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Home , Moment of inertia

Moments of

Suppose a body is moving on a circular path and reminds us of the analogous equation for with constant . Let’s consider two quan- linear tities: the body’s L about the p= mv , center of the circle, and its kinetic T. How are these quantities related to its angular which is of the form ω ? linear momentum = H linear velocity .

LP , ωP The of the body is

1 Pv T = mv2 ,

2 Pr and is of the form kinetic energy = (1/2) mass H linear velocity2 . The angular momentum about the center of the circle has magnitude Substituting in for the velocity, we find L = mvr , 1 T= mr2 ω 2 and the velocity has magnitude 2 , v = r ω . which is of the form The first relation we seek is therefore kinetic energy = (1/2) constant H angular velocity2 . L= mr 2 ω . The “constant” in these equations is the rotational analog of mass. It is called the of inertia This is of the form of the body about the point at the center of the circle, and is symbolized by I: angular momentum = constant H , 2 1 2 I = 3 m r L= I ω and T = I ω . i i . 2 . i

In this particular case, the value of the moment of In the case of a continuous body, the sum inertia is becomes an integral. We will see some cases of this below. I = mr2 . Example: moment of inertia of a ring about its Now let’s suppose that we have a system center consisting of several bodies held rigidly together in a single plane, and that this composite body is Suppose that instead of a single body moving in a rotating about a point in the plane of the body: circular path, we have a thin ring spinning around

on its axis like this:

Pω Pω

v r P 1 P 1 R

Pr 2

Pv 2 The mass of the ring is m and its radius is R. Then each piece of the whole thing goes around in a circle, and they all have the same angular What is the moment of inertia of the ring about velocity. The angular momentum of the system its center? is just the sum of the angular momenta of the We imagine the ring split up into tiny pieces. All parts: are at the same R from the center of the 2 circle. The expression for the moment of inertia L= ä 3 m r ë ω , å i i ì simplifies, becoming ã i í I = ä m ë R2 . so the moment of inertia of the system is the sum å 3 i ì of the individual moments of inertia: ã i í The sum of the of the pieces is just m, so ring, of radius r and (infinitesimal) thickness dr: we get the result I = mR 2 .

This is the same as the moment of inertia of a single body at distance R from the center. r

Example: moment of inertia of a disc about its dr center Now let’s soup up our example even more. R Suppose we have a circular disc spinning around on its axis:

ω (The thickness has been exaggerated in the P figure.) What’s the mass of this ring? Well, the mass will be the mass of the whole disc, the ratio of the of the ring to the area of the disc. The area of the ring is just its length times its thickness, or 2π r dr . So the mass is . 2π r dr 2m Let’s let the mass of the disc be m, and its radius m = r dr . πR 2 R 2 be R. We’ll also suppose that the mass is uniformly distributed over the disc. What is the The moment of inertia of the ring is its mass moment of inertia of the disc about its center? times the of its radius r, and we want to Well, we can think of the disc as being made up add these up for all radii from 0 to R. This means of a bunch of thin rings. We can “add up” the we have to do an integral R R moments of inertia of all the rings using , ä 2m ìë 2m 2m 1 = 2 å ì = 3 = 4 and the result will be the moment of inertia of the I I r å 2 r drì 2 I r dr 2 R . R R R 4 disc. 0 ã í 0 Let’s see how this works. Consider a typical The result is multiply their mass times the square of their 1 I = m R 2 distance from the point of . We then add 2 . these up over the whole ring. Warning! This is the kind of calculation that This is an interesting result. If all the mass were looks very easy when done by someone else, but concentrated at the edge (that is, if the plate were can appear nearly impossible when you try it actually a ring), then the moment of inertia would 2 yourself! Don’t be fooled! Practice on bodies of be mR . The moment of inertia of the disc is other shapes yourself. actually only half as big as this, because the rings nearer to the center contribute less than they One of the biggest challenges when doing a would if they were right at the edge. calculation of this kind is the very first step: the choice of suitable coordinates. Here is a good Example illustrating dependence of moment of choice for the present case:

inertia on the point of rotation R dθ Suppose we return to our thin ring of mass m and

radius R, and we constrain it to rotate about a r θ/2 θ point on the ring. Here is a view: R d

calculating θ moment of θ/2 inertia about this point

rotates about this point The mass of the arc subtended by the dθ, shown in blue, is the mass of the ring times the What’s the moment of inertia now? ratio of the length of the arc to the circumference Well, the principle is no different from before. of the ring, or Only the details of its application change. We Rdθ m m = d θ . just take all the little pieces of the ring and 2 πR 2 π The distance of the little arc to the point about has mass m, again uniformly distributed. which we are calculating the moment of inertia is θ =

r 2R cos . 2 a Multiplying the mass times the square of the

distance, and integrating over the whole ring, b gives for the moment of inertia

2π 2π ä θ ë ä m ë 2m R 2 1 = å 2 2 ì å θ ì = + θ θ I I å 4R cos ì å π d ì π I (1 cos )d 0 ã 2 í ã 2 í 0 2 Show that the moment of inertia of the wheel The value of the integral is π, so the result is about its axle is 1 4 I = 2 mR 2 I = M( a 2 + b 2 ) + m b 2 . . 2 3 Notice that this is the same body for which we Hint: this exercise is designed to be rather earlier calculated the moment of inertia to be half complicated. You have to break up the wheel as large! That’s because the two moments of into separate parts, calculate their moments of inertia are taken about different points. The inertia individually, and add them up in the end. moment of inertia is not an intrinsic property of Example of use of energy of rotation: body the body, but rather depends on the choice of the down slope point around which the body rotates. Here is a famous example. Consider a circular Exercise: moment of inertia of a wagon wheel body which rolls without slipping down an about its center inclined plane. Let’s not make any assumptions Consider a wagon wheel made up of a thick rim about the nature of the body, except that its mass and four thin spokes. The rim’s outer radius is a is m and its moment of inertia about its central and inner radius is b, and its mass M is uniformly axis is I. What is the along the distributed. The spokes have length b and each slope? ω = v

. ω R

Substituting, we find x θ v x sin R θ 1 ä I ìë 2 E = m å 1 + ì v − mgx sin θ . 2 ã mR2 í We know that this is constant in , so we may You could tackle this by trying to figure out all set its time derivative to zero: the involved, but by far the easiest way is ä I ìë to use the method of energy. 0 = mv å 1 + ì a − mgv sin θ . ã mR2 í Let the distance travelled along the slope be x. Then the vertical distance travelled in the Solving for the acceleration, we find downward direction is x sin θ. g sin θ The total energy is a = I . 1 + 1 1 2 E = mv2 + Iω 2 − mgx sin θ . mR 2 2 Notice that this is less than the acceleration of a The first term is the kinetic energy due to the of the and the is body down a frictionless plane at the the kinetic energy due to rotation about the center same angle. Why is this? Well, in the present of mass. (We used a result of an earlier section case the of pushes uphill along the to split up the kinetic energy in this way.) The slope, causing the body to rotate as it moves. third term is the gravitational . This extra frictional force slows down the body, compared with a body sliding down a frictionless A little reflection will convince you that if no plane. slipping occurs, then the angular of rotation is related to the speed along the plane by Exercise - practice with forces Derive the above result by considering the forces involved. Exercise - how to measure the moment of inertia LP = r m v = m r (ω r ) . 3 Pi H i Pi 3 iP i H P H P i Think of a way in which you could use the above i i result to measure the moment of inertia of a body We then make use of the standard relation for the with circular cross-section. triple

P P P = P P P − P P P The inertia A H (B H C) ( A A C) B (A A B) C . We find the result Up until now, we have been considering only the LP = m ä r 2 ω − (r ω) r ë . 3 i ã i P Pi A P P i í rotation of two-dimensional planar bodies about i axes perpendicular to the body. But you can This is a nice compact form of the result, but easily see that this is only a very small subset of there’s another form that gives more physical all possible of a . So now insight into the relation between the angular let’s move on to the more general case of the momentum and the angular velocity. This other rotation of a three-dimensional rigid body about form is what we get when we re-write the above an arbitrary axis. We will assume only that the equation in notation. direction of the axis of rotation does not change. Let’s begin by looking at the components of the Let’s let the angular velocity vector of the body last equation. We write the vector as be ωP . This is the same for every piece of the r = xˆ x + yˆ y + zˆ z , body. As we learned in our section on rotational Pi i i i motion, the velocity of any particular piece is and obtain for the x-component of the angular v = ω r . Pi P H P i momentum L = m ä r 2 ω − (x ω + y ω + z ω )x ë , Let’s ask the question: what is the relation x 3 i ã i x i x i y i z i í between the total angular momentum of the body i and the angular velocity vector? We begin by with similar equations for the other components. writing down the expression for the angular momentum and substituting for the velocity: We can re-write this as a linear combination of the three components of the angular velocity: L = ä m (r 2 − x 2 ) ë ω − ä m x y ë ω − ä m x z ë ω x å 3 i i i ì x å 3 i i i ì y å 3 i i i ì z ã i í ã i í ã i í We recognize the right-hand side as the top component of the product of a matrix and the angular velocity vector:

2 2 ω ä m (r − x ) ë ä − m x y ë ä − m x z ë L å 3 i i i ì å 3 i i i ì å 3 i i i ì x x ã i í ã i í ã i í = ω í í í í y í ω í í í z (For clarity, the other components have been temporarily suppressed.) This matrix is called the inertia tensor. We will symbolize it by ¯. Written out in full, it is ä m (r 2 − x 2 ) ë ä − m x y ë ä − m x z ë å 3 i i i ì å 3 i i i ì å 3 i i i ì ã i í ã i í ã i í ä − 3 m y x ë ä 3 m (r 2 − y 2 ) ë ä − 3 m y z ë ¯ = å i i i ì å i i i ì å i i i ì ã i í ã i í ã i í ä − m z x ë ä − m z y ë ä m (r 2 − z 2 ) ë å 3 i i i ì å 3 i i i ì å 3 i i i ì ã i í ã i í ã i í and the relation between the angular momentum and the angular velocity is written compactly as

LP = ¯ ωP .