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ROTATIONAL MOTION
7
Q7.1. Reason: Looking down from above the player runs around the bases in a counterclockwise direction, hence
the angular velocity is positive.
Assess: Note that looking from below (from under the grass) the motion would be clockwise and the angular
velocity would be negative. We assumed the bird’s eye view because it is standard to do so, and it is difficult to view
the game from below the ground. This is akin to setting up a coordinate system with the positive x-axis pointing left
and the positive y-axis pointing down; the real-life physics wouldn’t change any, and the calculations of measurable
quantities would produce the same results, and it might even be occasionally convenient. But unless there is a clear
reason to do otherwise, the usual conventions should be your first thought.
Q7.2. Reason: The earth rotates such that the sun rises in the east. A diagram of the sun-earth system as seen from
above the north pole follows.
The North Pole is represented as the point at the center of the circle in the figure. The directions of east and west are
indicated for a person standing at the equator. The person at this position would be seeing the sun just above the
horizon in the east. In order for the sun to rise in the east, the earth must be rotating counterclockwise as seen from
above the north pole. A point on the earth’s equator would have a positive angular velocity as seen from this position.
Assess: Angular velocity is positive for counterclockwise rotations.
Q7.3. Reason: By convention, clockwise rotations are negative and counterclockwise rotations are positive. As a
result, an angular acceleration that decreases/increases a negative angular velocity is positive/negative. In like
manner, an angular acceleration that decreases/increases a positive angular velocity is negative/positive. Knowing
this we can establish the situation for each figure. Figure (a) the pulley is rotating clockwise (ω = −), however since
the large mass is on the left it is decelerating (α = + ).
Figure (b) the pulley is rotating counterclockwise (ω = +) and since the large mass is on the left it is accelerating
(α = + ).
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7-1
7-2
Chapter 7
Figure (c) the pulley is rotating clockwise (ω = −) and since the large mass is on the right it is accelerating (α = −).
Figure (d) the pulley is rotating counterclockwise (ω = + ), however since the large mass is on the right it is
decelerating (α = −).
Assess: It is important to know the sign convention for all physical quantities that are vectors. This is especially
important when working with rotational motion.
Q7.4. Reason: Since we are dealing with a stubborn nut, we want a large torque. For a given force the torque is
increased by increasing the moment arm. Slipping a length of pipe over the wrench handle increases the moment arm
and hence the applied torque.
Assess: You can increase the torque by either increasing the force or the moment arm. In this case we chose to
increase the torque by increasing the moment arm.
Q7.5. Reason: The question properly identified where the torques are computed about (the hinge). Torques that
tend to make the door rotate counterclockwise in the diagram are positive by convention (general agreement) and
torques that tend to make the door rotate clockwise are negative.
(a) +
(b) −
(c) +
(d) −
(e) 0
Assess: Looking at the diagram we see that F a and F c are parallel and are both creating a negative or
counterclockwise torque. But since F c is farther from the hinge, its torque will be greater. A similar argument can be
made for F b and F d . Note that F e causes no torque since it has no moment arm.
Q7.6. Reason: The screwdriver with a thick handle has a larger radius and so provides a larger moment arm for the
force you exert when turning the driver. The larger moment arm leads to a larger torque on the screw. This can be
seen from Equation 7.11.
Assess: For a given force a larger moment arm leads to a larger torque.
Q7.7. Reason: The reason for large-diameter steering wheels in trucks is that more torque is needed to turn the
wheels due to the greater mass of the truck. Making the steering wheel larger means that more torque is exerted on
the steering shaft for the same force from the driver’s hands.
Assess: Most light cars employ a rack-and-pinion steering system, while larger SUVs and trucks often employ a
recirculating-ball steering system; however both systems can be assisted by pressurized hydraulic fluid (power
steering), so steering, even in trucks, can be much easier than in the old days.
Q7.8. Reason: When put farther from the hinge of the door, the torque exerted on the door by the doorstop is larger
than when the doorstop is placed nearer the hinge. This is due to the larger moment arm for the force the doorstop
exerts on the door.
Assess: Torque is proportional to the moment arm of a force, from Equation 7.18.
Q7.9. Reason: The torque the student exerts turns the ball and rod in a clockwise direction as seen from the top.
Clockwise torques are negative. The student exerts a negative torque.
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Rotational Motion
7-3
Assess: Note that gravity does produce a torque on the ball tending to rotate the ball in the vertical direction, though
this torque is balanced by the vertical torque of the rod and pivot keeping the ball moving in a horizontal plane.
Q7.10. Reason: Any object free to rotate about a pivot will come to rest with its center of gravity directly below the
pivot. When the center of gravity is below the pivot there are no torques to rotate it. That means the center of gravity
is somewhere on the blue and, for pivot 2, somewhere on the red line. Only the intersection of the two lines satisfies
both requirements.
Assess: A clever way to find the center of gravity of an irregularly shaped object.
Q7.11. Reason: As suggested by the figure, we will assume that the larger sphere is more massive. Then the center
of gravity would be at point 1, because if we suspend the dumbbell from point 1 the counterclockwise torque due to
the large sphere (large weight times small lever arm) will be equal to the clockwise torque due to the small sphere
(small weight times large lever arm).
Assess: Look at the figure and mentally balance the dumbbell on your finger; your finger would have to be at
point 1.
The sun-earth system is similar to this except that the sun’s mass is so much greater than the earth’s that the center of
mass (called the barycenter for astronomical objects orbiting each other) is only 450 km from the center of the sun.
Q7.12. Reason: When the mass is distributed close to the axis of rotation the moment of inertia is smaller and the
object is easier to move and get rotating. Swinging the hammer by its light handle puts the heavy mass farther from
the rotational axis and it is therefore harder to wave it rapidly.
Assess: The rapid waving involves angular acceleration.
Q7.13. Reason: Spin them. Because they would have different moments of inertia (2/5MR 2 for the solid sphere
and 2/3MR 2 for a thin-walled hollow sphere, with something in between if the wall of the hollow sphere is not
particularly thin) their angular accelerations would be different with the same torque acting on them (remember the
rotational version of Newton’s second law: α = τ net /I ). The solid one (with the smaller moment of inertia) would
accelerate quicker given the same torque.
Assess: The two balls would have the same linear inertia (having the same mass) so dropping them in free fall, for
example, and observing their linear accelerations would not distinguish between them; but the rotational inertia
(moment of inertia) is different, and so exerting a net torque on them and measuring the resultant angular acceleration
can distinguish between them.
Q7.14. Reason: The more distant an object’s mass is from an axis of rotation, the larger the moment of inertia of the
object. Mass is distributed farther from the axis of rotation for a rod rotated about one end than a rod rotated about its center.
Assess: From Equation 7.21, the contribution to moment of inertia is proportional to the square of the distance from
the axis of rotation.
Q7.15 Reason: A review of Section 7.7 in the text will prepare you to answer this question. Let’s separate the
rolling motion into translation and rotation. Figure (a) below shows the translational velocity of the points of interest.
Figure (b) below shows the velocity due to rotation at all points of interest. Finally, Figure (c) below combines the
velocity vectors due to both translation and rotation.
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7-4
Chapter 7
Looking at the magnitude of the resulting velocity vectors we can write the following:
v1 > v2 > v3 > v4 > v5 = 0
Assess: We know that the magnitudes of v2 and v3 are greater than the magnitude of v4 because their horizontal
component is equal to the magnitude of v4 .
Q7.16 Reason: When the barrel is full it has a much greater moment of inertia and resistance to changes in its
rotation, so it will be a more stable moving platform than the empty barrel with its small moment of inertia.
Assess: The full barrel will take more torque to change its angular velocity.
Q7.17. Reason: Assuming all the forces have the same magnitude, the largest torque will be exerted on the nut by
choice C. Choice A has a lesser torque than C because the moment arm of the force is smaller than in choice C. This
can be seen from Equation 7.11. Choice D has a lesser torque than choice C because the component of the force
perpendicular to the radial line is smaller. This can be seen from Equation 7.10. In choice B there is no torque exerted
on the nut since the radial line and the force are parallel. The torque in this case is zero from Equation 7.12.
Assess: The torque created by a given force can be increased by increasing the moment arm of a force or by
increasing the component of the force perpendicular to the radial line.
Q7.18. Reason: We are given that r = 0.15 m and that sin θ = 1.
Solve Equation 7.12 for F.
τ = rF sin θ
F=
τ
r sin θ
=
20 N ⋅ m
= 133 N ≈ 130 N
(0.15 m)(sin 90 °)
The correct answer is C.
Assess: This force is not unreasonable; it is like lifting 30 lbs. Notice the units worked out in the equation.
Q7.19. Reason: Since the center of gravity of piece 2 is to the right of the center of gravity of piece 1, the
horizontal position of the center of gravity of the two pieces should be between the center of gravity of the two
pieces. The same argument applies to the vertical position of the center of gravity of the pieces. The only point that is
located between the two centers of gravity is point D.
Assess: Our solution to the problem is based on the fact that we can replace piece 2 with a single mass point (with
the same mass as piece 2) located at the center of gravity of piece 2 and we can replace piece 1 with a single mass
point (with the same mass as piece 1) located at the center of gravity of piece 1. We now need to find the center of
gravity of these two mass points and our knowledge of the physics involved makes us aware that it must be
somewhere on the line connecting them. Only point D satisfies this condition.
Q7.20. Reason: We look up the formula for the moment of inertia of a disk about its axis of symmetry in Table
7.1. Neglect the small hole in the center. The radius is half the diameter.
I = 12 MR 2 = 12 (0.015 kg)(0.060 m) 2 = 2.7 × 10−5 kg ⋅ m 2
The correct answer is A.
Assess: The answer is a small number, but CDs are small and light. An old vinyl LP record would be both larger and
more massive so it would have a larger moment of inertia, but modeled as a disk would still be I = 12 MR 2 .
Q7.21. Reason: Since the diameter of the disk increases by a factor of two, the radius of the disk increases by a
factor of two. Assume that the density of the material used is the same and that the hole in the center of the disk is
negligible. The total mass of the disk goes up by a factor of four since the volume of the disk increases by a factor of
four. The moment of inertia of a disk is given in Table 7.1 and is proportional to the mass of the disk and the square
of the radius of the disk. Since the mass goes up by a factor of four and the radius goes up by a factor of two, the
moment of inertia of the disk goes up by a total factor of sixteen. The correct choice is D.
Assess: Note that the increase of the total mass of the disk needed to be included in the calculation.
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Rotational Motion
7-5
Q7.22. Reason: Assume the rods are both uniform. Then the downward gravitational force acts as if concentrated
at the center of the rod. The moment of inertia for a rod swinging from an axis through one end is I = 13 ML2 . Use
ratios so lots of factors cancel out.
τ2
( L)(2Mg )
r2 F2
2
1
( 3 (2M )(2 L) 2 ) 8 1
α2 I2
I2
=
=
=
= =
r1F1
( L / 2)( Mg )
1 2
α1 τ 1
I1
I1
( 13 ML2 )
2
Because the ratio of the angular acceleration for rod 2 to rod 1 is less than 1 then the angular acceleration of rod 1
must be greater. The correct choice is A.
Assess: Note that the increase of the total mass of the disk needed to be included in the calculation.
Q7.23. Reason: Consider the angular acceleration of the bat as it is just about to tip, as in Example 7.16. The center of
gravity of the bat will be nearer the barrel because most of the bat’s mass is concentrated in the barrel. With the barrel
up, the moment arm for the gravitational torque is larger than with the barrel down. The moment of inertia of the bat is
also larger when the barrel is up. Torque is proportional to the length of the moment arm of the applied force. Moment
of inertia depends on the square of the distance to the particles that make up the object. The moment of inertia of the bat
with the barrel up is much larger than with the handle down, though the gravitational torque is only slightly larger. The
bat is easier to balance with the barrel up, so the correct choice is A.
Assess: This result makes sense.
Q7.24. Reason: A particle on the positive y-axis has an angle of θ = π /2 rad. An equivalent angle can be obtained
by adding or subtracting 2π rad any number of times. In choice A, we have the basic angle, π /2 rad. In choice C,
2π rad has been added and in part D, 2π rad has been subtracted. Choice B, however, is π rad, which is a
completely different angle, so the answer is B.
Assess: Two angles may look different numerically, but if their difference is a multiple of 2π rad, then they are the
same angle.
Q7.25. Reason: The skater turns one-and-a-half revolutions in 0.5 s. One-and-a-half revolutions is 3π radians. Her
angular velocity is
ω=
Δθ 3π rad
=
= 20 rad/s
Δt
0.5 s
The correct choice is D.
Assess: This result is reasonable. She makes three revolutions in one second, which is 6π radians per second.
Q7.26. Reason: We first find ω (in rad/s) and then the centripetal acceleration.
Δθ 1.5 rev ⎛ 2π rad ⎞
ω=
=
⎜
⎟ = 18.8 rad/s
0.5 s ⎝ 1 rev ⎠
Δt
Let’s make an estimate (one significant figure) for how far the hand is from the center of the body (the axis of
rotation). Either think of a meter stick, or get one. From the center of your body to your hand is about 0.8 m for an
average person. Now use Equation 6.1:
a = ω 2 r = (18.8 rad/s) 2 (0.8 m) = 284 m/s 2 ≈ 300 m/s 2
The correct choice is C.
Assess: We kept an extra “guard digit” in the intermediate calculations, but rounded the final result to one
significant figure since that is all the data justifies.
The result of 300 m/s 2 is large—about 30g. If your whole body experienced 30g for more than a few seconds you
would black out. However, this is only the skater’s hand and only for a short time.
Q7.27. Reason: In Question Q7.25 we found that the angular velocity of the skater is 20 rad/s. Estimating the
length of her arm to be about 0.75 m from the center of rotation, we can calculate the speed of her hand.
v = ω r = (20 rad/s)(0.75 m) = 15 m/s
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7-6
Chapter 7
The correct choice is D.
Assess: This is actually a high velocity, over 30 mph.
Problems
P7.1. Prepare: The position of the minute hand is determined by the number after the colon. There are
60 minutes in an hour so the number of minutes after the hour, when divided by 60, gives the fraction of a circle
which has been covered by the minute hand. Also, the minute hand starts at π /2 rad and travels clockwise, thus
decreasing the angle. If we get a negative angle, we can make it positive by adding 2π rad.
Solve: (a) The angle is calculated as described above. Since the number after the colon is 0, we subtract nothing
from π /2 rad, so θ = π /2.
(b) We subtract 15/60 of 2π rad from the starting angle, so we have:
π ⎛ 15 ⎞
θ = − ⎜ ⎟ (2π ) = 0
2 ⎝ 60 ⎠
(c) As before, the angle is given by:
π ⎛ 35 ⎞
2
θ = − ⎜ ⎟ (2π ) = − π
2 ⎝ 60 ⎠
3
Since this angle is negative, we can add 2π rad to obtain: θ = − 2π /3 + 2π = 4π /3.
Assess: The first two parts make sense from our experience with clocks. In part (a), the minute hand is straight up. In
part (b), it points to the right.
P7.2. Prepare: This problem gives us a period of rotation and a time interval. If we find the angular velocity, we
can use Δ θ = ω Δt.
Solve: Since he or she completes one revolution in 3.0 s, his or her angular velocity is given by:
ω = Δθ /Δ t = (2π rad) /(3.0 s) = 2.09 rad/s. We include an extra digit beyond the significant two digits since this is
an intermediate step. Now we can find his or her angular displacement as described previously:
Δ θ = (2.09 rad/s)(1.0 s) = 2.1 rad
Assess: This makes sense because if he or she completes one revolution in 3.0 s, then he or she completes one third
of a revolution in 1.0 s.
P7.3. Prepare: To compute the angular speed ω we use the equation in the text and convert to rad/s. The minute
hand takes an hour to complete one revolution.
Solve:
Δθ 1.0 rev ⎛ 2π rad ⎞ ⎛ 1 min ⎞
ω=
=
= 0.0017 rad/s = 1.7 × 10−3 rad/s
Δ t 60 min ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠
Assess: This answer applies not just to the tip, but the whole minute hand. The answer is small, but the minute hand
moves quite slowly.
The second hand moves 60 times faster, or 0.10 rad/s. This too seems reasonable.
P7.4. Prepare: Every part of the record (on a turntable) is rotating at 45 rpm. We note that 1 rev = 2π radians and
1 min = 60 s. To find the period, we will use the equation in the text.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotational Motion
Solve: (a) The angular speed is
ω = 45 rpm ×
(b) The period is
T=
7-7
1 min 2π rad
×
= 1.5 π rad/s
60 s
1 rev
2π rad
ω
=
2π rad
= 1.3 s
1.5 π rad/s
P7.5. Prepare: The airplane is to be treated as a particle in uniform circular motion on the equator around the
center of the earth. We show the following pictorial representation of the problem and a list of values. To convert
radians into degrees, we note that 2π rad = 360°.
Solve: (a) The angle you turn through is
s 5000 mi
180 °
= 1.25 rad = 1.25 rad ×
= 71.6°
θ f − θi = =
π rad
r 4000 mi
So, the angle is 1.3 rad or 72°.
(b) The plane’s angular speed is
θ − θ 1.25 rad
rad
1h
= 0.139 rad/h = 0.139
×
= 3.9 × 10−5 rad/s
ω= f i =
t f − ti
9h
h 3600 s
Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the
separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference.
P7.6. Prepare: Since the angular velocity of the Ferris wheel is positive, the wheel is rotating counterclockwise.
Solve: Refer to the following diagram.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7-8
Chapter 7
Since all angles in this problem are measured from the top of the Ferris wheel, we set the x-axis to be along the line
from the center of the wheel to the top of the wheel. Seth starts out at the top of the wheel, which is at an angle of
θi = 0 rad.
θ f = θi + ω Δt = (0.036 rad/s)(180 s) = 6.48 rad
An additional significant figure has been kept in this intermediate result. Converting to degrees
⎛ 360° ⎞
θ f = (6.48 rad) ⎜
⎟ = 370°
⎝ 2π rad ⎠
Note that since the answer is larger than 360 °, Seth has made more than one complete rotation on the Ferris wheel.
Since one rotation is 360°, Seth’s angular position is actually only 370° − 360° = 10° counterclockwise from the
top of the wheel.
Assess: Choosing the coordinate system aligned with the top of the wheel simplified the calculations.
P7.7. Prepare: We’ll use the equation in the text to compute the angular displacement. We are given θi = 0.45 rad
and that Δ t = 8.0 s − 0 s = 8.0 s.
We’ll do a preliminary calculation to convert ω = 78 rpm into rad/s:
78 rpm = 78
rev ⎛ 2π rad ⎞ ⎛ 1 min ⎞
= 8.17 rad/s
min ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠
Solve: Solve the equation for θ f :
θ f = θi + ω Δt = 0.45 rad + (8.17 rad/s)(8.0 s) = 65.8 rad = 10.474 × 2π rad
= 10 × 2π rad + 0.474 × 2π rad = 10 × 2π rad + 2.98 rad
So the speck completed almost ten and a half revolutions. An observer would say the angular position is 3.0 rad (to
two significant figures) at t = 8.0 s.
Assess: Ask your grandparents if they remember the old records that turned at 78 rpm. They turned quite fast and so
the music didn’t last long before it was time to turn the record over.
Singles came on smaller records that turned at 45 rpm, and later “long play” (LP) records turned at 33 rpm.
CDs don’t have a constant angular velocity, instead they are designed to have constant linear velocity, so the motor
has to change speeds. For the old vinyl records the recording had to take into account the changing linear velocity
because they had constant angular velocity.
P7.8. Prepare: We can use an equation from the text.
Solve: One revolution is 2π rad, so the superhero has run (5.5)(2π rad) = 11π rad. Her angular speed is
11π rad
= 12 rad/s
ω=
3.0 s
Assess: Note that the diameter of the track was not needed in the calculation. Angular speed is independent of radius
in uniform circular motion.
P7.9. Prepare: To find angular displacement we simply subtract the initial value of the angle from the final value
of the angle: Δ θ = θ F − θ I .
Solve: (a) At t = 5 s, the angle is 100 rad and at t = 15 s, the angle is 125 rad. Thus
Δ θ = 125 rad − 100 rad = 25 rad
(b) The angular velocity of the wheel at 15 s is the slope of the θ vs t graph at 15 s. We can find this slope by
comparing the angle at 10 s and the angle at 20 s:
θ − θ 100 rad − 150 rad
= − 5 rad/s
ω= F I =
20 s − 10 s
tF − tI
Assess: The angular velocity is negative at t = 15 s because the angle is decreasing, that is, the wheel is rotating
clockwise.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotational Motion
7-9
P7.10. Prepare: The angular displacement of a rotating object equals the area under the angular velocity graph.
We need to convert the angular velocities from rpm to rad/s. The two which are nonzero are 100 rpm and
250 rpm:
100
rev ⎛
rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ 10 rad
and
= 100
= π
min ⎜⎝
min ⎟⎠ ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠ 3
s
rev ⎛
rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ 25 rad
= 250
=
π
min ⎜⎝
min ⎟⎠ ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠ 3
s
Solve: (a) The area under the angular velocity graph from t = 0 s to t = 20 s is the sum of the area from 0 s to 5 s
and from 5 s to 15 s and from 15 s to 20 s:
⎛ 25 rad ⎞
⎛ 10 rad ⎞
Δ θ = (0 rad/s)(5 s) + ⎜ π
(10 s)+ ⎜ π
(5 s) = 100π rad
⎟
s ⎠
s ⎟⎠
⎝ 3
⎝ 3
If θ = 0 rad at t = 0 s, the angle at t = 20 s is 100π rad. We can subtract 2π from this angle 50 times to obtain an
angle between 0 and 2π . The result is 0 rad.
(b) The blade does not turn during the first five seconds. It turns (250 rev/min)(10 s)(1 min/60 s) = 42 rev during
the interval from 5 s to 15 s. Thus the 10 full revolutions are completed during that time interval. We can solve the
equation ω = Δ θ / Δt to obtain Δ t = Δ θ / ω. The amount of time the blade must turn at 250 rev/min to turn 10
revolutions is given by:
10 rev
1
⎛ 1
⎞⎛ 60 s ⎞
min = ⎜
min ⎟⎜
Δt =
=
⎟ = 2.4 s
250 rev/min 25
⎝ 25
⎠⎝ 1 min ⎠
The blade has turned 10 revolutions 2.4 s into the second time interval at t = 7.4 s.
Assess: If we convert 250 rpm to units of rev/s we have about 4 rev/s. So it makes sense that we have to wait
about 2.5 s for the blade to complete 10 revolutions.
250
P7.11. Prepare: The smooth rotation assures constant angular velocity.
Solve: (a) The second hand has a rotational speed of one revolution per minute, or 1 rpm. We simply need to convert
this to SI units.
rev ⎛ rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞
rad
1
= 1
= 0.105
min ⎜⎝ min ⎟⎠ ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠
s
(b) The tip of the hand has speed
v = ω R = (0.105 rad/s)(0.0100 m) = 0.00105 m/s
Assess: This is about a millimeter per second, which is about right.
P7.12. Prepare: Ignore the earth’s orbit around the sun.
Solve: (a) The angular speed is 1 rev/24 hr, but we should convert this to SI units.
1 rev ⎛ 1 rev ⎞⎛ 2π rad ⎞⎛ 1 h ⎞⎛ 1 min ⎞
rad
=⎜
= 7.3 × 10−5
⎟⎜
⎟⎜
⎟⎜
⎟
24 h ⎝ 24 h ⎠⎝ 1 rev ⎠⎝ 60 min ⎠⎝ 60 s ⎠
s
(b) To get the sun to “rise” in the east and “set” in the west the earth must turn counterclockwise as seen from above
the north pole, so the angular velocity is positive.
(c) A point on the equator has speed
v = ω R = (7.3 × 10−5 rad/s)(6.38 × 106 m) = 460 m/s
(d)
A
point
cos 45° Rearth =
halfway
between
the
equator
and
the
north
pole
has
an
orbital
radius
of
2
Rearth = 0.707(6.38 × 106 m) = 4.51 × 106 m and a speed of
2
v = ω R = (7.3 × 10−5 rad/s)(4.51× 106 m) = 330 m/s
Assess: The speed of a point on the equator is about 1000 mi/h.
P7.13. Prepare: We’ll assume a constant acceleration during the one revolution. We’ll use the second and third
equations for circular motion in Synthesis 7.1., the third to find α and then the second to find ωf .
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7-10
Chapter 7
Known
r = 12 D = .90 m
Δ t = 1.0 s
Δ θ = 1.0 rev = 2π rad
Find
vf = rωf
Solve: Using Synthesis 7.1 allows us to solve for α . That ω0 = 0 makes it easier.
1
Δθ = α (Δt )2
2
2Δθ 2(2π rad)
=
= 12.6 rad/s 2
α=
(Δt )2
(1.0 s) 2
Synthesis 7.1 gives ωf :
ωf = ω0 + α Δt = 0 rad/s + (12.6 rad/s 2)(1.0 s) = 12.6 rad/s
Finally, we compute vf = rωf = (.90 m)(12.6 rad/s) = 11 m/s.
Assess: This speed seems reasonable, about 1/4 of a baseball fast pitch. The hammer throw is similar to the discus,
but the weight is on a wire so the radius of the circular motion is a bit longer than the arm and the release speed is a
bit larger, hence the distance it goes before landing is a few meters more.
P7.14. Prepare: The motion has two parts: the disk accelerates for some time until it reaches a constant angular
velocity and then rotates at this constant angular velocity for the remainder of the time. We can use the equations in
Synthesis 7.1 to calculate the number of turns the disk has made.
Solve: One additional significant figure has been kept in each of the intermediate results. The disk starts from rest,
so ωi = 0 rad/s. The disk’s final angular velocity converted to rad/s is
⎛ 7200 rev ⎞⎛ 2π rad ⎞ ⎛ 1 min ⎞
⎟ = 754 rad/s
⎟⎜
⎟⎜
⎝ min ⎠⎝ rev ⎠ ⎝ 60 s ⎠
ωf = 7200 rpm = ⎜
We can find the time the disk takes to accelerate to this angular velocity using Synthesis 7.1.
ω − ωi 754 rad/s − 0 rad/s
Δ taccelerating = f
=
= 3.97 s
α
190 rad/s 2
Using Synthesis 7.1., the total angular displacement of the disk during this time is
1
1
Δ θ accelerating = α ( Δ taccelerating ) 2 = (190 rad/s 2)(3.97 s) 2 = 1500 rad
2
2
After it has reached its final angular velocity ωf , the disk spins with that angular velocity for the remainder of the
time. The angular displacement during this time is given by Synthesis 7.1.
Δ θ constant ω = ω f Δ tconstant ω = (754 rad/s)(10.0 s − 3.97 s) = 4550 rad
The total angular displacement of the disk is Δ θ = 1500 rad + 4550 rad = 6050 rad.
Converting this result to revolutions, we have Δ θ = 960 revolutions.
Assess: It’s important to realize that there are two different parts to this motion.
P7.15. Prepare: We assume constant angular acceleration; then we can use Synthesis 7.1.
Known
Δ t = 2.0 s
ω0 = 0
ωf = 3000 rpm
Find
α
Δθ
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Rotational Motion
7-11
Convert Δ ω to rad/s.
Δ ω = ωf −ω0 = 3000
Solve: (a)
α=
rev ⎛ 2π rad ⎞⎛1 min ⎞
⎜
⎟⎜
⎟ = 314 rad/s
min ⎝ 1 rev ⎠⎝ 60 s ⎠
Δ ω 314 rad/s
=
≈ 160 rad/s 2
Δt
2.0 s
(b) We’ll use Synthesis 7.1.2, using ω0 = 0.
1
1
Δ θ = ω0 Δ t + α (Δt ) 2 = (157 rad/s 2 )(2.0 s) 2 = 314 rad
2
2
Finally, convert to revolutions:
⎛ 1 rev ⎞
314 rad = 314 rad ⎜
⎟ = 50 rev
⎝ 2π rad ⎠
Assess: 50 rev seems like a lot in 2 s, but it is reasonable with the large angular acceleration and the final angular
velocity of 3000 rpm.
P7.16. Prepare: We can use Equation 7.10 to calculate the torque in this case.
Solve: Consider the diagram in Example 7.10 with the radius of the wrench increased to 35 cm.
The torque Luis exerts is given by τ = − rF⊥ = − rF sinφ. The angle and the torque Luis exerts are the same as in the
example. Solving for the force required for a wrench with a longer handle,
−τ
− (− 17 N ⋅ m)
F=
=
= 56 N
r sin φ (0.35 m)sin 120 °
Assess: This result makes sense. The radius has been increased by almost a factor of two, so only about one half of
the force is required to exert the same torque.
P7.17. Prepare: The magnitude of the torque in each case is τ = rF because sin φ = 1.
Solve: τ 1 = rF
τ 2 = r 2F
τ 3 = 2rF
τ 4 = 2r 2 F
Examining the above we see that τ 1 < τ 2 = τ 3 < τ 4 . Since for each case τ = rF (because sin φ = 1 ), in order to
determine the torque we have just kept track of each force (F), the magnitude of the position vector (r) which locates
the point of application of the force, and finally the product (rF).
Assess: As expected, both the force and the lever arm contribute to the torque. Larger forces and larger lever arms
make larger torques. Case 4 has both the largest force and the largest lever arm, hence the largest torque.
P7.18. Prepare: Refer to Figure P7.6. We will calculate the torques τ1 through τ6 about the hinge using Equation
7.5, that is, τ = rFsinφ.
Solve: τ1 = (L/4)(2F) sin 90° = LF/2; τ2 = (L/2)(F) sin 45° =
2
2
τ1; τ3 = (L/2)F sin 90° = τ 1 ; τ4 = (L/2)F sin 135° = τ 2 ;
τ5 = L(2F) sin 90° = 4τ1; and τ6 = LF sin 0° = 0. So, from smallest to largest, the ranking order of the torques is
τ 6 < τ 2 = τ 4 < τ1 = τ 3 < τ 5.
Assess: Note that φ is the angle between r and F .
P7.19. Prepare: Torque by a force is defined as τ = Fr sin φ [Equation 7.12], where φ is measured
counterclockwise from the r vector to the F vector. The radial line passing through the axis of rotation is shown
below by the broken line. We see that the 20 N force makes an angle of + 90° relative to the radius vector r2 , but the
30 N force makes an angle of − 90° relative to r1.
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7-12
Chapter 7
Solve: The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N
force:
(30 N)r1 sin φ1 + (20 N) r2 sin φ2 = (30 N)(0.02 m)sin(− 90°) + (20 N)(0.02 m)sin (90°)
= ( − 0.60 N ⋅ m) + (0.40 N ⋅ m) = − 0.20 N ⋅ m
Assess: A negative torque will cause a clockwise rotation of the pulley.
P7.20. Prepare: Torque by a force is defined through Equation 7.12. The radial line passing through the axis of
rotation is shown below by the broken line. We see below that the force F applied by your pull makes an angle of
− 120° relative to the radial line.
Solve: The net torque on the spark plug is
τ = Fr sin φ = − 38 N ⋅ m = F (0.25 m)sin(− 120 °) ⇒ F = 180 N
That is, you must pull with a force of 180 N to tighten the spark plug.
Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for
the net torque.
P7.21. Prepare: The height, thickness, and mass of the door are all irrelevant (for this problem, but the mass is
important for Problem 7.42). If the door closer exerts a torque of 5.2 N ⋅ m, then you need to also apply a torque of
5.2 N ⋅ m in the opposite direction. The way to do that with the least force is to make r as big as possible (the entire
width of the door), and make sure the angle φ = 90 °.
Solve: From τ = rF sin φ solve for F . Then we see that the needed torque is produced with the smallest force by
maximizing r and sin φ .
τ
5.2 N ⋅ m
F=
=
= 5.7 N
r sin φ (0.91 m) sin 90°
Assess: It is good to have problems where more than the required information is given. Part of learning to solve realworld problems is knowing (or learning) which quantities are significant, which are irrelevant, and which are
negligible. Of course, in this case the mass is used later in Problem 7.42.
The answer of 5.7 N seems like a reasonable amount of force, which might be supplied, say, by a doorstop. If your
doorstop is a simple wedge of wood inserted under the door (as they are at my college), you can see that it should be
positioned near the outside edge of the door so the friction force will produce enough torque to keep the door open.
P7.22. Prepare: We can use Equations 7.12 and 7.10 to calculate the torques due to each force and then set the net
torque to zero to solve for F2 .
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Rotational Motion
7-13
Solve: Using Equation 7.11 to calculate the torque due to F1 with φ1 = 45 °,
τ1 = + r1F1 sin φ1 = r1 (20.0 N) sin 45°
At this point, we can’t calculate the numerical value of this torque, since r1 is not given. Since F2 is perpendicular to
the radial line the torque due to F2 is
⎛r ⎞
τ 2 = − r2 ( F2 ) = −r2 F2 = − ⎜ 1 ⎟ F2
2
⎝ ⎠
A negative sign has been inserted since F2 tends to turn the rod clockwise.
Setting the net torque to zero gives the equation
⎛r⎞
τ1 + τ 2 = r1 (20.0 N) sin 45 ° − ⎜ 1 ⎟ F2 = 0
⎝ 2⎠
Solving for F2 gives
F2 = 2(20.0 N) sin 45° = 28.3 N
Assess: This answer makes sense. F1 is twice as far from the pivot but its contribution to the torque is diminished by
almost a factor of two due to the angle it makes with the radial line. The force F2 should be a larger than the force
F1.
P7.23. Prepare: For both Tom and Jerry r = 1.5 m (or half the diameter).
Compute the magnitude of each torque.
τ Tom = (1.50 m)(50.0 N) sin 60° = 64.95 N ⋅ m
τ Jerry = (1.50 m)(35.0 N) sin 80° = 51.70 N ⋅ m
We keep extra significant figures in the intermediate calculations because we will be subtracting two nearly equal
numbers.
Solve: (a) The torque due to Tom will be positive because it will tend to produce counterclockwise rotation, while
the torque due to Jerry will be negative because it will tend to produce clockwise rotation.
τ = τ Tom + τ Jerry = 64.95 N ⋅ m + (− 51.70 N ⋅ m) = 13.2 N ⋅ m
(b) With both Tom’s and Jerry’s torques tending to produce counterclockwise rotation, they are both positive and so
the net torque would be their sum:
τ = τ Tom + τ Jerry = 64.95 N ⋅ m + 51.70 N ⋅ m = 116.6 N ⋅ m
This rounds to 117 N ⋅ m to 3 significant figures.
Assess: As the difference between the two parts of the problem demonstrates, the direction of the forces really
matters. But we can produce torques of the same magnitude with forces of different magnitude by adjusting the
angles—even with the same r.
P7.24. Prepare: Assume the bar is weightless. We can calculate the torques due to both forces using Equation 7.10.
Solve: Refer to the diagram below.
For F1
τ 1 = r1 ( F1 ) ⊥ = (0.75 m)(10 N) = 7.5 N ⋅ m
This torque is positive since F1 tends to turn the bar counterclockwise.
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7-14
Chapter 7
For F2
τ 2 = r2 ( F2 ) ⊥ = (0.75 m)(0 N) = 0 N ⋅ m
The net torque about the pivot is τ 1 + τ 2 = 7.5 N ⋅ m.
Assess: The torque due to F2 is zero because it has no component perpendicular to the radial line.
P7.25. Prepare: Knowing that torque may be determined by τ = rF⊥ , that counterclockwise torque is positive and
clockwise torque negative we can determine the net torque acting on the bar.
\
Solve:
τ clockwise = − (0.25 m)(8.0 N) = − 2.0 N ⋅ m
τ counterclockwise = (0.75 m)(10 N) = 7.5 N ⋅ m
τ net = τ counterclockwise + τ clockwise = 7.5 N ⋅ m − 2.0 N ⋅ m = 5.5 N ⋅ m
Since the net torque is + 5.5 N ⋅ m, the bar will rotate in the counterclockwise direction around the dot.
Assess: The counterclockwise torque had both the larger r and the larger F, so the net torque was also
counterclockwise. The numbers also seem reasonable, and the units work out.
P7.26. Prepare: We can use Equation 7.10 to calculate the net torque on the bar. Recall that clockwise torques are
negative and counterclockwise torques are positive.
Solve: Refer to the diagram below.
Calculate the torque associated with F1
τ 1 = r1 ( F1 ) ⊥ = r1F1 sin(θ1 ) = (0.75 m)(10 N)sin(30 °) = 3.75 N ⋅ m
An additional significant figure has been kept in the intermediate result above.
Calculate the torque associated with F2
τ 2 = − r2 ( F2 ) ⊥ = − r2 F2 sin(θ 2) = −(0.25 m)(8.0 N)sin(40 °) = − 1.29 N ⋅ m
The net torque is
τ 1 + τ 2 = 3.75 N ⋅ m − 1.29 N ⋅ m = +2.46 N ⋅ m
Since all information is given to two significant figures, the result should be reported to two significant figures as
τ net = 2.5 N ⋅ m
Assess: This result makes sense. Force F1 is at a much larger distance from the axis than F2 , while both forces
are close in magnitude and in the angle they make with the radial line. The net torque should act to turn the
bar counterclockwise.
P7.27. Prepare: We’ll set up the coordinate system so the barbell lies along the x-axis with the origin at the left end
and use Equation 7.14 to determine the x-coordinate of the center of gravity. We’ll model each weight (and the
barbell itself in part (b)) as a particle, as if the mass were concentrated at each item’s own center of gravity.
Solve: (a)
xm +x m
(0.0 m)(35 kg) + (1.7 m)(20 kg)
xcg = 1 1 2 2 =
= 0.62 m
m1 + m2
35 kg + 20 kg
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Rotational Motion
7-15
(b) In this part we include the barbell (call it particle 3), whose own center of gravity is at its geometrical center
( x3 = 1.7 m/2 = 0.85 m), since we assume it has uniform density.
x1m1 + x2 m2 + x3m3 (0.0 m)(35 kg) + (1.7 m)(20 kg) + (0.85 m)(8.0 kg)
=
= 0.65 m
m1 + m2 + m3
35 kg + 20 kg + 8.0 kg
Assess: To two significant figures the answers to both parts are close. Taking the barbell into account didn’t move
the center of gravity much for two reasons: It wasn’t very massive, and its center of mass was already near the center
of mass of the system.
xcg =
P7.28. Prepare: The procedure in Tactics Box 7.1 can be used to calculate the center of gravity of the coins.
Solve: Refer to the figure below.
The coordinates for the three masses are
x1 = 0 m
y1 = 0 m
x2 = 0 m
y2 = 0.100 m
x3 = 0.100 m
y3 = 0 m
All the coins have the same mass since they are identical. The x- and y -coordinates of the center of gravity of the
coins may be determined by
xm +x m +x m
x m + x2 m + x3m x3m 1
1
xcg = 1 1 2 2 3 3 = 1
=
= x3 = (0.100 m) = 0.0333 m
m1 + m2 + m3
m+m+m
3m 3
3
y1m1 + y2 m2 + y3m3 y1m + y2 m + y3m y2 m 1
1
=
=
= y2 = (0.100 m) = 0.0333 m
m1 + m2 + m3
m+m+m
3m 3
3
Assess: Most of the mass lies near the x- and y -axes, so this answer makes sense.
ycg =
P7.29. Prepare: How will you estimate the mass of your arm? Of course different people’s arms have different
masses, but even different methods of estimating the mass of your specific arm will produce slightly different results.
But it is a good exercise, even if we have only one significant figure of precision. One way would be to make some
rough measurements. Because the density of your arm is about the density of water, you could fill a garbage can to
the brim with water, insert your arm, and weigh the water that overflows. Or you could look at Figure 7.31, which
indicates that a guess of m = 4.0 kg for a person whose total mass is 80 kg is good.
The gravitational force on a 4.0 kg object is w = mg(4.0 kg)(10 m/s 2 ) = 40 N.
I know it is about 1 yd from the tip of my outstretched arm to my forward-facing nose, but a meter is a bit bigger than
a yard, so I would estimate that from fingertip to shoulder would be about 0.7 m. If I model my arm as a uniform
cylinder of uniform density then the center of gravity would be at its center—0.35 m from the shoulder joint (the
“hinge”). If I further refine the model of my arm, it is heavier nearer the shoulder and lighter toward the hand, so I
will round the location of its center of gravity down to 0.30 m from the shoulder.
Solve: Since the arm is held horizontally sin φ = 1.
τ = rF sin φ = (0.30 m)(40 N)(1) = 12 N ⋅ m
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7-16
Chapter 7
Assess: Your assumptions and estimates (and, indeed, your arm) might be different, but your answer will probably
be in the same order of magnitude; you probably won’t end up with an answer 10 times bigger or 10 times smaller.
P7.30. Prepare: The procedure in Tactics Box 7.1 provides a guide for determining the center of gravity of an
object and/or combination of objects. It is important to note that the x-component of the center of gravity is on the
line of symmetry.
Solve: The y-component of the center of gravity is determined by:
y m + y2 m2 (2.5 cm)(800 g) + (7.5 cm)(400 g)
ycm = 1 1
=
= 4.2 cm
m1 + m2
800 g + 400 g
Assess: Note that it was not necessary to convert the units to the MKS system. Since the mass units cancel, any
appropriate unit may be used. Since the number is small, cm is an appropriate unit. Note that the center of gravity of
the two objects is inside the cube and beneath the cylinder. Since they have the same height but the cube has twice
the mass, this is a reasonable answer.
P7.31. Prepare: First let’s divide the object into two parts. Let’s call part #1 the part to the left of the point of
interest and part #2 the part to the right of the point of interest. Next using our sense of center of gravity, we know
the center of mass of part #1 is at 12.5 m and the center of mass of part #2 is at +37.5 cm. We also know the mass of
part #1 is one fourth the total mass of the object and the mass of part #2 is three fourths the total mass of the object.
Finally, we can determine the gravitational torque of each part using any of the three expressions for torque as shown
below:
Equation 7.3 τ = rF⊥ is straightforward to use because the forces are perpendicular to the position vectors, which
locate the point of application of the force.
Equation 7.4 τ = r⊥ F is straightforward to use because the position vectors that locate the point of application of the
forces are also the moment arms for the forces.
Equation 7.4 τ = rF sin φ is straightforward to use because the angles are either 90 or 270 .
Solve: Using Equation 7.4 we obtain the following
τ net = τ 1 + τ 2 = m1gr1sin(90 ) + m2 gr2sin(−90 )
= (0.5 kg)(9.8 m / s )(−0.125 m)(1) + (0.75 kg)(9.8 m / s2 )(0.375 m)(−1) = −2.1 N ⋅ m
Assess: According to this answer, if released the object should rotate in a clockwise direction. Looking at the figure
this is exactly what we would expect to happen. It might be easier to consider the whole rod as if acting at its center
of mass and then the distance would be just 25 cm and there would be only one torque calculation.
2
P7.32. Prepare: The beam is a solid rigid body. The steel beam experiences a torque due to the construction
worker’s weight wC and the beam’s weight wB. The normal force exerts no torque since the net torque is calculated
about the point where the beam is bolted into place. The net torque on the steel beam about point O is the sum of the
torque due to wC and the torque due to wB. The weight of the beam acts at the center of mass.
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Rotational Motion
Solve:
7-17
τ = (wC )(4.0 m) sin ( − 90°) + (wB )(2.0 m) sin ( − 90°)
= − (70 kg)(9.80 m/s 2 )(4.0 m) − (500 kg)(9.80 m/s 2 )(2.0 m)= − 12,000 N ⋅ m
The magnitude of the gravitational torque is 12,000 N ⋅ m.
Assess: The negative torque means these forces would cause the beam to rotate clockwise, as you can see without
doing any calculations.
P7.33. Prepare: Model the arm as a uniform rigid rod. Its mass acts at the center of mass. The force in each case is
the weight, w = mg .
Solve: (a) The torque is due both to the weight of the ball and the weight of the arm.
τ = τ ball + τ arm = rb ( mb g )sin 90 ° + ra ( ma g )sin 90 °
= (0.70 m)(3.0 kg)(9.80 m/s 2) + (0.35 m)(4.0 kg) (9.80 m/s 2 ) = 34 N ⋅ m
(b) The torque is reduced because the moment arms are reduced. Both forces act at φ = 45° from the radial line, so
τ = τ ball + τ arm = rb (mb g )sin 45° + ra (ma g )sin 45°
= (0.70 m)(3.0 kg)(9.80 m/s 2)(0.707) + (0.35 m)(4.0 kg) (9.80 m/s 2 )(0.707) = 24 N ⋅ m
Assess: This problem could also have been done by first finding the center of mass of the arm-ball system and
computing the torque due to the combined weight acting at that point. The final answers would be the same.
P7.34. Prepare: We can use Equation 7.10 or 7.11 to calculate the torque. The weight of the beam acts at the
center of gravity of the beam.
Solve: The center of gravity of the beam is at the center of the beam, which is 1.0 m from the hinge. Each of the
torques is negative since gravity tends to turn the bar in a clockwise direction in each case. The weight of the beam is
w = mg = (15 kg)(9.80 m/s 2 ) = 147 N. An additional significant figure has been kept in this intermediate result.
(a) A diagram of the beam in the upper position is below.
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7-18
Chapter 7
We can calculate the torque using Equation 7.10. The perpendicular component of the weight is w cos(θ ). The
torque is
τ = −rw⊥ = − rw cos(θ ) = −(1.0 m)(147 N)cos(20°) = − 140 N ⋅ m
(b) A diagram of the beam in the middle position is below.
We can calculate the torque using Equation 7.12. The radial line is along the x-axis. The angle between the weight
and the radial line is 90 °.
τ = −rw sinφ = −(1.0 m)(147 N)sin(90°) = − 150 N ⋅ m
(c) A diagram of the beam in its lower position is below.
Equation 7.11 can be used to calculate the torque. The moment arm for the force is shown in the diagram. It is the
perpendicular distance from the hinge to the line of action of the force. The torque is
τ = −r⊥ w = −(r cos θ ) w = −(1.0 m)cos(45°)(147 N) = − 100 N
Assess: Any one of Equations 7.10 through 7.12 can be used to calculate torque.
P7.35. Prepare: Set up a coordinate system along the beams with the origin at the left end. Assume each beam is
of uniform density so its own center of gravity is at its geometrical center. Then use Equation 7.14.
Solve: (a)
xm +x m
(0.500 m)(10.0 kg) + (1.00 m + 1.00 m)(40.0 kg)
xcg = 1 1 2 2 =
= 1.70 m
m1 + m2
10.0 kg + 40.0 kg
(b) The gravitational torque on the two-beam system is the total weight acting at the center of gravity of the system:
w = mg = (10.0 kg + 40.0 kg)(9.80 m/s 2 ) = 490 N.
τ = rF sin φ = (1.70 m)(490 N)sin 90 ° = 833 N ⋅ m
Because this torque is in the clockwise direction we report it as a negative torque: τ = −833 N ⋅ m.
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Rotational Motion
7-19
Assess: This problem could also be done by computing the gravitational torque individually on each beam and
adding them up. The final answer would be the same. For the beams to remain in equilibrium some other object must
supply an equal torque in the opposite direction.
P7.36. Prepare: Equations 7.15 and 7.16 can be used to calculate the center of gravity of the joined beams. To
calculate the gravitational torque we can use the fact that the weight of the entire structure acts on its center of
gravity.
Solve: (a) The center of gravity of each beam is at its center. Refer to the diagram below.
Using Equations 7.16 and 7.17, the position of the center of gravity is
m x + mh xh cg
mx
(25.0 kg)(1.00 m)
xcg = v v cg
= h h cg =
= 0.625 m
mv + mh
mv + mh (25.0 kg) + (15.0 kg)
ycg =
mv yv cg + mh yh cg
mv + mh
=
mv yv cg
mv + mh
=
(15.0 kg)(0.500 m)
= 0.188 m
(25.0 kg) + (15.0 kg)
(b) The gravitational torque about the origin is
τ = −r⊥ w = −(0.625 m)(15.0 kg + 25.0 kg)(9.80 m/s 2 ) = −245 N ⋅ m
Assess: This result makes sense. The center of gravity should be close to the vertical and horizontal axes.
P7.37. Prepare: A table tennis ball is a spherical shell, and we look that up in Table 7.1. The radius is half the
diameter.
Solve:
I = 32 MR 2 = 32 (0.0027 kg)(0.020 m) 2 = 7.2 × 10−7 kg ⋅ m 2
Assess: The answer is small, but then again, it isn’t hard to start a table tennis ball rotating or stop it from doing so.
By the way, this calculation can be done in one’s head without a calculator by writing the data in scientific notation
and mentally keeping track of the significant figures:
2
2
27
⋅ 2 ⋅ 22 × 10−4 × 10−4 kg ⋅ m 2
I = (2.7 × 10−3 kg)(2.0 × 10−2 m) 2 = (27 × 10−4 kg)(2 × 10−2 m) 2 =
3
3
3
= 9 ⋅ 8 × 10−8 kg ⋅ m 2 = 72 × 10−8 kg ⋅ m 2 = 7.2 × 10−7 kg ⋅ m 2
P7.38. Prepare: We will calculate the moments of inertia I1 through I 3 about axes through the centers of the rods
using Equation 7.21, that is, I = m1r 12 + m2 r 22 .
Solve:
I1 = m( R/2) 2 + m( R/2) 2 = mR 2/ 2
I 2 = 2m( R/4) 2 + m( R/4) 2 = 3mR 2/16 = 3I1 / 8
and
I 3 = ( m/2) R 2 + ( m/2) R 2 = mR 2 = 2 I1
So, from smallest to largest, the ranking order of the moments of inertia is I 2 < I1 < I 3 .
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7-20
Chapter 7
Assess: The moment of inertia is linearly proportional to mass and also to the second power of distance from the
axis. Although the masses in case 3 are one-half the masses in case 1, the distance of the masses from the axis for
case 3 is twice the distance for case 1. As a result I 3 is larger than I1.
P7.39. Prepare: When the problem says the connecting rods are “very light” we know to ignore them (especially
since we aren’t given their mass).
Since all of the mass (seats plus children) is equally far from the axis of rotation we simply use I = MR 2 where M is
the total mass.
Solve:
I = MR 2 = [4(5.0 kg) + 15 kg + 20 kg](1.5 m)2 = 120 kg ⋅ m 2
Assess: When all of the mass is the same distance R from the axis of rotation, the calculation of I is relatively simple.
P7.40. Prepare: We can obtain expressions for the moment of inertia for a cylinder and a sphere from the table in
the text. Given that the masses and the moments of inertia are the same, we can obtain a relationship between their
radii and then determine the radius of the sphere.
Solve: Knowing that the moment of inertia of the cylinder is equal to the moment of inertia of the sphere we may
write
I cylinder = I sphere
Inserting expressions for the moment of inertia from Table 7.1.
M c Rc2 / 2 = 2 M s Rs2 / 5
Cancel the masses (recall they are equal) and solve for Rs
Rs = (5/4)1/ 2 Rc = 4.5 cm
Assess: Knowing that the masses are equal and that the coefficient for the cylinder is 0.50 and for the sphere 0.40,
we should expect the radius of the sphere to be greater than the radius of the cylinder in order for them to have the
same moment of inertia.
P7.41. Prepare: Treat the bicycle rim as a hoop, and use the expression given in Table 7.1 for the moment of
inertia of a hoop. Manipulate this expression to obtain the mass.
Solve: The mass of the rim is determined by
m = I/ R 2 = 0.19 kg ⋅ m 2 /(0.65 m/2) 2 = 1.8 kg
Assess: Note that the units reduce to kg as expected. This amount seems a little heavy, but since we are not told what
type of bicycle it is not an unreasonable amount.
P7.42. Prepare: We will assume we are to compute I about an edge (the hinge of the door), so from Table 7.1 we
have I = 13 Ma 2 where a = 0.91 m and M = 25 kg.
Solve: (a)
I = 13 Ma 2 = 13 (25 kg)(0.91 m) 2 = 6.9 kg ⋅ m 2
(b) After we let go of the door (or remove the doorstop) the net torque is the torque due to the door closer, which
Problem 7.21 says is 5.2 N ⋅ m.
τ net
5.2 N ⋅ m
=
= 0.75 rad/s 2
I
6.9 kg ⋅ m 2
Assess: The results seem to be in a reasonable range. As long as the net torque of 0.52 N ⋅ m continues to act, the
α=
door will continue to accelerate. Usually there are dampeners to exert a slowing torque so that the door doesn’t slam.
P7.43. Prepare: We can calculate the angular acceleration using Equation 7.22.
Solve: Using Equation 7.22 we find
τ = Iα = (14.0 × 10−5 kg ⋅ m 2 )(150 rad/s 2 ) = 6.0 × 10−3 N ⋅ m
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Rotational Motion
7-21
Assess: This result is reasonable. The moment of inertia of the grinding wheel is small.
P7.44. Prepare: We will assume that the torque produced by the frictional force of the floor on your foot is the net
torque (i.e., we will ignore frictional and other torques). We know that F = 7.0 N, r = 0.40 m, and α = 1.8 rad/s 2 .
We are also indirectly told (“in the direction that causes the greatest angular acceleration”) that φ = 90°. We will
solve for I from the rotational version of Newton’s second law: α = τ net / I .
Solve:
I=
τ net rF sin φ (0.40 m)(7.0 N)(sin 90°)
=
=
= 1.6 kg ⋅ m 2
α
α
1.8 rad/s 2
Assess: The answer is not particularly large, but your mass is distributed quite close to the axis of rotation, so I is
small. Carefully observe the units work out since N = kg ⋅ m/s 2 .
P7.45. Prepare: We are given the moment of inertia I and the angular acceleration α of the object. Since τ = Iα
is the rotational analog of Newtons second law F = ma, we can use this relation to find the net torque on the object.
Solve: τ = (2.0 kg ⋅ m 2 )(4.0 rad/s 2 ) = 8.0 kg ⋅ m 2 /s 2 = 8.0 N ⋅ m
Assess: The units are correct and the relatively small torque magnitude is consistent with the size of the moment of
inertia and the angular acceleration.
P7.46. Prepare: A circular plastic disk rotating on an axle through its center is a rigid body. Assume the axis is
perpendicular to the disk. Since τ = Iα is the rotational analog of Newton’s second law F = ma, we can use this
relation to find the net torque on the object. To determine the torque (τ ) needed to take the plastic disk from
ωi = 0 rad/s to ωf = 1800 rpm = (1800)(2π ) / 60 rad/s = 60π rad/s in tf − ti = 4.0 s, we need to determine the
angular acceleration (α ) and the disk’s moment of inertia (I) about the axle in its center. The radius of the disk is
R = 10.0 cm.
Solve: We have
I = 12 MR 2 = 12 (0.200 kg)(0.10 m) 2 = 1.0 × 10−3 kg ⋅ m 2
ωf = ωi + α (tf − ti ) ⇒ α =
ωf − ωi
tf − ti
=
60π rad/s − 0 rad/s
= 15π rad/s 2
4.0 s
Thus, τ = Iα = (1.0 × 10 kg ⋅ m )(15 π rad/s ) = 0.047 N ⋅ m.
−3
2
2
Assess: The solution to this problem required a knowledge of torque, moment of inertia, rotational dynamics and
rotational kinematics. You should consider it an accomplishment to have mastered these concepts and then combined
them to solve a problem.
P7.47. Prepare: Equation 7.22 can be used to find the initial angular acceleration once the torque is known. The
torque can be calculated with Equation 7.11.
Solve: Refer to the diagram below.
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7-22
Chapter 7
Gravity is the only force acting on the object. The object’s weight has a moment arm equal to r⊥ = r cos θ . The
gravitational torque on the object is
τ = −r⊥ w = − rw cos θ = −(0.17 m)(2.5 kg)(9.80 m/s 2 )cos(25°) = − 3.77 N ⋅ m
An additional significant figure has been kept in this intermediate calculation.
The angular acceleration is then
τ
−3.77 N ⋅ m
= − 44 rad/s 2
α= =
I 0.085 kg ⋅ m 2
Assess: Any one of Equations 7.10 through 7.12 can be used to calculate the torque. In this problem, using Equation
7.11 avoids the need to determine the angle between the direction of the vectors r and W.
P7.48. Prepare: What causes angular accelerations? (Net) torques. We’ll apply the rotational version of Newton’s
second law. We’ll write the net torque as Σ τ to emphasize that we are summing the two given torques; the 12 N
force is producing a positive (counterclockwise) torque, while the 10 N force is producing a negative (clockwise)
torque. For each torque R = 0.30 m.
We will assume that the rope comes off tangent to the pulley on each side, so that φ = 90° and sin φ = 1.
Looking up the formula for I of a cylinder in Table 7.1, and using M = 0.80 kg, gives
I = 12 MR 2 = 12 (0.80 kg)(0.30 m) 2 = 0.036 kg ⋅ m 2
Solve:
α=
Σ τ (0.30 m)(12 N) − (0.30 m)(10 N) (0.30 m)(12 N − 10 N)
=
=
= 17 rad/s 2
I
0.036 kg ⋅ m 2
0.036 kg ⋅ m 2
Assess: This result answers the question. The proper units cancel to give α in rad/s 2 .
Notice that the specific angles the ropes make with the vertical do not matter, as long as they are exerting torques in
opposite directions and coming off of the pulley tangentially.
P7.49. Prepare: Equation 7.22 can be used to calculate the frictional torque once the angular acceleration is
known. Using Synthesis 7.1 we can calculate the angular acceleration.
Solve: The wheel stops spinning in 12 seconds from an initial angular velocity of 0.72 revolutions per second. The
initial angular velocity in radians per second is
⎛ 2 π rad ⎞
ωi = 0.72 rev/s = (0.72 rev/s) ⎜
⎟ = 4.5 rad/s
⎝ rev ⎠
The angular acceleration of the wheel is
Δω − 4.5 rad/s
=
= − 0.38 rad/s 2
α=
Δt
12 s
The only force acting on the wheel is friction. Using the results above in Equation 7.22 the magnitude of the
frictional torque is
τ = Iα = (0.30 kg ⋅ m 2 )(0.38 rad/s 2 ) = 0.11 N ⋅ m
Assess: This is a relatively large amount of friction. Assuming the torque of the wheel bearings acts at a radius of
around a centimeter the frictional force exerted by the bearings is over 10 N. This shows you just one of the problems
you will have if you do not properly maintain your bicycle.
P7.50. Prepare: Since the sphere starts from rest we know ωi = 0 rad/s so the kinematic equation is easier. The
moment of inertia for a solid sphere is I = 52 Mr 2 . We’ll also use τ = Iα .
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Rotational Motion
7-23
Solve:
Δθ = 12 α (Δt ) 2 ⇒
Δt =
2Δ θ
α
=
2Δ θ
=
τ /I
2(Δ θ ) I
τ
2( Δθ )( 52 Mr 2 )
=
rF
=
2(2π rad)( 52 (8200 kg)(0.90 m) 2 )
= 27 s
(0.90 m)(50 N)
Assess: The 27 s seems reasonable given the large moment of inertia of the sphere.
P7.51. Prepare: With a constant string tension we will have constant angular acceleration, so we can use the
equations in Synthesis 7.1., but it is also clear that we will need to do a preliminary calculation to get α .
The string will come off tangentially so that φ = 90 °. We will also interpret “to get the top spinning” as ω0 = 0. The
radius is half the diameter.
Known
r = 0.025 m
F = 0.30 N
φ = 90°
I = 3.0 × 10−5 kg ⋅ m 2
Δ θ = 5.0 rev=31.4 rad
ωo = 0
Find
α (Preliminary)
Δt
α=
τ net
I
=
rF sin φ (0.025 m)(0.30 N)(1)
=
= 250 rad/s 2
3.0 × 10−5 kg ⋅ m 2
I
Solve: Solve the angular displacement equation for circular motion Synthesis 7.1 (with ω0 = 0) for Δ t:
Δθ = 12 α (Δt ) 2
Δt =
2Δ θ
α
=
2(31.4 rad)
= 0.50 s
250 rad/s 2
Assess: As you may know from personal experience, it doesn’t take long for a toy top to complete five revolutions,
and this bears that out.
P7.52. Prepare: The rope is a constraint that makes the magnitudes of the accelerations of the blocks the same, and
if the rope doesn’t slip that further implies that α block = α pulley R. See Equation 7.23.
We will assume that the pulley is a solid cylinder, so that I pulley = 12 MR 2 . “Light rope” means we can assume that the
mass of the rope is zero. For the torque calculation r = R = 0.30 m.
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7-24
Chapter 7
Known
R = 0.30 m
m1 = 2.5 kg
m2 = 1.5 kg
M = 0.75 kg
(pulley)
φ = 90°
Find
a
In order to keep track of directions, draw and label a figure.
We will need to set up a system of three equations in three unknowns, because we do not know the tensions T1 and
T2 in the ropes (they are not equal to each other nor to the weights of the blocks).
Solve: Use ∑ F = ma for each block. Call up positive, so block 1 will have a negative (downward) acceleration
because it weighs more than block 2.
T1 − m1 g = − m1a
(1)
T2 − m2 g = m2 a
(2)
Also write an equation for the rotational form of Newton’s second law for the pulley. The forces producing torques
are T1 and T2 .
(T1 − T2 ) R = Iα
(3)
These are the three equations with three unknowns. Now substitute α = a / R and divide both sides by R so that
equation (3) becomes
a
(4)
T1 − T2 = I 2
R
Solve equations (1) and (2) for T1 and T2 and then subtract equation (2) from equation (1).
T1 − T2 = (m1 − m2 ) g − (m1 + m2 )a
(5)
Now set T1 − T2 from equations (4) and (5) equal to each other.
I
a
= (m1 − m2 ) g − (m1 + m2 ) a
R2
(6)
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Rotational Motion
7-25
All that remains is to solve equation (6) for a.
⎛ I
⎞
a ⎜ 2 + (m1 + m2 ) ⎟ = (m1 − m2 ) g
⎝R
⎠
(m − m2 ) g
a= 1 1
2 + ( m1 + m2 )
R
Substitute I = 12 MR 2 for the pulley.
(m1 − m2 ) g
(m1 − m2 ) g
(2.5 kg − 1.5 kg)(9.80 m/s 2 ) (1.0 kg)(9.80 m/s 2 )
=
=
=
= 2.2 m/s 2
2
0.75 kg
1
M
+
0.375
kg
4.0
kg
+
+
(2.5
kg
1.5
kg)
2 MR
2 + ( m1 + m2 )
2
+ (m1 + m2 )
R2
So block 2 (the lighter one) accelerates up at 2.2 m/s 2 .
a=
Assess: Because we had three unknowns, T1 , T2 , and a, we needed three equations.
We are glad to get an answer less than 9.8 m/s 2 since that is the limit that block 1 could accelerate down if m2 and
M were zero. In the other limit, a → 0 as (m1 − m2 ) → 0, as we would expect.
This set-up with blocks of unequal masses connected by a light rope over a pulley is called an Atwood Machine.
P7.53. Prepare: We can use the rolling constraint to find the speed of the dot and the angular speed of the tires.
Solve: (a) Since the tires are rolling without slipping the angular velocity of the tires is given by Equation 7.25.
v
5.6 m/s
ω = cm =
= 14 rad/s
R
0.40 m
(b) The blue dot is undergoing translation and rotation. At the top of the tire, it has a translational velocity equal to
the speed of the bike, and an additional velocity equal to ω R due to the rotation of the tire. See Figure 7.43. The
speed of the dot at this point is
v = vcm + ω R = 2vcm = 2(5.6 m/s) = 11 m/s
(c) See the diagram below.
The dot has a translational velocity equal to the velocity of the center of mass of the tire in the horizontal direction.
The tangential velocity of the dot is in the vertical direction and has a magnitude equal to ω R. The velocity of the
dot is equal to the sum of these two vectors. Since the tire is rolling without slipping, vcm = ω R.
See the vector diagram below.
The speed of the dot is equal to
v = (vcm ) 2 + (vcm ) 2 = 7.9 m/s
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7-26
Chapter 7
Assess: Note that the speed of the dot during the downward motion for part (c) is equal to the speed of the dot during
the upward motion shown in the diagram.
P7.54. Prepare: Both the slab and the rollers move to the right, but not at the same speed. The initial distance
between the center of gravity of the slab and the right-most roller is 1.0 m.
Solve: Because the speed of the top of a wheel is twice the speed of the center of mass, the slab moves at twice the
translational speed of the rollers. In the same time the slab moves 2.0 m the rollers on the right will move 1.0 m.
At that point the center of gravity will have caught up with the right-most roller. So the slab moves 2.0 m.
Assess: This matches experience.
P7.55. Prepare: The particle is moving in a circle and a motion of 2π radians corresponds to one full rotation.
Angular velocity at any given time is defined as ω = Δθ / Δ t , or the slope of the angular position-versus-time graph.
Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position + 4π rad to − 2π rad.
Therefore, Δ θ = − 2π − (+ 4π ) = − 6π rad in one sec, or ω = − 6π rad/s. From t = 1 s to t = 2 s, ω = 0 rad/s. From
t = 2 s to t = 4 s the particle rotates counterclockwise from the angular position − 2π rad to 0 rad. Thus
Δ θ = 0 − (− 2π ) = 2π rad rad and ω = +π rad/s.
(b)
Assess: Since we take positive angular displacements as counterclockwise and negative angular displacements
clockwise, we know the particle is traveling around the circle in a clockwise direction. Knowing that the slope of the
angular position-versus-time plot is the angular velocity, we can establish a plot of angular velocity-versus-time.
P7.56. Prepare: The crankshaft is a rotating rigid body. The crankshaft’s angular acceleration is given as
α = Δ ω /Δ t , or slope of the angular velocity-versus-time graph.
Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s,
maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up to 200 rad/s from t = 4 s to t = 7 s.
The angular acceleration (α ) graph is based on the fact that α is the slope of the ω -versus-t graph.
Assess: Knowing that the slope of the angular velocity-versus-time plot is the angular acceleration, we can establish
a plot of angular acceleration-versus-time.
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Rotational Motion
7-27
P7.57 Prepare: Knowing the kinematic equations and the fact that the distance the car travels is some number of
revolutions (or circumferences) of the tire, we can solve this problem.
Solve: The acceleration of the car may be obtained from the expression:
v − vo v
v = vo + at which gives a =
= since vo = 0 m/s
t
t
The distance traveled by the car during the time it is accelerating may be determined by:
v 2 − vo2 v 2
v2
vt
=
=
=
2aΔ x = v 2 − vo2 which gives Δ x =
2a
2a 2(v/ t ) 2
Finally, the number of times the tires rotate (i.e. the number of circumferences of the tires) may be determined by:
vt
Δx (vt /2)
(20 m/s)(10 s)
=
=
=
= 55 rotations
Δ x = N (2π r ) = N π d which gives N =
2π d 2[π (0.58 m)/rotation]
πd
πd
Assess: As with many kinematics problems, we can check our work by approaching the problem in a different
manner. For example, since the acceleration is constant, the distance the car travels is just the average velocity times
the time of travel. This may be expressed as follows:
⎛ v + vo ⎞ vt
Δ x = vavet = ⎜
⎟ t = 2 which is the same as the expression obtained above for the distance traveled.
⎝ 2 ⎠
Note also that the final units are in rotations and that 55 is a reasonable number of rotations for a tire in 10 seconds.
P7.58. Prepare: This problem requires a knowledge of one-dimensional kinematics and the fact that as the
cylinder goes around once, the elevator will advance an amount equal to the diameter of the cylinder.
Solve: Knowing that the elevator advances an amount equal to the circumference of the cylinder allows us to write
Δ y = 2π R
From one dimensional kinematics we have
v 2 − vo2 = 2aΔy
and
v = vo + at
Combine the first two equations, solve for a, and insert the result into the third equation to obtain
t = 4π R / vo = 4π (0.50 m)/1.6 m/s = 3.9 s
Assess: This is a rather long time for a passenger elevator but a reasonable time for a freight elevator.
P7.59. Prepare: The disk is a rotating rigid body and it rotates on an axle through its center. We will use Equation
7.12 to find the net torque.
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7-28
Chapter 7
Solve: The net torque on the axle is
τ = FA rA sin φA + FB rB sin φB + FC rC sin φC + FD rD sin φD
= (30 N)(0.10 m) sin( − 90°) + (20 N)(0.05 m) sin 90° + (30 N)(0.05 m) sin 135° + (20 N)(0.10 m) sin 0°
= − 3 N ⋅ m + 1 N ⋅ m + 1.0607 N ⋅ m = − 0.94 N ⋅ m
Assess: A negative net torque means a clockwise acceleration of the disk.
P7.60. Prepare: The torque is provided by the force of static friction between your fingers and the knob.
Solve: See the diagram below.
Since the knob has no velocity in the horizontal direction the net force in the horizontal direction is zero, which
means F = n. The maximum force of static friction is given by fs, max = μs n. The torque provided by the force of
friction is given by Equation 7.10, τ = r μs n. Since there are two torques acting in the same direction, the net torque
is
τ net = 2τ = 2r μs n = 2(0.0050 m)(0.12)(0.60 N) = 7.2 × 10−4 N ⋅ m
Assess: Note that though the net force on the knob is zero, the net torque on the knob is not zero.
P7.61. Prepare: Equation 7.16 tells us the center of gravity of a compound object. If we take all the rest of the body
other than the arms as one object (call it the trunk, even though it includes head and legs) then we can write
y m
+ 2 yarm marm
ycg = trunk trunk
M
Where M = mtrunk + marm = 70 kg (the mass of the whole body).
The language “by how much does he raise his center of gravity” makes us think of writing Δycg .
Since we have modeled the arm as a uniform cylinder 0.75 m long, its own center of gravity is at its geometric center,
0.375 m from the pivot point at the shoulder. So raising the arm from hanging down to straight up would change the
height of the center of gravity of the arm by twice the distance from the pivot to the center of gravity:
(Δ ycg )arm = 2(0.375 m) = 0.75 m.
Solve:
Δ ( ycg ) body = ( ycg ) with arms up − ( ycg ) with arms down =
( ycg ) trunk mtrunk + 2( ycg )arm, up marm
M
−
( ycg ) trunk mtrunk + 2( ycg )arm, down marm
M
2marm
2marm
2(3.5 kg)
=
(( ycg )arm, up − ( ycg )arm, down ) =
(Δ ycg )arm =
(0.75 m) = 0.075 m = 7.5 cm
M
M
70 kg
Assess: 7.5 cm seems like a reasonable amount, not a lot, but not too little. The trunk term subtracted out, which is
both expected and good because we didn’t know ( ycg ) trunk .
P7.62. Prepare: The three masses connected by massless rigid rods is a rigid body. Refer to Figure P7.62 for
masses and their distance from the rotation axis.
Solve: (a)
∑ mi xi (0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m)
=
= 0.060 m
xcg =
∑ mi
0.100 kg + 0.200 kg + 0.100 kg
(
)
2
2
∑ mi yi (0.100 kg)(0 m) + (0.200 kg) (0.10 m) − (0.06 m ) + (0.100 kg)(0 m)
=
= 0.040 m
ycg =
∑ mi
0.100 kg + 0.200 kg + 0.100 kg
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Rotational Motion
7-29
(b) The moment of inertia about an axis through A and perpendicular to the page is
I A = ∑ mi ri 2 = mB (0.10 m) 2 + mC (0.10 m) 2 = (0.100 kg)[(0.10 m) 2 + (0.10 m)2 ] = 0.0020 kg ⋅ m 2
(c) The moment of inertia about an axis that passes through B and C is
I BC = mA
(
(0.10 m) 2 − (0.06 m) 2
) = 0.0013 kg ⋅ m
2
2
Assess: Note that mass mA does not contribute to I A , and the masses mB and mC do not contribute to I BC .
P7.63. Prepare: We can use Equation 7.21, and Table 7.1.
Solve: Refer to the figure below.
(a) The moment of inertia of the skater will be the moment of inertia of her body plus the moment of inertia of her
arms. The center of mass of each arm is at her side, 20 cm from the axis of rotation. The mass of each arm is half of
one eighth of the mass of her body, which is 4 kg. The mass of her body is 64 kg − 8 kg = 56 kg. With her arms at
her sides, her total moment of inertia is
I = I body + I arm + I arm = 12 M body ( Rbody ) 2 + M arm ( Rarm ) 2 + M arm ( Rarm ) 2
= 12 (56 kg)(0.20 m) 2 + (4 kg)(0.20 m) 2 + (4 kg)(0.20 m) 2 = 1.4 kg ⋅ m 2
(b) With her arms outstretched, the center of mass of her arms is now 50 cm from the axis of rotation. Her moment of
inertia is now
I = I body + I arm + I arm = 12 M body ( Rbody ) 2 + M arm ( Rarm ) 2 + M arm ( Rarm ) 2
= 12 (56 kg)(0.20 m) 2 + (4 kg)(0.50 m) 2 + (4 kg)(0.50 m) 2 = 3.1 kg ⋅ m 2
Her moment of inertia has increased.
Assess: Her moment of inertia with arms outstretched is almost twice as large as with them at her side. This is
reasonable, since their distance from the axis of rotation is much larger.
P7.64. Prepare: The compact disk is a rigid body rotating about its center. The initial angular velocity is ωi = 0
and the final angular velocity is ωf = 2000 rpm, which must be converted to SI units of rad/s. Noting that 1
revolution corresponds to 2π radians, ωf = (2000 rpm)( 260π ) rad/s = 209.4 rad/s.
Solve: (a) The rotational kinematics equation ωf = ωi + α (tf − ti ) gives
209.4 rad/s = 0 rad + α (3.0 s − 0 s) ⇒ α = 69.81 rad/s 2
The torque needed to obtain this operating angular velocity is
τ = Iα = (2.5 × 10−5 kg ⋅ m 2 )(69.81 rad/s 2 ) = 1.8 × 10−3 N ⋅ m
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7-30
Chapter 7
(b) From the rotational kinematics equation,
θ f = θ i + ω i (tf − ti ) + 12 α (tf − ti ) 2 = 0 rad + 0 rad + 12 (69.81 rad/s 2 )(3.0 s − 0 s) 2 = 314.1 rad
314.1
revolutions = 50 rev
2π
Assess: Fifty revolutions in 3 seconds or 1000 rpm is a reasonable value.
=
P7.65. Prepare: This problem requires a knowledge of translational ( Fnet = ma ) and rotational (τ net = Iα )
dynamics. Notice that the counterclockwise torque is greater than the clockwise torque, hence the system will rotate
counterclockwise. Let’s agree to call any force that tends to accelerate the system positive and any force that tends to
decelerate the system negative. Also let’s agree to call the small disk M 1 and the large disk M 2 .
Solve:
Write Newton’s second law equation for m2 and m1 as follows:
m2 g − T2 = m2 a2 = m2 R2α or T2 = m2 g − m2 R2α
and
T1 − m1 g = m1a1 = m1R1α or T1 = m1 g + m1R1α
The net torque acting on the system may be determined by
τ = R2T2 − R1T1 = R2 (m2 g − m2α R2 ) − R1 (m1 g + m1α R1 )
The moment of inertia of the system is
I = I1 + I 2 = ( M 1R12 /2) + ( M 2 R22 /2)
Knowing
τ = Iα
we may combine the above to get
R2 (m2 g − m2α R2 ) − R1 (m1 g + m1α R1 ) = [( M 1R12 /2) + ( M 2 R22 /2)]α
which may be solved for α to obtain
( m2 R2 − m1R1 ) g
= 3.5 rad/s 2
α= 2
R2 (m2 + M 2 /2) + R12 (m1 + M 1/2)
Assess: This angular acceleration amounts to speeding up about a half revolution per second every second. That is
not an unreasonable amount.
P7.66. Prepare: The flywheel is a rigid body rotating about its central axis as shown below. The initial angular
speed will be taken as ωi = 0 and the final angular speed is ωmax = ωf = 1200 rpm, which must be converted to SI
units of rad/s. Noting that 1 revolution corresponds to 2π radians, ωf = 1200 rpm (2 π /60) rad/s = 40π rad/s.
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Rotational Motion
7-31
Solve: The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about the axis
of rotation is that of a disk:
I = 12 MR 2 = 12 (250 kg)(0.75 m) 2 = 70.31 kg ⋅ m 2
The angular acceleration is calculated as follows:
τ net = Iα ⇒ α = τ net / I = (50 N ⋅ m)/(70.31 kg ⋅ m 2 ) = 0.711 rad/s 2
Using the kinematics equation for angular velocity gives
ωf = ωi + α (tf − ti ) = 1200 rpm = 40 π rad/s = 0 rad/s + 0.711 rad/s 2 (tf − 0 s) ⇒ tf = 180 s
Assess: To solve this problem, you used your knowledge of moment of inertia, rotational dynamics, and rotational
kinematics. As you move farther into physics, you will find that you need mastery of multiple concepts in order to
solve the problems. You should enjoy this process.
P7.67. Prepare: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the
center of gravity. Assume that the size of the balls is small compared to 1 m. We have placed the origin of the
coordinate system on the 1.0 kg ball. Since τ = I about cgα , we need the moment of inertia and the angular acceleration
to be able to calculate the required torque.
Solve: The center of gravity and the moment of inertia are
(1.0 kg)(0 m) + (2.0 kg)(1.0 m)
xcm =
= 0.667 m and ycm = 0 m
(1.0 kg + 2.0 kg)
I about cm = ∑ mi ri 2 = (1.0 kg)(0.667 m) 2 + (2.0 kg)(0.333 m) 2 = 0.667 kg ⋅ m 2
We have ωf = 0 rad/s,
tf − ti = 5.0 s, and ωi = 20 rpm = 20(2π rad/60 s) = 32 π rad/s, so ωf = ωi + α (tf − ti )
becomes
2π
⎛ 2π
⎞
0 rad/s = ⎜
rad/s ⎟ + α (5.0 s) ⇒ α = −
rad/s 2
15
⎝ 3
⎠
Having found I and α , we can now find the torque τ that will bring the balls to a halt in 5.0 s:
4π
⎛2
⎞⎛ 2π
⎞
τ = I about cmα = ⎜ kg ⋅ m 2 ⎟⎜ −
rad/s 2 ⎟ = −
N ⋅ m = − 0.28 N ⋅ m
45
⎝3
⎠⎝ 15
⎠
Assess: The minus sign with the torque indicates that the torque acts clockwise.
P7.68. Prepare: Assume the rope has no mass and does not slip on the pulley. Equation 7.22 and Newton’s second
law for linear motion can be applied to find the acceleration of the mass. There is motion constraint on the rope and
pulley.
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7-32
Chapter 7
Solve: See the diagram below.
Choosing the upward direction as positive, Newton’s second law applied to the mass gives
ma y = T2 − mg
The normal force on the pulley and the pulley’s weight exert no torque. The moment arm for both tensions is the
radius of the pulley. For the rotational motion Equation 7.22 gives T2 R − T1R = Iα .
Since upward motion gives clockwise acceleration, the motion constraint is a y = −α R.
Using this constraint and solving for T2 in the torque equation we obtain T2 = T1 − Ia / R 2 .
Substituting this result into the equation for linear motion and solving for the acceleration gives
T − mg
ay = 1
m + I/R 2
The moment of inertia of a cylinder is I = 12 MR 2 . Substituting this into the acceleration equation gives
T1 − mg 10 N − (1.5 kg)(9.80 m/s 2 )
=
= − 1.88 m/s 2
m + M/2
1.5 kg + 1.0 kg
An additional significant figure has been kept in this intermediate result.
Note that the acceleration is negative. The block is moving downward.
Starting with zero initial velocity, the block moves Δ y = −0.30 m. Using Equation 2.12, the time it takes the block to
ay =
move 30 cm is
Δt =
2Δ y
2(− 0.30 m)
=
= 0.56 s
ay
− 1.88 m/s 2
Assess: The expression for the acceleration makes sense. If the mass of the block or the moment of inertia of the
pulley increases, the acceleration decreases. If the tension applied to the rope is smaller than the weight of the mass,
the acceleration is negative instead of positive. Note that the tensions acting on either side of the pulley can’t be equal
due to the moment of inertia of the pulley. If the tensions were equal there would be no torque on the pulley, and the
pulley would not rotate.
P7.69. Prepare: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a
disk and that the string does not slip. Because the pulley is not massless and frictionless, tension in the rope on both
sides of the pulley is not the same. We will have to be careful with the appropriate masses when we write below
Newton’s second law for the blocks and the pulley. A pictorial diagram of the problem and free-body diagrams for
the two blocks are shown. We have placed the origin of the coordinate system on the ground.
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Rotational Motion
7-33
Solve: Applying Newton’s second law to m1 , m2 , and the pulley yields the three equations:
T1 − w1 = m1a1
T2 R − T1R − 0.50 N ⋅ m = Iα
− w2 + T2 = m2 a2
Noting that − a2 = a1 = a, I = mp R , and α = a / R, the above equations simplify to
1
2
T1 = m1 g = m1a
2
m2 g − T2 = m2 a
T2 − T1 =
(
1
2
0.50 N ⋅ m
⎛ a ⎞ 1 0.50 N ⋅ m 1
mp R 2 ⎜ ⎟ +
= 2 mp a +
R
R
R
0.060 m
⎝ ⎠
)
Adding these three equations,
(
)
(m2 − m1 ) g = a m1 + m2 + 12 mp + 8.333 N ⇒ a
=
(m2 − m1 ) g − 8.333 N
m1 + m2 + 12 mp
(4.0 kg − 2.0 kg)(9.80 m/s 2 ) − 8.333 N
= 1.610 m/s 2
2.0 kg + 4.0 kg + (2.0 kg / 2)
We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor:
yf = yi + vi (tf − ti ) + 12 a2 (tf − ti ) 2 ⇒ 0 = 1.0 m + 0 + 12 (− 1.610 m/s 2 )(tf − 0 s) 2 ⇒ tf =
2(1.0 m)
= 1.1 s
(1.610 m/s 2 )
Assess: Compared to free fall where we would use a = − 9.80 m/s 2 , a = − 1.61 m/s 2 , and a time of 1.1 s for the
block to reach floor are reasonable.
P7.70. Prepare: The disk is a rigid spinning body. The initial angular velocity of 300 rpm in SI units is
(300)(2π ) / 60 = 10π rad/s. After 3.0 s the disk stops, so ωf = 0.
Solve: Using the kinematic equation for angular velocity,
ω − ωi (0 rad/s − 10 π rad/s) − 10π
ωf = ωi + α (tf − ti ) ⇒ α = f
=
=
rad/s 2
(3.0 s − 0 s)
3
tf − ti
Thus, the torque due to the force of friction that brings the disk to rest is
1
mR 2 α
π
Iα
⎛
⎞
2
=−
= − 12 (mR)α = 12 (2.0 kg)(0.15 m) ⎜ − 10 rad/s 2 ⎟ = 1.6 N
τ = Iα = − fR ⇒ f = −
R
R
3
⎝
⎠
Assess: The minus sign with τ = − fR indicates that the torque due to friction acts clockwise.
(
)
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7-34
Chapter 7
P7.71. Prepare: This is an excellent review problem. In order to solve this problem you will need a working
knowledge of rotational kinematics (ω f = ω o + α t ), moment of inertia of a cylinder ( I = MR 2 /2), rotational
dynamics (τ = Iα ), torque (τ = Rf k sin φ ), and kinetic friction ( f k = μ k N ).
Solve: First determine an expression for the angular acceleration
α = (ωf − ωo ) / Δt
Next obtain an expression for the moment of inertias of the grindstone
I = MR 2 / 2
Then obtain an expression for the torque acting on the grindstone
⎛ MR 2 ⎞ ⎛ ωf − wo ⎞
τ = Iα = ⎜
⎟⎜
⎟
⎝ 2 ⎠ ⎝ Δt ⎠
Write a second expression for the torque in terms of the force of friction and then the normal force
τ = f k R = μ k NR
Finally, equate the last two expressions for the torque and solve for N (the force with which the man presses the knife
against the grindstone).
MR (ωf − ωo ) (28 kg)(0.15 m)(180 rev/min − 200 rev/min)(2π rad/rev)(min / 60 s)
N=
=
= 2.2 N
2μk Δt
2(0.2)(10 s)
Assess: This is a reasonable force (i.e. one that the man could easily exert and yet not grind the knife to a sliver in a
matter of minutes).
P7.72. Prepare: The sacs are constrained by the stamens.
Solve: Using Synthesis 7.1. to find the angular acceleration of a sac, and then use Equation 7.22 to find the
corresponding tangential acceleration.
α=
(
)
π rad
2(60 °) 2360°
2Δ θ
=
= 2.3 × 107 rad/s 2
Δt 2 (0.30 × 10−3 s) 2
The correct answer is D.
Assess: The acceleration is huge. This is due to the relatively low moment of inertia of the structure.
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Rotational Motion
7-35
P7.73. Prepare: The sacs are constrained by the stamens.
Solve: Using Synthesis 7.1 with the result from the previous problem gives
ωf = α Δ t = 7.0 × 103 rad/s
With the angular speed we can now compute the speed.
v = ω r = (7.0 ×103 rad/s)(0.001 m) = 7.0 m/s
The correct choice is B.
Assess: These results seem reasonable.
P7.74. Prepare: The sacs are constrained by the stamens.
Solve: The moment of inertia of the stamen and sac can be calculated using Equation 7.21.
I = 13 mstamen R2 + M sac R 2 = (1.0 ×10−6 kg)(5.0 ×10−4 m) 2 + (1.0 ×10−6 kg)(1.0 ×10−3 m) 2 = 1.25 ×10−12 kg ⋅ m 2
An additional significant figure has been kept in this intermediate result.
The “straightening torque” is
I = 13 mstamen R2 + M sac R 2 = (1.0 ×10−6 kg)(5.0 ×10−4 m) 2 + (1.0 ×10−6 kg)(1.0 ×10−3 m) 2 = 1.25 ×10−12 kg ⋅ m 2
The correct answer is B.
Assess: The accelerations are huge, while the torque is relatively small. This is due to the relatively low moment of
inertia of the structure.
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7-36
Chapter 7
P7.75. Prepare: The center of gravity must follow a parabolic path, so the answer isn’t C.
While it is possible to bend over and get one’s center of gravity outside the body (search the Web for the Fosbury flop
technique in the high jump), in this case the center of gravity of the dancer stays in the body, so the answer isn’t D.
Solve: The correct answer is B, with the reasoning being that, as shown in the diagram, the head can stay quite level
by bringing up the arms and legs and then lowering them again. This allows the center of gravity to follow the
parabola while the head stays about level.
Assess: The whole point of movement is to raise and lower the center of gravity, and this is done by raising and
lowering the arms and legs.
P7.76. Prepare: Your center of gravity will travel as if it were in free fall and so reaches the same height no matter
how you move your arms.
Solve: In order to get your head as high as possible, you should lower your center of gravity as much as possible.
You should have your arms at your side at the top of the leap. The correct answer is A.
Assess: Compare to the motion in a grand jeté. There, the dancer moves her center of gravity upward so her head
appears to move in a straight line.
P7.77. Prepare: While the dancer is in the air, the gravitational force on her acts as if her mass were concentrated
at her center of gravity.
Solve: The correct answer is B; there is no lever arm to create a torque because the gravitational force is directly
through the center of gravity. Even if the arms and legs move, the center of gravity moves and the gravitational force
is still through the center of gravity. So if the axis of rotation is through her center of gravity then there can be no
gravitational force.
Assess: Of course, computing the gravitational torque about some axis other than through the center of gravity could
produce a torque, but we were told to use an axis through the center of gravity.
P7.78. Prepare: The basic expression used to determine the moment of inertia is Equation 7.21. As indicated in
this expression, the moment of inertia depends on the mass and how the mass is distributed.
Solve: As the dancer moves her arms and legs outward, she is moving the center of gravity of her arms and legs
farther from the vertical axis through the center of her body and hence increasing her moment of inertia about that
axis. The correct answer is A.
Assess: Moment of inertia increases as the square of the distance of a mass from the axis of rotation.
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