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Moments of Inertia

Suppose a body is moving on a circular pathwith constant speed. Lets consider two quan-tities: the bodys angular momentumL about the

center of the circle, and its kinetic energy T. How

are these quantities related to its angular velocity?

r

P

v

P

L

P

, P

The angular momentum about the center of thecircle has magnitude

L=mvr,

and the velocity has magnitude

v=r .

The first relation we seek is therefore

L=mr2.

This is of the formangular momentum = constant H angular velocity ,

and reminds us of the analogous equation forlinear momentum

p=mv ,

which is of the form

linear momentum = mass H linear velocity .

The kinetic energy of the body is

T=1

2mv2 ,

and is of the form

kinetic energy = (1/2) mass H linear velocity2 .

Substituting in for the velocity, we find

T=1

2mr22

,

which is of the form

kinetic energy = (1/2) constant H angular velocity2 .

The constant in these equations is the rotationalanalog of mass. It is called themoment of inertia

of the body about the point at the center of thecircle, and is symbolized byI:

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L=I and T=1

2I2 .

.

In this particular case, the value of the moment ofinertia is

I=mr2 .

Now lets suppose that we have a systemconsisting of several bodies held rigidly togetherin a single plane, and that this composite body isrotating about a point in the plane of the body:

r

P

v

P

P

v

P

r

P

1

1

2

2

Then each piece of the whole thing goes aroundin a circle, and they all have the same angular

velocity. The angular momentum of the systemis just the sum of the angular momenta of theparts:

L=

3

i

mir2

i

,

so the moment of inertia of the system is the sumof the individual moments of inertia:

I=3i

mir

i

2

.

In the case of a continuous body, the sumbecomes an integral. We will see some cases ofthis below.

Example: moment of inertia of a ring about itscenter

Suppose that instead of a single body moving in acircular path, we have a thin ring spinning aroundon its axis like this:

P

R

The mass of the ring is m and its radius is R .

What is the moment of inertia of the ring about

its center?We imagine the ring split up into tiny pieces. Allare at the same distance R from the center of the

circle. The expression for the moment of inertiasimplifies, becoming

I=

3im

i

R2 .

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The sum of the masses of the pieces is just m, so

we get the result

I=mR2 .

This is the same as the moment of inertia of asingle body at distanceR from the center.

Example: moment of inertia of a disc about itscenter

Now lets soup up our example even more.Suppose we have a circular disc spinning aroundon its axis:

P

.

Lets let the mass of the disc be m, and its radius

be R. Well also suppose that the mass is

uniformly distributed over the disc. What is the

moment of inertia of the disc about its center?

Well, we can think of the disc as being made upof a bunch of thin rings. We can add up themoments of inertia of all the rings using calculus,and the result will be the moment of inertia of thedisc.

Lets see how this works. Consider a typical

ring, of radius rand (infinitesimal) thickness dr:

r

dr

R

(The thickness has been exaggerated in thefigure.) Whats the mass of this ring? Well, themass will be the mass of the whole disc, timesthe ratio of the area of the ring to the area of the

disc. The area of the ring is just its length timesits thickness, or 2r dr. So the mass is

2r dr

R2m =

2m

R2r dr .

The moment of inertia of the ring is its mass

times the square of its radius r, and we want toadd these up for all radii from 0 toR. This means

we have to do an integral

I=R

I0

r2

2m

R2r dr

=

2m

R2

R

I0

r3dr =2m

R21

4R4 .

The result is

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I=1

2m R2

.

This is an interesting result. If all the mass wereconcentrated at the edge (that is, if the plate wereactually a ring), then the moment of inertia wouldbe mR2. The moment of inertia of the disc is

actually only halfas big as this, because the rings

nearer to the center contribute less than theywould if they were right at the edge.

Example illustrating dependence of moment ofinertia on the point of rotation

Suppose we return to our thin ring of mass m and

radius R, and we constrain it to rotate about a

point on the ring. Here is a top view:

rotates aboutthis point

Whats the moment of inertia now?

Well, the principle is no different from before.

Only the details of its application change. We just take all the little pieces of the ring and

multiply their mass times the square of theirdistance from the point of rotation. We then addthese up over the whole ring.

Warning! This is the kind of calculation thatlooks very easy when done by someone else, butcan appear nearly impossible when you try ityourself! Dont be fooled! Practice on bodies ofother shapes yourself.

One of the biggest challenges when doing a

calculation of this kind is the very first step: thechoice of suitable coordinates. Here is a goodchoice for the present case:

/2

/2calculatingmoment ofinertia aboutthis point

d

R d

Rr

The mass of the arc subtended by the angle d,

shown in blue, is the mass of the ring times theratio of the length of the arc to the circumferenceof the ring, or

Rd2R

m = m2

d .

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The distance of the little arc to the point aboutwhich we are calculating the moment of inertia is

r=

2R cos

2 .

Multiplying the mass times the square of thedistance, and integrating over the whole ring,gives for the moment of inertia

I=

2

I0

4R

2

cos2

2

m

2 d

=2mR2

2

I0

1

2 (1 +cos )d

The value of the integral is , so the result is

I= 2mR2

.

Notice that this is the same body for which we

earlier calculated the moment of inertia to be halfas large! Thats because the two moments ofinertia are taken about different points. The

moment of inertia is not an intrinsic property ofthe body, but rather depends on the choice of the

point around which the body rotates.

Exercise: moment of inertia of a wagon wheelabout its center

Consider a wagon wheel made up of a thick rimand four thin spokes. The rims outer radius is a

and inner radius is b, and its massMis uniformlydistributed. The spokes have length b and each

has mass m, again uniformly distributed.

b

a

Show that the moment of inertia of the wheelabout its axle is

I=1

2M(a2 +b2) +

4

3m b2 .

Hint: this exercise is designed to be rathercomplicated. You have to break up the wheelinto separate parts, calculate their moments ofinertia individually, and add them up in the end.

Example of use of energy of rotation: body

rolling down slopeHere is a famous example. Consider a circularbody which rolls without slipping down an

inclined plane. Lets not make any assumptionsabout the nature of the body, except that its massis m and its moment of inertia about its central

axis is I. What is the acceleration along theslope?

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x sin

xv

R

You could tackle this by trying to figure out allthe forces involved, but by far the easiest way is

to use the method of energy.

Let the distance travelled along the slope be x.

Then the vertical distance travelled in thedownward direction isx sin .

The total energy is

E=1

2mv2 +

1

2I2 mgx sin .

The first term is the kinetic energy due to themotion of the center of mass and the second isthe kinetic energy due to rotation about the center

of mass. (We used a result of an earlier sectionto split up the kinetic energy in this way.) Thethird term is the gravitational potential energy.

A little reflection will convince you that if noslipping occurs, then the angular frequency ofrotation is related to the speed along the plane by

=v

R.

Substituting, we find

E=1

2m

1 +

I

mR2

v

2 mgx sin .

We know that this is constant in time, so we mayset its time derivative to zero:

0 =mv

1 + ImR2

a mgv sin .

Solving for the acceleration, we find

a =g sin

1 +

I

mR2.

Notice that this is less than the acceleration of a

body sliding down a frictionless plane at thesame angle. Why is this? Well, in the presentcase the force of friction pushes uphill along theslope, causing the body to rotate as it moves.This extra frictional force slows down the body,compared with a body sliding down a frictionlessplane.

Exercise - practice with forces

Derive the above result by considering the forcesinvolved.

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Exercise - how to measure the moment of inertia

Think of a way in which you could use the aboveresult to measure the moment of inertia of a body

with circular cross-section.

The inertia tensor

Up until now, we have been considering only the

rotation of two-dimensional planar bodies aboutaxes perpendicular to the body. But you caneasily see that this is only a very small subset ofall possible rotations of a rigid body. So nowlets move on to the more general case of therotation of a three-dimensional rigid body aboutan arbitrary axis. We will assume only that the

direction of the axis of rotation does not change.

Lets let the angular velocity vector of the bodybe P . This is the same for every piece of thebody. As we learned in our section on rotationalmotion, the velocity of any particular piece is

vPi = PH rPi .

Lets ask the question: what is the relationbetween the total angular momentum of the bodyand the angular velocity vector? We begin bywriting down the expression for the angular

momentum and substituting for the velocity:

LP= 3i

rPiHm

ivP

i= 3

i

mirP

iH(P HrP

i) .

We then make use of the standard relation for the

triple cross product

AP H (BP HCP) = (AP ACP)BP(AP ABP)CP .

We find the result

LP=3i

mir

2

iP (rP

iAP)rP

i .

This is a nice compact form of the result, buttheres another form that gives more physicalinsight into the relation between the angularmomentum and the angular velocity. This otherform is what we get when we re-write the above

equation in matrix notation.Lets begin by looking at the components of thelast equation. We write the position vector as

rPi

=xxi+yy

i+zz

i,

and obtain for the x-component of the angular

momentum

Lx

= 3i

mir

2

i

x(x

i

x+y

i

y+z

i

z)x

i ,

with similar equations for the other components.

We can re-write this as a linear combination ofthe three components of the angular velocity:

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Lx

=

3

i

mi(r2

ix2

i)

x

3

i

mix

iy

i

y

3

i

mix

iz

i

z

We recognize the right-hand side as the top

component of the product of a matrix and theangular velocity vector:

Lx

=

3

i

mi(r2

ix2

i)

3

i

mix

iy

i

3

i

mix

iz

i

x

y

z

(For clarity, the other components have beentemporarily suppressed.) This matrix is called the

inertia tensor. We will symbolize it by .Written out in full, it is

=

3

im i(r

2

i x2

i )

3

i

miy

ix

i

3

i

miz

ix

i

3

im ixiyi

3

i

mi(r2

iy2

i)

3

i

miz

iy

i

3

im ixizi

3

i

miy

iz

i

3

i

mi(r2

iz2

i)

and the relation between the angular momentumand the angular velocity is written compactly as

LP= P.