1
Physics 221
Fall 2008
Homework #8 Solutions
Ch. 9:4-6; Ch. 10:1-4
Due Tues, Oct 21, 2008
8.1
N
identical point-like particles are equally spaced around the rim of a circular massless wheel of
radius
R
.
The total mass of all the particles is
M
.
Derive the formula for the moment of inertia of
the wheel for rotation about an axis through the center of the wheel and perpendicular to the plane
of the wheel.
Why is this formula for the moment of inertia the same as for the thin-walled cylinder
in Table 9.2(g) on p. 299 of the text?
I
=
m
i
r
i
2
i
=
1
N
±
=
R
2
m
i
i
=
1
N
±
=
MR
2
This expression is the same as for a thin-walled cylinder because all of the mass is a distance
R
from the axis of rotation, just as for a thin-walled cylinder.
8.2 State in words what the Parallel-Axis Theorem says.
Why is the Parallel-Axis Theorem a useful
theorem?
The Parallel-Axis Theorem states that if the axis of rotation of a rigid body is a distance
d
from a
parallel axis passing through the center of mass, then the moment of inertia of the rigid body about
the axis of rotation is equal to the moment of inertia of the body about the axis through the center of
mass, plus the mass of the body times
d
2
.
This theorem is very useful because one need only do one detailed and possibly complicated
calculation of the moment of inertia about an axis through the center of mass in order to simply
calculate the moment of inertia about any axis parallel to that axis.
8.3 A solid sphere of mass 1.00 kg and radius
R
= 0.100 m rotates at angular speed
20.0 rad/s about an axis that touches the edge of the sphere (see figure).
(a) What is the value of the moment of inertia of the sphere about its axis of
rotation?
According to Table 9.2(h), the moment of inertia of the sphere about an axis
through its center is
I
cm
= (2/5)
MR
2
.
Then using the Parallel Axis Theorem,
we get
I
=
I
cm
+
M d
2
=
(2/5)
MR
2
+
MR
2
=
(7/5)
MR
2
=
(7/5)(1.00 kg)(0.100 m)
2
=
0.0140 kg m
2
.
(b) What is the rotational kinetic energy of the sphere for rotation of the sphere about its axis of
rotation?
K
rot
=
(1/2)
I
±
2
=
(1/2)(0.0140 kg m
2
)(20.0 rad/s)
2
=
2.80 J .
R

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