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Rotational Motion
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Looking Ahead
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Looking Back
The goal of Chapter 7 is to understand the physics of rotating objects. In this
chapter you will learn to:
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Understand what a rigid body is.
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Calculate torque from two perspectives.
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Determine the center of gravity of an object, and calculate the
gravitational torque.
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Understand how torque affects rotational motion, and the role of the
moment of inertia in rotational motion.
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Apply concepts of torque and moment of inertia to the rotation of an
object about a fixed axis.
Rotational motion will revisit many of the major themes introduced in
previous chapters, especially the properties of circular motion. Please review:
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Section 4.6 Newton’s second law.
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Section 6.2 Describing angular motion.
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Section 6.3 Velocity and acceleration in circular motion.
Photo suggestion: Starting an old-fashioned airplane propeller
by pulling on it.
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Not all motion can be described as that of a particle. Rotation requires the idea of an
extended object.
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To get any object rotating, such as this airplane propeller, you have apply a
force away from the axis of rotation. If the pilot were to pull down right at
the propeller’s shaft, no rotation would occur at all. The application of a
force at a distance from an object’s rotation axis produces a torque that leads
to rotational motion of the object.
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7.1 Torque
Our goal in this chapter is to understand rotational motion. We will focus our
attention on what are called rigid bodies. Wheels, axles, and gyroscopes are
examples of rigid bodies that rotate. Divers, gymnasts, and ice skaters also
rotate, although the fact that they are not rigid bodies makes their motions
more complex. Even so, we will be able to understand many aspects of their
motion by modeling them as rigid bodies.
You will quickly discover that the physics of rotational motion is analogous
to the physics of linear motion that you studied in earlier chapters. For
example, the new concepts of torque and angular acceleration are the
rotational analogs of force and acceleration, and we’ll find a new version of
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Newton’s second law that is the rotational equivalent of F  ma
Thus far, our study of physics has focused almost exclusively on the particle
model in which an object is represented as a mass at a single point in space.
The particle model is a perfectly good description of the physics in a vast
number of situations, but there are other situations for which we need to
consider the motion of an extended object—a system of particles for which
the size and shape do make a difference and cannot be neglected.
A rigid body is an extended object whose size and shape do not change as it
moves. For example, a bicycle wheel can be thought of as a rigid body.
Figure 7.1 shows a rigid body as a collection of atoms held together by the
rigid “massless rods” of molecular bonds.
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Figure 7.1 The rigid-body model of an extended object.
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Real molecular bonds are, of course, not perfectly rigid. That’s why an object
seemingly as rigid as a bicycle wheel can flex and bend. Thus Figure 7.1 is
really a simplified model of an extended object, the rigid-body model. The
rigid-body model is a very good approximation of many real objects of
practical interest, such as wheels and axles. Even nonrigid objects can often
be modeled as a rigid body during parts of their motion. For example, a diver
is well described as a rotating rigid body while she’s in the tuck position.
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Torque
Newton’s genius, summarized in his second law of motion, was to recognize
force as the cause of acceleration. But what about angular acceleration?
What do Newton’s laws have to tell us about rotational motion? To begin our
study of rotational motion, we’ll need to find a rotational equivalent of force.
Consider the common experience of pushing open a door. Figure 7.2 is a top
view of a door that is hinged on the left. Four pushing forces are shown, all
of equal strength. Which of these will be most effective at opening the door?
Figure 7.2 The four forces are the same strength, but they have different effects on
the swinging door.
Force F1 will open the door, but force F2 , which pushes straight at the hinge,
will not. Force F3 will open the door, but not as easily as F1 What about
F4? It is perpendicular to the door, it has the same magnitude as F1 , but you
know from experience that pushing close to the hinge is not as effective as
pushing at the outer edge of the door.
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The ability of a force to cause a rotation or a twisting motion thus depends on
three factors:
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We can incorporate these three observations into a single quantity called the
torque  (Greek tau). Loosely speaking,  measures the “effectiveness” of
the force at causing an object to rotate about a pivot. Torque is the
rotational equivalent of force.
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To make these ideas specific, Figure 7.3 shows a force F applied at one
point on a rigid body that can rotate about a pivot or axis. For example, a
string might be pulling on the object at that point, in which case the force
would be a tension force. Figure 7.3 defines the distance r from the pivot to
the point at which the force is applied; the radial line, the line starting at the
pivot and extending through the point of force application; and the angle 
(Greek phi) measured between the force direction and the radial line.
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Figure 7.3 Force
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1. The magnitude F of the force.
2. The distance r from the point of application to the pivot.
3. The angle at which the force is applied.
F
exerts a torque about the pivot point.
Torque can be interpreted from two perspectives. The first can be understood
by considering the door in Figure 7.2, where we saw that force F1 was quite
effective in opening the door, but force F2 had no effect on its rotation. In
general, the component of a force FP that is parallel to the radial line—that
is, the component that points either directly towards or away from the
pivot—has no effect on an object’s rotation, and thus contributes nothing to
the torque. Only the component of the force F that is perpendicular to the
radial line has an effect on the object’s rotation, and so it is this component
of the force that determines the torque. This perspective is shown in Figure
7.4.
Figure 7.4 Torque is due to the perpendicular component of the force.
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In our example of the door we’ve also seen that the larger the distance r from
the pivot to the point of application of the force, the greater the effect on
rotation, so we would expect a larger value of r also leads to a greater torque.
Putting this together with our observation that only F contributes to the
torque leads to our first expression for torque:
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  rF
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Torque is the productof the distance to pivot r and the perpendicular force
component F
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(7.1)
From this equation, we see that the SI units of torque are newton-meters,
abbreviated N m.
A second interpretation of torque is illustrated in Figure 7.5, which shows an
object, free to rotate on a pivot, subject to a force F. The force F acts at
point A, at an angle to the radial line; we’ve seen that because of this it is less
effective at producing a torque on the object. Consider now what would
happen if the same force acted at point B, which is on the line of action—the
line passing through the point of application of the force (point A) and
parallel to the direction of the force. You can see that at point B the force
would be more perpendicular to its radial line, and therefore more effective at
producing a torque—but its distance rB from the pivot is less, which is less
effective at producing a torque. It seems plausible, then, that these two
factors cancel, and the torque generated by the force acting at B is the same
as when it acted at A. This is in fact the case; indeed, the torque produced
by a given force acting anywhere along the line of action is the same.
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Figure 7.5 Torque the the product of the force and the moment arm.
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Thus, if the force acted at point C, which is also on the line of action, it
would produce the same torque as it did at the original point A. But point C
is special: here, the force F and the radial line are perpendicular to each
other. Further, the distance from the pivot to the line of action is as short as
possible. This distance is called the moment arm and, since it is
perpendicular to F and the line of action we give it the symbol r .
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Because the torque is the same whether the force acts at A or C, we can find
the torque by calculating it at C, where it has a simple expression. The force
at C is perpendicular to the radial line, and hence has only a perpendicular
component. Thus F  F. Then, using Equation (7.1) for the torque we have
  rF or, since here r  r and F  F, the torque can be written as
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  r F
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Torque is the product of the moment arm r and the magnitude of the force
F
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These two interpretations of torque can be written in one equivalent form, as
shown in Figure 7.6. From this figure we see that F  F sin  , so that
Equation (7.1) for the torque can be written   rF  rF sin  . We also note
that the moment arm can be written as r  r sin  , so that Equation (7.2) for
the torque becomes   r F  (r sin  )F  rF sin  . Thus our two
interpretations of torque are indeed equivalent, and we have the general
result that
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  r F  rF  rF sin 
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Torque in terms of r, force magnitude F, and the angle  between F and the
radial line
(7.2)
(7.3)
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Figure 7.6 Finding an expression for torque.
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Returning to the door of Figure 7.2, you can see that F1 is most effective at
opening the door because F1 exerts the largest torque about the pivot point:
it has the largest distance r from the pivot, and acts at   90, so that
sin   1, its maximum value. F3 has equal magnitude, but it is applied at an
angle less than 90 and thus exerts less torque. F2 , pushing straight at the
hinge with   0 and sin   0, exerts no torque at all. And F4 , with a
smaller value for r, exerts less torque than F1
Torque, like force, has a sign. A torque that rotates the object in a ccw
direction is positive while a negative torque gives a cw rotation. Figure
Figure 7.7 summarizes the signs. Notice that a force pushing straight toward
the pivot or pulling straight out from the pivot exerts no torque.
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Figure 7.7 Signs and strengths of the torque.
Equations (7.1)–(7.3) give only the magnitude of the torque;
the sign has to be supplied by observing the direction in which the
torque acts, as in Figure 7.7. b
NOTE c
Torque differs from force in a very important way. Torque
is calculated or measured about a particular point. To say that a
torque is 20 N m is meaningless. You need to say that the torque is
20 N m about a particular point. Torque can be calculated about any
point, but its value depends on the point chosen, because this choice
determines r. In practice, we usually calculate torques about a point
about which the object rotates, such as a pivot or an axle. b
NOTE c
EXAMPLE 7.1 Applying a torque
Luis uses a 20-cm-long wrench to turn a nut. The wrench handle is tilted 30
above the horizontal, and Luis pulls straight down on the end with a force of
100 N. How much torque does Luis exert on the nut?
Prepare Figure 7.8 shows the situation. We’ve drawn one figure
corresponding to each of our two interpretations of torque.
Figure 7.8 A wrench being used to turn a nut.
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Solve According to our first interpretation of torque, the torque is given as
  rF . From Figure 7.8a we see that perpendicular component of the force
is
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The torque is then
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F  F cos 30  (100 N)cos 30  866 N
  rF   020 m 866 N   173 N m
We put in the minus sign because the torque is negative—it tries to rotate the
nut in a clockwise direction.
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Our second interpretation of torque is shown in Figure 7.8b. The moment
arm r is the shortest distance from the pivot to the line of action. From the
figure we see that
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Then the torque is
r  r cos 30  (0.20 m)(cos30)  0.173 m
  r F   0173 m 100 N   173 N m
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Again, we insert the minus sign because the torque acts to give a cw rotation.
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  (0.20 m)(100 N)sin120  17.3 N m
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Finally, we can also calculate the torque from Equation (7.3),   rF sin .
From Figure 7.8b,   30  90  120, so we have
as before.
Assess In using any of these methods to find the torque, remember to add in
the minus sign if the torque acts to rotate the object in a clockwise direction.
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CONCEPTUAL EXAMPLE 7.1 Starting a bike
If you try to start your bike with the pedal you’re pushing nearly at the top, it
can be hard to get going. Why is this?
Reason Aided by the weight of the body, the greatest force can be applied to
the pedal straight down. But with the pedal at the top, this force is directed
almost directly towards the pivot, leading to only a small torque. In terms of
our two interpretations of torque, we would say either that the perpendicular
component of the force is small, or that the moment arm is small. Either
leads to a small torque.
Assess If you’ve ever climbed a steep hill while standing on the pedals, you
know that you get the greatest forward motion when you stand hard on the
pedal with its crank parallel to the ground. In this case the force you apply is
entirely perpendicular to the radial line, and the moment arm is as long as it
can be.
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STOP TO THINK 7.1 Rank in order, from largest to smallest, the five
torques  A to  E  The rods all have the same length and are pivoted at the
dot.
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Net Torque
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r r
Figure 7.9 shows forces F1 , F2 , F3 , applied to an extended object. The
object is free to rotate about the axle, but the axle prevents the object from
having any translational motion. It does so by exerting force Faxle on the
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object to balance the other forces and keep Fnet  0
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Figure 7.9 The forces exert a net torque about the pivot point.
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r r
Forces F1 , F2 , F3 , exert torques 1 ,  2 ,  3 , on the object, but Faxle does
not exert a torque because it is applied at the pivot point and has zero
moment arm. Thus the net torque about the axle is the sum of the torques due
to the applied forces:
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net  1  2   3 L  
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In practice, usually only a small number of forces exert torques. For example,
the only torque we considered in Example 7.1 was due to Luis’s pull.
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EXAMPLE 7.2 Turning a capstan
(7.4)
A capstan is a device used on sailing ships to raise the anchor. A sailor
pushes the long lever, turning the capstan and winding up the anchor rope.
Assume a tension in the rope due to the anchor of 1500 N. If the distance
from the axis of the capstan to where the sailor pushes is seven times the
radius of the capstan around which the rope is wound, with what force must
the sailor push if the net torque on the capstan is to be zero?
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
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Prepare Shown in Figure 7.10 is a view looking down from above the
capstan. The rope pulls with a tension force T , acting at a distance R from
the axis of rotation. The sailor pushes with a force F at a distance 7R from
the axis. Both forces are perpendicular to their radial lines, so  in Equation
(7.3) is 90.
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Figure 7.10 A sailor trying to turn a capstan.
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Solve The torque due to the tension in the rope is given by
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Since we don’t know the capstan radius, we’ll just leave it as R for now. This
torque is positive, since it tends to turn the capstan counterclockwise. The
torque due to the sailor is
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We put the minus sign in because this torque acts in the clockwise (negative)
direction. The net torque is zero, so we have  T   S  0, or
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R(1500 N)  7.0RF  0
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 T  RT sin 90  RT  R(1500 N)
 S  (7.0R)F sin 90  7.0RF
Note that the radius R cancels, leaving
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1500 N
 214 N
7.0
Assess We’ve found that the force the sailor must exert is seven times less
than the force the rope exerts: the long lever helps him lift the heavy anchor.
In the HMS Warrior, built in 1860, it took 200 men turning the capstan to lift
the huge anchor that weighed close to 55,000 N!
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Note that in Figure 7.10 the two forces F and T point in very different
directions. The torques we calculate from these forces depend not on the
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F
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relative directions of the two forces with respect to each other, but their
directions with respect to their own radial lines. If the sailor continue to
circle the capstan, so his force pointed in an entirely different direction, the
torque he would need to exert in order to make the net torque zero would be
the same.
7.2 Gravitational Torque and the Center of Gravity
It is common for gravity to exert a torque on an object. If the car hood in
Figure 7.11 is released, a torque due to the force of gravity will cause it to
close by rotating about its hinges. To calculate the torque due to gravity, we
need to know at what point on the object the gravitational force acts. As
shown in Figure 7.12a, the force of gravity on an object—that is, its
weight—actually acts on every particle making up the object. But Figure
7.12b shows that for the purpose of calculating the torque due to gravity, the
weight can be considered as acting at a special point of the object—its center
of gravity (symbol ).
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Figure 7.11 The torque due to gravity will cause this hood to close.
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Figure 7.12 The center of gravity is the point where the weight appears to act.
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Finding the Center of Gravity
There is a simple experimental method for finding the center of gravity of
any object, based on the observation that any object free to rotate about a
pivot will come to rest with its center of gravity directly below the pivot.
To see this, consider the punching bag of Figure 7.13, swinging back and
forth on its pivot. When the bag is to the left, the gravitational torque pulls it
to the right; when it’s to the right, the torque pulls it back to the left. But
when the bag’s center of mass is directly below the pivot, the weight force
pulls directly away from the pivot and the torque is zero. So only here can
the bag can remain at rest. Eventually, then, as friction slows the swinging,
the bag ends up with its center of gravity directly below the pivot.
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Figure 7.13 An object comes to rest with its center of gravity below the pivot.
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So when an object free to rotate comes to rest, we know that its center of
gravity must be below the pivot point. But we don’t know how far below the
pivot the center of gravity is. To find out, we need to suspend the object from
a second pivot, and again let it come to rest. To see how this works, consider
an oddly-shaped object, such as the cutout map of the United States shown in
Figure 7.14. If it is free to rotate about pivot 1 as in Figure 7.14a, we know
that it will come to rest with its center of gravity directly below the pivot,
that is, with its center of gravity somewhere along the blue line. We then
suspend it from a second pivot 2; the map’s center of gravity must also end
up directly below this pivot, along the red line in Figure 7.14b. Since the
center of gravity must lie on both the blue and red lines, it must be at their
intersection as shown. Interestingly, the geographical center of the United
States is defined in just this way, as the center of gravity of a map of the
United States. This point is one mile northwest of Lebanon, Kansas.
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Figure 7.14 Method for finding the center of gravity of an object.
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The geographical center of the United States is near Lebanon, Kansas.
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A balancing bird Because this toy bird is balanced, its center of gravity
must lie somewhere below the pivot point at its beak. But this means its
center of gravity lies outside of the body of the bird itself! For the center of
gravity to lie outside an object is actually quite common; this is often the
case for the human body, depending on its pose.
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For a simple symmetrical object, such as a rod, sphere, or cube made from a
uniform material, the above considerations show that the center of gravity
of a symmetrical object lies at its center. As shown in Figure 7.15, such an
object can always be hung from several pivots such that half of the object is
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on each side of a vertical line descending from the pivot. Such lines must
intersect at the object’s center. A particularly simple case of this is a point
particle, whose center of gravity lies at the position of the particle.
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Figure 7.15 The center of gravity of a symmetrical object lies at its center.
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Finally, let’s examine the most general case of an object made up of a
combination of particles and other objects whose center of gravity positions
are known. Figure 7.16a shows such an object, a mallet with a handle of
mass m1 and a head of mass m2 . Let’s imagine that the mallet is free to pivot
from an axle on its left end; we’ll measure all distances from this pivot. The
x-coordinate of the handle’s center of gravity is x1 , and that of the head is
x2 . The center of mass of the combined object, shown in Figure 7.16b, is at
x-coordinate xcg ; this is the coordinate we want to find.
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Figure 7.16 Finding the center of gravity of a combination of objects.
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We use the basic definition of the center of gravity: it is the point at which
gravity appears to act for the purpose of calculating the gravitational torque.
From Figure 7.16a the net torque from the two weights w1 and w2 is
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 net  x1w1  x2 w2  x1 m1g  x1 m2 g  x1 m1  x1 m2 g
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But in Figure 7.16b, the net torque must also be given as
 net  xcg w  xcg mg  xcg (m1  m2 )g
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Setting these two expressions for  net equal gives
x1 m1  x2 m2 g  xcg (m1  m2 )g
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so that
xcg 
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x1 m1  x1 m2
m1  m2
This gives the x-position of the center of gravity of the combined object in
terms of the center of gravity positions and masses of the objects making it
up. If more than two objects are combined, then the x-position of the
combined center of gravity is
xcg 
x1 m1  x2 m2  x3 m3 L
m1  m2  m3 L
(7.5)
If the centers of gravity don’t all line up along a horizontal line, then we’ll
need the y-position of the center of gravity as well. Using the same kind of
argument gives
ycg 
y1 m1  y2 m2  y3 m3 L
m1  m2  m3 L
(7.6)
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Because the center of gravity position depends on products such as x 1m1 ,
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EXAMPLE 7.3 Finding the center of gravity of a carpenter’s square
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objects that have large masses count more in the sum. This means that the
center of gravity tends to lie closer to the heavier objects or particles that
make up the entire object.
A carpenter’s square has the dimensions and masses shown in Figure 7.17.
The blade extends into the handle, where it is held by three rivets. Where is
the square’s center of gravity?
Figure 7.17 A carpenter's square.
Prepare We’ll assume that the blade and handle are both uniform rectangles,
so that their centers of gravity are at their centers. We’ll ignore the rivets, and
the fact that the handle is slotted to accept the blade. We’ll make all
measurements from an origin at the lower-left corner, although any other
choice would work as well. Then the x- and y-positions of the handle (h) and
blade’s (b) centers of gravity are
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xh
yh
xb
yb
Solve From Equation (7.5) we have
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2.0 cm
5.5 cm
10.0 cm
9.5 cm
xcg 
xh mh  xb mb (2.0 cm)(40.0 g)  (10.0 cm)(80.0 g)

 7.3 cm
mb  mb
40.0 g  80.0 g
ycg 
yh mh  yb mb (5.5 cm)(40.0 g)  (9.5 cm)(80.0 g)

 8.2 cm
mb  mb
40.0 g  80.0 g
and
Assess The calculated center of gravity position of the entire square is
shown in Figure 7.17. Note that the center of gravity of the square is closer to
that of the heavier blade than to that of the lighter handle. Also, the center of
gravity of the square lies along a line connecting the centers of gravity of the
blade and the handle. Since the center of gravity can be considered as the
point where all the mass of an object is concentrated, the handle and the
blade act as two particles positioned at their centers of gravity. The combined
center of gravity of two particles must clearly lie on a line connecting them.
18
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28
EXAMPLE 7.4 Calculating the torque on a flag pole
A flag pole consists of a 2.0-m-long pole, having a mass of 5.0 kg, with a
3.0 kg decorative ball on its end. The pole is fixed at an angle of 25 from
the horizontal. What is the gravitational torque on the flag pole, calculated
about an axis at the fixed end of the pole?
Prepare For the purpose of calculating the gravitational torque we can
consider the 7.0 kg total weight of the flag pole to act at its center of gravity.
Then the torque is the moment arm r times the total weight. From Figure
7.18 we see that r , the distance from the pivot to the line of action, is equal
to the x-component of the pole’s center of gravity (COG).
29
30
Figure 7.18 A flag and flag pole.
31
32
Solve We use Equation (7.5) to find the position of the center of gravity. We
have
xcg 
1
2
3
4
5
6
7
8
9
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11
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28
mp x p  mb x b
mp  mb
where xp and xb are the x-positions of the centers of gravity of the pole and
ball, respectively. Now the center of gravity of the pole alone is at its center,
a distance L / 2 up the pole from the pivot, where L = 2.0 m is the pole’s
length. Then from the figure we can see that
L
cos  (1.0 m)cos25  0.91 m
2
Similarly, the center of gravity is at its center, and from the figure its x
position is twice that of the pole’s center of gravity, so
xb  (2.0 m)cos25  1.81 m. Then we have for the overall center of gravity
of the pole
xp 
xcg 
(5.0 kg)(0.91 m)  (3.0 kg)(1.81 m)
 1.25 m
(5.0 kg)  (3.0 kg)
Since in this example xcg  r , we have
  r F  xcg w  xcg mg  (1.25 m)(8.0 kg)(9.8 m/s2 )  98.0 N m
This result is just the magnitude of the torque. Since the torque here tends to
rotate the flag pole clockwise, the sign of the torque is negative, so we have
  98.0 N m
as the complete value for the torque.
Assess There is another way to find the net gravitational torque in this
example. We could find the torques due to the weight of the pole and the
weight of the ball individually, and then take their sum. This would require
finding the moment arms for each weight.
STOP TO THINK 7.2 The balls are connected by very lightweight rods. The
pivot point is indicated by a dot. Rank in order, from least to greatest, the
gravitational torque about the pivot for each arrangement of balls.
1
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3
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5
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7.3 Rotational Dynamics and Moment of Inertia
We’ve found that torque is the rotational equivalent of force. Now we need
to learn what torque does. Let’s start by examining a single particle subject
to a torque. Figure 7.19 shows a particle of mass m attached to a lightweight,
rigid rod of length r, so that the particle is constrained to move in a circle.
The particle is subject to an external force F directed tangentially to the
circle. Thus there is a torque on the particle of magnitude
  rF  rF
Figure 7.19 A force F exerts a torque on the particle and causes an angular
acceleration.
Because there is a force F acting on the particle, it will undergo an
acceleration given by Newton’s second law given by
r
F
at 
m
We call this the tangential acceleration at because its direction is tangent
to the circle. The tangential acceleration measures the rate at which the speed
v of the particle is changing. Thus, just as for ordinary linear acceleration, we
can write for the magnitude of the tangential acceleraton
at 
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21
22
23
24
25
26
As long as we apply the tangential force F, the particle will continue to have
a tangential acceleration and its speed will increase. As we found in Section
6.2, a particle that is speeding up (or slowing down) as it moves in a circle is
also undergoing an angular acceleration . To find the relationship between
 and the magnitude at of the tangential acceleration, we can use Equation
6.10, the relationship v  r between the particle’s speed and its angular
velocity  . We then have
at 
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28
v
t
so that
v   r 


r
 r
t
t
t
1
at   r
2
3
4
5
Now we’re ready to find out what happens when a torque is applied to a
single particle such as the one in Figure 7.19. We can write Newton’s second
law for the particle as
6
7
8
9
10
11
F  mat  mr
Multiplying both sides of this equaton by r gives
rF  mr 2
But rF is the torque  on the particle. Hence Newton’s second law for a
single particle is
  mr 2
(7.7)
We can rewrite Equation (7.7) in the more suggestive form

12

(7.8)
mr 2
13
14
15
16
17
18
At the beginning of Section 7.1 we asked the question, “What do Newton’s
laws have to tell us about rotational motion?” We now have the first part of
an answer: A torque causes an angular acceleration. This is the rotational
equivalent of our prior discovery, for motion along a line, that a force causes
an acceleration. Now all that remains is to expand this idea from a single
particle to an extended object.
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29
Newton’s Second Law for Rotational Motion
We’ve found that torque causes an angular acceleration for a single particle
constrained to move in a circle. Next we wish to understand how a rigid
body, free to rotate about a fixed axis, responds to a torque caused by an
external force F.
Figure 7.20 shows a rigid body that undergoes such motion about a fixed and
unmoving axis. Some representative particles 1–4 are shown. As the external
torque is applied, the object will undergo an angular acceleration . Since all
the particles that make up the object rotate together, each particle has this
same angular acceleration . So for each particle we can apply Equation
(7.7), with the same . For particle 1 we have
1  m1r12
30
31
and for particles 2, 3, 4,… we have
32
 2  m2 r22
33
 3  m3r32
34
 4  m4 r42
35
36
and so on for every particle in the object. If we add up all these torques, we
find that the net torque on the object is
37
 net  1   2   3 L  m1r12  m2 r22  m3r32 L
38
  m1r12  m2 r22  m3r32 L    mr 2


(7.9)
1
2
3
Note that by factoring  out of the sum, we’re making explicit use of the
fact that every particle in a rotating rigid body has the same angular
acceleration 
4
5
Figure 7.20 The forces on a rigid body exert a torque about the rotation axis.
6
7
8
9
10
11
12
13
14
15
16
17
Note that the torque on a particle such as particle 2 is due only to internal
forces due to its neighboring particles, such as particle 3, pushing or pulling
on it. Figure 7.21 shows a closeup of these two particles. By Newton’s third
law the forces F2 on 3 and F3 on 2 form an action/reaction pair, and are thus
equal in magnitude but point in opposite directions. Further, since they both
act at the same point—the point of contact between the particles—both
forces have the same moment arm. This means that the torques that the two
forces exert on the two particles are equal in magnitude but opposite in sign:
their combined contibution to the net torque is zero. In fact, every internal
torque acting on a particle in the object is exactly canceled by a second
internal torque acting on an adjacent particle. We are thus left with the result
that the net torque due to internal torques is zero.
18
19
Figure 7.21 The torques due to internal forces cancel.
20
21
22
23
24
25
26
27
Does this mean that  net in Equation (7.9) is zero? No, because the total net
torque on the objects also includes the external torque  ext due to any
external forces that act. These don’t cancel in the sum in Equation (7.9).
Since the sum of the internal torques is zero, we are left with the important
result that the net torque  net in Equation (7.9) is the net torque due to
external forces only. This is an key simplification, since we generally know
the external forces and torques, while the internal torques would be very
difficult to calculate.
28
29
The quantity  mr 2 , which is the proportionality constant between angular
acceleration and net torque, is called the object’s moment of inertia I:
30
I  m1r12  m2 r2 2  m3r32 L   mr 2
(7.10)
1
Moment of inertia of a collection of particles
2
3
4
5
6
7
8
9
10
11
The units of moment of inertia are kg m2  An object’s moment of inertia,
like torque, depends on the axis of rotation. Once the axis is specified,
allowing the values of r1 , r2, r3 ,L to be determined, the moment of inertia
about that axis can be calculated from Equation (7.10).
12

13
14
15
16
17
18
19
20
21
22
23
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30
31
32
33
The moment in moment of inertia and moment arm has
nothing to do with time. These terms stem from the Latin momentum,
meaning “motion.” b
NOTE c
Substituting the moment of inertia into Equation (7.9) puts the final piece of
the puzzle into place. An object that experiences a net torque  net about the
axis of rotation undergoes an angular acceleration
 net
I
Newton’s second law for rotational motion
(7.11)
where I is the moment of inertia of the object about the rotation axis. This
result, which is Newton’s second law for rotation, is the fundamental
equation of rigid-body dynamics.
In practice we often write  net  I , but Equation (7.11) better conveys the
idea that torque is the cause of angular acceleration. In the absence of a
net torque  net  0 , the object either does not rotate   0  or rotates with
constant angular velocity   constant 
This large granite ball floats with nearly zero friction on a thin layer of pressurized
water. As the girl pushes on it, she exerts a torque on the ball. Because of the ball’s
large moment of inertia, the torque causes only a small angular acceleration, so that
the ball’s angular velocity increases only slowly as she pushes.
Interpreting the Moment of Inertia
Before rushing to calculate moments of inertia, let’s get a better
understanding of its meaning. First, notice that moment of inertia is the
rotational equivalent of mass. It plays the same role in Equation (7.11) as
r
mass m in the now-familiar a  Fnet /m Recall that objects with larger mass
have a larger inertia, meaning that they’re harder to accelerate. Similarly, an
object with a larger moment of inertia is harder to rotate. Loosely speaking, it
takes a larger torque to spin an object with a larger moment of inertia than an
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
object with a smaller moment of inertia. The fact that moment of inertia
retains the word inertia reminds us of this.
But why does the moment of inertia depend on the distances r from the
rotation axis? Think about the two wheels shown in Figure 7.22. They have
the same total mass M and the same radius R. As you probably know from
experience, it’s much easier to spin the wheel whose mass is concentrated at
the center than to spin the one whose mass is concentrated around the rim.
This is because having the mass near the center (smaller values of r ) lowers
the moment of inertia.
Figure 7.22 Moment of inertia depends both on the mass and how the mass is
distributed.
Thus an object’s moment of inertia depends not only on the object’s mass but
on how the mass is distributed around the rotation axis. This is well known to
bicycle racers. Every time a cyclist accelerates, she has to “spin up” the
wheels and tires. The larger the moment of inertia, the more effort it takes
and the slower her acceleration. For this reason, racers use the lightest
possible tires, and they put those tires on wheels that have been designed to
keep the mass as close as possible to the center without sacrificing the
necessary strength and rigidity.
Table 7.1 summarizes the analogies between linear and rotational dynamics.
Rotational dynamics
Linear dynamics
force
Fnet
mass
m
angular acceleration
net
I

acceleration
second law
   net /I
second law
a
r
a  Fnet /m
torque
moment of inertia
22
Table 7.1 Rotational and linear dynamics
23
24
25
26
27
Example 7.5 Finding the moment of inertia of point particles
An object consists of three small, heavy spheres attached by very lightweight
10-cm-long rods as shown in Figure 8.23. The spheres have masses
m1  1.0 kg, m2  1.5 kg, and m3  1.0 kg. What is the object’s moment of
inertia if it is rotated about axis a? About axis b?
1
2
Figure 8.23 Three point particles separated by lightweight rods.
3
Prepare We’ll use Equation (7.10) for the moment of inertia:
4
I  m1r12  m2 r2 2  m3r32
5
6
7
8
9
10
11
Recall that in this expression, r1 , r2 , and r3 are the distances from each
particle to the axis of rotation, so that they depend on the axis chosen.
Particle 1 lies directly on both axes, so r1  0 cm in both cases. Particle 2 lies
10 cm (0.10 m) from both axes. But Particle 3 is 10 cm from axis a, but
further from axis b. We can find r3 for axis b by using the Pythagorean
12
13
Solve For each axis, we can prepare a table of the values of r, m, and mr 2
for each particle. For axis a we have
theorem. We have r32  (10 cm)2  (10 cm)2 , which gives r3  14.1 cm.
These distances are indicated in the figure.
Particle
r
m
1
0m
1.0 kg
0 kg m 2
2
0.10 m
1.5 kg
0.015 kg m 2
3
0.10 m
1.0 kg
0.010 kg m 2
I   mr 2
14
r
m
mr 2
1
0m
1.0 kg
0 kg m 2
2
0.10 m
1.5 kg
0.015 kg m 2
3
0.14 m
1.0 kg
0.020 kg m 2
I   mr 2
18
0.025 kg m 2
For axis b we have
Particle
15
16
17
mr 2
0.035 kg m 2
Assess We’ve already noted that the moment of inertia of an object is
higher when its mass is distributed farther from the axis of rotation. Here, m 3
is farther from axis b, leading to a higher moment of inertia about that axis.
TRY IT YOURSELF
Waving a hammer Most of the mass of a hammer is in its head, so its
moment of inertia is large when calculated about an axis passing through the
end of the handle (far from the head), but small when calculated through an
axis passing through the head itself. You can feel this difference by
attempting to wave a hammer back and forth about the handle end and the
head end. It’s much harder to do about the handle end, because the large
moment of inertia keeps the angular acceleration small.
1
2
3
4
5
6
7
8
9
10
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15
Calculating the Moment of Inertia
Newton’s second law for rotational motion is easy to write, but we can’t
make use of it without knowing an object’s moment of inertia. Unlike mass,
we can’t measure moment of inertia by putting an object on a scale. And
while we can guess that the center of mass of a symmetrical object is at the
physical center of the object, we can not guess the moment of inertia of even
a simple object. To find I, we really must carry through the calculation.
For an object consisting of only a few point particles connected by massless
rods, we can use Equation (7.10) to directly calculate I. But such an object is
pretty unrealistic. All real objects are made up of solid material that is itself
comprised of countless atoms. To calculate the moment of inertia of even a
simple such object requires integral calculus, and is beyond the scope of this
text. However, moments of inertia for all simple objects have been
calculated; a short list of common moments of inertia is given in Table 7.2.
Table 7.2 Moments of inertia of objects with uniform density
Object and
axis
Picture
Thin rod,
about center
1
12
Thin rod,
about end
1
3
Plane or slab,
about center
1
12
I
Object and axis
Picture
I
ML2
Cylinder or disk,
about center
1
2
ML2
Cylindrical hoop,
about center
MR 2
Ma 2
Solid sphere,
about diameter
2
5
MR 2
MR 2
Plane or slab,
about edge
1
3
Ma 2
Spherical shell,
about diameter
1
2
3
4
5
Even though we can’t calculate most of the moments of inertia of Table 7.2,
we can make some general observations. For instance, the cylindrical hoop is
composed of particles that are all the same distance R from the axis. Thus
each particle of mass m makes the same contribution mR2 to the hoop’s
moment of inertia. Adding up all these contributions gives
6
I  m1 R2  m2 R2  m3 R2 L  m1  m2  m3 L R2  MR2
7
8
9
10
11
12
13
14
15
as given in the table. The solid cylinder of the same mass and radius has a
lower moment of inertia than the hoop, since much of the cylinder’s mass is
nearer its center, where the contribution to the moment of inertia is less. In
the same way we can see why a slab rotated about its center has a lower
moment of inertia than the same slab rotated about its edge: in the latter case
some of the mass is twice as far from the center as the furthest mass in the
former case. This means that those particles contribute four times as much to
the moment of inertia, leading to an overall larger moment of inertia for the
slab rotated about its edge.
Novel golf clubs The latest craze in golf putters is the High Moment of
Inertia head. By adding much of the mass rather far from the pivot, the
moment of inertia about that pivot can be greatly increased. The idea is that if
the ball is accidently struck off center, the large moment of inertia of the
head will keep its angular acceleration small—reducing unwanted rotation of
the head and allowing a truer putt.
16
17
2
3
MR 2
1
2
3
4
5
6
Stop to Think 7.3 Four very lightweight disks each have three identical heavy
marbles glued to them as shown. Rank in order, from smallest to largest, the
moments of inertia of the disks, when they are rotated about the indicated
axes.
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10
11
12
13
14
7.4
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23
PROBLEM-SOLVING STRATEGY 7.1 Rotational dynamics problems
In this section we’ll look at several examples of rotational dynamics for rigid
bodies that rotate about a fixed axis. The restriction to a fixed axis avoids
complications that arise for an object undergoing a combination of rotational
and translational motion. The problem-solving strategy for rotational
dynamics is very similar to that of linear dynamics.
Prepare Model the object as a simple shape. Draw a pictorial representation
to clarify the situation, define coordinates and symbols, and list known
information.

Identify the axis about which the object rotates.

Identify forces and determine their distance from the axis.

Identify the torques caused by the forces and the signs of the torques.
Solve The mathematical representation is based on Newton’s second law for
rotational motion
net  I
24
25
26
27
28
29
30
Rotation About a Fixed Axis
or

net
I

Find the moment of inertia either by direct calculation using Equation
(7.10) for point particles, or from Table 7.2 for common shapes of
objects.

Use rotational kinematics to find angles and angular velocities.
Assess Check that your result has the correct units, is reasonable, and
answers the question.
31
32
33
34
35
EXAMPLE 7.6 Starting an airplane engine
The engine in a small airplane is specified to have a torque of 500 Nm. This
engine drives a 2.0-m-long, 40 kg propeller. On start-up, how long does it
take the propeller to reach 2000 rpm?
1
2
3
Model The propeller can be modeled as a rod that rotates about its center.
The engine exerts a torque on the propeller. Figure 7.24 shows the propeller
and the rotation axis.
4
5
Figure 7.24 A rotating airplane propeller.
6
7
Solve The moment of inertia of a rod rotating about its center is found from
Table 7.2:
8
9
1
1
2
ML2  40 kg 20 m   133 kg m2
12
12
The 500 Nm torque of the engine causes an angular acceleration
I

10
11
12
13
14
15
16
17
18

I

500 N m
 37.5 rad/s 2
133 kg m 2
The time needed to reach  f  2000 rpm  33.3 rev/s  209 rad/s is
t 



 f   i 209 rad/s  0 rad/s

 5.7 s

37.5 rad/s2
Solve We’ve assumed a constant angular acceleration, which is reasonable
for the first few seconds while the propeller is still turning slowly.
Eventually, air resistance and friction will cause opposing torques and the
angular acceleration will decrease. At full speed, the negative torque due to
air resistance and friction cancels the torque of the engine. Then  net  0 and
the propeller turns at constant angular velocity with no angular acceleration.
19
20
21
22
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25
26
27
28
29
30
31
32
EXAMPLE 7.7 A falling pole
A 7.0-m-tall telephone pole with a mass of 260 kg is hit by lightning at its
base, nearly severing the pole from its base. The pole begins to fall, rotating
about the part still connected to the base. Estimate the pole’s angular
acceleration when it has fallen by 25 from the vertical.
Prepare The situation is shown in Figure 7.25, where we define our symbols
and list the known information. There are two forces acting one the pole: the
pole’s weight w, and the force of the base on the pole (not shown). If we
choose our axis of rotation as shown in the figure, this second force will exert
no torque since it acts at the axis of rotation. Thus we can ignore this force
for the purpose of calculating the torque. The torque on the pole is then due
only to gravity. From the figure we see that this torque tends to rotate the
pole in a counterclockwise direction, so that the torque is positive.
1
2
3
4
5
Figure 7.25 A falling telephone pole undergoes an angular acceleration due to a
gravitational torque.
Solve We’ll model the pole as a thin rod, so that its center of gravity is at its
L
center, a distance d  from the axis. Then the torque is
2
 L
2
 net  Fd sin  w   sin 
 
6
7
8
9
10
11
12
13
14
15
16
mgL
sin
2
From Table 7.2, the moment of inertia of a thin rod rotated about its end is
1
I  mL2 . Thus from Newton’s second law for rotational motion we can
3
find the angular acceleration as
1
mgL sin 3gsin 3(9.8 m/s2 )sin 25

2


 0.31 rad/s2
(7.12)
1 2
I
2L
2(20
m)
mL
3
Assess The problem asked for an estimate of the angular acceleration. This
is usually a hint that you should make some simplifying assumptions, as we
did here in modeling the tree as a thin rod. Note also that  is inversely
proportional to L, but doesn’t depend on the mass m. This means that taller
objects fall more slowly: compare the fall of a pencil balanced on its tip to
that of a tall tree when cut.
 net
17
18
19
20
21
CONCEPTUAL EXAMPLE 7.2 Balancing a meter stick
You’ve probably tried balancing rod-shaped objects vertically on your
fingertip. If the object is very long, like a meter stick or a baseball bat, it’s
not too hard. But if it’s short, like a pencil, it’s impossible. Why is this?
1
2
3
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5
6
7
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23
This juggler can balance the long stick an teapot on his head because long rods fall
slowly.
Reason Suppose you’ve managed to balance a meter stick on your finger,
but then it starts to fall. You’ll need to quickly adjust your finger to bring the
stick back into balance. But from Equation (7.12) the angular acceleration 
of a thin rod is inversely proportional to L. Thus a long object like a meter
stick topples much more slowly than a short one like a pencil. Your reaction
time is fast enough to correct for a slowly-falling meter stick, but not for a
rapidly-falling pencil.
Assess If we double the length of a rod, its mass doubles and its center of
gravity is twice as high, so the gravitatonal torque  on it is four times as
1
much. But since a rod’s moment of inertia is I  ML2 , the longer rod’s
3
moment of inertia will be eight times greater.
Constraints Due to Ropes and Pulleys
Many important applications of rotational dynamics involve objects, such as
pulleys, that are connected via ropes or belts to other objects. Figure 7.26
shows a rope passing over a pulley and connected to an object in linear
motion. If the rope turns on the pulley without slipping, then the rope’s speed
vrope must exactly match the speed of the rim of the pulley, which is
vrim   R If the pulley has an angular acceleration, the rope’s acceleration
arope must match the tangential acceleration of the rim of the pulley,
at   R
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Figure 7.26 The rope's motion must match the motion of the rim of the pulley.
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The object attached to the other end of the rope has the same speed and
acceleration as the rope. Consequently, an object connected to a pulley of
radius R by a rope that does not slip must obey the constraints
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vobj   R
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aobj   R
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Motion constraints for a nonslipping rope
(7.13)
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These constraints are very similar to the acceleration constraints introduced
in Chapter 5 for two objects connected by a string or rope.
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EXAMPLE 7.8 Lowering a block
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The constraints are given as magnitudes. Specific problems
will need to introduce signs that depend on the direction of motion
and on the choice of coordinate system. b
NOTE c
A 2.0 kg block is attached to a massless string that is wrapped around a 1.0
kg, 4.0-cm-diameter cylinder, as shown in Figure 7.27a. The cylinder rotates
on an axle through the center. The block is released from rest 1.0 m above
the floor. How long does it take to reach the floor?
Figure 7.27 The falling block turns the cylinder.
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Prepare Assume the string does not slip. Figure 7.27b shows the free-body
diagrams for the cylinder and the block. The string tension exerts an upward
force on the block and a downward force on the outer edge of the cylinder.
The string is massless, so these two tension forces act as if they are an
action/reaction pair, so Tb  Tc  Since these two tensions have equal
magnitudes, we’ll call both their magnitudes T.
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where, as usual, the y-axis points upward. What about the cylinder? There is
a normal force n on the cylinder due to the axle. However, n does not exert
a torque because it passes through the rotation axis. The only torque comes
from the string tension. The moment arm for the tension is d  R, and the
torque is positive because the string turns the cylinder ccw. Thus  string  TR
Solve Newton’s second law applied to the linear motion of the block is
may  T  mg
and Newton’s second law for the rotational motion is

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 net
I

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TR
2T

MR 2 MR
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The moment of inertia of a cylinder rotating about a center axis was taken
from Table 7.2.
The last piece of information we need is the constraint due to the fact that the
string doesn’t slip. Equation (7.13) relates only the magnitudes of the linear
and angular accelerations, but in this problem  is positive (ccw
acceleration) while ay is negative (downward acceleration). Hence
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ay   R
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Using  from the cylinder’s equation in the constraint, we find
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ay   R  
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Thus the tension is T   12 May  If we use this value of the tension in the
block’s equation, we can solve for the acceleration:
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may   May  mg
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ay  
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g
 784 m/s2
1 M / 2m
The time to fall through y  10 m is found from kinematics:
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y  ay t 
2
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2T
2T
R
MR
M
t 
2 10 m 
2y

 0505 s
ay
784 m/s2
Assess The expression for the acceleration gives ay  g if M  0 This
makes sense because the block would be in free fall if there were no cylinder.
When the cylinder has mass, the downward force of gravity on the block has
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to accelerate the block and spin the cylinder. Consequently, the acceleration
is reduced and the block takes longer to fall.