Algebra and Trigonometry based Physics Serway

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Chapter 8

Page 8.1

8

Rotational Equilibrium and Rotational Dynamics

CLICKER QUESTIONS

Question E1.01

Description: Reasoning with rotational inertia.

Question

The rotational inertia of the dumbbell (see figure) about axis A is twice the rotational inertia about axis B. The unknown mass

is:

1. 4/7 kg

2. 2 kg

3. 4 kg

4. 5 kg

5. 7 kg

6. 8 kg

7. 10 kg

8. None of the above

9. Cannot be determined

10. The rotational inertia cannot be different about different axes.

Commentary

Purpose: To practice problem solving with rotational motion ideas.

Discussion: Rotational inertia is not an intrinsic property of an object; its value depends on the axis about which it is

calculated. In this case, the object’s rotational inertia is twice as large about one axis as another.



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Page 8.2

Let’s avoid numerical computations, even though we are given specific values for all known quantities. Let d = 10 cm, m = 2

kg, and M = the unknown mass. We’ll solve for M in terms of the given quantities.

The rotational inertia about axis A is 2

2 22 4 A

I md M d m M d . The rotational inertia about axis B is

2

2 22 4 B I m d Md m M d . (We are treating each dumbbell as a point mass.)

We are told that the rotational inertia about axis A is twice as large as that about axis B ( I A = 2 I B), so m + 4 M = 2(4m + M ).

(Note that d has cancelled out, and does not affect the answer.) Therefore, M = 7/2 m. Since m = 2 kg, M = 7 kg.

Key Points:

• An object's rotational inertia (“moment of inertia”) depends on the axis you choose to calculate it about.

• The rotational of inertia of a point-like object about an axis is md 2, where m is the object’s mass and d is its distance from

the axis.

• The rotational of inertia of an object composed of multiple subobjects is just the sum of their individual rotational inertias

(about the same axis).

• Avoid putting numbers into calculations until the very end. Instead, define variables and work with those. (Often, some

will cancel out, simplifying your calculations.)

For Instructors Only

Students can get bogged down in unnecessary calculations and computations. This problem presents a good opportunity to

discuss problem solving procedures.

Many students will translate the given relationship between rotational inertias incorrectly, and will interpret “The rot ational

inertia . . . about axis A is twice the rotational inertia about axis B” as 2 I A = I B. Those who do this will get M = 4/7 kg. (This

is a well-documented error in translating verbal to algebraic representations.)

Students should be encouraged to decide on qualitative grounds which mass is larger, so that they can check their answers for

reasonability.

Question E1.02a

Description: Integrating linear and rotational dynamics ideas.

Question

A disk, with radius 0.25 m and mass 4 kg, lies flat on a smooth horizontal tabletop. A string wound about the disk is pulled



Chapter 8

Page 8.4

Key Points:

• The acceleration of a body as a whole depends on the net force acting on the body, period. It does not depend on where on

the body the force acts or whether the body spins in addition to accelerating.

• Newton’s second law is true for bodies that spin or rotate as well as those that don’t. F = ma and I

are both true,always.


This is the first in a two-question set exploring linear and angular acceleration from a force that exerts a torque.

The most prevalent misunderstanding to confront here is students’ belief that somehow I replaces F = ma, rather than

augmenting it.

Students may assume that the question asks for angular acceleration. They should be cautioned against jumping to

conclusions based on the superficial features of a question!

Question E1.02b

Description: Integrating linear and rotational dynamics ideas.

Question

A disk, with radius 0.25 m and mass 4 kg, lies flat on a smooth horizontal tabletop. A string wound about the disk is pulled

with a force of 8 N. What is the angular acceleration of the disk?

1. 0

2. 64 rad/s2

3. 8 rad/s2

4. 4 rad/s2

5. 12 rad/s2




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Commentary

Purpose: To explore the angular acceleration of an object experiencing both angular and linear acceleration due to a force

that also exerts a torque.

Discussion: A force exerted a distance R from an axis of rotation causes a torque sin RF , where is the angle

between the direction of the force and the vector from the axis to the point of application. In this case, the angle i s 90°, so the

torque exerted about the center of mass is 2 N · m. The rotational equivalent of Newton’s second law, I , relates a

body’s angular acceleration to the net torque it experiences. The moment of inertia I of the disk is 2 20.25 kg m MR .

Therefore, the angular acceleration of the disk about its center of mass is 28 N kg m 8 rad s . (This is true even if

the disk translates.)

A common mistake is to calculate the angular acceleration from the linear acceleration (found in the previous problem) via

a r . This relationship between linear and angular acceleration is not generally true; it describes a physical (“geometric”)

constraint that only applies in special circumstances, such as when a round object is rolling without slipping, or when it is

rotating about a fixed axis through its center and a refers to the acceleration of a point on its rim.

Key Points:

• I describes the relationship between torque and angular acceleration the way Newton’s second law describes that

between force and (translational) acceleration, and is always valid.

• The relationship a r between translational and rotational acceleration is only valid in special cases. (The same is true

of r v and s r .)


This is the second in a two-question set exploring linear and angular acceleration from a force that exerts a torque.

Some students will get the correct answer by misunderstanding the problem and thinking that the center of the disk is fixed in

place. The question “Is there friction at the pivot?” indicates such a misunderstanding.

Other students will give an answer less than 8 rad/s2, thinking that the translational motion somehow reduces the torque or its

effect.



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Page 8.6

Question E1.03a

Description: Linking force and torque ideas in the context of mechanical advantage.

Question

A 100 kg crate is attached to a rope wrapped around the inner disk as shown. A person pulls on another rope wrapped around

the outer disk with force F to lift the crate.

What force F is needed to lift the crate 2 m?

1. about 20 N

2. about 50 N

3. about 100 N

4. about 200 N

5. about 500 N

6. about 1 000 N

7. about 2 000 N

8. about 5 000 N

9. Impossible to determine without knowing the radii

10. Impossible to determine for some other reason(s)

Commentary

Purpose: To link force and torque ideas in the context of mechanical advantage.

Discussion: Consider a static situation in which the crate is held motionless in the air. To support the crate, the tension in the

rope attached to it must be 1 000 N (using g = 10 N/kg). The tension in the other rope is equal to F . If the disk arrangement is

stationary, the torques exerted on it by the two ropes must balance.

We do not know the radii of the pulleys, so let’s use r for the smaller disk and R for the larger. The rope supporting the crate

is tangential to its disk, so the torque exerted by this rope is 1 000 r clockwise. The other rope is also tangential to its disk, so



Chapter 8

Page 8.7

it exerts a torque of FR counter-clockwise. The force of the pivot holding up the disks exerts zero torque, and we will assume

the axle is frictionless.

For the torques to balance each other, 1000 N FR r , or = 1000 N F r R . In other words, the force exerted by

the person is a fraction of the weight of the crate, and the fraction depends on the ratio of the disk radii.

We do not know the exact ratio r/R, but we can estimate it from the diagram. It looks to be about 1/5, so the force needed is

about 200 N.

That’s the force required to hold the crate stationary in the air, or to lift or lower it with constant speed (no acceleratio n). In

order to start the crate moving from rest, a slightly larger force is necessary, but it can be infinitesimally larger. (If there were

friction in the pivot, the force to get it moving would have to be enough larger to support the weight of the crate and

overcome static friction.)

The angle at which F is applied does not matter, as long as it acts tangentially to the disk. (It would, however, affect the force

exerted on and by the pivot axle.)

Key Points:

• Two different forces can exert the same torque on an object.

• “Mechanical advantage” is gained by having the app lied force act farther from the pivot point than the force acting on the

object to be move.

• The closer to the pivot point a force acts, the larger it must be to balance other torque-causing forces.


This is the first of two questions exploring the concepts of force, torque, work, and energy in this mechanical advantage

situation.

Many students may say that F is impossible to determine, either because they are not given the radii and don’t assume the

diagram is to scale or they are not told if friction can be neglected. These are defensible responses.

Other students may say that the answer is impossible to determine for another reason, such as not knowing the speed of the

crate or the angle of the rope. These are not valid reasons.

Some students will ignore units and treat 100 kg as the “weight,” and therefore say that F is “about 20.” Some will invert the

ratio, thinking the applied force is 5 000 N, even though this does not agree with experience.

Some students, not understanding mechanical advantage, will think that a force of 1 000 N must be applied to lift the crate no

matter what the radii are.



Chapter 8

.8

Question E1.03b

Description: Linking force, torque, work, and energy ideas in the context of mechanical advantage.

Question

A 100-kg crate is attached to a rope wrapped around the inner disk as shown. A person pulls on another rope wrapped around

the outer disk with force F to lift the crate.

How much work is done by the person to lift the crate 2 m?

1. about 400 J

2. slightly less than 2 000 J

3. exactly 2 000 J

4. slightly more than 2 000 J

5. much more than 2 000 J

6. Impossible to determine without knowing F

7. Impossible to determine without knowing the radii

8. Impossible to determine without knowing the mass of the pulley

9. Impossible to determine for two or more of the reasons given in 6, 7, and 8 above

10. Impossible to determine for some other reason(s)

Commentary

Purpose: To link force, torque, work, and energy ideas in the context of mechanical advantage.

Discussion: The only ambiguity here is whether or not to ignore frictional effects at the axle where the two disks are

attached. If friction can be ignored, then energy conservation demands that the work done by the person is exactly equal to

the work done on the crate, which is = 2 000 Jmg y .

If friction is not ignored, then the work done by the person must be larger than the work done on the crate.

Note that even though the force F is much smaller than the weight of the crate, it acts through a much larger displacement

than the crate travels. If, for example, the ratio of the disk diameters is 1 to 5, then the force F would be about 200 N and the



Chapter 8

Page 8.9

displacement of the end of the rope would be about 10 m, even though the crate only moves up by 2 m.

Key Points:

• The presence of mechanical advantage in a system does not invalidate the work – energy theorem.

• A smaller force can do as much work as a larger one if it the smaller one acts through a longer distance.


This is the second of two questions exploring the concepts of force, torque, work, and energy in this mechanical advantage

situation.

Some students will say that the work done is impossible to determine, because they are not told if friction can be neglected.

This is a defensible response.

Other students will say that it is impossible to determine for another reason, such as not knowing the speed of the crate, the

angle of the rope, the force F , the radii of the disks, or whether the diagram is to scale. These are not valid responses, since

none of that information is required to answer the question.

Some students will use g = 9.8 N/kg, and get a value “slightly less than 2 000 J.”

Students who are including friction are likely to choose “slightly more than 2 000 J,” perhaps assuming that friction is small.

If the coefficient of friction is large enough, a valid response would be “much more than 2 000 J,” though students might

choose this response for other reasons as well.

Question E1.04a

Description: Reasoning and problem solving with linear and rotational forms of Newtons’ laws in the context of rolling

without slipping.

Question

A spool has string wrapped around its center axle and is sitting on a horizontal surface. If the string is pulled in the horizontal

direction when tangent to the top of the axle, the spool will:



Chapter 8

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1. Roll to the right

2. Not roll, only slide to the right

3. Spin and slip, without moving left or right

4. Roll to the left


6. The motion cannot be determined.

Commentary

Purpose: To reason about a rotational system using the linear and rotational forms of Newton’s second law.

Discussion: There are four forces on the spool: (1) gravitation, down; (2) normal, up; (3) tension, right; and (4) friction, left

or right. Gravitation is balanced by the normal force, and their torques balance about any origin. We don’t know yet which

direction friction will point.

First, imagine that the surface is frictionless. The net force is then due exclusively to the tension, producing an acceleration to

the right. The net torque about the spool’s center is also due exclusively to the tension, producing a clockwise angular

acceleration. The spool will start to move to the right and also rotate clockwise. It will be rolling to the right, and perhaps

slipping at the same time. (An object only rolls without slipping when its rate of rotation and translation are just right so the

contact point has zero velocity.)

Now, add friction back in. A little bit of static friction will prevent the spool from slipping and cause it to roll only; this is

answer (1). (If the contact point starts to slip, the friction force will oppose that, exerting a countering torque.)

If F is very large, the spool will not be able to roll without slipping, since the (static) friction force has a maximum possible

value. In that case, the spool will slide to the right while rotating slightly: not an available answer.

Answer (2) is impossible, since the net torque on the spool cannot be zero. (If the spool slips at all, the torque from the

friction force will supplement, not counteract, the torque from the string.)

Answer (3) is impossible, since the net force on the spool cannot be zero. If the net force were zero, the friction force would

have to point to the left and have the same magnitude as the tension. But if the spool spins, the bottom surface slides to the

left and the friction force must point to the right.

Answer (4) is impossible, since it would require a net force to the left and a counterclockwise net torque, which cannot both

exist. (What direction would the friction force point?)

Key Points:

• When two forces are balanced (same strength, opposite directions) and colinear (having the same line of action), their



Chapter 8

Page 8.12

4. Roll to the left


6. The motion cannot be determined.

Commentary

Purpose: To explore the choice of origin and its effect on the torque.

Discussion: Intuitively, it may not be obvious what will happen here. Pulling on the string seems like it might cause the spool

to unwind, thus rotating counter clockwise and perhaps rolling to the left. On the other hand, the string pulls to the right, so

perhaps it will cause the spool to roll to the right (clockwise) along the surface. A more careful analysis is required.

There are four forces on the spool: (1) gravitation, down; (2) normal, up; (3) tension, right; and (4) friction, left or right.

Gravitation is balanced by the normal force. Because they are balanced and colinear, their torques also balance about any

origin.

If the spool rolls to the left without slipping, the spool’s center of mass accelerates to the left. Since the tension acts t o the

right, the static friction force must act to the left and must have a larger magnitude so that the net force acts to the left.

However, if that were true the net torque about the spool’s center would be clockwise, causing the spool to rotate to the right.

Contradiction!

Since we don’t know what direction the friction force points, let’s choose an origin about which the friction force exerts no

torque: the point of contact between the spool and surface. For this origin, the only force exerting a nonzero torque is the

tension force, so the net torque is clockwise and the spool rotates to the right relative to the contact point. This means it rolls

to the right. There is no reason the spool must necessarily slip or slide. If we pull gently enough, there will be enough static

friction so that the spool rolls without slipping.

Why doesn’t the spool unroll to the left? Because although the string applies a torque in the counter -clockwise direction, the

static friction force exerts a larger torque in the clockwise direction. (If we yank hard enough on the string, the spool will

overcome static friction and slide to the right as it spins counter-clockwise. This is not the intent, so it is not any of the

answers provided.)

Key Points:

• By choosing your origin carefully, you can avoid dealing with torques due to an unknown force.

• The torque exerted by a force depends on the origin you calculate it about.

• You do not need to choose the center of the object as the origin.

• I is true, valid, and useful in addition to, not instead of, m aF . In other words, Newton’s second law in its



Chapter 8

Page 8.14

Question

A spool has string wrapped around its center axle and is sitting on a horizontal surface. If the string is pulled at an angle to

the horizontal when drawn from the bottom of the axle, the spool will:

1. Roll to the right

2. Not roll, only slide to the right

3. Spin and slip, without moving left or right

4. Roll to the left


6. It is impossible to determine the motion.

Commentary

Purpose: To explore the choice of origin and its effect on the torque.

Discussion: There are four forces on the spool: (1) gravitation, down; (2) normal, up; (3) tension, as shown; and (4) friction.

Gravitation and the normal force exert zero torques about any origin along a vertical line through the center of the spool.

(They no longer exert balancing torques, because they are no longer equal to each other in strength. The normal force is

smaller than the weight, because the tension force has a component “up.”) Thus, whether we choose the center of the spool or

the contact point, the torques are zero.

Tension exerts a counter-clockwise torque about the center of the spool. It is hard to predict what direction the friction force

will point, and therefore what direction its torque about the center of the spool will be. So, from this analysis it is not obvious

what will happen.

As with the previous problem, we can choose our origin to be at the point of contact between the spool and surface, so that

friction exerts zero torque. About this origin, the net torque is exerted exclusively by the tension in the string. But in what

direction is that torque?

To figure that out, we need to know the “line of action” of the tension force. The line of action is a straight line having the



Chapter 8

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same orientation as the force and passing through the point of application of the force, as shown. The torque due to F can be

found by treating it as though it is applied anywhere along the line of action. Thus, the way the given diagram is drawn, the

torque is clockwise. This means the spool will roll to the right.

However, the angle is left unspecified. If we don’t assume the drawing is to scale, the angle could be anything. There is an

angle for which the line of action passes through the chosen origin, in which case the net torque on the spool is zero, the net

force is to the right, and the spool will slide to the right without spinning.

If the angle is even larger than this, so that the line of action passes to the right of the origin/contact point, the net torque is

counter-clockwise. This would cause the spool to unroll to the left. If the tension is vertical, the spool will also unroll to the

left.

Demonstrations can confirm all of these outcomes.

Thus, the motion of the spool depends upon the angle . The angle drawn in the figure will cause the spool to roll to the right.

Key Points:

• The “line of action” is a useful concept for determining the direction of the torque exerted by a force. Each force has its

own line of action. We can treat the torque as though the force is applied anywhere along the line of action.

• A clever choice of origin can make torque problems much easier to analyze.


This is the last of three questions using this situation. This question drives home the idea that the choice of origin should be

done with some strategic thinking and goal. It also shows the utility of the “line of action” concept.

A physical demonstration is extremely valuable here, so that students can see that the angle of the tension force critically

affects the motion of the spool.

A family of diagrams showing the line of action for different angles of the tension force can help students understand the

analysis and come to a better understanding of torques and lines of action.



Chapter 8

Page 8.16

Question E1.06a

Description: Reasoning with force, energy, and torque ideas in the context of mechanical advantage.

Question

Two blocks hang from strings wound around different parts of a double pulley as shown. Assuming the system is not in

equilibrium, what happens to the system’s potential energy when it is released from rest?

1. It remains the same.

2. It decreases.

3. It increases.

4. It is impossible to determine without knowing the radii of the two pulleys

5. It is impossible to determine without knowing the ratio of the radii of the two pulleys

6. It is impossible to determine for some other reason

Commentary

Purpose: To develop qualitative reasoning and problem-solving skills by applying energy ideas in a rotational dynamics

context.

Discussion: We do not know the radii of the two pulleys, or even their ratio, so we cannot predict which block will fall and

which will rise when the system is released from rest. To answer the question, however, we do not need to know any of these

features.

We do know that the system is “not in equilibrium,” which means that one block will start to fall and the other will start to

rise, and the double pulley will begin to rotate. Thus, the kinetic energy of the system will rise.

Where will this energy come from? There are no external forces doing work on the system, so it can only come from the

potential energy of the gravitational interaction between the blocks and the Earth. If the kinetic energy is increasing, the

potential energy must decrease so that total mechanical energy will be conserved.



Chapter 8

Page 8.18

1. F N = 20 N

2. 20 N < F N < 27 N

3. F N = 27 N

4. F N > 27 N

5. It is impossible to predict what the normal force on the double pulley will be.

Commentary

Purpose: To develop qualitative reasoning and problem-solving skills by applying force ideas in a rotational dynamics

context.

Discussion: We do not know the radii of the two pulleys, or even their ratio, so we cannot predict which block will fall and

which will rise when the system is released from rest. To answer the question, however, we do not need to know any of these

features.

We do know that the center of mass of the system is falling, and in fact is accelerating downward. This means that a net force

in the downward direction must be acting on the system. The only (external) forces are gravitation and the normal force

exerted by the pivot, which means the normal force must be smaller than the total weight of the system (27 N).

Further, the double pulley is not accelerating, so the net force on it must be zero. There is tension in both strings pulling

down, so the normal force must be larger than the weight of the double pulley (20 N).

Note that the tensions in the strings are not 2 N and 5 N, the weights of the blocks. For the falling block (whichever that turns

out to be), the tension will be slightly smaller than the weight; for the rising block, the tension will be slightly larger than theweight. This is needed to satisfy Newton’s second law applied to each hanging mass.

In the previous question, we considered the Earth to be part of the “system” we were analyzing. In this question, it is more

convenient not to, but rather to treat the gravitational force as an external force acting upon a system comprised of the double

pulley, ropes, and two blocks. Reasoning about the center of mass motion of the system, if the Earth were included in that

system, would be difficult!



Chapter 8

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Key Points:

• Many questions can be answered through qualitative reasoning from general principles, without numerical calculations or

solving for anything.

• If the center of mass of a body or system is accelerating — even if part of it is held in place — there must be a nonzero net

external force acting on one or more components of system.

• If the center of mass of a body or system is not accelerating, all external forces on that body or system must sum to zero.


This is the second of two questions on this situation. This one explores qualitative reasoning with force ideas; the previous

was similar but applied energy ideas. Although the “topics” of these two questions are conservation of energy and forces,

they are useful as broader “integrating” questions that teach students to use their inventory of basic physics principles for

reasoning about various situations. Revisiting old ideas in new contexts is valuable: it enriches the new context and helps

students cross-link new and old ideas.

Students may have difficulty focusing on the pivot and the forces it exerts. They are not accustomed to applying New ton’s

second law (linear) to situations involving pulleys and torque.

Some students will say that the force supporting the pulley is equal to the pulley’s weight, 20 N, ignoring the tensions pulling

down. Others will say that the tensions are 2 N and 5 N, so the force supporting the system is 27 N.

This set of two questions presents an excellent opportunity to hold a higher-level discussion about choosing a “system” as

part of strategic problem solving: for example, why one would decide to include the Earth as part of the system sometimes

but not others.

Note that treating the diagram as a scale drawing will not help students determine which mass falls and which rises, since the

ratio of the radii is the same as the ratio of the hanging masses (2:5).

Question E1.07

Description: Developing problem solving skills by choosing an origin for statics problems.

Question

A uniform rod of length L, mass M , is suspended by two thin strings. Which of the following statements is true regarding the

tensions in the strings?



Chapter 8

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1. T 2 = T 1

2. T 2 = 2.5 T 1

3. T 2 = 0.6 T 1

4. T 2 = 0.8 T 1


6. Not enough information to determine

Commentary

Purpose: To practice determining torques in static situations, once again making the point that a good choice of pivot point

can make a problem easy.

Discussion: A consideration of forces tells us that the two tensions must add up to the bar’s weight, in order to have zero net

force in the y-direction. Thus, we must turn to torques to answer this. There must be zero net torque about any point on the

bar if the bar is to remain static. The question is, what choice of pivot point will make the problem easiest?

Since the question doesn’t require us to know the weight, choosing the center of mass as the pivot point is advantageous:

gravity exerts no torque about that point. Then, our equation that states the sum of the torques equals zero will relate the two

tensions, providing the answer we seek.

We need to use the markings on the rod to determine where the center of mass is and how far each string is from it. We don’t

know the units — how long each segment is —but it doesn’t matter, since we’re looking for a ratio between T 1 and T 2.

Counting segments, we see that T 2 acts 5 units from the center and T 1 acts 3 units away, so T 2 must be 3/5 of T 1. Thus, answer

(3) is appropriate.

Key Points:

• Look for the most convenient origin about which to calculate torques.

• Use all the information provided in a question, including the diagram.

• If you think you need a quantity that is not given, define a variable for it and proceed. The variable will often cancel out.


If any students choose “Not enough information,” we suggest asking them what it is they would need to know to make the

question answerable, and why they need it. It’s likely they want to know physical dimensions for the locations of the string.

They may not realize they can count segments to find relative distances, or they may not be comfortable working with ratios

rather than actual distances.



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Question E1.08

Description: Developing problem-solving skills by working with forces and torques in a nontrivial statics situation.

Question

A uniform rod is hinged to a wall and held at a 30° angle by a thin string that is attached to the ceiling and makes a 90° angle

to rod. Which statement(s) must be true? (At least one of them is true and at least one is false.)

1. The hinge force is purely vertical.

2. The hinge force is purely horizontal.

3. The string tension is equal to the hinge force.

4. The string tension is smaller than the rod’s weight.

5. 1 and 3 are true.





10. Three of the statements are true.

Commentary

Purpose: To help you learn to reason using forces and torques.

Discussion: The rod is at rest, so the net force on it must be zero, and the net torque about any origin must also be zero. This

yields many possible relationships, all of which are valid, but only some of which bring out relevant features of this situation.

In other words, we do not need to write down every valid equation or relationship to answer this question. Rather, thoughtful

choices about how to proceed will yield efficient results.

It is useful to assume nothing about the hinge force and to think of it as having a vertical and a horizontal component. These

components can be treated as independent forces. (We often separate one force into two separate component forces, for

example when treating the contact force between two surfaces as a normal force and a friction force.)

Let’s focus on each statement and determine its truth or falsehood.



Chapter 8

Page 8.22

The hinge force is purely vertical. This statement is false, because the net force in the horizontal direction must be zero. The

tension force has a component pulling to the right, so the hinge must pull to the left with an equal force.

The hinge force is purely horizontal. This statement is false, because the net torque about the center of mass must be zero.

Both the tension force and the horizontal component of the hinge force exert counter clockwise torques about the center of

mass of the rod. Therefore, the hinge must have a vertical component to provide a balancing clockwise torque.

The string tension is equal to the hinge force. This statement is false, since the two forces have different directions, with the

hinge pulling up and to the left and the tension pulling up and to the right. (It turns out that the two forces do have the same

magnitude.)

The string tension is smaller than the rod’s weight. This statement is true, since the net torque about the left end of the rod

must be zero. The torque exerted by the hinge is zero (about this point), so the tension must balance the weight. The moment

arm is larger for the tension, so the force must be smaller.

Key Points:

• For an object at rest, all components of the net force and the net torque about any origin are zero.

• Strategic choices of relationships and origins can make analysis and reasoning particularly efficient.

• To isolate an unknown, choose an origin such that the torque due to the other unknown(s) is zero. You can consider the

situation using several different origins if you want; I must be true for all of them.


This question provides an excellent opportunity to explore problem-solving approaches and strategies. The equations are

simple; finding the most efficient use of those equations is more difficult.

Students can be either unaware of or overwhelmed by the decision making needed to solve statics problems. It all looks so

easy when the instructor does it, yet when students are doing homework or exams, it becomes impossible, largely because

they have not practiced the skill of strategic thinking.

The key to evaluating statements 1 and 2 is to focus on the other component to determine its validity. That is, to determine if

the hinge force is purely vertical, one must find out whether the horizontal component is zero. (Students sometimes think that

hinges only exert vertical forces.)

Some good students might figure out that the hinge force and the tension force have the same magnitude, since their

horizontal components balance and their vertical components each support half the weight of the rod. They might not realize,

however, that statement 3 is about vector equality, requiring magnitudes and directions to be the same. (This falls into the

category of students giving the right answer to the wrong question.)



Chapter 8

Page 8.24

In other words, each string supports half the weight of the rod. The lengths of the strings does not matter.

Another way to solve the problem is to choose the origin to be at the right end of the rod. This is also a strategic choice,

because then the torque due to the tension in the right string is zero. Two forces on the rod exert nonzero torques: the tension

in the left string, which is the desired unknown, and gravitation. The moment arm for the tension is twice as large as the

moment arm for gravitation. (Remember, gravitation acts “as though” the force is exerted at the center of mass.) As before,

since the rod is at rest, the two torques must balance each other. Therefore, since the moment arm for the tension is twice as

large, the tension must be half as large as the weight of the rod.

Key Points:

• For a body in static equilibrium (i.e., one that is stationary), the net force on the body and the net torque on the body about

any point must both be zero.

• Any origin may be chosen to analyze and solve a problem, but thoughtful, strategic choices of origin can make the analysis

much simpler.

• The “moment arm” of a force about an origin, used to determine the torque it exerts, is the shortest distance from the or igin

to the imaginary line you get by extending the line along which the force acts to infinity in both directions. It is not

necessarily the distance from the origin to the point at which the force acts on the object.

• Gravitation acts “as though” the force is exerted at the center of mass.


This is one of the simplest static situations we can create, yet students often cannot sort out the relevant features because of

the angle of the rod. Many conclude intuitively that the left string exerts the larger force, and thoughts of torque, center of

mass, and moment arms are often neglected.

Students often have difficulty applying the concept of “moment arm” to a situation in which forces are exerted at an a ngle

relative to the rod or object. A diagram may help many sort this out.

Students can be flustered by having to choose which point is the origin. They frequently get hung up on making the “right”

choice, not realizing that all choices are correct but some are easier to work with than others. It is useful to give students

opportunities to think about strategic choices of origin, and also to have students solve the problem two or more times with

different origins, as this will encourage comparison of approaches. If students only see one choice for any given situation,

they will (reasonably but incorrectly) conclude that there is one best (and therefore “right”) choice to any problem.



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Question E1.10a

Description: Developing problem-solving skills by choosing an origin for torque problems.

Question

A uniform disk with mass M and radius R sits at rest on an incline 30° to the horizontal. A string is wound around the disk

and attached to the top of the incline as shown. The string is parallel to incline. What is the tension in the string?

1. Mg

2. Mg /2

3. 2 Mg /5

4. Mg /4



Commentary

Purpose: To develop your problem-solving skills by considering multiple approaches to a question, and explore the

significance of where you choose your origin for torque calculations.

Discussion: There are often multiple ways to solve a problem, and part of learning to “do” physics well is learning to select

the easiest approach to a given problem. This statics question is conceptually straightforward to answer, but the algebra can

be relatively simple or complicated depending on the coordinate system you choose.

The unknown forces acting on the disk are the tension force of the string that we seek, and the normal and friction forces of

the plane. We know the direction of each of these and the location at which it acts on the disk, but not its magnitude. Gravity,

whose magnitude and direction we know, also acts on the disk. So, we have three unknowns: the magnitudes of the tension,

normal, and friction forces. We therefore need three independent equations in order to solve for them. Since the disk is

stationary, we know it is not accelerating in the x- or y-directions, and that it is not rotating about any pivot point we choose

to consider. This means that the net force in the x-direction is zero, the net force in the y direction is zero, and the net torque

about any point is zero. By writing each of these statements in terms of the actual forces (and trigonometric functions of the

incline angle), we get three equations, and all we have to do is eliminate the variables we don’t care about and solve f or the

tension. This is a classic statics problem.



Chapter 8

Page 8.26

The question is, in order to find the tension, what choice of coordinate system and pivot point is best? Any choice will work ,

but some will lead to rather ugly algebra. How to choose?

If we choose the pivot point for our torque equation to be at the point of contact between the disk and incline, then neither the

normal nor friction forces exert any torque (since both act at the pivot point itself). So, the only two torques are due to gravity

and the string tension; the only unknown is the magnitude of the tension, and we can solve this one equation for the tension.

Simple! We don’t need to use the two force equations or solve for the friction or normal forces at all.

The tension exerts a torque of 2 RT out of the page. The weight acts at the center of the disk, and the component of the weight

perpendicular to the vector from pivot point to the center of the disk is Mg sin (the component parallel to the plane), so the

weight exerts a torque of Mg R sin into the page. Thus, 2 sin RT MgR , and 2 sinT Mg . The sine of 30° is 1/2,

so answer (4) is correct.

Had we chosen a different pivot point, the system of equations we’d have to solve would be significantly more complex.

Key Points:

• For a body in static equilibrium, the forces on it must add to zero (vector sum, i.e., along all axes), and the torques must

add to zero about any pivot point you choose.

• Choosing the orientation of your coordinate axes and the pivot point for your torque equation can make the algebra of a

problem easier or harder.

• It’s generally wise to choose a pivot point where as many forces as possible, especially unknown ones you aren’t interested

in solving for, exert no torque: that is, the forces act at that point, or along a line that passes through the point.


In this question, the issue is not so much which answer is right as what approach provides the easiest path to it. Many

students will place their pivot point at the center of the disk or perhaps where the string attaches to the disk, and then dive

into two (or three) equations in two (or three) unknowns (depending on their choice of coordinate system). When the simple

solution described above is revealed during discussion, a good deal of forehead-smacking occurs, and the point is taken to

heart.

We find that talking explicitly about problem-solving strategies and providing questions with multiple approaches, some

clearly superior to others, helps students to become more effective, efficient, and aware problem solvers.



Chapter 8

Page 8.27

Question E1.10b

Description: Developing problem-solving skills by choosing an origin for torque problems.

Question

A uniform disk with mass M and radius R sits at rest on an incline 30° to the horizontal. A string is wound around disk and

attached to top of incline as shown. The string is parallel to incline. The friction force acting at the contact point is:

1. Mg /2, down the incline

2. Mg /2, up the incline

3. Mg /4, up the incline

4. Mg /0.86, down the incline



Commentary

Purpose: To develop your problem-solving skills by considering multiple approaches to a question, and to explore the

significance of where you choose your origin for torque calculations.

Discussion: This question is almost identical to the last one (19a), except it asks for the magnitude of the friction force rather

than the tension. It can be solved just as easily as that problem, by choosing a pivot point where the string meets the disk (so

the tension and normal forces exert no torque), setting up the torque equation, and solving for the friction force.

However, having answered the previous question, we can solve this one even more simply. We know that the net force on the

disk is zero. Consider forces acting parallel to the incline. The weight component down the plane is Mg sin =Mg /2, and the

tension force up the plane is one-half that ( Mg/ 4), as we found last question. So, if there is to be no net force along the plane,

friction must exert a force equal to tension ( Mg/ 4) up the plane, so that together tension and friction can balance the parallel

component of the weight. Using what we found last question, we can answer this one in our heads with no significant algebra

at all.

It might bother some students that the friction force acts upward along the plane. Imagine what would happen if the plane

were frictionless: the bottom of the disk would slip forward down the plane, sliding “out from under” the point attached to t he

string. Thus, friction must oppose this motion by pointing up the plane.



Chapter 8

Page 8.28

Key Points:

• Again, think carefully before diving into a calculation. You might already know enough, or be able to reason enough, to

answer a question without getting into the algebra. “Work smarter, not harder.”

• The moral of these questions isn’t just to choose the best pivot point for a statics problem. It’s to use all your information

and knowledge to choose the easiest way to answer a question.


This question is meant to follow 19a. We suggest thoroughly discussing 19a before beginning this one. Some students will

not have taken the point of 19a, and will once again wade into multiple equations in multiple unknowns. Some will have

taken the specific moral about choosing a pivot point carefully, and will repeat that approach here — better, but not ideal. Few

are likely to use what they found last question to reason the answer, as outlined above.

We strongly suggest drawing a free-body diagram of the disk to support the argument above. Not only does this help some

students grasp the argument, but it communicates by example that graphical representations such as free-body diagrams are

useful problem-solving tools that should be part of students’ working toolkit.

Question E2.01

Description: Developing understanding of angular momentum for linear, circular, and spinning motion.

Question

Which situation has the least magnitude of angular momentum about the origin?

A. A 2-kg mass travels along the line y = 3 m with speed 1.5 m/s.

B. A 1-kg mass travels in a circle of r = 4.5 m about the origin with speed 2 m/s.

C. A disk with I = 3 kg · m2 rotates about its center (on origin) with = 3 rad/s.

1. A

2. B

3. C

4. Both A and B

5. Both A and C

6. Both B and C

7. All have the same magnitude angular momentum.



Chapter 8

Page 8.29

Commentary

Purpose: To hone your understanding of angular momentum and confront a common misconception that objects traveling in

a straight line must have zero angular momentum.

Discussion: The angular momentum of a point-like object is defined by L r p , where r is the vector from the origin

(about which angular momentum is being determined) to the object, and p is the object’s momentum. For a rigid object

rotating about an axis, we can use this to derive an expression for the object’s total angular momentum: L I , where I is

the moment of inertia of the extended object about the rotation axis and is its angular velocity about that axis.

If we simply apply the first form to situation A we find that the angular momentum has a magnitude of 2 29 kg · m /s . Note

that an object does not need to be rotating or traveling in a curve to have nonzero angular momentum; it merely needs a

nonzero velocity that isn’t purely radial (towards or away from the origin).

If this bothers you, imagine that the mass in situation A strikes and sticks to a stationary disk free to rotate about an axis at

the origin. What will happen? The disk with the mass stuck to it will begin to rotate about the axis. The final situation clearly

has nonzero angular momentum. For the principle of “conservation of angular momentum” to have any meaning, the initial

situation — the mass moving in a straight line — must also have nonzero angular momentum.

Similarly, we can apply the first form to situation B to find an angular momentum of 2 29 kg · m /s . Alternatively, we can use

the second form determining the angular velocity from the speed and the circle’s circumference, and using 2 I mR as the

moment of inertia of a point mass a distance R from the axis. We will get the same answer.

Applying the second form to situation C also yields an angular momentum of 2 29 kg · m /s . Thus, the best answer is (7).

Key Points:

• Angular momentum is defined by L r p for a point mass.

• The total angular momentum of a rigid, rotating object can be determined using L I .

• A mass does not need to be rotating or spinning to have nonzero angular momentum. Translating past the origin a nonzero

distance away is sufficient.


Answer (1) is common, and reveals the prevalent misconception that objects traveling along a straight line have no angular

momentum.

Answer (4) reveals the less prevalent misconception that only nonpoint, rotating objects can have a nonzero angular

momentum.



Chapter 8

Page 8.30

The other incorrect answers are merely distractors.

This is a good problem for stressing that the angular momentum of an object depends on one’s choice of origin. The mass

moving linearly in situation A would have no angular momentum if one chose an origin directly along its path, rather than off

to the side.

Question E2.02a

Description: Exploring student thinking about rotational motion and angular momentum conservation.

Question

A child is standing at the rim of a rotating disk holding a rock. The disk rotates freely without friction. If the rock is dropped

at the instant shown, which of the indicated paths most nearly represents the path of the rock as seen from above the disk?

1. Path (1)

2. Path (2)

3. Path (3)

4. Path (4)

5. Path (5)


Commentary

Purpose: To check your understanding of Newton’s first law in a rotational context.

Discussion: We interpret the word “drop” in the question statement to mean that the rock is released without delivering any

impulse to it.

At the instant shown, the rock’s velocity is straight forward, tangential to the disk: direction (2). When the rock is r eleased,

the only forces acting on the rock are gravity and air resistance. Gravity will accelerate it downwards (into the page, as seen



Chapter 8

Page 8.31

from above), and air resistance will tend to slow it down but not change its direction. Thus, there are no forces that provide an

acceleration to the left or right, and the rock must continue along path (2) as seen from above. (Seen from the side, the rock

would follow a parabola as it continues traveling forward while accelerating downward.) This is a consequence of Newton’s

first law: an object maintains its existing velocity — speed and direction — unless an external force acts upon it.

You may think that the “velocity” of the rock before it is dropped is curved. We often use a curved arrow to represent the

angular velocity of something, but this does not mean that the (linear) velocity is curved. Velocity is always a vector

representing the speed and direction of an object at one point in time, and can be indicated by a (straight) arrow. If the object

follows a curved path, the velocity at any point is tangent to the curve.

Path (3) might be what the trajectory of the rock would look like to the child who dropped it. In other words, the child is

moving away from the path of the rock, and therefore, the rock looks like it is curving away. This is an illusion, however,

because the child is accelerating (moving in a circle); the child’s frame of reference is a noninertial (invalid) frame.

You might think that the rock travels along path (1), that the velocity changes direction because the rock experiences a

“centripetal acceleration,” and that this acceleration is caused by the “centripetal force.” Centripetal force is not a real force.

Rather, it is a component of the net force. The “centripetal force” is whatever component points radially inward when all the

actual forces, caused by interactions with other objects, are summed. In this situation, the only forces acting on the rock after

it is released are gravity and air resistance, so the net force has no component in the radial directio n. Thus, the “centripetal

force” is zero.

You might think that the rock travels along paths (3), (4), or (5) because it is acted upon by the “centrifugal force.” There is

no such force. “Centrifugal force” is an illusion experienced within an accelerated (noninertial) reference frame. To the child,

it feels like something is pulling the rock outward from her hand; that is the illusory centrifugal force. In real ity, the rock’sinertia is just carrying it in a straight line according to Newton’s first law, an d the child must exert a force radially inward on

it to make it travel in a circle (until she drops it).

Key Points:

• Newton's first law says that an object travels with constant velocity unless a force acts upon it to change its speed and/or

direction.

• An object travels along a circular path only if some interaction with another object pulls it towards the center of that circle;

when that interaction ceases, the object stops moving in a curve and continues in a straight line.

• “Centripetal force” is not a real force like tension, gravity, and the like, but is a way of talking about one component of the

sum of all forces.

• “Centrifugal force” is an illusion experienced in an accelerating reference frame, not a real force.




Chapter 8

Page 8.32

This is the first of three questions about this situation. Later questions explore conservation of angular momentum and

energy; this question sets those up by clarifying the situation, establishing students’ understanding of the rock’s actual

trajectory. It also provides a valuable refresher on Newton’s first law and reveals stubborn misconceptions about circular

motion.

You may be surprised by how many misconceptions this question can reveal, and how tenaciously students cling to them.

This question, and others integrating rotational and linear ideas, deserve extended discussion time.

To stimulate productive discussion, you may wish to pose questions such as: What path would the child see? What is the

velocity of the rock just before and just after it is dropped? What would the path of the rock have been if the child continued

to hold it? Do you expect the path to be the same or different when the child drops it?

Question E2.02b

Description: Exploring student thinking about rotational motion and angular momentum conservation.

Question

A child is standing at the rim of a freely rotating disk holding a rock. The disk rotates without friction. The rock is dropped at

the instant shown. As a result of dropping the rock, what happens to the angular velocity of the child and disk?

1. Increases

2. Stays the same

3. Decreases


Commentary

Purpose: To check your understanding of angular momentum and rotational inertia.

Discussion: The angular momentum of a system is conserved when the system experiences no net torque. When the child

drops the rock, no external forces are acting on the child and disk that could exert a net torque, so the angular momentum of

the child and disk must remain constant. (Internal forces cannot exert a net torque.) Thus, their angular velocity cannot



Chapter 8

Page 8.33

change.

To put it another way, in the process of dropping the rock, no angular impulse is delivered to the child and disk, so no change

in angular momentum occurs.

Key Points:

• If a system experiences no net torque, its angular momentum will remain constant.

• Only forces external to a system can exert a net force or a net torque on it.

• If a system’s angular momentum does not change and its mass distribution remains constant, its angular velocity must also

remain constant.


This is the second of three questions about this situation. The first question established the trajectory of the rock following its

release; this one establishes the behavior of the child and disk when the rock is dropped.

This question makes a good context for discussing the concept of “angular impulse.”

Students may confuse themselves on this question by applying angular momentum conservation incorrectly. They may, for

example, believe that the angular momentum of the disk-child-rock system must be constant, and that the rock has no angular

momentum after it is released, so the angular momentum of the disk and child must increase. This confusion may be

addressed in the context of discussion about the next question in the set.

Question E2.02c

Description: Linking and relating energy and angular momentum conservation for rotational motion.

Question

A child is standing at the rim of a freely rotating disk holding a rock. The disk rotates without friction. The rock is dropped at

the instant shown. Which of the following statements is true about the process of dropping the rock?



Chapter 8

Page 8.34

1. Angular momentum is conserved, mechanical energy increases.

2. Angular momentum is conserved, mechanical energy decreases.

3. Angular momentum increases, mechanical energy is conserved.

4. Angular momentum decreases, mechanical energy is conserved.

5. Both angular momentum and mechanical energy are conserved.

6. Both angular momentum and mechanical energy increase.

7. Angular momentum decreases, mechanical energy increases.

8. Angular momentum increases, mechanical energy decreases.

9. Both angular momentum and mechanical energy decrease.

10. The conserved quantities cannot be determined.

Commentary

Purpose: To develop your understanding of energy and angular momentum conservation in a rotational context.

Discussion: The angular momentum of a system is conserved when no net torque acts on it. A net torque must come from

external forces, since internal interactions cannot apply a net force or a net torque.

The mechanical energy of a system is conserved when no external forces or nonconservative internal forces do work on it. (If

conservative internal forces do work on it, kinetic energy is exchanged for potential energy, but the total mechanical energy

remains constant. If conservative external forces do work, kinetic energy is exchanged for potential, but the potential energy

gained or lost is not part of the “system.”)

If we take the system to be the rock, the child, and the merry-go- round, and we interpret “drop” to mean that the child

releases the rock without doing any work on it, then there is no net torque on the system and also no work done by

nonconservative forces. Thus, both angular momentum and mechanical energy are conserved.

In the previous question, we established that the angular momentum and angular velocity of the child and disk do not change

when the rock is dropped. If the total angular momentum of the disk-child-rock system is also conserved, then the rock itself

must have the same angular momentum before and after it is released. This is possible because an object moving in a straight

line can have nonzero angular momentum about a point that is not on that line. (Refer to the definition of angular momentum

in terms of linear momentum to see why.)

Note that as the rock is falling, its kinetic energy is increasing due to the gravitational force of the Earth. Therefore, if the

Earth is not part of the “system” under consideration, the system’s mechanical energy does increase while the rock is falling

towards the ground. However, this is after the “process” the question asks about.



Chapter 8

Page 8.35

Key Points:

• The angular momentum of a system is conserved whenever no net torque is exerted upon it.

• Only external forces can exert a net torque.

• An object moving in a straight line can have angular momentum about a point not on that line.

• The mechanical energy of a system is conserved whenever no external or nonconservative internal forces do work on it.

• Whether angular momentum or mechanical energy is conserved for a system depends what one includes in the “system.”


This is the last of three questions involving this situation. Questions 69a – c ask about similar situations in which the child

throws the rock radially or tangentially rather than dropping it. By using these two question sets together, you can draw

students’ attention to the significance of that facet.

Some students will say that mechanical energy and/or angular momentum is lost when the rock is dropped because it is no

longer part of the system.

Students generally understand angular momentum as “spinning” or “rotating,” and have great difficulty understanding how

something moving in a straight line can have nonzero angular momentum. Reconciling their intuition and concept of angular

momentum with L r p (the definition of angular momentum) will be difficult. It may be helpful to make a connection by

consider ing the constituent “particles” of an extended, rigid, rotating body.

To stimulate productive discussion, you may wish to pose questions such as: What is the “system” before the rock is

dropped? How about afterwards? What exactly is the “process” we are looking at? When does it start and end? Does the rock

have angular momentum (or mechanical energy) just before it is dropped? How about just afterwards? If energy is “lost,”

what happens to it? If angular momentum changes, what torques act on the system to change it?

Question E2.03a

Description: Developing understanding of angular momentum and energy in rotational motion (set-up: addressing velocity

vector addition).

Question

A child is standing at the rim of a disk holding a rock. The disk rotates freely without friction. At the instant shown, the child

throws the rock radially outward . Which of the indicated paths most nearly represents the trajectory of the rock as seen from

above?



Chapter 8

Page 8.36

1. Path (1)

2. Path (2)

3. Path (3)

4. Path (4)

5. Path (5)



Commentary

Purpose: To revisit velocity and vector addition in the context of rotational motion.

Discussion: Just before the rock is thrown, it is moving in a circle, and its instantaneous linear velocity points directly up the

page (tangential to the circle of motion). If the child merely dropped the rock, it would continue in a straight line along path

(2). However, when the child throws the rock radially outwards, she delivers an impulse in direction (5), so it now has a

velocity component in that direction as well as its original velocity component in direction (2). As a result, the rock moves in

direction (4).

After it has been released, the rock experiences no forces except gravity and air resistance, and neither of those act in a

direction that would change its direction left or right. So, seen from above, the rock continues along path (4).

You may be tempted to put yourself in the frame of the child, to imagine throwing or dropping the rock. This is dangerous,

because the frame of the child is accelerating and is not a proper inertial reference frame. Newton’s laws do not hold in a

noninertial frame, so it is best to avoid using one to analyze situations.

Key Points:

• At any point in time, an object traveling along a curved path has a linear velocity tangential to the curve.

• If an object is moving in one direction and receives an impulse in another direction, its resulting motion will be a

combination of those two directions. (The exact direction can be found from the vector addition of its original momentum

and the applied impulse.)

• An object does not continue moving along a curve when the force causing it to follow that curve has ceased.



Chapter 8

Page 8.37


This is the first of three related questions. This set is similar to Questions 68a – c and is intended to follow them.

Some students may answer (5) because they interpret the question to mean that the rock is thrown so that it travels directly

outward. In discussing and resolving this ambiguity, students can learn more than if the misinterpretation had been prevented

by careful problem wording.

Students may also choose (5) because they think of the rock as “at rest” before it is thrown, looking at it from the child’s

frame.

Presenting students with an analogous situation may be helpful. For example, imagine a ball rolling along the floor in one

direction and receiving a kick sideways. What direction does it roll after the kick? (This is suitable for a demonstration.)

Additional questions to ask during discussion: What is the radial component of the velocity if the rock follows path (2)? Is it

possible to throw the rock in such a way that it follows path (5)?

Question E2.03b

Description: Developing understanding of angular momentum and energy in rotational motion.

Question

A child is standing at the rim of a rotating disk, and throws a rock radially outward at the instant shown. The disk rotatesfreely without friction. Which of the following statements is correct about the disk-child-rock system as the rock is thrown?

1. Angular momentum is conserved; mechanical energy increases.

2. Angular momentum is conserved; mechanical energy decreases.

3. Angular momentum increases; mechanical energy is conserved.

4. Angular momentum decreases; mechanical energy is conserved.

5. Both angular momentum and mechanical energy are conserved.



Chapter 8

Page 8.38

6. Both angular momentum and mechanical energy increase.

7. Angular momentum decreases; mechanical energy increases.

8. Angular momentum increases; mechanical energy decreases.

9. Both angular momentum and mechanical energy decrease.

10. The conserved quantities cannot be determined.

Commentary

Purpose: To hone your understanding of energy and angular momentum conservation in a rotational context.


external forces, since internal interactions cannot apply a net force or a net torque.

The mechanical energy of a system is conserved when no external forces or nonconservative internal forces do work on it. (If

conservative internal forces do work on it, kinetic energy is exchanged for potential energy, but the total mechanical energy

remains constant. If conservative external forces do work, kinetic energy is exchanged for potential, but the potential energy

gained or lost is not part of the “system.”)

Any forces between the child and rock during the act of throwing are internal to the system, and therefore cannot exert a

torque on the system. No other forces are present that can exert a torque, so no net torque exists, and the system’s ang ular

momentum is conserved. (After the rock is thrown, it still has angular momentum even though it is traveling in a straight

line.)

Internal forces between the child and rock are doing work, however. The kinetic energy of the rock increases. Furthermore,

these are nonconservative forces, and so the total mechanical energy of the system increases during the throwing.

Key Points:


• Only external forces can exert a net torque.

• The mechanical energy of a system is conserved whenever no external or nonconservative internal forces do work on it.

• Nonconservative internal forces can do work on and increase the mechanical energy of a system.


This is the second of three related questions. It is parallel to Question E2.02c, but for a thrown rather than dropped rock.

Students who answer that mechanical energy is conserved – (3), (4), or (5) – may be expressing their belief that “energy is

always conserved,” without appreciating that this does not r equire the mechanical energy of a particular system to always be



Chapter 8

Page 8.40

1. Increases

2. Remains the same

3. Decreases

4. Impossible to determine

Commentary

Purpose: To hone and relate the concepts of angular velocity, angular momentum, angular impulse, and torque.


external forces, since internal interactions cannot apply a net force or a net torque. For the system consisting of the disk,

child, and rock, any forces between the child and rock during the act of throwing are internal and cannot exert a net torque on

the system (as a whole). If no net torque exists, no angular impulse is delivered, so the angular momentum of the system

cannot change.

However, this does not tell us whether the angular velocity of the disk and child changes. Let’s consider the “system”

consisting of the disk and child, but not the rock. During the act of throwing, the child must exert a force on the rock in the

direction the rock is being thrown: direction (2). According to Newton’s third law, the rock must be exerting a force of equal

magnitude on the child’s hand, pointing back in the opposite direction. As far as the child and disk are concerned, this is an

external force, and exerts a torque in the clockwise direction. This torque causes an angular impulse, changing the angular

momentum of the child and disk. Since the torque is in the opposite direction of the rotation, the angular velocity of the disk

and child decrease.

How is it possible for the angular momentum of the child and disk to decrease if the angular momentum of the child, disk,and rock together is conserved? Because the angular momentum of the rock increases as it is thrown. Recall that an object

traveling in a straight line can have nonzero angular momentum about a point not on that line, according to the definition of

angular momentum L pr . The rock’s linear momentum increases as it is thrown, so its angular momentum about the

center of the disk does as well.

Key Points:


• Only external forces can exert a net torque on a system.

• An object traveling in a straight line has nonzero angular momentum about a point not on that line.

• Choosing your “system” wisely makes analyzing situations and answering questions easier.




Chapter 8

Page 8.41

This is the third of three related questions. This question differs from the others in that the rock is thrown tangentially rather

than radially, drawing students’ attention to the significance of that change.

This question is useful for relating several different concepts: force, torque, momentum, angular momentum, impulse,

angular impulse, velocity, and angular velocity. We encourage you to lead a discussion that explores the situation and

question from many angles, helping students to fit these ideas together and resolve apparent contradictions.

For example, many students will still most likely have difficulty with the idea that an object traveling in a straight line can

have angular momentum, but that is crucial to understanding how the system as a whole can conserve angular momentum

when the disk and child alone do not.

Many students may choose the correct answer based on intuition or vague heuristic reasoning, and should be encouraged to

formalize or defend this with rigorous analysis.

Note that the kinetic energy of the system has increased, even though the angular speed of the disk-child system has

decreased.

Question E2.04

Description: Relating angular momentum to kinetic energy for rotation.

Question

An ice skater begins a spin in the middle of a large rink, but then starts to spin faster by pulling her arms in. Which of the

following statements is true?

1. Both kinetic energy and angular momentum are conserved.

2. Kinetic energy is conserved; angular momentum increases.

3. Kinetic energy is conserved; angular momentum decreases.

4. Kinetic energy increases; angular momentum is conserved.

5. Kinetic energy decreases; angular momentum is conserved.

6. Both kinetic energy and angular momentum increase.

7. Kinetic energy increases; angular momentum decreases.

8. Kinetic energy decreases; angular momentum increases.

9. Both kinetic energy and angular momentum decrease.

10. Impossible to determine



Chapter 8

Page 8.42

Commentary

Purpose: To improve your understanding of conservation laws and internal forces.

Discussion: If we treat the ice as frictionless, only two external forces act on the skater: gravitation (down) due to the Earth

and a normal force (up) due to the ice. Neither of these exerts a torque about the skater’s center of mass, so her angular

momentum is conserved. By pulling in her arms, however, she has decreased her rotational inertia, so her angular velocity

must increase: L I .

Her kinetic energy, however, must increase. If angular momentum is conserved, 1 1 2 2 I I . This means that if her

rotational inertia I has decreased by a factor we’ll call f (so that I 2 = I 1/ f ), her angular speed must increase by that same factor:

2 1 f . Her kinetic energy must therefore increase by a factor of f : 2 2

2 2 2 1 1 12 2 K I f I fK .

Where does the additional energy come from? As the skater pulls her arms in, she is applying a force through a displacement

and thus doing work on the system. She is converting chemical energy stored in her body to kinetic energy.

Key Points:

• If a system experiences no net external torque, its angular momentum does not change.

• Internal forces can do work on a system and increase its kinetic energy.

• Angular momentum can remain constant while angular speed increases (or decreases), if rotational inertia decreases (or

increases) by the same factor.

• If angular speed changes but angular momentum does not, kinetic energy from rotation must also change.


Students who indicate that angular momentum decreases – (3), (7), or (9) – may be taking friction into account. They are not

incorrect; they are merely not making the “expected” assumption.

Students may claim that angular momentum increases – (2), (6), or (8) – because they associate larger angular speed with

larger angular momentum, ignoring the role of rotational inertia or not realizing that it decreases sufficiently in this situation.

Some students may correctly believe that kinetic energy increases, but without understanding how this arises from the

interplay between I and .

Students indicating that kinetic energy decreases – (5), (8), or (9) – may think that kinetic energy would be conserved except

for the effect of friction.



Chapter 8

Page 8.43

Question F1.03a

Description: Integrating energy and angular momentum ideas by considering projectile motion in a universal gravitation

context.

Question

Two masses m1 and m2, having m1 > m2, are launched with the same speed in the direction A. Which mass reaches the

greatest height?

1. m1

2. m2

3. Both reach the same height.

Commentary

Purpose: To extend your understanding of energy conservation and projectile motion into the realm of “universal

gravitation.”

Discussion: Under the assumptions of “local gravity,” the maximum height a projectile reaches does not depend on the

ob ject’s mass, only on its launch direction and speed. This problem inquires whether that is also true under universal

gravitation, thus stimulating you to generalize your understanding of projectile motion and work with universal gravitation

ideas.

Assuming each projectile’s launch speed is less than its escape velocity, it will rise straight up, gradually slow, stop

momentarily, and fall back down. Its maximum height thus occurs when the kinetic energy is zero. Initially, a projectile has

some positive kinetic energy and some negative potential energy. (Under universal gravitation, all gravitational potential

energy is negative; the closer one is to the source of attraction, the more negative it becomes.) At the highest point, it has only

negative potential energy. Since all forces are conservative, total mechanical energy is conserved, initial energy is equal to

final energy, and by writing the appropriate expressions for kinetic and potential energy, we can solve for the pro jectile’s

height.



Chapter 8

Page 8.44

We don’t, however, actually have to solve for the height here. Note that the expressions for kinetic and potential energy are

both proportional to the projectile’s mass m: K = mv2/2 and U = – GmM Earth/r . Thus, the mass cancels out of the equation,

meaning that the maximum height achieved is independent of the mass. (The same argument applies to the analogous

problem in local gravitation, though the equation for potential energy is different.)

It may be tempting to argue qualitatively that the heavier object has more initial kinetic energy, and therefore will have more

final potential energy, and thus must go higher; this argument overlooks the fact that the heavier object also requires more

energy to reach a given height.

Key Points:

• Object masses often drop out of gravitational problems, for both local and universal gravitation.

• Understanding the reasoning behind facts and derived results helps you to know whether they are valid in new

circumstances (e.g., universal gravitation vs. local gravitation).


This question is an example of the “extend the context” pattern, presenting a familiar question in a novel circumstance: in t his

case, a common projectile-motion question in a situation requiring universal rather than local gravitation.

Do not stop after finding out whether students provide the correct answer; persevere to find out their reasoning. Some

students will get the right answer by erroneously applying “the square root of 2 gh” or similar rules memorized during local

gravitation, and this must be detected and challenged.

Question F1.03b

Description: Integrating energy and angular momentum ideas by considering projectile motion in a universal gravitation

context.

Question

Two masses m1 and m2, having m1 > m2, are launched with the same speed in the direction B. Which mass reaches the

greatest height?



Chapter 8

Page 8.45

1. m1

2. m2

3. Both reach the same height.

Commentary

Purpose: To help you integrate angular momentum and energy ideas in universal gravitation problems.

Discussion: In the previous question, a projectile launched directly upwards reached its apogee (maximum height) when its

speed was zero. In this question, the projectiles are launched at an angle and therefore have a nonzero speed even at apogee.

In order to apply the principle of conservation of energy to find the maximum height, we need to know how much kinetic

energy each projectile will have at the top.

If we were doing this problem with local gravity (a “flat earth” with constant gravitational force), the trajectory of a pr ojectile

would be a parabola, the horizontal component of the velocity would be constant, and by finding the initial horizontal

velocity we would know it at all times and could find the kinetic energy at apogee (when the vertical component of the

velocity is zero). With this problem, however, the projectile will follow an elliptical orbit rather than a parabola, and the

horizontal velocity (in a coordinate system fixed at the launch point) is not constant.

What is constant? The angular momentum of the projectile around the center of the earth is conserved, since the only force

acting upon it is central (points toward the center of the earth) and thus exerts no torque. We can relate the angular

momentum to the “horizontal” (tangential) velocity component vt as follows:t

m mr L r p r v v . Since the

kinetic energy at apogee K a is2 2t

mv (since the radial component of velocity vr is zero), we can write K a in terms of the

angular momentum: 2 22a a

K L mr (where r a is the radius of the projectile — its height above the center of the earth — at

apogee).

Since the angular momentum is constant, we can find its value at the moment of launch: 0 cos

t E L mr mR v v where

is the launch angle (above horizontal) and v0 the launch speed. Now, if we substitute this into our expression for K a, we get

2 2 2 20

cos 2a E

K mR r v . In other words, we find that the kinetic energy at apogee is propor tional to the projectile’s

mass.

Since kinetic energy at launch and potential energy at launch and apogee are all also proportional to the mass (as discussed in

the previous question), mass cancels out of the conservation of energy equation, and we find that the maximum height does

not depend on mass. Thus, both projectiles must reach the same maximum height. In other words, the same argument applies

to this question as to the previous one, but in this case we had to do a little more work to demonstrate that the argument does

apply.

Intuition might suggest to you that the mass does not matter, but part of physics is learning to back up your intuition with a



Chapter 8

Page 8.48

contradict the reasoning they’ve just used for this question. Exploring and resolving the apparent contradiction pushes

students to further refine their understanding of escape velocity and the relationship between angular momentum and energy

in orbit problems.

QUICK QUIZZES

1. (d). A larger torque is needed to turn the screw. Increasing the radius of the screwdriver handle provides a greater

lever arm and hence an increased torque.

2. (b). Since the object has a constant net torque acting on it, it will experience a constant angular acceleration. Thus,

the angular velocity will change at a constant rate.

3. (b). The hollow cylinder has the larger moment of inertia, so it will be given the smaller angular acceleration and

take longer to stop.

4. (a). The hollow sphere has the larger moment of inertia, so it will have the higher rotational kinetic energy.

5. (c). Apply conservation of angular momentum to the system (the two disks) before and after the second disk is

added to get the result: 1 1 1 2 I I I .

6. (a). Earth already bulges slightly at the Equator, and is slightly flat at the poles. If more mass moved towards the

Equator, it would essentially move the mass to a greater distance from the axis of rotation, and increase the moment

of inertia. Because conservation of angular momentum requires that constant z z I , an increase in the moment

of inertia would decrease the angular velocity, and slow down the spinning of Earth. Thus, the length of each day

would increase.

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. sin 0.500 m 80.0 N sin 60.0° 36.4 N mrF which is choice (a).

2. Using the left end of the plank as a pivot and requiring that

0 gives 22.00 m 3.00 m 0mg F , or

2

2

2 20.0 kg 9.80 m s2131 N

3 3

mg F

so choice (d) is the correct response.



Chapter 8

Page 8.49

3. Assuming a uniform, solid disk, its moment of inertia about a perpendicular axis through its center is 2 2 I MR ,

so I gives

2

22

2 40.0 N m25.00 rad s

25.0 kg 0.800 m MR

and the correct answer is (b).

4. For a uniform, solid sphere,

22 5 I MR and 5

4

2 rad 1 d7.27 10 rad s

1 d 8.64 10 s E

so

224 6

22 5

2 5.98 10 kg 6.38 10 m1 17.27 10 rad s

2 2 5r E

KE I

yielding 293 10 Jr KE , making (a) the correct choice.

5. In order for an object to be in equilibrium, it must be in both translational equilibrium and rotational equilibrium.

Thus, it must meet two conditions of equilibrium, namely net 0F and net 0 . The correct answer is therefore

choice (d).

6. In a rigid, rotating body, all points in that body rotate about the axis at the same rate (or have the same angular

velocity). The centripetal acceleration, tangential acceleration, linear velocity, and total acceleration of a point in

the body all vary with the distance that point is from the axis of rotation. Thus, the only correct choice is (b).

7. The moment of inertia of a body is determined by its mass and the way that mass is distributed about the rotation

axis. Also, the location of the body’s center of mass is determined by how its mass is distributed. As long as these

qualities do not change, both the moment of inertia and the center of mass are constant. From = I , we see that

when a body experiences a constant, nonzero torque, it will have a constant, nonzero angular acceleration.

However, since the angular acceleration is nonzero, the angular velocity (and hence the angular momentum, L =

I ) will vary in time. The correct responses to this question are then (b) and (e).

8. When objects travel down ramps of the same length, the one with the greatest translational kinetic energy at the

bottom will have the greatest final translational speed (and, hence, greatest average translational speed). This means



Chapter 8

Page 8.50

that it will require less time to travel the length of the ramp. Of the objects listed, all will have the same total

kinetic energy at the bottom, since they have the same decrease in gravitational potential energy (due to the ramps

having the same vertical drop) and no energy has been spent overcoming friction. All of the block’s kinetic energy

is in the form of translational kinetic energy. Of the rolling bodies, the fraction of their total kinetic energy that is in

the translational form is

1 22

1 1 22 2 22 2

1 1

11

t

t r

M KE f

KE KE M I I MR I M

v

v v

Since the ratio I / MR2equals 2/5 for a solid ball and 2/3 for a hollow sphere, the ball has the larger translational

kinetic energy at the bottom and will arrive before the hollow sphere. The correct rankings of arrival times, from

shortest to longest, is then block, ball, sphere, and choice (e) is the correct response.

9. Please read the answer to Question 8 above, since most of what is said there also applies to this question. The total

kinetic energy of either the disk or the hoop at the bottom of the ramp will be KE total = Mgh, where M is the mass of

the body in question and h is the vertical drop of the ramp. The translational kinetic energy of this body will then be

KE t = f KE total = fMgh, where f is the fraction discussed in Question 8. Hence, M 2 2 f M v gh and the

translational speed at the bottom is 2 . fghv

Since f = 1/(1+1/2) = 2/3 for the disk and f = 1/(1+1) = 1/2 for the hoop, we see that the disk will have the greater

translational speed at the bottom, and hence, will arrive first. Notice that both the mass and radius of the object has

canceled in the calculation. Our conclusion is then independent of the object’s mass and/or radius. Therefore, the

only correct response is choice (d)

10. The ratio of rotational kinetic energy to the total kinetic energy for an object that rolls without slipping is

1 22

1 1 2 22 2total 2 2

1 1

11

r r

t r

I KE KE

MR KE KE KE M I M

I I

v v

For a solid cylinder, I = MR2/2 and this ratio becomes

total

1 1

2 1 3

r KE

KE

so the correct answer is (c).

11. If a car is to reach the bottom of the hill in the shortest time, it must have the greatest translational speed at the



Chapter 8

Page 8.51

bottom (and hence, greatest average speed for the trip). To maximize its final translational speed, the car should be

designed so as much as possible of the car’s total kinetic energy is in the form of translational kin etic energy. This

means that the rotating parts of the car (i.e., the wheels) should have as little kinetic energy as possible. Therefore,

the mass of these parts should be kept small, and the mass they do have should be concentrated near the axle in

order to keep the moment of inertia as small as possible. The correct response to this question is (e).

12. Please review the answers given above for questions 8 and 9. In the answer to question 9, it is shown that the

translational speed at the bottom of the hill of an object that rolls without slipping is 2 fghv where h is the

vertical drop of the hill and f is the ratio of the translational kinetic energy to the total kinetic energy of the rolling

body. For a solid sphere, 22 5 I MR , so the ratio f is

2

1 1 1

1 2 5 1.41 f

I MR

and the translational speed at the bottom of the hill is 2 1.4 ghv Notice that this result is the same for all

uniform, solid, spheres. Thus, the two spheres have the same translational speed at the bottom of the hill. This also

means that they have the same average speed for the trip, and hence, both make the trip in the same time. The

correct answer to this question is (d).

13. Since the axle of the turntable is frictionless, no external agent exerts a torque about this vertical axis of the mouse-

turntable system. This means that the total angular momentum of the mouse-turntable system will remain constant

at its initial value of zero. Thus, as the mouse starts walking around the axis (and developing an angular

momentum,mouse m m

L I , in the direction of its angular velocity), the turntable must start to turn in the opposite

direction so it will possess an angular momentum, table t t L I , such that

total mouse table 0

m m t t L L L I I . Thus, the angular velocity of the table will be t m t m

I I .

The negative sign means that if the mouse is walking around the axis in a clockwise direction, the turntable will be

rotating in the opposite direction, or counterclockwise. The correct choice for this question is (d).

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. If the bar is, say, seven feet above the ground, a high jumper has to lift his center of gravity approximately to a

height of seven feet in order to clear the bar. A tall person already has his center of gravity higher than that of a

short person. Thus, the taller athlete has to raise his center of gravity through a smaller distance.

4. The lever arm of a particular force is found with respect to some reference point. Thus, an origin for calculating

torques must be specified. However, for an object in equilibrium, the calculation of the total torque is independent

of the location of the origin.



Chapter 8

Page 8.52

6. The critical factor is the total torque being exerted about the line of the hinges. For simplicity, we assume that the

paleontologist and the botanist exert equal magnitude forces. The free body diagram of the original situation is

shown on the left and that for the modified situation is shown on the right in the sketches below:

In order for the torque exerted on the door in the modified situation to equal that of the original situation, it is

necessary that Fd = Fd 0 + F (8 cm) or d = d 0 + 8 cm. Thus, the paleontologist would need to relocate about 8 cm

farther from the hinge.

8.

10. After the head crosses the bar, the jumper should arch his back so the head and legs are lower than the midsection

of the body. In this position, the center of gravity may pass under the bar while the midsection of the body is still

above the bar. As the feet approach the bar, the legs should be straightened to avoid hitting the bar.

12. (a) Consider two people, at the ends of a long table, pushing with equal magnitude forces directed in opposite

directions perpendicular to the length of the table. The net force will be zero, yet the net torque is not zero.

(b) Consider a falling body. The net force acting on it is its weight, yet the net torque about the center of gravity

is zero.

14. As the cat falls, angular momentum must be conserved. Thus, if the upper half of the body twists in one direction,

something must get an equal angular momentum in the opposite direction. Rotating the lower half of the body in

the opposite direction satisfies the law of conservation of angular momentum.

PROBLEM SOLUTIONS



Chapter 8

Page 8.53

8.1 Since the friction force is tangential to a point on the rim of the wheel, it is perpendicular to the radius line

connecting this point with the center of the wheel. The torque of this force about the axis through the center of the

wheel is then = rf sin 90.0º = rf , and the friction force is

76.0 N m

217 N0.350 m f r

8.2 The torque of the applied force is = rF sin . Thus, if r = 0.330 m, = 75.0º, and the torque has the maximum

allowed value of max = 65.0 N m, the applied force is

max 65.0 N m

204 Nsin 0.330 m sin 75.0

F r

8.3 First resolve all of the forces shown in Figure P8.3 into components parallel to and perpendicular to the beam as

shown in the sketch below.

(a) 25 N cos 30 2.0 m 10 N sin 20 4.0 m 30 N mO

or 0 = 30 N m counterclockwise

(b) 30 N sin 45 2.0 m 10 N sin 20 2.0 m 36 N mC

or c = 30 N m counterclockwise

8.4 The lever arm is 2 31.20 10 m cos 48.0 8.03 10 md , and the torque is

380.0 N 8.03 10 m 0.642 N m counterclockwise Fd



Chapter 8

Page 8.54

8.5 (a) sin g F lever arm mg

23.0 kg 9.8 m s 2.0 m sin 5.0 5.1 N m

(b) The magnitude of the torque is proportional to the sin , where is the angle between the direction of the

force and the line from the pivot to the point where the force acts. Note from the sketch that this is the same as the

an gle the pendulum string makes with the vertical.

Since sin increases as increases, the torque also increases with the angle.

8.6 The object is in both translational and rotational equilibrium. Thus, we may write:

0 0 x x x F F R

0 0 y y y g F F R F

and

0 cos sin cos 02

O y x g F F F

8.7

Requiring that = 0, using the shoulder joint at point O as a pivot, gives

sin12.0 0.080 m 41.5 N 0.290 m 0t F or 724 Nt F

Then 0 724 N sin 12.0 41.5 N = 0 y sy F F , yielding 109 N sy F



Chapter 8

Page 8.56

19.6 N 33.0 cm

312 N8.00 cm cos 75.0

B F

8.10 Since the bare meter stick balances at the

49.7 cm mark when placed on the fulcrum, the

center of gravity of the meter stick is located 49.7 cm

from the zero end. Thus, the entire weight of the meter

stick may be considered to be concentrated at this point.

The free-body diagram of the stick when it is balanced

with the 50.0-g mass attached at the 10.0 cm mark is as

given at the right.

Requiring that the sum of the torques about point O be zero yields

50.0 g g 39.2 cm 10.0 cm M g 49.7 cm 39.2 cm 0

or

39.2 cm 10.0 cm

50.0 g 139 g49.7 cm 39.2 cm

M

8.11 Consider the remaining plywood to consist of two

parts: A1 is a 4.00-ft-by-4.00-ft section with center of gravity

located at (2.00 ft, 2.00 ft), while A2 is a 2.00-ft-by-4.00-ft section with

center of gravity at (6.00 ft, 1.00 ft). Since the plywood is uniform, its

mass per area is constant and the mass of a section having

area A is m = A. The center of gravity of the remaining

plywood has coordinates given by

cgi i

i

m x x

m

1 1 A x 2 2 A x

1 A

2 2

2 22

16.0 ft 2.00 ft 8.00 ft 6.00 ft3.33 ft

16.0 ft 8.00 ft A

and

cgi i

i

m y y

m

1 1 A y 2 2 A y

1 A

2 2

2 22

16.0 ft 2.00 ft 8.00 ft 1.00 ft1.67 ft

16.0 ft 8.00 ft A



Chapter 8

Page 8.57

8.12 (a)

290.0 kg 9.80 m s 882 N Mg 255.0 kg 9.80 m s 539 Nmg

(b) The woman is at x = 0 when n1 is greatest. With this location of the woman, the counterclockwise torque

about the center of the beam is a maximum. Thus, n1 must be exerting its maximum clockwise torque about

the center to hold the beam in rotational equilibrium.

(c) n1 = 0 As the woman walks to the right along the beam, she will eventually reach a point where the beam will

start to rotate clockwise about the rightmost pivot. At this point, the beam is starting to lift up off of the left

most pivot and the normal force exerted by that pivot will have diminished to zero.

(d) When the beam is about to tip, n1 = 0 and F y = 0, gives 0+n2 – Mg – mg = 0, or

32 882 N 539 N 1.42 10 Nn Mg mg

(e) Requiring that rightmost pivot

0 when the beam is about to tip (n1 = 0) gives

4.00 m 4.00 m 3.00 m 0 x mg Mg

or 1.00 m 4.00 mmg x Mg mg , and

1.00 m 4.00 m M

xm

Thus,

90.0 kg

1.00 m 4.00 m 5.64 m55.0 kg

x

(f) When n1 = 0 and n2 = 1.42 × 103 N, requiring that leftend

0 gives



Chapter 8

Page 8.58

30 539 N 882 N 3.00 m 1.42 10 N 4.00 m 0 x

or

33.03 10 N m5.62 N

539 N

x

which, within limits of rounding errors, is the same as the answer to part (e).

8.13 Requiring that cg 0i i i x m x m gives

5.0 kg 0 3.0 kg 0 4.0 kg 3.0 m 8.0 kg0

5.0 3.0 4.0 8.0 kg

x

or 8.0 x + 12 m = 0 which yields x = – 1.5 m

Also, requiring that cg 0i i i y m y m gives

5.0 kg 0 3.0 kg 4.0 m 4.0 kg 0 8.0 kg0

5.0 3.0 4.0 8.0 kg

y

or 8.0 y + 12 m = 0 yielding y = – 1.5 m

Thus, the 8.0-kg object should be placed at coordinates ( – 1.5 m, – 1.5m) .

8.14 (a) As the woman walks to the right along the beam,

she will eventually reach a point where the beam will start

to rotate clockwise about the rightmost pivot. At this point,

the beam is starting to lift up off of the leftmost pivot and the

normal force, n1, exerted by that pivot will have diminished to

zero.

Then, 20 0 0 y F mg Mg n , or

2n m M g



Chapter 8

Page 8.59

(b) When n1 = 0 and n2 = (m + M ) g , requiring that leftend

0 gives

0 02

Lmg x Mg mg Mg or 1

2

M M x L

m m

(c) If the woman is to just reach the right end of the beam ( x = L) when n1 = 0 and n2 = (m+M ) g (i.e., the beam is

ready to tip), then the result from Part (b) requires that

12

M M L L

m m

or 1 2 2

1

M m m M L

M m m M

8.15 In each case, the distance from the bar to the center of mass of the body is given by

arms arms torso torso thighs thighs legs legs

cgarms torso thighs legs

i i

i

m x m x m x m xm x x

m m m m m

where the distance x for any body part is the distance from the bar to the center of gravity of that body part. In each

case, we shall take the positive direction for distances to run from the bar toward the location of the head.

Note that 6.87 33.57 14.07 7.54 kg 62.05 kgim .

With the body positioned as shown in Figure P8.15b, the distances x for each body part is computed using the

sketch given below:

arms cgarms

0.239 m x r

torso arms cgtorso

0.548 m +0.337 m 0.885 m x r

thighs arms torso cgthighs

0.548+ 0.601+0.151 m 1.30 m x r



Chapter 8

Page 8.60

legs arms torso thighs cglegs

0.548+ 0.601+0.374 +0.227 m 1.75 m x r

With these distances and the given masses we find

cg 62.8 kg m 1.01 m62.05 kg

x

With the body positioned as shown in Figure P8.15c, we use the following sketch to determine the distance x for

each body part:

arms cgarms

0.239 m x r

torso arms cgtorso

0.548 m 0.337 m 0.211 m x r

thighs arms torso cgthighs

0.548 0.601 0.151 m 0.204 m x r

legs arms torso thighs cglegs

0.548 0.601 0.374 0.227 m 0.654 m x r

With these distances, the location (relative to the bar) of the center of gravity of the body is

cg

0.924 kg m0.015 m 0.015 m towards the head

62.05 kg x

8.16 With the coordinate system shown below, the coordinates of the center of gravity of each body part may be

computed:



Chapter 8

Page 8.61

cg,arms 0 x cg,arms arms cgarms

0.309 m y r

cg,torso cgtorso

0.337 m x r cg,torso 0 y

cg,thighs torso cgthighs

cos60.0 0.676 m x r cg,thighs cgthighs

sin 60.0 0.131 m y r

cg,legs torso thighs cglegs

cos 60.0 1.02 m x r cg,legs thighs sin60.0 0.324 m y

With these coordinates for individual body parts and the masses given in Problem 8.15, the coordinates of the

center of mass for the entire body are found to be

arms cg,arms torso cg, torso th ighs cg, thighs l egs cg, legs


28.5 kg m0.459 m

62.05 kg

m x m x m x m x

x

m m m m

and

arms cg ,arms torso cg ,torso th ighs cg, th ighs legs cg, legs


6.41 kg m0.103 m

62.05 kg

m y m y m y m y

ym m m m

8.17 The free-body diagram for the spine is shown below.

When the spine is in rotational equilibrium, the sum of the torques about the left end (point O) must be zero. Thus,

2

350 N 200 N 03 2 y

L L

T L

Yielding sin 12.0 562 N yT T .

The tension in the back muscle is then 3562 N

= 2.71 10 N 2.71 kNsin12.0

T

.



Chapter 8

Page 8.62

The spine is also in translational equilibrium, so 0 0 x x x

F R T and the compression force in the

spine is

cos 12.0 = 2.71 kN cos 12.0 2.65 kN x x

R T T

8.18 In the free-body diagram of the foot

given at the right, note that the force

R (exerted on the foot by the tibia) has been

replaced by its horizontal and vertical

components. Employing both conditions of

equilibrium (using point O as the pivot point)

gives the following three equations:

0 sin15.0 sin 0 x F R T

or

sin

sin15.0

T R

[1]

0 700 N cos15.0 cos 0 y F R T [2]

0 700 N 18.0 cm cos 25.0 cm 18.0 cm 0O

T

or

T = (1 800 N) cos [3]

Substituting Equation [3] into Equation [1] gives

1 800 Nsin cos

sin15.0 R

[4]

Substituting Equations [3] and [4] into Equation [2] yields

21 800 N

sin cos 1 800 N cos 700 Ntan15.0

which reduces to: sin cos = (tan 15.0º) cos2 + 0.104 2.



Chapter 8

Page 8.63

Squaring this result and using the identity 2 2sin 1 cos gives

2

2 4 2tan 15.0 1 cos 2 tan15.0 0.104 2 1 cos 0.104 2 0

In this last result, let u = cos2

and evaluate the constants to obtain the quadratic equation

21.071 8 0.944 2 0.010 9 0u u

The quadratic formula yields the solutions u = 0.869 3 and u = 1.011 7.

Thus,

1cos 0.869 3 21.2 or 1cos 0.011 7 83.8

We ignore the second solution since it is physically impossible for the human foot to stand with

the sole inclined at 83.8° to the floor. We are the left with: = 21.2º.

Equation [3] then yields

31 800 N cos 21.2 1.68 10 NT

and Equation [1] gives

3 31.68 10 N sin 21.2 2.34 10 Nsin15.0

R

8.19 Consider the torques about an axis perpendicular to the page through the left end of the rod.

100 N 3.00 m 500 N 4.00 m0

6.00 m cos30.0T



Chapter 8

Page 8.64

443 NT

0 sin 30.0 443 N sin 30.0 x x F R T

R x = 221 N toward the right

0 cos 30.0 100 N 500 N 0 y y F R T

R y = 600 N (443 N) cos 30.0 = 217 N upward

8.20 Consider the torques about an axis perpendicular to the page through the left end of the scaffold.

1 20 0 700 N 1.00 m 200 N 1.50 m 3.00 m 0T T

From which, T 2 = 333 N.

Then, from F y = 0, we have

T 1+T 2 – 700 N – 200 N = 0

or

T 1 = 900 N – T 2 = 900 N – 333 N = 567 N

8.21 Consider the torques about an axis

perpendicular to the page and through

the left end of the plank.

= 0 gives

1700 N 0.500 m 294 N 1.00 m sin 40.0 2 .00 m 0T

or T1 = 501 N.

Then, 0 x F gives 3 1 cos 4 0.0° 0T T , or

3 501 N cos 40.0 384 NT

From 0 y F , 2 1994 N sin 40.0 0T T ,

or 2 994 N 501 N sin 40.0 672 NT .



Chapter 8

Page 8.65

8.22 (a) See the diagram below

(b) If x = 1.00 m, then

left end

0 700 N 1.00 m 200 N 3.00 m

80.0 N 6.00 m sin 60.0° 6.00 m 0T

giving T = 434 N.

Then, 0 cos 60.0° 0 x

F H T , or 343 N cos 60.0° 172 N H

and 0 980 N + 343 N sin 60.0° 0 y F V , or V = 683 N.

(c) When the wire is on the verge of breaking, T = 900 N and

maxleft end

700 N 200 N 3.00 m

80.0 N 6.00 m 900 N sin 60.0° 6.00 m 0

x

which gives xmax = 5.14 m

8.23 The required dimensions are as follows:

1 4.00 m cos 50.0° 2 .57 md

2 cos 50.0° 0.643d d d

3 8.00 m sin 50.0 6.13 md

0 y F yields 1 200 N 800 N = 0 F , or F 1 = 1.00 103 N.

When the ladder is on the verge of slipping,



Chapter 8

Page 8.66

1max s s s f f n F or 30.600 1.00 10 N 600 N f

Then, F x = 0 gives F 2 = 600 N to the left.

Finally, using an axis perpendicular to the page and through the lower end of the ladder = 0, gives

– (200 N)(2.57 m) – (800 N)(0.643)d +(600 N)(6.13 m) = 0

or

33.68 10 550 N m6.15 m

0.643 800 Nd

when the ladder is ready to slip

8.24 (a)

(b) The point of intersection of two unknown forces is always a good choice as the pivot point in a torque calcu-

lation. Doing this eliminates these two unknowns from the calculation (since they have zero lever arms about

the chosen pivot) and makes it .easier to solve the resulting equilibrium equation .

(c) hinge

0 0 0 cos sin 02

Lmg T L

(d) Solving the above result for the tension in the cable gives

2 cos

sin 2 tan

mg L mg T

L

or



Chapter 8

Page 8.67

216.0 kg 9.80 m s136 N

2 tan 30.0T

(e) 0 0 x x F F T and 0 0 y y F F mg

(f) Solving the above results for the components of the hinge force gives

F x = T = 136 N and 216.0 kg 9.80 m s 157 N y F mg

(g) Attaching the cable higher up would allow the cable to bear some of the weight, thereby reducing the stress

on the hinge. It would also reduce the tension in the cable.

8.25 Consider the free-body diagram of the

material making up the center point inthe rope given at the right. Since this

material is in equilibrium, it is necessary

to have F x = 0 and F y = 0, giving

0: x

F 2 1sin sin 0T T

or T 2 = T 1, meaning that the rope has a uniform tension T throughout its length.

0: y F cos cos 475 N 0T T where

2 2

0.500 mcos

6.00 m 0.500 m

and the tension in the rope (force applied to the car) is

2 2

3475 N 6.00 m 0.500 m475 N

2.86 10 N 2.86 kN2 cos 2 0.500 m

T

8.26 (a)



Chapter 8

Page 8.68

(b) lower

end

0 0 0 cos sin 02

Lmg T L

or

cos

cot2 sin 2

mg mg

T

(c) 0 0 x s F T n or sT n [1]

0 0 y F n mg n mg [2]

Substitute Equation [2] into [1] to obtain T = smg .

(d) Equate the results of parts (b) and (c) to obtain s = cot /2

This result is valid only at the critical angle where the beam is on the verge of slipping

(i.e., where f s = ( f s)max is valid.)

(e) At angles below the critical angle (where max s s f f is valid), the beam will slip. At larger angles, the

static friction force is reduced below the maximum value, and it is no longer appropriate to use s

in the

calculation.

8.27 Consider the torques about an axis perpendicular

to the page and through the point where the force

T acts on the jawbone.

0 (50.0 N) (7.50 cm) (3.50 cm) 0 R , which

yields 107 N R .

Then, 0 50.0 N + 107 N 0 y F T , or 157 NT .

8.28 Observe that the cable is perpendicular to the boom. Then, using = 0 for an axis perpendicular to the page and

through the lower end of the boom gives

3

1.20 kN cos 65° 2.00 kN cos 65° 02 4

LT L L

or .T = 1.47 kN



Chapter 8

Page 8.69

From 0 x F , cos25° 1.33 kN to the right H T

and F y = 0 gives

V = 3.20 kN – T sin 25 = 2.58 kN upward

8.29 First, we resolve all forces into components parallel to and

perpendicular to the tibia, as shown. Note that = 40.0 and

w y = (30.0 N) sin 40.0 = 19.3 N

F y = (12.5 N) sin 40.0 = 8.03 N

and

T y = T sin 25.0

Using = 0 for an axis perpendicular to the page and

through the upper end of the tibia gives

sin 25.0° 19.3 N 8.03 N 05 2

d d T d

or T = 209 N.

8.30 When x = xmin , the rod is on the verge of slipping, so

max

0.50 s s f f n n

From 0 x F , cos37° 0, or 0.80n T n T . Thus,

0.50 0.80 0.40 f T T

From 0 y F , sin 37 2 0, or f T w

0.40 0.60 2 0T T w , giving 2T w



Chapter 8

Page 8.70

Using = 0 for an axis perpendicular to the page and through the left end of the beam gives

min 2.0 m 2 sin 37° 4.0 m 0w x w w , which reduces to xmin = 2.8 m.

8.31 The moment of inertia for rotations about an axis is 2i i

I m r , where r i is the distance mass mi is from that axis.

(a) For rotation about the x-axis,

2 2

2 2 2

3.00 kg 3.00 m 2 .00 kg 3.00 m

2.00 kg 3.00 m 4.00 kg 3.00 m 99.0 kg m

x I

(b) When rotating about the y-axis,

2 2

2 2 2

3.00 kg 2 .00 m 2 .00 kg 2 .00 m

2.00 kg 2.00 m 4.00 kg 2.00 m 44.0 kg m

y I

(c) For rotations about an axis perpendicular to the page through point O, the distance r i for each mass is

2 2

2 .00 m 3.00 m 13.0 mi

r

Thus,

2 23.00 2.00 2.00 4.00 kg 13.0 m 143 kg mO

I

8.32 The required torque in each case is = I . Thus,

2 299.0 kg m 1.50 rad s 149 N m x x I

2 244.0 kg m 1.50 rad s 66.0 N m y y I

and

2 2143 kg m 1.50 rad s 215 N mO O I



Chapter 8

Page 8.71

8.33 (a) net 2

net 2

0.330 m 250 Nsin 90 87.8 kg m

0.940 rad s

rF I I

(b) For a solid cylinder, I = Mr 2/2, so

2

322

2 87.8 kg m21.61 10 kg

0.330 m

I M

r

(c) 20

0 0.940 rad s 5.00 s 4.70 rad st

8.34 (a) 2 2disk cylinder 2 2 2 2 I I I MR mr or 2 2 2 I MR mr

(b) g = 0 Since the line of action of the gravitational force passes through the rotation axis, it has zero lever arm

about this axis and zero torque.

(c) The torque due to the tension force is positive. Imagine gripping the cylinder with your right hand so

your fingers on the front side of the cylinder point upward in the direction of the tension force. The thumb of

your right hand then points toward the left (positive direction) along the rotation axis. Because I , the

torque and angular acceleration have the same direction. Thus, a positive torque produces a

positive angular acceleration. When released, the center of mass of the yoyo drops downward, in the

negative direction. The translational acceleration is negative.

(d) Since, with the chosen sign convention, the translational acceleration is negative when the angular accelera-

tion is positive, we must include a negative sign in the proportionality between these two quantities. Thus, we

write: a = r or = – a/r

(e) Translation:

total 2 2 y F m a T M m g M m a [1]

(f) Rotational:

sin 90 or I rT I rT I [2]

(g) Substitute the results of (d) and (a) into Equation [2] to obtain



Chapter 8

Page 8.72

2

2

22

a r mr aT I I MR

r r r

or

2

2

R mT M as

r

[3]

Substituting Equation [3] into [1] yields

2

2 2 2 M R r m a M m g M m a or

2

2

2 3 2

M m g a

M M R r m

(h)

2

22

2 2.00 kg 1.00 kg 9.80 m s2.72 m s

2 2.00 kg 2.00 kg 10.0 4.00 3 1.00 kg 2a

(i) From Equation [1], 2 22 5.00 kg 9.80 m s 2.72 m s 35.4 NT M m g a .

( j)

22

2 2 1.00 m0 2 0.857 s

2.72 m s

y y t at t

a

8.35 (a) Consider the free-body diagrams of the cylinder and

man given at the right. Note that we shall adopt a sign

convention with clockwise and downward as the positive

directions. Thus, both a and are positive in the indicated

directions and a = r . We apply the appropriate form of Newton’s

second law to each diagram to obtain the following:

Rotation of Cylinder: sin 90° , or , I rT I T I r

so

21 12

aT Mr r r

and 1

2T Ma [1]

Translation of man:

y F ma mg T ma or T m g a [2]



Chapter 8

Page 8.73

Equating Equations [1] and [2] gives 1

2 Ma m g a , or

2

275.0 kg 9.80 m s

3.92 m s2 75.0 kg+ 225 kg 2

mg a

m M

(b) From a = r , we have

2

23.92 m s

9.80 rad s0.400 m

a

r

(c) As the rope leaves the cylinder, the mass of the cylinder decreases, thereby decreasing the moment of inertia.

At the same time, the weight of the rope leaving the cylinder would increase the downward force acting tan-

gential to the cylinder, and hence increase the torque exerted on the cylinder. Both of these effects will cause

the acceleration of the system to increase with time. (The increase would be slight in this case, given the

large mass of the cylinder.)

8.36 The angular acceleration is ( ) ( ) f i it t since 0 f

Thus, the torque is ( )i I I t . But, the torque is also = – fr , so the magnitude of the required

friction force is

212 kg m 50 rev min 2 rad 1 min21 N

0.50 m 6.0 s 1 rev 60 s

i I f

r t

Therefore, the coefficient of friction is

21 N0.30

70 Nk

f

n

8.37 (a) 0.800 N 30.0 m 24.0 N m F r

(b)

222

24.0 N m0.0356 rad s

0.750 kg 30.0 m I m r



Chapter 8

Page 8.74

(c) 2 230.0 m 0.0356 rad s 1.07 m st a r

8.38 22 21.80 kg 0.320 m 0.184 kg m I MR

net applied resistive I , or F r f R I

yielding

I f R F

r

(a) 2 2

2

0.184 kg m 4.50 rad s 120 N 0.320 m872 N

4.50 10 m F

(b) 2 2

2

0.184 kg m 4.50 rad s 120 N 0.320 m1.40 kN

2.80 10 m F

8.39 22 2

1 1150 kg 1.50 m 169 kg m

2 2 I MR

and

2

0.500 rev s 0 2 rad rad s

2 .00 s 1 rev 2

f i

t

Thus, F r I gives

2 2169 kg m rad s2

177 N1.50 m

I F

r

8.40 (a) It is necessary that the tensions T 1 and T 2 be different in

order to provide a net torque about the axis of the pulley

and produce and angular acceleration of the pulley.

Since intuition tells us that the system will accelerate in

the directions shown in the diagrams at the right

when m2 > m1, it is necessary that T 2 > T 1.



Chapter 8

Page 8.75

(b) We adopt a sign convention for each object with the positive

direction being the indicated direction of the acceleration of that

object in the diagrams at the right. Then, ap ply Newton’s

second law to each object:

For 1 1 1 1 1: ym F m a T m g m a or 1 1T m g a [1]

For 2 2 2 2 2: ym F m a m g T m a m2 or 2 2T m g a [2]

For 2 1: M I rT rT I or 2 1

T T I r [3]

Substitute Equations [1] and [2], along with the relations 2 2 and I Mr a r , into Equation [3] to obtain

2

2 12 2

Mr a Mam g a m g ar r

or 1 2 2 1

2

M m m a m m g

and

22 1 2

1 2

20.0 kg 10.0 kg 9.80 m s2.88 m s

2 20.0 kg 10.0 kg+ 8.00 kg 2

m m g a

m m M

(c) From Equation [1]: 2 2

1 10.0 kg 9.80 m s 2.88 m s 127 NT

.

From Equation [2]: 2 22 20.0 kg 9.80 m s 2.88 m s 138 NT .

8.41 The initial angular velocity of the wheel is zero, and the final angular velocity is

50.0 m s40.0 rad s

1.25 m f

r

v

Hence, the angular acceleration is

240.0 rad s 0

83.3 rad s0.480 s

f i

t



Chapter 8

Page 8.76

The torque acting on the wheel is k f r , so I gives

2 2

3110 kg m 83.3 rad s

7.33 10 N1.25 m

k

I f

r

Thus, the coefficient of friction is

3

4

7.33 10 N0.524

1.40 10 N

k k

f

n

8.42 (a) The moment of inertia of the flywheel is

22 3 2

1 1

500 kg 2 .00 m 1.00 10 kg m2 2 I MR

and the angular velocity is

rev 2 rad 1 min5000 524 rad s

min 1 rev 60 s

Therefore, the stored kinetic energy is

22 3 2 8

stored

1 11.00 10 kg m 524 rad s 1.37 10 J

2 2 KE I

(b) A 10.0-hp motor supplies energy at the rate of

3746 W

10.0 hp 7.46 10 J s1 hp

P

The time the flywheel could supply energy at this rate is

8stored 4

3

1.37 10 J1.84 10 s 5.10 h

7.46 10 J s

KE t

P



Chapter 8

Page 8.77

8.43 The moment of inertia of the cylinder is

22 2 2

2

1 1 1 800 N1.50 m 91.8 kg m

2 2 2 9.80 m s

w I MR R

g

The angular acceleration is given by

2

2

50.0 N 1.50 m0.817 rad s

91.8 kg m

F R

I I

At t = 3.00 s, the angular velocity is

20 0.817 rad s 3.00 s 2 .45 rad si t

and the kinetic energy is

22 2

rot

1 191.8 kg m 2.45 rad s 276 J

2 2 KE I

8.44 (a) Hoop: 22 24.80 kg 0.230 m 0.254 kg m I MR

Solid Cylinder: 22 2

1 14.80 kg 0.230 m 0.127 kg m

2 2 I MR

Solid Sphere: 22 2

2 24.80 kg 0.230 m 0.102 kg m

5 5 I MR

Thin, Spherical, Shell: 22 2

2 24.80 kg 0.230 m 0.169 kg m

3 3 I MR

(b) When different objects of mass M and radius R roll

without slipping a R down a ramp, the one with the largest

translational acceleration a will have the highest

translational speed at the bottom. To determine the

translational acceleration for the various objects,

consider the free-body diagram at the right:



Chapter 8

Page 8.78

sin x F Ma Mg f Ma [1]

2 or I f R I a R f Ia R [2]

Substitute Equation [2] into [1] to obtain

2sin Mg Ia R Ma or2

sin Mg a

M I R

Since M , R, g are the same for all of the objects, we see that the translational acceleration (and hence the

translational speed) increases as the moment of inertia decreases. Thus, the proper rankings from highest to

lowest by translational speed will be:

Solid sphere; solid cylinder; thin, spherical, shell; and hoop

(c) When an object rolls down the ramp without slipping, the friction force does no work and mechanical energy

is conserved. Then, the total kinetic energy gained equals the gravitational potential energy given up:

r t g KE KE PE Mgh and 1 2

2r KE Mgh M v , where h is the vertical drop of the ramp and

is the translational speed at the bottom. Since M , g , and h are the same for all of the objects, the rotational

kinetic energy decreases as the translational speed increases. Using this fact, along with the result of Part (b),

we rank the object’s final rotational kinetic energies, from highest to lowest, as :

hoop; thin, spherical, shell; solid cylinder; and solid sphere

8.45 (a) Treating the particles on the ends of the rod as point masses, the total moment of inertia of the rotating

system is 2 2 2rod 3 4 rod 3 412 2 2( ) ( ) I I I I m L m L m L . If the mass of the rod can be ignored,

this reduces to

2

2 23 40 3.00 kg 4.00 kg 0.500 m 1.75 kg m

2

L I m m

and the rotational kinetic energy is

22 2

1 11.75 kg m 2.50 rad s 5.47 J

2 2r

KE I

(b) If the rod has mass rod 2.00 kgm

2 2 2

12.00 kg 1.00 m 1.75 kg m 1.92 kg m

12 I



Chapter 8

Page 8.79

and

22 2

1 11.92 kg m 2.50 rad s 6.00 J

2 2r

KE I

8.46 Using conservation of mechanical energy,

trans rot trans rot g g f i

KE KE PE KE KE PE

or

2 21 1

0 0 0 sin2 2

t M I M g L v

Since 22 5 I M R for a solid sphere and t R v when rolling without slipping, this becomes

2 2 2 21 1

sin2 5

M R M R M g L

and reduces to

2

22

10 9.8 m s 6.0 m sin 3710 sin36 rad s

7 7 0.20 m

gL

R

8.47 (a) Assuming the disk rolls without slipping, the friction force between the disk and the ramp does no work. In

this case, the total mechanical energy of the disk is constant with the value

0 sin( )i g i E KE PE Mgh MgL . When the disk gets to the bottom of the ramp, 0 g PE and

sin f t r KE KE KE E MgL . Also, since the disk does not slip, R v and

2

2 2 21 1 1 1 1 1

2 2 2 2 2 2r t

KE I MR M KE R

v

v

Then,

total

1sin

2t t KE KE KE E MgL or

3 1

2 2 M 2 M

v sin gL



Chapter 8

Page 8.81

8.49 Using 1 22

0net f i f W KE KE I , we have

4 2

2 5.57 N 0.800 m2 2149 rad s

4.00 10 kg m

net f

W F s

I I

8.50 The work done on the grindstone is net W F s F r F r .

Thus, 1 12 22 2net f iW I I , or

222 rad 1

25.0 N m 15.0 rev 0.130 kg m 0

1 rev 2

f

This yields

rad 1 rev190 30.3 rev s

s 2 rad f

8.51 (a) 22

1 110.0 kg 10.0 m s 500 J

2 2trans t KE mv

(b)

22 2

2

22

1 1 1

2 2 2

1 110.0 kg 10.0 m s 250 J

4 4

t

rot

t

v KE I m R

R

mv

(c) 750 Jtotal trans rot

KE KE KE

8.52 As the bucket drops, it loses gravitational potential energy. The spool

gains rotational kinetic energy and the bucket gains translational kinetic

energy. Since the string does not slip on the spool, r where r is the



Chapter 8

Page 8.82

radius of the spool. The moment of inertia of the spool is 1 2

2 I Mr ,

where M is the mass of the spool. Conservation of energy gives

t r g t r g f i

KE KE PE KE KE PE

2 21 1

0 02 2

f im I mgy mgy

or

22 2

1 1 1

2 2 2 i f m r Mr mg y y

This gives

2

21 2 12 2

2 2 3.00 kg 9.80 m s 4.00 m10.9 rad s

3.00 kg+ 5.00 kg 0.600 m

i f mg y y

m M r

8.53 (a) The arm consists of a uniform rod of 10.0 m length and the mass of the seats at the lower end is negligible.

The center of gravity of this system is then located at the

geometric center of the arm, located 5.00 m from the upper end.

From the sketch at the right, the height of the center of gravity above the zero level is

10.0 m 5.00 m coscg y .

(b) When 45.0 , 10.0 m 5.00 m cos 45.0 6.46 mcg y

and

2 4365 kg 9.80 m s 6.46 m 2.31 10 J g cg PE mgy

(c) In the vertical orientation, = 0 and cos = 1, giving

10.0 m 5.00 m 5.00 mcg y . Then,

2 4365 kg 9.80 m s 5.00 m 1.79 10 J g cg PE mgy



Chapter 8

Page 8.83

(d) Using conservation of mechanical energy as the arm starts

from rest in the 45° orientation and rotates about the upper

end to the vertical orientation gives

21

02

end f cg cg f i

I mg y mg y or 2 cg cg i f

f end

mg y y

I

[1]

For a long, thin rod: 2 3end

I mL . Equation [1] then becomes

2

f

m

cg cg

i f g y y

m

22

2

2

6

3

6 9.80 m s 6.46 m 5.00 m 0.927 rad s

10.0 m

cg cg i f

g y y

L L

Then, from r , the translational speed of the seats at the lower end of the rod is

10.0 m 0.927 rad s 9.27 m s

8.54 (a)

22 2

2.40 kg 0.180 m 35.0 rad s 2.72 kg m s L I MR

(b) 22 2

1 12.40 kg 0.180 m 35.0 rad s 1.36 kg m s

2 2 L I MR

(c) 22 2

2 22.40 kg 0.180 m 35.0 rad s 1.09 kg m s

5 5 L I MR

(d) 22 22 2

2.40 kg 0.180 m 35.0 rad s 1.81 kg m s3 3

L I MR

8.55 (a) The rotational speed of Earth is



Chapter 8

Page 8.84

5

4

2 rad 1 d7.27 10 rad s

1 d 8.64 10 s E

2

224 6 5 33

2

52

5.98 10 kg 6.38 10 m 7.27 10 rad s 7.08 10 J s5

spin spher e E E E E L I M R

(b) For Earth’s orbital motion,

77

2 rad 1 y1.99 10 rad s

1 y 3.156 10 s

orbit

and using data from Table 7.3, we find

2

224 11 7 405.98 10 kg 1.496 10 m 1.99 10 rad s 2.67 10 J s

orbit point orbit E orbit orbit L I M R

8.56 (a) Yes, the bullet has angular momentum about an axis through the

hinges of the door before the collision. Consider the sketch at

the right, showing the bullet the instant before it hits the door. The

physical situation is identical to that of a point mass m g moving in

a circular path of radius r with tangential speed t = i. For that

situation the angular momentum is

2 i

i i i B B i L I m r m r

r

and this is also the angular momentum of the bullet about the axis

through the hinge at the instant just before impact.

(b) No, mechanical energy is not conserved in the collision. The bullet embeds itself in the door with the two

moving as a unit after impact. This is a perfectly inelastic collision in which a significant amount of mechanic

cal energy is converted to other forms, notably thermal energy.



Chapter 8

Page 8.85

(c) Apply conservation of angular momentum with i B i L m r as discussed in part (a). After impact,

1 2 2

2 f f f door bullet f door B f L I I I M L m r where L = 1.00 m = the width of the door

and r = L – 10.0 cm = 0.900 m. Then,

3

2 22 2

0.005 kg 0.900 m 1.00 10 m s

1 118.0 kg 1.00 m 0.005 kg 0.900 m

3 3

B i f i f

door B

m r L L

M L m r

yielding 0.749 rad s f .

(d) The kinetic energy of the door-bullet system immediately after impact is

2 2 22

1 1 118.0 kg 1.00 m 0.005 kg 0.900 m 0.749 rad s

2 2 3 f f f KE I

or 1.68 J f KE .

The kinetic energy (of the bullet) just before impact was

2

2 3 31 1

0.005 kg 1.00 10 m s 2.50 10 J2 2

i B i KE m

8.57 Each mass moves in a circular path of radius r = 0.500 m/s about the center of the connecting rod. Their angular

speed is

5.00 m s10.0 m s

0.500 mr

Neglecting the moment of inertia of the light connecting rod, the angular momentum of this rotating system is

22 2

1 2 4.00 kg 3.00 kg 0.500 m 10.0 rad s 17.5 J s L I m r m r

8.58 Using conservation of angular momentum, aphelion perihelion L L .

Thus, 2 2a a p pmr mr . Since = 1/r at both aphelion and perihelion, this is equivalent to

2 2a a a p p pmr r mr r , giving



Chapter 8

Page 8.86

0.59 A.U.

54 km s 0.91 km s35 A.U.

p

a p

a

r

r

8.59 The initial moment of inertia of the system is

22 24 1.0 m 4.0 mi i i I m r M M

The moment of inertia of the system after the spokes are shortened is

22 24 0.50 m 1.0 m f f f I m r M M

From conservation of angular momentum, I f f = I i i, or

4 2.0 rev s 8.0 rev si

f i f

I

I

8.60 From conservation of angular momentum: - - - -child m g r f child m g r i

f i I I I I

where 2- - 275 kg mm g r I is the constant moment of inertia of the merry-go-round.

Treating the child as a point object, 2child

I mr where r is the distance the child is from the rotation axis.

Conservation of angular momentum then gives

22 2- -

22 2- -

25.0 kg 1.00 m 275 kg m14.0 rev min

25.0 kg 2.00 m 275 kg m

i m g r

f i f m g r

mr I

mr I

or

rev 2 rad 1 min11.2 1.17 rad s

min 1 rev 60.0 s f

8.61 The moment of inertia of the cylinder before the putty arrives is

22 2

1 110.0 kg 1.00 m 5.00 kg m

2 2i I M R



Chapter 8

Page 8.87

After the putty sticks to the cylinder, the moment of inertia is

22 2 25.00 kg m 0.250 kg 0.900 m 5.20 kg m f i I I mr

Conservation of angular momentum gives I f f = I i i, or

2

2

5.00 kg m7.00 rad s 6.73 rad s

5.20 kg m

i f i

f

I

I

8.62 The total moment of inertia of the system is

2 22 3.0 kg mtotal masses student

I I I mr

Initially, r = 1.0 m, and 2 2 22 3.0 kg 1.0 m 3.0 kg m 9.0 kg mi I .

Afterward, r = 0.30 m, so

2 2 22 3.0 kg 0.30 m 3.0 kg m 3.5 kg m f I

(a) From conservation of angular momentum, I f f = I i i, or

2

2

9.0 kg m0.75 rad s 1.9 rad s3.5 kg m

i f i f

I

I

(b) 22 2

1 19.0 kg m 0.75 rad s 2.5 J

2 2i i i KE I

22 2

1 13.5 kg m 1.9 rad s 6.3 J

2 2 f f f KE I

8.63 The initial angular velocity of the puck is

0.800 m s rad2.00

0.400 m s

t i

i

ir

Since the tension in the string does not exert a torque about the axis of revolution, the angular



Chapter 8

Page 8.88

momentum of the puck is conserved, or I f f = I i i.

Thus,

22

2

0.400 m2 .00 rad s 5.12 rad s

0.250 mi i

f i i f f

I mr I mr

The net work done on the puck is

2 2 2 2 22 2 2 2 21 1 1

2 2 2 2net f i f f i i f f i i f f i i

mW KE KE I I mr mr r r

or

2 2 2 20.120 kg0.250 m 5.12 rad s 0.400 m 2.00 rad s

2net W

This yields 25.99 10 Jnet

W .

8.64 For one of the crew, c c F m a becomes 2 2/t in m v r mr . We require n = mg , so the initial angular

velocity must be /i g r . From conservation of angular momentum, f f i i I I or ( / ) f i f i I I . Thus,

the angular velocity of the station during the union meeting is

28 2

28 2

5.00 10 kg m 150 65.0 kg 100 m1.12

5.00 10 kg m 50 65.0 kg 100 m

i f

f

I g g g

I r r r

The centripetal acceleration experienced by the managers still on the rim is

2 22 2 2 212.3 m s 1.12 1.12 9.80 m s 12.3 m sc f

g a r r

r

8.65 (a) From conservation of angular momentum, I f f = I i i, so

1

1 2

i f i o

f

I I

I I I



Chapter 8

Page 8.89

(b) 2

1 1 12 2 21 2 1

1 2 1 2 1 2

1 1 1

2 2 2 f f f o o i

I I I KE I I I I KE

I I I I I I

or

1

1 2

f

i

KE I

KE I I

Since this is less than 1.0, kinetic energy was lost.

8.66 The initial angular velocity of the system is

rev 2 rad0.20 0.40 rad s

s 1 revi

The total moment of inertia is given by

22 2 2

1 180 kg 25 kg 2 .0 m

2 2man cyli nder I I I mr M R r

Initially, the man is at 2.0 mr from the axis, and this gives 2 23.7 10 kg mi

I . At the end, when r = 1.0,

the moment of inertia is 2 21.3 10 kg m f I .

(a) From conservation of angular momentum, I f f = I i i, or

2 2

2 2

3.7 10 kg m0.40 rad s 1.14 rad s 3.6 rad s

1.3 10 kg m

i f i

f

I

I

(b) The change in kinetic energy is1 12 22 2 f f f i KE I I , or

2 2

2 2 2 21 rad 1 rad

1.3 10 kg m 1.14 3.7 10 kg m 0.402 s 2 s

KE

or 25.4 10 J KE . The difference is the work done by the man as he walks inward.



Chapter 8

Page 8.90

8.67 (a) The table turns counterclockwise, opposite to the way the woman walks. Its angular momentum cancels that

of the woman so the total angular momentum maintains a constant value of 0total woman table

L L L .

Since the final angular momentum is 0total w w t t

L I I , we have

2w w w w

t w w

t t t

I m r m r

I I r I

or

2

60.0 kg 2 .00 m1.50 m s 0.360 rad s

500 kg mt

Hence, 0.360 rad s counterclockwisetable .

(b) 2 21 1

02 2

net f w t t W KE KE m I

2 22

1 160.0 kg 1.50 m s 500 kg m 0.360 rad s 99.9 J

2 2net

W

8.68 (a) In the sketch at the right, choose an axis perpendicular to

the page and passing through the indicated pivot. Then,

with = 30.0, the lever arm of the force P is observed to be

5.00 cm 5.00 cm5.77 cm

cos cos 30.0

and = 0 gives

5.77 cm 150 N 30.0 cm 0 P

so

150 N 30.0 cm780 N

5.77 cm P



Chapter 8

Page 8.91

(b) 0 cos 30.0 0 y F n P , giving

cos 30.0 780 N cos30.0 675 Nn P

0 sin 30.0 0 x F f F P , or

sin 30.0 780 N sin 30.0 150 N 240 N f P F

The resultant force exerted on the hammer at the pivot is

2 2

2 2 240 N 675 N 716 N R f n

at 1 1tan ( / ) tan (675 N/240 N) 70.4n f , or

716 N at 70.4 above the horizontal to the right R

8.69 (a) Since no horizontal force acts on the child-boat system, the center of gravity of this system will remain sta-

tionary, or

cg constantchild child boat boat

child boat

m x m x x

m m

The masses do not change, so this result becomes mchild xchild + m boat x boat = constant.

Thus, as the child walks to the right, the boat will move to the left .

(b) Measuring distances from the stationary pier, with away from the pier being positive, the child is initially at

( xchild)i = 3.00 m and the center of gravity of the boat is at ( x boat)i = 5.00. At the end, the child is at the right

end of the boat, so ( xchild) f = ( x boat) f + 2.00 m. Since the center of gravity of the system does not move,

we have child child boat boat child child boat boat f im x m x m x m x , or

2.00 m 3.00 m 5.00 mchild child boat child child boat f f m x m x m m

and



Chapter 8

Page 8.92

3.00 m 5.00 m 2.00 mchild boat

child f child boat

m m x

m m

40.0 kg 3.00 m 70.0 kg 5.00 m 2.00 m

5.55 m40.0 kg 70.0 kg

child f x

(c) When the child arrives at the right end of the boat, the greatest distance from the pier that he can reach is

max1.00 m 5.55 m 1.00 m 6.55 mchild f

x x This leaves him 0.45 m short of reaching the

turtle .

8.70 (a) Consider the free-body diagram of the block given at

the right. If the + x-axis is directed down the incline,

F x = ma x gives

sin37.0 t mg T m a , or sin 37.0 t T m g a

2 212.0 kg 9.80 m s sin 37.0 2.00 m s

46.8 N

T

(b) Now, consider the free-body diagram of the pulley.

Choose an axis perpendicular to the page and passing

through the center of the pulley, = I

gives

t a

T r I r

or

2

22

2

46.8 N 0.100 m0.234 kg m

2.00 m st

T r I

a

(c) 22.00 m s

0 2.00 s 40.0 rad s0.100 m

t

i

at t

r

8.71 If the ladder is on the verge of slipping, max s s f f n at both the



Chapter 8

Page 8.93

floor and the wall.

From F x = 0, we find f 1 – n2 = 0, or

n2 = sn1 [1]

Also, F x = 0 gives n1 – w + sn2.= 0

Using Equation [1], this becomes

1 1 0 s sn w n

or

1 2 0.800

1 1.25 s

w w

n w

[2]

Thus, Equation [1] gives

2 0.500 0.800 0.400n w w [3]

Choose an axis perpendicular to the page and passing through the lower end of the ladder. Then, = 0 yields

2 2cos sin cos 02

L

w n L f L

Making the substitutions n2 = 0.400 w and2 2

0.200 s f n w , this becomes

cos 0.400 sin 0.200 cos 02

Lw w L w L

and reduces to

0.500 0.200sin cos

0.400

Hence, tan = 0.750 and = 36.9 .

8.72



Chapter 8

Page 8.94

We treat each astronaut as a point object, m, moving at speed v in a circle of radius r = d /2. Then the total angular

momentum is

21 2 2 2 L I I mr m r r

(a) 2 2 75.0 kg 5.00 m s 5.00 mi i i L m r

3 23.75 10 kg m si L

(b) 2 2 21 1 2 2

1 1 12

2 2 2i i i i KE m m m

2 375.0 kg 5.00 m s 1.88 10 J= 1.88 kJi KE

(c) Angular momentum is conserved: 3 23.75 10 kg m s f i L L .

(d)

3 23.75 10 kg m s10.0 m s

2 75.0 kg 2.50 m2

f

f

f

L

mr

(e) 22

12 75.0 kg 10.0 m s 7.50 kJ

2 f f KE m

(f) 5.62 kJnet f iW KE KE

8.73 (a) 22

i

d L M M d

(b) 2 21

22

i i KE M M

(c) f i L L M d



Chapter 8

Page 8.95

(d)

22 42

f

f

f

L M d

M d Mr

(e) 2

2 2

1

2 2 42

f f KE M M M

(f) 23net f iW KE KE M

8.74 Choose an axis that is perpendicular to the page

and passing through the left end of the scaffold.

Then = 0gives

750 N 1.00 m 345 N 1.50 m

500 N 2 .00 m 1000 N 2 .50 m

3.00 m 0 RT

or

31.59 10 N= 1.59 kN RT

Then,

30 750 345 500 1000 N 1.59 10 N 1.01 kN y L F T

8.75 (a) From conservation of angular momentum, f f f i i i L I I L , or

2 22 2 9

5

2 32

5

1.50 10 m0.010 0 rev d

15.0 10 m

ii i f i i i

f f f

MR I R

I R MR

giving

8rev

1.00 10 f d

2 rad

1 rev

1 d

34

7.27 10 rad s8.64 10 s

(b) 3 3 815.0 10 m 7.27 10 rad s 1.09 10 m st f f f R (which is about one-third the



Chapter 8

Page 8.96

speed of light).

8.76 (a) Taking PE g = 0 at the level of the horizontal axis passing

through the center of the rod, the total energy of the rod

in the vertical position is

1 2 1 20

g E KE PE

m g L m g L m m gL

(b) In the rotated position of Figure P8.76b, the rod is

in motion and the total energy is

21 1 2 2

1

2r g total E KE PE I m gy m gy Figure P8.76

2 2 21 2 1 2

1sin sin

2m L m L m g L m g L

or

2 21 2

1 2 sin

2

m m L E m m gL

(c) In the absence of any nonconservative forces that do work on the rotating system, the total mechanical energy

of the system is constant. Thus, the results of parts (a) and (b) may be equated to yield an equation that can be

solved for the angular speed, , of the system as a function of angle .

(d) In the vertical position, the net torque acting on the system is zero, net = 0. This is because the lines of action

of both external gravitational forces (m1 g and m2 g )pass through the pivot and hence have zero lever arms

about the rotation axis. In the rotated position, the net torque (taking clockwise as positive) is

1 2 1 2cos cos cosnet m g L m g L m m gL

Note that the net torque is not constant as the system rotates. Thus, the angular acceleration of the rotating

system, given by = net/ I , will vary as a function of . Since a net torque of varying magnitude acts on the

system, the angular momentum of the system will change at a nonuniform rate.

(e) In the rotated position, the angular acceleration is



Chapter 8

Page 8.97

1 2 1 2

2 21 2 1 2

cos cosnet

m m gL m m g

I m L m L m m L

8.77 Let m p be the mass of the pulley, m1 be the mass of the sliding block, and m2 be the mass of the counterweight.

(a) The moment of inertia of the pulley is 21

2 p p I m R and its angular velocity at any time is, = /R p where

is the linear speed of the other objects. The friction force retarding the sliding block is

1k k k f n m g

Choose 0 g PE at the level of the counterweight when the sliding object reaches the second photogate.

Then, from the work – energy theorem,

nc trans rot g trans rot g f i

W KE KE PE KE KE PE

2

2 21 2 2

2

2 21 2 22

1 1 10

2 2 2

1 1 1

2 2 2

f

k f p p p

ii p p

p

f s m m m R R

m m m R m gs R

or

2 21 2 1 2 2 1

1 1 1 1

2 2 2 2 p f p i k m m m m m m m gs m g s

This reduces to

2 12

1 2

2

1

2

k

f i

p

m m gs

m m m

and yields

2 22 0.208 kg 9.80 m s 0.700 mm0.820 1.63 m s

s 1.45 kg f



Chapter 8

Page 8.98

(b)1.63 m s

54.2 rad s0.030 0 m

f

f p

v

R

8.78 (a) The frame and the center of each wheel moves forward at = 3.35 m/s and each wheel also turns at angular

speed = / R. The total kinetic energy of the bicycle is KE = KE t + KE r , or

2 2

2

2 2

2

1 12 2

2 2

1 12

2 2

frame wheel wheel

frame wheel wheel

KE m m I

m m m R R

This yields

2

2

132

18.44 kg 3 0.820 kg 3.35 m s 61.2 J

2

frame wheel KE m m v

(b) Since the block does not slip on the roller, its forward speed must

equal that of point A, the uppermost point on the rim of the roller.

That is, AE v where

AEv is the velocity of A relative to Earth.

Since the roller does not slip on the ground, the velocity of point O

(the roller center) must have the same magnitude as the tangential

speed of point B (the point on the roller rim in contact with the

ground). That is, OE O R v . Also, note that the velocity of point A relative

to the roller center has a magnitude equal to the tangential speed R , orAO O

R v .

From the discussion of relative velocities in Chapter 3, we know thatAE AO OE

v v v . Since all of these

velocities are in the same direction, we may add their magnitudes gettingAE AO OE

v v v , or

2 2O O O

R .

The total kinetic energy is KE = KE t + KE r , or



Chapter 8

Page 8.99

2

2 2

2

2 2

2

1 1 12 2

2 2 2 2

1 1 1

2 4 2 4

stone tree tree

ston e tree tree

KE m m I

m m m R R

This gives 21 3

2 4 stone tree KE m m

, or

21 3

844 kg 82.0 kg 0.335 m s 50.8 J2 4

KE

8.79 We neglect the weight of the board and assume that

the woman’s feet are directly above the point of support

by the rightmost scale. Then, the free-body diagram

for the situation is as shown at the right.

From 0 y F , we have 1 2 0 g g F F w , or 380 N 320 N=700 Nw .

Choose an axis perpendicular to the page and passing through point P.

Then 0 gives 12 .00 m 0 g w x F or

1 2.00 m 380 N 2.00 m1.09 m

700 N

g F x

w

8.80 Choose PE g = 0 at the level of the base of the ramp. Then, conservation of mechanical energy gives

trans rot g trans rot g f i

KE KE PE KE KE PE

2

2 21 1

0 0 sin 02 2

iimg s m mR

R

or

2 22 2 2

2

3.0 m 3.0 rad s24 m

sin sin 9.80 m s sin 20

i i R s

g g



Chapter 8

Page 8.100

8.81 Choose an axis perpendicular to the page and passing

through the center of the cylinder. Then, applying = I

to the cylinder gives

2 21 122 2

t aT R M R M R R

or 1

4 t

T M a [1]

Now F y = ma y apply to the falling objects to obtain

2 2 2 t m g T m a or t

T a g

m [2]

(a) Substituting Equation [2] into [1] yields

4 4

M g M T T

m

which reduces to4

M mg T

M m

(b) From Equation [2] above,

1 4

4 4 4t

M mg M g mg a g g

m M m M m M m

8.82 (a) A smooth (that is, frictionless) wall cannot exert a force parallel to its surface. Thus, the only force the verti-

cal wall can exert on the upper end of the ladder is a horizontal normal force.

(b) Consider the free-body diagram of the ladder given at the

right. If the rotation axis is perpendicular to the page and

passing through the lower end of the ladder, the lever

arm of the normal force2n that the wall exerts on the

upper end of the ladder is

d2 = L sin



Chapter 8

Page 8.101

(c) The lever arm of the force of gravity, m g , acting on the ladder is

2 cos cos 2d L L

(d) Refer to the free-body diagram given in part (b) of this solution and make use of the fact that the ladder is in

both translational and rotational equilibrium.

10 0 y p F n m g m g or 1 p

n m m g

When the ladder is on the verge of slipping, 1 1 1max s s p f f n m m g .

Then 2 10 x F n f , or 2 s p

n m m g .

Finally, 20 sin ( /2) cos cos 0 pn L m g L m gx where x is the maximum distance

the painter can go up the ladder before it will start to slip. Solving for x gives

2 sin cos

21 tan

cos 2 s

p p p

Ln L m g

m m x L L

m g m m

and using the given numerical data, we find

30 kg 30 kg

0.45 1 4.0 m tan 53 4.0 m 2.5 m80 kg 2 80 kg

x

8.83 The large mass (m1 = 60.0 kg) moves in

a circular path of radius r 1 = 0.140 m, while the

radius of the path for the small mass (m2 = 0.120 kg) is

2 1

3.00 m 0.140 m 2.86 m

r r

The system has maximum angular speed

when the rod is in the vertical position as

shown at the right.

We take PE g = 0 at the level of the horizontal



Chapter 8

Page 8.102

rotation axis and use conservation of energy to find:

2 21 max 2 max 2 2 1 1

1 1 0 0

2 2 f g i g

f i KE PE KE PE I I m gr m gr

Approximating the two objects as point masses, we have2 2

1 1 1 2 2 2and I m r I m r . The energy conservation

equation then becomes 1 2 2 21 1 2 2 max 1 1 2 22

m r m r m r m r g and yields

2

1 1 2 2max 2 22 2

1 1 2 2

2 60.0 kg 0.140 m 0.120 kg 2.86 m 9.80 m s2

60.0 kg 0.140 m 0.120 kg 2.86 m

m r m r g

m r m r

or max8.56 rad s . The maximum linear speed of the small mass object is then

2 2 maxmax2.86 m 8.56 rad s 24.5 m sr

8.84 (a) Note that the cylinder has both translational and rotational

motion. The center of gravity accelerates downward while

the cylinder rotates around the center of gravity. Thus, we

apply both the translational and the rotational forms of

Newton’s second law to the cylinder:

y y F ma T mg m a

or

T m g a [1]

I Tr I a r

For a uniform, solid cylinder, 21

2 I mr so our last result becomes

2

2

mr a

Tr r

or

2T

a m [2]

Substituting Equation [2] into Equation [1] gives T = mg – 2T , and solving for T yields T = mg /3.

(b) From Equation [2] above,



Chapter 8

Page 8.103

2 22 3

3

T mg a g

m m

(c) Considering the translational motion of the center of gravity, 2 20

2 y y y

a y gives

2

0 2 4 33

y

g h gh

Using conservation of energy with PE g at the final level of the cylinder gives

t r g t r g f i

KE KE PE KE KE PE or1 12 2

2 20 0 0 ym I mgh

Since1 2

2and y r I mr , this becomes 2 2

1 1 1

2 2 2 ym m r

2

2

y

r

mgh

, or23

4

ym mgh

yielding 4 3 y gh .

8.85 Considering the shoulder joint as the pivot, the second

condition of equilibrium gives

0 70 cm sin 45 4.0 cm 02

m

w F

or

70 cm12.4

2 4.0 cm sin 4 5m

w F w

Recall that this is the total force exerted on the arm by a set of two muscles. If we approximate that the two muscles

of this pair exert equal magnitude forces, the force exerted by each muscle is



Chapter 8

Page 8.104

312.4

6.2 6.2 750 N 4.6 10 N 4.6 kN2 2

m

each

muscle

F w F w

8.86 Observe that since the torque opposing the rotational motion of the gymnast is constant, the work done by

nonconservative forces as the gymnast goes from position 1 to position 2 (an angular displacement of /2 rad) will

be the same as that done while the gymnast goes from position 2 to position 3 (another angular displacement of /2

rad).

Choose PE g = 0 at the level of the bar, and let the distance from the bar to the center of gravity of the outstretched

body be r eg. Applying the work – energy theorem, nc g g f i

W KE PE KE PE , to the rotation from

position 1 to position 2 gives

1 22 cg212

0 0ncW I mgr or 1 22 cg212ncW I mgr [1]

Now, apply the work – energy theorem to the rotation from position 2 to position 3 to obtain

1 12 23 cg 22 223

0ncW I mg r I or 1 12 2

3 2 cg2 223ncW I I mgr [2]

Since the frictional torque is constant and these two segments of the motion involve equal angular

displacements, 23 12nc nc

W W . Thus, equating Equation [2] to Equation [1] gives

1 12 23 2 cg

2 2

I I mgr 1 2

2 cg2

I mgr

which yields 2 23 2

2 , or 3 22 2 4.0 rad s 5.7 rad s .

8.87 (a) Free-body diagrams for each block and the

pulley are given at the right. Observe that

the angular acceleration of the pulley will be

clockwise in direction and has been given a

negative sign. Since I , the positive

sense for torques and angular acceleration

must be the same (counterclockwise).

For m1: 1 1 1 y y F ma T m g m a



Chapter 8

Page 8.105

1 1T m g a [1]

For m2:

2 2 x x F ma T m a [2]

For the pulley: 2 1 I T r T r I a r or

1 2 2

I T T a

r

[3]

Substitute Equations [1] and [2] into Equation [3] and solve for a to obtain

1

21 2

m g a I r m m

or

2

222

4.00 kg 9.80 m s3.12 m s

0.500 kg m 0.300 m 4.00 kg 3.00 kga

(b) Equation [1] above gives: 2 21 4.00 kg 9.80 m s 3.12 m s 26.7 NT ,

and Equation [2] yields: 22 3.00 kg 3.12 m s 9.37 NT .

8.88 (a)

(b) 0 120 N 0 y F monkey F n m g

2120 N 10.0 kg 9.80 m s 218 N

F

n

(c) When x = 2 L/3, we consider the bottom end of the ladder as our pivot and obtain

0 120 Nbottomend

L

2cos 60.0 98.0 N

2

L

Wcos60.0

3n L

sin 60.0 0

or



Chapter 8

W

60.0 N 196 3 N cos 60.072.4 N

sin 60.0n

Then,

W0 0

x F T n or

W 72.4 NT n

(d) When the rope is ready to break, W80.0 NT n . Then 0bottom

end

yields

120 N cos 60.0 98.0 N cos 60.0 80.0 N sin 60.0 02

L x L

or

80.0 N sin 60.0 60.0 N cos 60.00.802 0.802 3.00 m 2.41 m

98.0 N cos60.0

L

x L

(e) If the horizontal surface were rough and the rope removed, a horizontal static friction force directed toward

the wall would act on the bottom end of the ladder. Otherwise, the analysis would be much as what is done

above. The maximum distance the monkey could climb would correspond to the condition that the friction

force have its maximum value, F sn , so you would need to know the coefficient of static friction to solve

part (d).

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