 5.00g).

Insight: The same centripetal acceleration could be achieved at lower angular speeds by using platform radii corresponding to curves 1 and 2.

115. Picture the Problem: A human-powered centrifuge platform rotates


Strategy: By the reasoning presented in problem 113 above, we expect the radius of platform 4 to be smaller than the 6.25 ft of platform 3. The centripetal acceleration of a point on a rotating platform is given by 2

cp .a r ω= Solve this expression for r and use data points on the graph to determine the radius of platform 4.

Solution: Curve 4 passes through the point (55 rpm, 3.0g). Use these values to find r:

2cp2 2

3.0 9.81 m/s 0.89 m 3.281 ft/m 2.9 ft55 rev 2 rad 1 minmin rev 60 s

ar

ω π×

= = = × =⎛ ⎞× ×⎜ ⎟⎝ ⎠

Insight: This platform would not be practical unless the astronauts were less than 3 ft tall!



10 – 39

116. Picture the Problem: The disk and the hollow sphere roll

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. Because the sphere and the disk roll without slipping, the equation v rω = can be used to write the expression in terms of linear velocity alone. Use the resulting equations to find the speeds of each object at the bottom of the 0.75 m high ramp.

Solution: 1. (a) The expression for the speed of the disk and the spherical shell are independent of mass and radius, but the different moments of inertia result in a faster speed for the disk than for the basketball. The disk wins the race.

2. (b) Set i fE E= and solve for fK : i i f f

f

f

0 0U K U Kmgh K

K mgh

+ = ++ = +

=

3. Set fK equal to t rK K+ for the disk: ( )( )22 2 2 21 1 1 1 1f 2 2 2 2 2

2 2 231 12 4 4


ω= + = +

= + =

4. Solve the equation in step 3 for diskv : ( )( )2f 4disk 3

4 4 9.81 m/s 0.75 m 3.1 m/s3 3K mghvm m

= = = =

5.(c) Set fK equal to t rK K+ for the hollow shell: ( )( )22 2 2 21 1 1 1 2f 2 2 2 2 3

2 2 251 12 3 6


ω= + = +

= + =

6. Solve the equation in step 3 for shellv : ( )( )2f 6shell 5

6 6 9.81 m/s 0.75 m 3.0 m/s5 5K mghvm m

= = = =

Insight: The disk always wins a race with a hollow sphere, regardless of their mass or radii. Try it! In this case the disk has a larger moment of inertia than the basketball, but it also has more mass and therefore more energy.

117. Picture the Problem: The objects roll down the ramp without slipping, gaining both translational and rotational kinetic

energy.

Strategy: The object with the smallest moment of inertia will store the smallest fraction of its total energy in rotational kinetic energy, and will therefore have the greatest translational kinetic energy.

Solution: 1. (a) Compare the moments of inertia: 21disk 2

22solid 5

22shell 3

I MR

I MR

I MR

=

=

=

2. The solid sphere has the smallest moment of inertia and therefore the highest translational speed. The solid sphere will win the race, followed by the disk and then the hollow sphere.

3. (b) In each case the total kinetic energy of the object at the bottom of the ramp equals the gravitational potential energy it had at the top of the ramp. Since all three objects have the same gravitational potential energy at the top of the ramp, all the kinetic energies are equal at the bottom of the ramp.

Insight: The results of the race are independent of mass and radius. The solid sphere will always win.



10 – 40

118. Picture the Problem: The yo-yo falls straight down, increasing both its

translational and rotational kinetic energy. The Active Example 10-3 indicates that the mass of the yo-yo is 0.056 kg and its moment of inertia is 5 22.9 10 kg m−× ⋅ . In this problem the radius r of the axle the string wraps around is 0.0075 m.

Strategy: Use conservation of energy, including both translational and rotational kinetic energy, to determine the speed of the yo-yo after falling through a height h = 0.50 m.

Solution: 1. (a) Upon inspection of the expression given in the example,

2

21

ghvI mr

=+

, we can see that increasing r decreases the denominator and

increases the speed. The speed of the yo-yo will increase.

2. (b) Set i fE E= and solve for v: ( ) ( )

( )( )( ) ( )( )

i i f f22 2 2 2 21 1 1 1 1

2 2 2 2 2

22 2

2

2–5 2

02 2

1

2 9.81 m/s 0.50 m

1 2.9 10 kg m 0.056 kg 0.0075 m

0.98 m/s

K U K Umgh mv I mv I v r v m I r

mgh ghvm I r I mr

v

ω+ = +

+ = + = + = +

= =+ +

=+ × ⋅

=

Insight: As predicted, the 0.98 m/s speed is faster than the 0.85 m/s that corresponded to the 0.0064 m radius. 119. Picture the Problem: The yo-yo falls straight down, increasing both its

translational and rotational kinetic energy. The Active Example 10-3 indicates that the mass of the yo-yo is 0.056 kg and the radius r is 0.0064 m.

Strategy: Use conservation of energy, including both translational and rotational kinetic energy, to determine the moment of inertia of the yo-yo. Let the yo-yo fall through a height h = 0.50 m and attain a speed of 0.64 m/s.

Solution: 1. (a) If more mass is concentrated near the rim of the yo-yo, its moment of inertia will be greater than that of the original yo-yo.

2. (b) Set i fE E= and solve for I: ( )

( )

( )( )( )( )

( )

i i f f22 2 21 1 1 1

2 2 2 22 21

2 22 2

22

2

5 2

02 2 1

2 9.81 m/s 0.50 m0.056 kg 0.0064 m 1

0.64 m/s

5.3 10 kg m

K U K Umgh mv I mv I v r

r mgh mv ghI mrv v

I

ω

−

+ = +

+ = + = +

− ⎛ ⎞= = −⎜ ⎟⎝ ⎠

⎡ ⎤⎢ ⎥= −⎢ ⎥⎣ ⎦

= × ⋅

Insight: As predicted, the 5 25.3 10 kg m−× ⋅ moment of inertia is larger than the 5 22.9 10 kg m−× ⋅ given in Active Example 10-3.

physics chapter 10 answers

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