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Holt Physics
Problem 8C
NEWTON’S SECOND LAW FOR ROTATION
PROBLEM
The giant sequoia General Sherman in California has a mass of about 2.00
106 kg, making it the most massive tree in the world. Its height of 83.0 m
is also impressive. Imagine a uniform bar with the same mass and length as
the tree. If this bar is rotated about an axis that is perpendicular to and
passes through the bar’s midpoint, how large an angular acceleration would
result from a torque of 4.60 107 N•m? (Note: Assume the bar is thin.)
SOLUTION
Given:
M = 2.00 × 106 kg
l = 83.0 m
t = 4.60 × 107 N•m
Unknown:
a×?
Calculate the bar’s moment of inertia using the formula for a thin rod with the
axis of rotation at its center.
1
1
I = 12 M l 2 = 12 (2.00 × 106 kg)(83.0 m)2 = 1.15 × 109 kg•m2
Now use the equation for Newton’s second law for rotating objects. Rearrange the
equation to solve for angular acceleration.
t = Ia
Copyright © by Holt, Rinehart and Winston. All rights reserved.
t
(4.60 × 107 N•m)
a =  = 
= 4.00 × 10−2 rad/s2
I (1.15 × 109 kg•m2)
ADDITIONAL PRACTICE
1. One of the largest Ferris wheels currently in existence is in Yokohama,
Japan. The wheel has a radius of 50.0 m and a mass of 1.20 × 106 kg. If
a torque of 1.0 × 109 N•m is needed to turn the wheel from a state of
rest, what would the wheel’s angular acceleration be? Treat the wheel as
a thin hoop.
2. In 1992, Jacky Vranken from Belgium attained a speed of more than
250 km/h on just the back wheel of a motorcycle. Assume that all of the
back wheel’s mass is located at its outer edge. If the wheel has a mass of
22 kg and a radius of 0.36 m, what is the wheel’s angular acceleration
when a torque of 5.7 N•m acts on the wheel?
3. In 1995, a fully functional pencil with a mass of 24 kg and a length of
2.74 m was made. Suppose this pencil is suspended at its midpoint and
a force of 1.8 N is applied perpendicular to its end, causing it to rotate.
What is the angular acceleration of the pencil?
Problem 8C
91
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4. The turbines at the Grand Coulee Third Power Plant in the state of
Washington have rotors with a mass of 4.07 × 105 kg and a radius of
5.0 m each. What angular acceleration would one of these rotors have if
a torque of 5.0 × 104 N•m were applied? Assume the rotor is a uniform
disk.
5. J. C. Payne of Texas amassed a ball of string that had a radius of 2.00 m.
Suppose a force of 208 N was applied tangentially to the ball’s surface
in order to give the ball an angular acceleration of 3.20 × 10–2 rad/s2.
What was the ball’s moment of inertia?
6. The heaviest member of British Parliament ever was Sir Cyril Smith.
Calculate his peak mass by finding first his moment of inertia from the
following situation. If Sir Cyril were to have ridden on a merry-goround with a radius of 8.0 m, a torque of 7.3 × 103 N•m would have
been needed to provide him with an angular acceleration of 0.60 rad/s2.
7. In 1975, a centrifuge at a research center in England made a carbonfiber rod spin about its center so fast that the tangential speed of the
rod’s tips was about 2.0 km/s. The length of the rod was 15.0 cm. If it
took 80.0 s for a torque of 0.20 N•m to bring the rod to rest from its
maximum speed, what was the rod’s moment of inertia?
9. In 1990, a cherry pie with a radius of 3.00 m and a mass of 17 × 103 kg
was baked in Canada. Suppose the pie was placed on a light rotating
platform attached to a motor. If this motor brought the angular speed
of the pie from 0 rad/s to 3.46 rad/s in 12 s, what was the torque the
motor must have produced? Assume the mass of the platform was negligible and the pie was a uniformly solid disk.
10. In just over a month in 1962, a shaft almost 4.00 × 108 m deep and with
a radius of 4.0 m was drilled in South Africa. The mass of the soil taken
out was about 1.0 × 108 kg. Imagine a rigid cylinder with a mass, radius, and length equal to these values. If this cylinder rotates about its
symmetry axis so that it undergoes a constant angular acceleration
from 0 rad/s to 0.080 rad/s in 60.0 s, how large a torque must act on the
cylinder?
11. In 1993, a bowl in Canada was filled with strawberries. The mass of the
bowl and strawberries combined was 2390 kg, and the moment of inertia about the symmetry axis was estimated to be 2.40 × 103 kg•m2. Suppose a constant angular acceleration was applied to the bowl so that it
made its first two complete rotations in 6.00 s. How large was the
torque that acted on the bowl?
92
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. The largest tricycle ever built had rear wheels that were almost 1.70 m
in diameter. Neglecting the mass of the spokes, the moment of inertia
of one of these wheels is equal to that of a thin hoop rotated about its
symmetry axis. Find the wheel’s moment of inertia and its mass if a
torque of 125 N•m is applied to the wheel so that in 2.0 s the wheel’s
angular speed increases from 0 rad/s to 12 rad/s.
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12. A steel ax with a mass of 7.0 × 103 kg and a length of 18.3 m was made
in Canada. If Paul Bunyan were to take a swing with such an ax, what
torque would he have to produce in order for the blade to have a tangential acceleration of 25 m/s2? Assume that the blade follows a circle
with a radius equal to the ax handle’s length and that nearly all of the
mass is concentrated in the blade.
Problem 8C
93
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Givens
Solutions
7. m = 449 kg
l
Apply the first condition of equilibrium to solve for F2.
F1 + F2 − mg = 0
= 5.0 m
3
F1 = 2.70 × 10 N
2
g = 9.81 m/s
F2 = mg − F1
F2 = (449 kg)(9.81 m/s2) − 2.70 × 103 N = 4.40 × 103 N − 2.70 × 103 N = 1.70 × 103 N
Apply the second condition of equilibrium, using the left end of the platform as the
pivot point.
F2 l − m g d = 0
F2 l (1.70 × 103 N)(5.0 m)
d =  = 
m g (449 kg)(9.81 m/s2)
d = 1.9 m from the platform’s left end
8. m1 = 414 kg
l
Apply the first condition of equilibrium to solve for F2.
F1 + F2 − m1 g − m2 g = 0
= 5.00 m
m2 = 40.0 kg
F2 = m1 g + m2 g − F1 = (m1 + m2) g − F1
F1 = 50.0 N
F2 = (414 kg + 40.0 kg)(9.81 m/s2) − 50.0 N = (454 kg)(9.81 m/s2) − 50.0 N = 4.45 ×
103 N − 50.0 N
2
g = 9.81 m/s
II
F2 = 4.40 × 103 N
Apply the second condition of equilibrium, using the supported end (F1) of the stick
as the rotation axis.
2 − m g l = 0
m
414 kg
 + m g l
 + 40.0 kg (9.81 m/s )(5.0 m)
2
2


d =
=
F2 d − m1 g
l
1
2
2
2
4.40 × 103 N
F2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(207 kg + 40.0 kg)(9.81 m/s2)(5.00 m)
(247 kg)(9.81 m/s2)(5.00 m)
d = 
= 
3
4.40 × 10 N
4.40 × 103 N
d = 2.75 m from the supported end
Additional Practice 8C
M = 1.20 × 106 kg
1.0 × 109 N • m
t
t
a =  = 2 = 
(1.20 × 106 kg)(50.02)
I MR
t = 1.0 × 109 N • m
a = 0.33 rad/s2
1. R = 50.0 m
2. M = 22 kg
R = 0.36 m
5.7 N•m
t
t
a =  = 2 = 2
(22 kg)(0.36 m)
I MR
t = 5.7 N • m
a = 2.0 rad/s2
Section Two—Problem Workbook Solutions
II Ch. 8–5
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Givens
Solutions
3. M = 24 kg
l
The force is applied perpendicular to the lever arm, which is half the pencil’s length.
= 2.74 m
Therefore,
F = 1.8 N
t = F d (sin q) = F
F
t
a =  =
I
2
l

1
 M l 2
12
2
l
2.74 m
(1.8 N) 
2
= 
1
 (24 kg)(2.74 m)2
12
a = 0.16 rad/s2
4. M = 4.07 × 105 kg
t
t
a =  = 
1
I
MR2
2
R = 5.0 m
t = 5.0 × 104 N • m
(5.0 × 104 N • m)
a = 
1
(4.07 × 105 kg)(5.0 m)2
2
a = 9.8 × 10−3 rad/s2
II
5. R = 2.00 m
The force is applied perpendicular to the lever arm, which is the ball’s radius.
F = 208 N
Therefore,
−2
a = 3.20 × 10
2
rad/s
t = F d (sin q) = F R
(208 N)(2.00 m)
t FR
T =  =  = 
3.20 × 10−2 rad/s2
a a
6. r = 8.0 m
3
t = 7.3 × 10 N• m
a = 0.60 rad/s2
t
I =  = mr 2
a
7.3 × 103 N•m
I = 
= 1.2 × 104 kg • m2
0.60 rad/s2
1.2 × 104 kg•m2
I
m = 2 = 
= 1.9 × 102 kg
(8.0 m)2
r
7. vt,i = 2.0 km/s
l
= 15.0 cm
∆t = 80.0 s
t = −0.20 N•m
vt,f = 0 m/s
t
vt,f − vt,i
t
t
I =  =  = 
a
wf − wi

d2 ∆t
∆t
l
I =
−0.20 N•m
3
0 m/s − 2.0 × 10 m/s

0.150 m
 (80.0 s)
2
I = 6.0 × 10−4 kg • m2
II Ch. 8–6
Holt Physics Solution Manual
=
−0.20 N•m
−2.0 × 103 m/s

(0.075 m)(80.0 s)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
I = 1.30 × 104 kg • m2
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Givens
Solutions
1.70 m
8. R =  = 0.85 m
2
t = 125 N • m
t
I =  = MR2
a
wi = 0 rad/s
125 N•m
125 N • m
t
t
I =  =  =  = 2
6.0
rad/s
a
12 rad/s − 0 rad/s
wf − wi


2.0 s
∆t
wf = 12 rad/s
I = 21 kg • m2
∆t = 2.0 s
21 kg•m2
I
M = 2 = 2 = 29 kg
(0.85 m)
R
9. R = 3.00 m
M = 17 × 103 kg
wi = 0 rad/s
wf = 3.46 rad/s
wf − wi
1

t = I a = 2 MR2 
∆t
(17 × 103 kg)(3.00 m)2(3.46 rad/s − 0 rad/s)
t =  = 2.2 × 104 N•m
(2)(12 s)
∆t = 12 s
10. R = 4.0 m
8
M = 1.0 × 10 kg
wi = 0 rad/s
wf = 0.080 rad/s
wf − wi
1

t = Ia = 2 MR2 
∆t
II
(1.0 × 108 kg)(4.0 m)2 (0.080 rad/s − 0 rad/s)
t =  = 1.1 × 106 N •m
(2)(60.0 s)
∆t = 60.0 s
11. I = 2.40 × 103 kg•m2
∆q = 2(2p rad) = 4p rad
∆t = 6.00 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
wi = 0 rad/s
1
∆q = wi ∆t + 2 a ∆t2
Because wi = 0,
1
∆q = 2 a∆t2
2∆q
a = 2
∆t
3
2
2 I ∆q (2)(2.40 × 10 kg•m )(4p rad)

t = Ia = 
2
2 =
(6.00 s)
∆t
t = 1.68 × 103 N • m
12. m = 7.0 × 103 kg
r = 18.3 m
at = 25 m/s2
a
t = Ia = (mr 2)t = mrat
r
t = (7.0 × 103 kg)(18.3 m)(25 m/s2)
t = 3.2 × 106 N • m
Section Two—Problem Workbook Solutions
II Ch. 8–7