Phys 2101
Homework 7 Solution Spring `11
These solutions use the parameter values from the problems printed in the book, not those that
appear in your personal homework assignment. The logic used to get to the answer is the same,
however.
1. The initial angular speed is
= (280 rev/min)(2
/60) = 29.3 rad/s.
(a) Since the rotational inertia is (Table 102(a))
2
2
(32 kg)(1.2 m)
46.1 kg
m
I
, the work done
is
2
2
2
4
1
1
0
(46.1kg
m
)(29.3 rad/s)
1.98
10
J
2
2
W
K
I
.
(b) The average power (in absolute value) is therefore



P
W
t

19.8
10
1.32
10
W.
3
3
15
2. (a) We use the parallelaxis theorem to find the rotational inertia:
2
2
2
2
2
2
com
1
1
20 kg
0.10 m
20 kg
0.50 m
0.15 kg
m .
2
2
I
I
Mh
MR
Mh
(b) Conservation of energy requires that
Mgh
I
1
2
2
, where
is the angular speed of the cylinder
as it passes through the lowest position. Therefore,
2
2
2
2(20 kg)(9.8 m/s
)(0.050 m)
11 rad/s.
0.15 kg m
Mgh
I
3. From Table 102, the rotational inertia of the spherical shell is 2
MR
2
/3, so the kinetic energy
(after the object has descended distance
h
) is
.
2
1
2
1
3
2
2
1
2
2
2
.
2
mv
I
MR
K
pulley
shell
sph
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Since it started from rest, then this energy must be equal (in the absence of friction) to the potential
energy
mgh
with which the system started. We substitute
v
/
r
for the pulley’s angular speed and
v
/
R
for that of the sphere and solve for
v
.
2
2
1
1
2
2
3
3
2
2
1
( /
)
(2
/3
)
2(9.8)(0.82)
1.4 m/s.
1
3.0
10
/((0.60)(0.050) )
2(4.5)/3(0.60)
I
M
r
mgh
gh
v
m
I
mr
M
m
4. By Eq. 1052, the work required to stop the hoop is the negative of the initial kinetic energy of
the hoop. The initial kinetic energy is
K
I
mv
1
2
2
1
2
2
(Eq. 115), where
I = mR
2
is its rotational
inertia about the center of mass,
m
= 140 kg, and
v
= 0.150 m/s is the speed of its center of mass.
Equation 112 relates the angular speed to the speed of the center of mass:
=
v
/
R
. Thus,
2
2
2
2
2
2
1
1
140 kg
0.150 m/s
3.15 J
2
2
v
K
mR
mv
mv
R
which implies that the work required is
0
3.15 J
3.15 J
W
K
.
5. (a) We find its angular speed as it leaves the roof using conservation of energy. Its initial kinetic
energy is
K
i
= 0 and its initial potential energy is
U
i
= Mgh
where
6.0sin30
3.0 m
h
(we are
using the edge of the roof as our reference level for computing
U
). Its final kinetic energy (as it
leaves the roof) is (Eq. 115)
K
Mv
I
f
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 Spring '07
 GROUPTEST
 Energy, Force, Mass, Work

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