Chapter 8
pages 869–870
1. The rotational velocity of a merry-goround is increased at a constant rate from
1.5 rad/s to 3.5 rad/s in a time of 9.5 s.
What is the rotational acceleration of
the merry-go-round?
6. A 92-kg man uses a 3.05-m board to attempt
to move a boulder, as shown in the diagram
below. He pulls the end of the board with a
force equal to his weight and is able to move
it to 45° from the perpendicular. Calculate
the torque applied.
(3.5 rad/s) (1.5 rad/s)
t
9.5 s
0.21 rad/s2
45°
2. A record player’s needle is 6.5 cm from the
center of a 45-rpm record. What is the
velocity of the needle?
First convert rpm to rad/s.
45 rev 2 rad 1 min
4.71 rad/s
1 min 1 rev 60 s 1m
v r (6.5 cm) (4.71 rad/s)
100 cm
0.31 m/s
3. Suppose a baseball rolls 3.2 m across the
floor. If the ball’s angular displacement is
82 rad, what is the circumference of the ball?
d r
82 rad
c 2r 2(0.039 m) 0.25 m
4. A painter uses a 25.8-cm long screwdriver to
pry the lid off of a can of paint. If a force of
85 N is applied to move the screwdriver 60.0°
from the perpendicular, calculate the torque.
Fr sin (85 N)(0.258 m)(sin 60.0°)
19 Nm
5. A force of 25 N is applied vertically at the
end of a wrench handle that is 45 cm long
to tighten a bolt in the clockwise direction.
What torque is needed by the bolt to keep
the wrench from turning?
In order to keep the wrench from turning, the torque on the bolt must be equal
in magnitude but opposite in direction of
the torque applied by the wrench.
Fr sin (25 N)(0.45 m)(sin 90.0°)
11 Nm counterclockwise
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Solutions Manual
mgr sin (92 kg)(9.80 m/s2)(3.05 m)(sin 45°)
1.9103 Nm
7. If a 25-kg child tries to apply the same torque
as in the previous question using only his or
her weight for the applied force, what would
the length of the lever arm need to be?
L r sin 1.9103 Nm
2
F
mg
(25 kg)(9.80 m/s )
7.8 m
8. Logan, whose mass is 18 kg, sits 1.20 m
from the center of a seesaw. If Shiro must
sit 0.80 m from the center to balance
Logan, what is Shiro’s mass?
FLrL FSrS
mLgrL mSgrS
m r
rS
L L
mS (18 kg)(1.20 m)
0.80 m
27 kg
9. Two forces—55-N clockwise and 35-N
counterclockwise—are applied to a merrygo-round with a diameter of 4.5 m. What is
the net torque?
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
d
3.2 m
r 0.039 m
Fr sin Chapter 8 continued
12. A merry-go-round in the park has a radius
of 2.6 m and a moment of inertia of
1773 kgm2. What is the mass of the
merry-go-round?
m
4.5
1
I mr 2
2
(2)(1773 kgm2)
(2.6 m)
2I
m 2 2
35 N
55 N
r
5.2102 kg
The net torque is the sum of the
individual torques.
net ccw cw
I
(35 N 55 N) (sin 90.0°)
4.5 m
2
45 Nm
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
45 Nm clockwise
10. A student sits on a stool holding a 5.0-kg
dumbbell in each hand. He extends his arms
such that each dumbbell is 0.60 m from the
axis of rotation. The student’s moment of
inertia is 5.0 kgm2. What is the moment of
inertia of the student and the dumbbells?
Isingle dumbbell mr 2 (5.0 kg)(0.60 m)2
Itotal 2Isingle dumbbell I student
kgm2
8.6 kgm2
11. A basketball player spins a basketball with a
radius of 15 cm on his finger. The mass of
the ball is 0.75 kg. What is the moment of
inertia about the basketball?
2
2
I mr 2 (0.75 kg)(0.15 m)2
5
5
6.8103 kgm2
(53 N)(2.6 m)(sin 90.0°)
2
1773 kgm
7.8102 rad/s2
14. What is the angular velocity of the merrygo-round described in problems 12 and 13
after 85 s, if it started from rest?
f
i
t
t
i 0 since it started from rest
f t
(7.8102 rad/s2)(85 s)
6.6 rad/s
1.8 kgm2
5.0
Fr sin 1
(Fccw Fcw)r sin (2)(1.8
net
Fccwr sin Fcwr sin kgm2)
13. The merry-go-round described in the previous problem is pushed with a constant force
of 53 N. What is the angular acceleration?
15. An ice-skater with a moment of inertia of
1.1 kgm2 begins to spin with her arms
extended. After 25 s, she has an angular
velocity of 15 rev/s. What is the net torque
acting on the skater?
net I
I( )
f
i
t
(1.1 kgm2)(15 rev/s 0 rev/s)
255
2 rad
rev 4.1 Nm
Physics: Principles and Problems
Solutions Manual
619
Chapter 8 continued
16. A board that is 1.5 m long is supported in two places. If the force exerted by the
first support is 25 N and the forced exerted by the second is 62 N, what is the
mass of the board?
Fnet F1 F2 (Fg)
Since the system is in equilibrium,
Fnet 0.
0 F1 F2 Fg
Fg F1 F2
mg F1 F2
F1 F2
25 N 62 N
m 2
g
9.80 m/s
8.9 kg
17. A child begins to build a house of cards by laying an 8.5-cm-long playing card
with a mass of 0.75 g across two other playing cards: support card A and support
card B. If support card A is 2.0 cm from the end and exerts a force of 1.5103 N,
how far from the end is support card B located? Let the axis of rotation be at the
point support card A comes in contact with the top card.
Fnet FA FB (Fg)
Since the system is in equilibrium,
Fnet 0.
0 FA FB Fg
FB Fg FA
Since the axis of rotation is about support card A, A 0.
so net B g
The system is in equilibrium, so net 0.
0 B g
B g
B rBFB and g rgFg
rBFB rgFg
rgFg
rgmg
FB
mg FA
rB 12(0.085 m) 0.020 m(7.5104 kg)(9.80 m/s2)
(7.5104 kg)(9.80 m/s2) 1.5103 N
2.8102 m
28 cm
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Solutions Manual
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
mg FA
Chapter 8 continued
18. If support card A in the previous problem was moved so that it now is 2.5 cm
from the end, how far from the other end does support card B need to be to
reestablish equilibrium?
r mg
mg FA
g
rB 12(0.085 m) 0.025 m(7.5104 kg)(9.80 m/s2)
(7.5104 kg)(9.80 m/s2) 1.5103 N
2.2102 m
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
22 cm
Physics: Principles and Problems
Solutions Manual
621
Chapter 9
pages 870–871
1. A ball with an initial momentum of 6.00 kgm/s bounces off a wall and travels in
the opposite direction with a momentum of 4.00 kgm/s. What is the magnitude
of the impulse acting on the ball?
Choose the direction away from the wall to be positive.
pf 4.00 kgm/s
pi 6.00 kgm/s
Impulse pf pi (4.00 kgm/s) (6.00 kgm/s)
10.0 kgm/s
2. If the ball in the previous problem interacts with the wall for a time interval of
0.22 s, what is the average force exerted on the wall?
Impulse Ft
Impulse
10.0 kgm/s
0.22 s
F 45 N
t
3. A 42.0-kg skateboarder traveling at 1.50 m/s hits a wall and bounces off of it.
If the magnitude of the impulse is 150.0 kg·m/s, calculate the final velocity of the
skateboarder.
Choose the direction away from the wall to be positive.
Impulse pf pi mvf (mvi)
m(vf vi)
150.0 kgm/s
42.0 kg
1.50 m/s
2.07 m/s
4. A 50.0-g toy car traveling with a velocity of 3.00 m/s due north collides head-on
with an 180.0-g fire truck traveling with a velocity of 0.50 m/s due south. The
toys stick together after the collision. What are the magnitude and direction of
their velocity after the collision?
Choose north to be positive.
pi pf
pci pti pf
mcvci mtvti (mc mt)vf
m v mv
mc mt
c ci
t ti
vf (50.0 g)(3.00 m/s) (180.0 g)(0.50 m/s)
50.0 g 180.0 g
0.26 m/s, due north
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Solutions Manual
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Impulse
m
vf vi
Chapter 9 continued
5. A 0.040-kg bullet is fired into a 3.50-kg
block of wood, which was initially at rest.
The bullet remains embedded within the
block of wood after the collision. The
bullet and the block of wood move at a
velocity of 7.40 m/s. What was the original
velocity of the bullet?
pi pf
pbi pwi pf
mbvbi mwvwi (mb mw)vf
where vwi 0. Thus,
(0.040 kg 3.50 kg)(7.40 m/s)
vbi 0.040 kg
6.5102 m/s
6. Ball A, with a mass of 0.20 kg, strikes ball B,
with a mass of 0.30 kg. The initial velocity
of ball A is 0.95 m/s. Ball B is initially at
rest. What are the final speed and direction
of ball A and B after the collision if they
stick together?
pi pf
mAvAi (mA mB)vf
m v
mA mB
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
A Ai
vf (0.20 kg)(0.95 m/s)
0.20 kg 0.30 kg
0.38 m/s in the same direction as
ball A’s initial velocity
7. An ice-skater with a mass of 75.0 kg pushes
off against a second skater with a mass of
42.0 kg. Both skaters are initially at rest.
After the push, the larger skater moves off
with a speed of 0.75 m/s eastward. What
is the velocity (magnitude and direction) of
the smaller skater after the push?
pi pf
m1vf1 m2vf2
m v
m2
1 f1
vf2 (75.0 kg)(0.75 m/s)
42.0 kg
1.3 m/s
The second skater moves west with a
velocity of 1.3 m/s.
8. Suppose a 55.0-kg ice-skater, who was
initially at rest, fires a 2.50-kg gun. The
0.045-kg bullet leaves the gun at a velocity
of 565.0 m/s. What is the velocity of the
ice-skater after she fires the gun?
pi pf
0 pfs pfb
(ms mg)vfs mbvfb
because the final mass of the skater
includes the mass of the gun held by
the skater. Then,
m v
ms mg
b fb
vfs (0.045 kg)(565.0 m/s)
55.0 kg 2.50 kg
0.44 m/s
9. A 1200-kg cannon is placed at rest on
an ice rink. A 95.0-kg cannonball is shot
from the cannon. If the cannon recoils at
a speed of 6.80 m/s, what is the speed of
the cannonball?
pi pf
Since the cannon and cannonball are at
rest before the blast, pi 0.00 kgm/s.
So pfc pfb
mcvfc mbvfb
m v
mb
c fc
vfb (1200 kg)(6.80 m/s)
95.0 kg
86 m/s
10. An 82-kg receiver, moving 0.75 m/s north,
is tackled by a 110.0-kg defensive lineman
moving 0.15 m/s east. The football players
hit the ground together. Calculate their final
velocity (magnitude and direction).
pri mrvri, y (82 kg)(0.75 m/s)
62 kgm/s north
pdi mdvdi, x (110.0 kg)(0.15 m/s)
16 kgm/s, east
Physics: Principles and Problems
Solutions Manual
623
Chapter 9 continued
Law of conservation of momentum states
pi pf
pf, x pi, x 16 kgm/s
pf, y pi, y 62 kgm/s
1
pf (pf, x 2 pf, y 2) 2
1
((16 kgm/s)2 (62 kgm/s)2) 2
64 kgm/s
p
64 kgm/s
f
vf 82 kg 110.0 kg
mr md
0.33 m/s
pf, y
tan1 pf, x
mrvri, y
mdvdi, x
tan1 (82 kg)(0.75 m/s)
(110.0 kg)(0.15 m/s)
tan1 75°
11. A 985-kg car traveling south at
29.0 m/s hits a truck traveling
18.0 m/s west, as shown in the figure
below. After the collision, the vehicles
stick together and travel with a final
momentum of 4.0104 kgm/s at
an angle of 45°. What is the mass
of the truck?
29.0 m/s
pf2 pfx 2 pfy2
pf2 pix 2 piy2
pf2 mc2vci2 mt2vti2
pf2 mc2vci2
mt vti2
1
2
p 4.0104 kgm/s
(4.0104 kgm/s) (985 kg)2(29.0 m/s)2
(18.0 m/s)
1
2
2
1.6103 kg
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Solutions Manual
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
18.0 m/s
45°
Chapter 9 continued
12. A 77.0-kg woman is walking 0.10 m/s east in the gym. A man throws a 15.0-kg
ball south and accidentally hits the woman. The woman and the ball move
together with a velocity of 0.085 m/s. Calculate the direction the woman and
the ball move.
p
pf
fx
cos1 p
pf
ix
cos1 m v
(mw mb)vf
w iw
cos1 (77.0 kg)(0.10 m/s)
cos1 (77.0
kg 15.0 kg)(0.085 m/s)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
1.0101 degrees south of east
Physics: Principles and Problems
Solutions Manual
625