The following are some examples of explicit evaluations
of the integral in Eq. (25).
EXAMPLE 6. Find the moment of inertia of a disk of uniform
density of mass M and radius R_{0}
rotating about its axis.
SOLUTION: Figure
12.14 shows the disk with the z axis along the axis of rotation.
Suppose the disk has a thickness l. The disk can be regarded as
made up of a large number of thin concentric hoops fitting one around
another. Figure 12.14 shows one such hoop of radius R, width dR,
and thickness l. The volume of the hoop is dV = 2p
Rl dR; with this, Eq. (25) gives
The total mass of the disk is the volume l
multiplied by the density r,
Hence rl = M/,
which, when substituted into Eq. (27), yields
for the moment of inertia of the disk.
COMMENTS AND SUGGESTIONS: This formula also gives the
moment of inertia of a cylinder, since such a body is simply a very thick
disk. The above calculation is equally valid for thin and for thick disks,
and the final formula is independent of the thickness l of the
disk.
EXAMPLE 7. Find the moment of inertia of a uniform thin
rod of length l and mass M rotating about an axis perpendicular
to the rod and through its center.
SOLUTION: Figure
12.15 shows the rod lying along the x axis; the axis of rotation
is the z axis. Consider a small slice dx of the rod. If
the crosssectional area of the rod is A, then dV = A dx and
Eq. (25) becomes
Since Al is the volume of the rod, rAl
is its total mass and therefore the moment of inertia can be written
EXAMPLE 8. Repeat the calculation of the preceding example
for an axis through one end of the rod.
SOLUTION: Figure
12.16 shows the rod and the axis of rotation. The rod extends from x
= 0 to x = l; hence instead of Eq. (29) we obtain
and instead of Eq. (30), we obtain
Comparison of Eqs. (30) and (32) makes it very clear
that the value of the moment of inertia depends on the location of the
axis of rotation. The moment of inertia is small if the axis passes through
the center of mass, and large if it passes through the end of the rod.
We can prove a general theorem, the parallelaxis
theorem, that relates the moment of inertia I_{CM}
about an axis through the center of mass to the moment of inertia I
about a parallel axis through some other point. The theorem asserts that
where M is the total mass of the body and d
the distance between the two axes. It is a corollary of Eq. (33) that
the moment of inertia about an axis through the center of mass is always
less than that about any other parallel axis.
A simple and instructive proof of the theorem is as follows:
Figure 12.17 shows
the body rotating about a fixed axis and also shows an alternative, parallel
axis through the center of mass. The kinetic energy of the body rotating
about the fixed axis is
However, we can express the kinetic energy in an alternative
form. According to Eq. (10.44) the kinetic energy of an arbitrary system
of particles is the sum of the translational energy of motion of the center
of mass and the "internal" energy of motion relative to the center of
mass,
In Figure 12.17 the center of mass moves along a circle
of radius d around the axis. Hence
Let us next look at the second term on the right side
of Eq. (35). Although the rotational motion of the body in Figure 12.17
is most simply described as a rotation about the fixed axis, it can also
be described as a rotation about the moving axis through the center of
mass. Each time the body completes one revolution about the fixed axis,
it also completes one revolution about this centerofmass axis; consequently,
the angular velocity of the rotation around the centerofmass axis is
also w. The kinetic energy of the rotational
motion relative to the center of mass is then
The values given by Eqs. (34) and (39) must be equal;
this implies that
which is the result we wanted to obtain.
We can also prove another theorem, the perpendicularaxis
theorem, which relates the moments of inertia of a thin flat plate
(for example, a sheet of metal) about three mutually perpendicular axes.
Figure 12.18 shows
a flat plate of arbitrary shape in the xy plane. The plate may
rotate either about an axis perpendicular to the plate (the z axis),
or about an axis in the plane of the plate (the x axis or the y
axis). We will call the moments of inertia for rotation about these alternative
axes I_{z}, I_{x},
and I_{y}, respectively. Then the
perpendicularaxis theorem asserts that
The proof of the theorem is very simple: For rotation
about the z axis, the distance from the axis of rotation to a particle
in the plate is
Comparison of these three formulas makes it obvious that
the sum of I_{x} and I_{y},
equals I_{z}.
Table 12.1 lists the moments of inertia of a variety
of rigid bodies about axes through their centers of mass; all the bodies
are assumed to have uniform density. The parallelaxis theorem can be
used to find the moment of inertia about any axis parallel to the centerofmass
axis. Furthermore, by means of combinations of the formulas of Table 12.1,
it is possible to find the moments of inertia of some other bodies.
EXAMPLE 9. Find the moment of inertia of a spherical
shell, of inner radius R_{1}
and outer radius R_{2} about
an axis through the center (Figure
12.19).
SOLUTION: According to Table 12.1, the moment of inertia
of a solid sphere of radius R_{2}
is
The shell of Figure 12.19 is a solid sphere with a concentric
spherical hole. Correspondingly, the inertia is that of a solid sphere
of radius R_{2} minus that of a
concentric sphere of radius R_{1}:
The volume of the shell is
and hence the density is
where M is now the mass of the shell. This leads to
for the inertia of the spherical shell.
