Physics 2
Chapter 10 problems
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10.6 A machinist is using a wrench to loosen a nut. The wrench is 25cm
long, and he exerts a 17-N force at the end of the handle.
a) What torque does the machinist exert about the center of the nut?
b) What is the maximum torque he could exert with this force?
Use the definition of torque.
The direction of the torque will
be out of the page.



 
r F
 
r F sin( )
(0.25m) (17N) sin(37 )
2.56N m
Maximum torque will always occur when the
radius and force are perpendicular to each other

m ax
(0.25m) (17N) sin(90 )
4.25N m
Prepared by Vince Zaccone
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10.12 A stone is suspended from the free end of a wire that is wrapped around
the outer rim of a pulley. The pulley is a uniform disk with mass 10kg and
radius 50cm and turns on frictionless bearings. You measure that the stone
travels 12.6m in the first 3 seconds, starting from rest.
Find a) the mass of the stone and b) the tension in the wire.
y
y0
v0,y t
12.6m
ay
0
0
1
2
ay t2
1
2
ay (3s)2
2.8 sm2
Positive = downward
First we can use kinematics to find
the acceleration of the falling stone:
R
T
M
T
Next use the torque on the pulley to
find the tension:
T R
I
a
T R ( 12 MR ) ( )
R
1
T 2 (10kg)(2.8 sm2 )
2
mg
T
1
2
M a
14N
Then we can use the forces on the
stone to find its mass:
F mg T ma
mg ma T
m(g a) T
T
14N
m
2kg
m
g a 9.8 s2 2.8 sm2
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10.20 A string is wrapped several times around the rim
of a small hoop with radius 8cm and mass 0.18kg. The
free end of the string is held in place and the hoop is
released from rest. After the hoop has descended 75cm,
calculate
a) the angular speed of the rotating hoop and
b) the speed of its center.
Use conservation of energy:
Eto p
Ebo tto m
mgh
1
2
mv2cm
1
2
I
2
mgh
1
2
mv2cm
1
2
(mr2 )(
mgh
1
2
mv2cm
1
2
mv2cm
mgh
mv2cm
vcm
gh
vcm 2
)
r
2.7 ms
Divide by the radius to get the angular speed:
vcm
r
33.9 ras d
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10.26 A bowling ball rolls without slipping up a ramp that slopes upward at
an angle to the horizontal. Treat the ball as a uniform, solid sphere,
ignoring the finger holes.
a) Draw the free-body diagram for the ball. Explain why friction is uphill.
b) What is the acceleration of the center of mass of the ball?
c) What minimum coefficient of static friction is needed to prevent slipping?
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10.44 A solid wood door 1m wide and 2m high is hinged along one side and has
a total mass of 40kg. Initially open and at rest, the door is struck at its center by
a handful of sticky mud with mass 5kg, traveling perpendicular to the door at
12m/s just before impact. Find the final angular speed of the door. Does the mud
make a significant contribution to the moment of inertia?
10.62 A large 16kg roll of paper with radius R=18cm rests
against the wall and is held in place by a bracket attached to
the rod through the center of the roll. The rod turns without
friction in the bracket, and the moment of inertia of the paper
and rod about the axis is 0.26kg-m2. The other end of the
bracket is attached by a frictionless hinge to the wall such
that the bracket makes and angle of 30° with the wall. The
weight of the bracket is negligible. The coefficient of kinetic
friction between the paper and the wall is μk=0.25. A
constant vertical force F=40N is applied to the paper, and the
paper unrolls.
a) What is the magnitude of the force the rod exerts on the
paper as it unrolls?
b) What is the angular acceleration of the roll?
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10.75 A solid, uniform ball roll without slipping up a
hill, as shown. At the top of the hill, it is moving
horizontally, and then it goes over the vertical cliff.
a) How far from the foot of the cliff does the ball land,
and how fast is it moving just before it lands?
b) Notice that when the ball lands, it has a larger
translational speed than when it was at the bottom of
the hill. Does this mean that the ball somehow gained
energy? Explain!
First we need to use conservation of energy to find the linear speed of the ball
when it reaches the top of the cliff, then we will use kinematics to find the
distance and final speed.
Eto p
Ebo tto m
1
2
mv2
1
2
1
2
mv2
1
2
7
10
mv2
7
10
v2
I
2
mgh
1
2
mv20
v
( 25 mr2 )( )2 mgh
r
7
mgh 10
mv20
(9.8 sm2 )(28m)
7
10
2
1
2
I
1
2
mv20
(25 ms )2
0
1
2
v to p
( 25 mr2 )(
v0 2
)
r
15.26 ms
Use free-fall to find the time it takes to fall, then multiply to get distance:
1
2
gt2
28m
1
2
y
(9.8 sm2 )t2
x
(15.26 ms )(2.4s)
vy
(9.8 sm2 )(2.4s)
t
2.4s
36.6m
23.5 ms
We need the Pythagorean theorem to get the final speed:

v
v2x
v2y
(15.26 ms )2
(23.5 ms )2
28 ms
Notice that this is larger than the initial speed! The ball is spinning more slowly,
so more energy is available for linear motion.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB