Content check — College Physics, Chapter 10: Rotational Motion and Angular Momentum

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 10: Rotational Motion and Angular Momentum

25 problems · 3 figures

At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?

Answer

ω =

0 . 737 rev/s

10-3 §10.1 source ↗

Integrated Concepts

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest?

Answer

(a) $- 0 . {26 rad/s}^{2}$

(b) $27 rev$

10-5 §10.2 source ↗

With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s.

(a) What is its angular acceleration in rad/s²?

(b) How many revolutions does it go through in the process?

Answer

(a) $80 {rad/s}^{2}$

(b) 1.0 rev

10-7 §10.2 source ↗

A gyroscope slows from an initial rate of 32.0 rad/s at a rate of $0.700 {rad/s}^{2}$ .

(a) How long does it take to come to rest?

(b) How many revolutions does it make before stopping?

Answer

(a) 45.7 s

(b) 116 rev

10-9 §10.2 source ↗

Everyday application: Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled.

(a) If the string is stationary and the yo-yo accelerates away from it at a rate of $1.50 {m/s}^{2}$ , what is the angular acceleration of the yo-yo?

(b) What is the angular velocity after 0.750 s if it starts from rest?

Answer

a) $6 {00 rad/s}^{2}$

b) 450 rad/s^s

c) 21.0 m/s^s

10-10 §10.3 source ↗

This problem considers additional aspects of example Calculating the Effect of Mass Distribution on a Merry-Go-Round. (a) Consider part b of that example, when the child is sitting on the merry-go-round. How long does it take the father to give the merry-go-round an angular velocity of 1.50 rad/s? (b) How many revolutions must he go through to generate this velocity? (c) If he exerts a slowing force of 300 N at a radius of 1.35 m, how long would it take him to stop them?

Answer

(a) 0.338 s

(b) 0.0403 rev

10-12 §10.3 source ↗

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of

2.00 \times 10^{3} N

with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of

120 {rad/s}^{2}

. What is the moment of inertia of the boxer’s forearm?

Answer

0.50 kg \cdot m^{2}

10-14 §10.3 source ↗

Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk).

(a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

Answer

(a) $50.4 N \cdot m$

(b) $17.1 {rad/s}^{2}$

10-16 §10.3 source ↗

Zorch, an archenemy of Superman, decides to slow Earth’s rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of

4.00 \times 10^{7} N

(a little greater than a Saturn V rocket’s thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.) Explicitly show how you follow the steps found in Problem-Solving Strategy for Rotational Dynamics.

Answer

$3 . 96 \times 10^{18} s$

or $1.26 \times 10^{11} y$

10-18 §10.3 source ↗

Starting with the formula for the moment of inertia of a rod rotated around an axis through one end perpendicular to its length

(I = {Mℓ}^{2} / 3)

, prove that the moment of inertia of a rod rotated about an axis through its center perpendicular to its length is

I =

{Mℓ}^{2} / 12

. You will find the graphics in Figure 10.11 useful in visualizing these rotations.

Illustrations of ten different objects accompanied by their rotational inertias. — Figure 10.11 in the book ↗

Answer

\begin{matrix} I_{end} = I_{center} + m {\frac{l}{2}}^{2} \\ Thus, I_{center} = I_{end} - \frac{1}{4} {ml}^{2} = \frac{1}{3} {ml}^{2} - \frac{1}{4} {ml}^{2} = \frac{1}{12} {ml}^{2} \end{matrix}

10-19 §10.3 source ↗

Unreasonable Results

A gymnast doing a forward flip lands on the mat and exerts a 500-N ∙ m torque to slow their angular velocity to zero. Their initial angular velocity is 10.0 rad/s, and their moment of inertia is $0.050 kg \cdot m^{2}$ . (a) What time is required for the gymnast to exactly stop their spin? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?

Answer

(a) 2.0 ms

(b) The time interval is too short.

(c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of $500 N \cdot m$ is reasonable.

10-20 §10.3 source ↗

Unreasonable Results

An advertisement claims that an 800-kg car is aided by its 20.0-kg flywheel, which can accelerate the car from rest to a speed of 30.0 m/s. The flywheel is a disk with a 0.150-m radius. (a) Calculate the angular velocity the flywheel must have if 95.0% of its rotational energy is used to get the car up to speed. (b) What is unreasonable about the result? (c) Which premise is unreasonable or which premises are inconsistent?

Answer

(a) 17,500 rpm

(b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is > 50,000 gs.

10-21 §10.4 source ↗

This problem considers energy and work aspects of Example 10.7—use data from that example as needed. (a) Calculate the rotational kinetic energy in the merry-go-round plus child when they have an angular velocity of 20.0 rpm. (b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. (c) Again, using energy considerations, calculate the force the father must exert to stop the merry-go-round in two revolutions

Example 10.7 View it in the book ↗

Answer

(a) 185 J

(b) 0.0785 rev

10-23 §10.4 source ↗

(a) Calculate the rotational kinetic energy of Earth on its axis. (b) What is the rotational kinetic energy of Earth in its orbit around the Sun?

Answer

(a) $2.57 \times 10^{29} J$

(b) ${KE}_{rot} =$ $2.65 \times 10^{33} J$

10-25 §10.4 source ↗

A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 20.0 m/s at a distance of 0.480 m from the joint and the moment of inertia of the forearm is

0.500 kg \cdot m^{2}

, what is the rotational kinetic energy of the forearm?

Answer

{KE}_{rot} = 434 J

10-27 §10.4 source ↗

A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy. (b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m/s at the top of the hill? Explicitly show how you follow the steps in the Problem-Solving Strategy for Rotational Energy.

Answer

(a) $128 rad/s$

(b) $19.9 m$

10-29 §10.4 source ↗

While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produced given the mass lifted is 10.0 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is

0.900 kg \cdot m^{2}

, the muscle force is 1500 N, and its effective perpendicular lever arm is 3.00 cm. (b) How much work is done if the leg rotates through an angle of

20.0º

with a constant force exerted by the muscle?

Answer

(a) $10. {4 rad/s}^{2}$

(b) $net W = 6. 11 J$

10-34 §10.4 source ↗

In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. (a) Find the rotational kinetic energy of the pitcher’s arm and ball together given that the arm's moment of inertia is

0.720 kg \cdot m^{2}

and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder. (b) What force did the muscles exert to cause the arm to rotate if their effective perpendicular lever arm is 4.00 cm and the ball is 0.156 kg?

Answer

(a) 1.49 kJ

(b) $2.52 \times 10^{4} N$

10-36 §10.5 source ↗

(a) Calculate the angular momentum of the Earth in its orbit around the Sun.

(b) Compare this angular momentum with the angular momentum of Earth on its axis.

Answer

(a) $2 . 66 \times 10^{40} kg \cdot m^{2} /s$

(b) $7 . 07 \times 10^{33} kg \cdot m^{2} /s$

The angular momentum of the Earth in its orbit around the Sun is $3 . 77 \times 10^{6}$ times larger than the angular momentum of the Earth around its axis.

10-38 §10.5 source ↗

Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What angular momentum is given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time?

Answer

22 . 5 kg \cdot m^{2} /s

10-40 §10.5 source ↗

Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

Answer

25.3 rpm

10-43 §10.6 source ↗

Repeat Example 10.15 in which the disk strikes and adheres to the stick 0.100 m from the nail.

Example 10.15 View it in the book ↗

Answer

(a) $0.156 rad/s$

(b) $1 . 17 \times 10^{- 2} J$

10-45 §10.6 source ↗

Twin skaters approach one another as shown in Figure 10.37 and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy.

Figure a shows two skaters from the top view approaching each other from opposite directions with velocity v. In figure b two skaters then lock their right hands and start to spin in the clockwise direction with angular velocity omega. — Figure 10.37 in the book ↗

Answer

(a) 3.13 rad/s

(b) Initial KE = 438 J, final KE = 438 J

10-47 §10.6 source ↗

Repeat Example 10.15 in which the stick is free to have translational motion as well as rotational motion.

Example 10.15 View it in the book ↗

Answer

(a) 1.70 rad/s

(b) Initial KE = 22.5 J, final KE = 2.04 J

10-48 §10.7 source ↗

Integrated Concepts

The axis of Earth makes a 23.5° angle with a direction perpendicular to the plane of Earth’s orbit. As shown in Figure 10.39, this axis precesses, making one complete rotation in 25,780 y.

(a) Calculate the change in angular momentum in half this time.

(b) What is the average torque producing this change in angular momentum?

(c) If this torque were created by a single force (it is not) acting at the most effective point on the equator, what would its magnitude be?

In the figure, the Earth’s image is shown. There are two vectors inclined at an angle of twenty three point five degree to the vertical, starting from the centre of the Earth. At the heads of the two vectors there is a circular shape, directed in counter clockwise direction. An angular momentum vector, directed toward left, along its diameter, is shown. The plane of the Earth’s orbit is shown as a horizontal line through its center. — Figure 10.39 in the book ↗

Answer

(a) $5 . 64 \times 10^{33} kg \cdot m^{2} /s$

(b) $1 . 39 \times 10^{22} N \cdot m$

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Problems and answers are adapted from College Physics 2e by Paul Peter Urone, Roger Hinrichs, Kim Dirks and Manjula Sharma, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.