262	Week 5: Torque and Rotation in One Dimension

pivot

M

R

Figure 77: A hoop of mass M and radius R is pivoted on the side – think of it as being hung on a nail from a barn door.

is the moment of inertia of the hoop about this new axis parallel to the one through the center of mass that we used before?

It’s so simple:

Iside pivot = M R2 + Icm = M R2 + M R2 = 2M R2

(558)

and we’re done!

For your homework, you get to evaluate the moment of inertia of a rod about an axis through its center of mass and about one end of the rod and compare the two, both using direct integration and using the parallel axis theorem. Good luck!

5.9: Perpendicular Axis Theorem

In the last section we saw how a bit of geometry and math allowed us to prove a very useful theorem

– useful because we can now learn a short table of moments of inertia about a given axis through the center of mass and then easily extend them to find the moments of inertia of these same shapes when they uniformly rotate around a parallel axis.

In this section we will similarly derive a theorem that is very useful for relating moments of inertia of planar distributions of mass (only) around axes that are perpendicular to one another

– the Perpendicular Axis Theorem. Here’s how it goes.

y
		dm
		M
y	r
	x	x
z

Figure 78: A planar blob of mass and the geometry needed to prove the Perpendicular Axis Theorem

Week 5: Torque and Rotation in One Dimension

263

Suppose that we wish to evaluate Ix, the moment of inertia of the plane mass distribution M

shown in figure 78. That’s quite easy:

Z

	Ix =	y2dm		(559)
Similarly,	Iy = Z
We add them, and presto chango!		x2dm		(560)
Ix + Iy = Z	y2dm + Z	x2dm = Z	r2dm = Iz	(561)

This is it, the Perpendicular Axis Theorem:

Iz = Ix + Iy

(562)

I’ll give a single example of its use. Let’s find the moment of inertia of a hoop about an axis through the center in the plane of the loop!

264	Week 5: Torque and Rotation in One Dimension

Example 5.9.1: Moment of Inertia of Hoop for Planar Axis

y

M

R

x

z

Figure 79: A hoop of mass M and radius R is drawn. What is the moment of inertia about the x-axis?

This one is really very, very easy. We use the Perpendicular Axis theorem backwards to get the answer. In this case we know Iz = M R2, and want to find Ix. We observe that from symmetry, Ix = Iy so that:

	Iz = M R2 = Ix + Iy = 2Ix						(563)
	or	1			1
	Ix =		Iz =			M R2	(564)
		2		2