Example 8-8 Earring Moment of Inertia
A woman’s earring is a thin, uniform disk that has a mass M and a radius R. The earring hangs from the earring post by a
small hole near the edge of the disk and is free to rotate. Find the moment of inertia of the disk around this rotation axis.
Set Up
A solid disk is an example of a solid cylinder like that
shown in Table 8-1. This table shows that the moment
of inertia of such a disk for an axis perpendicular to
the plane of the disk and passing through its center is
I = MR2 >2. (Unlike the cylinder shown in the table,
the length L of the earring is much less than its radius
R. But the length has no effect on the value of I.)
Because the disk is uniform, its geometrical center
is its center of mass, so MR2 >2 equals ICM. The axis we
want is parallel to the axis through the disk’s center, so
we can use the parallel-axis theorem, Equation 8-8, to
find the moment of inertia around the axis at the rim.
Solve
The edge of the disk is a distance h = R from the
center of mass of the disk. Substitute this value and
ICM = MR2 >2 into the parallel-axis theorem.
Reflect
If the wearer of these earrings nods her head up and
down while dancing at a club, she can make the
disks rotate back and forth around the posts in an
eye-catching way. The more massive the disks and
the larger their radius, the greater their moment of
inertia and the more effort will be required to get them
moving (recall that rotational kinetic energy is given
by K rotational = 12Iv2). Making the earrings small and
lightweight will require less effort from the wearer!
Parallel-axis theorem:
I = ICM + Mh2
(8-8)
rotation axis
disk, mass M
R
I =
MR2
+ MR2
2
or
I =
3MR2
2
center of mass