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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Chapter 8
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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
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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.
For Instructors Only
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
6. None of the above
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Chapter 8
Page 8.5
7. Cannot be determined
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 .)
For Instructors Only
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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Chapter 8
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
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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.
For Instructors Only
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.
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Chapter 8
Page 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
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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.
For Instructors Only
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:
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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
5. None of the above
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
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Chapter 8
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4. Roll to the left
5. None of the above
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
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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
5. None of the above
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
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Chapter 8
Page 8.15
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.
For Instructors Only
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.
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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.
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Chapter 8
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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!
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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.
For Instructors Only
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?
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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
5. None of the above
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.
For Instructors Only
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.
6. 2 and 3 are true.
7. 1 and 4 are true.
8. 2 and 4 are true.
9. 3 and 4 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.
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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.
For Instructors Only
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.)
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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.
For Instructors Only
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
5. None of the above
6. Cannot be determined
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.
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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.
For Instructors Only
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.
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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
5. None of the above
6. Cannot be determined
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.
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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.
For Instructors Only
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.
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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.
For Instructors Only
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.
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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)
6. Cannot be determined
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
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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.
For Instructors Only
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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
4. Cannot be determined
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
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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.
For Instructors Only
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?
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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.
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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.”
For Instructors Only
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?
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1. Path (1)
2. Path (2)
3. Path (3)
4. Path (4)
5. Path (5)
6. None of the above
7. Cannot be determined
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.
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For Instructors Only
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.
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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.
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.”)
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:
• The angular momentum of a system is conserved whenever no net torque is exerted upon it.
• 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.
For Instructors Only
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
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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.
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. 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:
• The angular momentum of a system is conserved whenever no net torque is exerted upon it.
• 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.
For Instructors Only
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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
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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.
For Instructors Only
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.
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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.
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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).
For Instructors Only
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?
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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
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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.
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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:
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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
cgarms torso thighs 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
cgarms torso thighs 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
.
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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.
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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
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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 .
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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,
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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
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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)
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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
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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
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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
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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
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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]
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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
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Chapter 8
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(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.
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Chapter 8
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(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
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Chapter 8
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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
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Chapter 8
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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:
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Chapter 8
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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
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Chapter 8
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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
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Chapter 8
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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
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Chapter 8
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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
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Chapter 8
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(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
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Chapter 8
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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.
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Chapter 8
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(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
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Chapter 8
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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
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Chapter 8
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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
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Chapter 8
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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
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Chapter 8
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(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.
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Chapter 8
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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
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Chapter 8
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(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
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Chapter 8
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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
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Chapter 8
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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
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Chapter 8
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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
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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
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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
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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
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Chapter 8
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(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
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Chapter 8
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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
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Chapter 8
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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
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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
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Chapter 8
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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,
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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
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Chapter 8
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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
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Chapter 8
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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
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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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