Week 5: Torque and Rotation in One Dimension

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5.3: The Moment of Inertia

We begin with a specific example to help smooth the way.

Example 5.3.1: The Moment of Inertia of a Rod Pivoted at One End

y
pivot	dm = λ dx
x	L	x
	dx

Figure 63: A solid rod of length L with a mass M uniformly pivoted about one end. One can think of such a rod as being the “massless rod” of the previous section with an infinite number of masses mi uniformly distributed along its length, that sum to the total mass M .

In figure 63 above a massive rod pivoted about one end is drawn. We would like to determine how this particular rod will rotationally accelerate when we (for example) attach a force to it and apply a torque. We therefore must characterize this rod as having a specific mass M , a specific length L, and we need to say something about the way the mass is distributed, because the rod could be made of aluminum (not very dense) at one end and tungsten (very dense indeed) at the other and still “look” the same. We will assume that this rod is uniformly distributed, and that it is very thin and symmetrical in cross-section – shaped like a piece of wire or perhaps a wooden dowel rod.

In a process that should be familiar to you from last week and from the previous section, we know that the moment of inertia of a sum of discrete point masses hung on a “massless” rod (that only serves to assemble them into a rigid structure) is just:

X
Itot = miri2	(477)
i

the sum of the moments of inertia of the point masses.

We can clearly approximate the moment of inertia of the continuous rod by dividing it up into N

	pieces, each of length x = L/N and mass			M = M/N , and treating each small piece as a “point
	mass” located at xi = i x:	N		µ			N
	I		M		i L	=		M x2	(478)
		rod ≈ i=1	N		N		i=1
						¶		i
		X					X

As before, the limit of this sum as N → ∞ is by definition the integral, and in this limit the sum exactly represents the moment of inertia of the rod.

We can easily evaluate this. To do so, we chant our ritual expression: “The mass of the chunk is the mass per unit length of the chunk times the length of the chunk”, or dm = λdx = ML dx, so:

Irod = Z0	x2dm = L Z0		x2dx =	3	(479)
L		M	L	M L2

5.3.1: Moment of Inertia of a General Rigid Body

This specific result can easily be generalized. If we consider a blob-shaped distribution of mass, the di erential moment of inertia of a tiny chunk of the mass in the distribution about some fixed axis

244	Week 5: Torque and Rotation in One Dimension

pivot axis

dm

r

Figure 64: A “blob-shaped” chunk of mass, perhaps a piece of modelling clay, constrained to rotate about an axis through the blob, perhaps a straight piece of nearly massless coat-hanger wire.

of rotation is clearly:

dI = r2dm

(480)

By now you should be getting the idea that summing up all of the little chunks that make up the

object is just integrating:

Z

Iblob = r2dm (481)

blob

where it is quite one thing to write down this formal expression, quite another to be able to actually do the integral over all of the chunks of mass that make up an object.

It isn’t too di cult to do this integral for certain simple distributions of mass, and we will need a certain “stock repertoire” of moments of inertia in order to solve problems. Which ones you should learn to do depends on the level of the course – math/physics majors should learn to integrate over spheres (and maybe engineers as well), but everybody else can probably get away learning to evaluate the moment of inertia of a disk. In practice, for any really complicated mass distribution (like the blob of clay pictured above) one would either measure the moment of inertia or use a computer to actually break the mass up into a very large number of discrete (but small/point-like) chunks and do the sum.

First let’s do an example that is even simpler than the rod.

Example 5.3.2: Moment of Inertia of a Ring

z

M

dθ R

ds = Rd θ

Figure 65: A ring of mass M and radius R in the x-y plane rotates freely about the z-axis.

We would like to find the moment of inertia of the ring of uniformly distributed mass M and radius R portrayed in figure 65 above. A di erential chunk of the ring has length ds = R dθ. It’s mass is thus (say the ritual words!):

	dm = λds =	M	R dθ =	M	dθ	(482)
		2πR		2π

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and its moment of inertia is very simple:

Iring = Z	r2dm = Z0	2π R2	dθ = M R2	(483)
		2π M

In fact, we could have guessed this. the axis of rotation, so its moment of and has no “vector” character) is just

All of the mass M in the ring is at the same distance R from inertia (which only depends on the mass times the distance M R2 just like a point mass at that distance.

Because it is so important, we will do the moment of inertia of a disk next. The disk will be many things to us – a massive pulley, a wheel or tire, a yo-yo, a weight on a grandfather clock (physical) pendulum. Here it is.

Example 5.3.3: Moment of Inertia of a Disk

dA = rdθdr

rdθ

M	dr

dθ

R r

Figure 66: A disk of mass M and radius R is pivoted to spin freely around an axis through its center.

In figure 66 a disk of uniformly distributed mass M and radius R is drawn. We would like to find its moment of inertia. Consider the small chunk of disk that is shaded of area dA. In plane polar coordinates (the only ones we could sanely hope to integrate over) the di erential area of this chunk is just its di erential height dr times the width of the arc at radius r subtended by the angle dθ, r dθ. The area is thus dA = r drdθ.

This little chunk was selected because the mass dm in it moves in a circle of radius r around the pivot axis. We need to find dm in units we can integrate to cover the disk. We use our litany to set:

	dm = σdA =		M		r drdθ	(484)
			πR2
	and then write down:
	dI = r2dm =		M	r3 drdθ		(485)
		πR2

We integrate both sides to get (note that the integrals are independent one dimensional integrals that precisely cover the disk):

Idisk =	πR2		Ã	0		r3 dr!		0		dθ
	M			Z	R			µZ	2π	¶
=	πR2		µ	4		¶	(2π)
	M			R4
=	1	M R2								(486)
	2

This is a very important and useful result, so keep it in mind.