The parallel axis theorem

Let us recall what we have learned about rigid bodies that are confined to lie and rotate in a plane, such as a rod or a sheet of some metal cut out into some arbitrary shape. The body has a mass M. It can translate and rotate, but for now we nail a point on it to the plane and let it rotate about that axis, with plans to bring in translations later on. A single angle θ, measured in radians, suffices to tell us what it is doing, because all it can do is rotate about the fixed point. This angleθ is to rotations what x was to translations. There is a corresponding angular velocityω =^dθ_dt and accelerationα =^d_dt^ω =^d2_dt2^θ. The novel attribute, owing to the object being extended and not point-like, is its moment of inertia I. If we imagine it as being made of point masses mi sitting at a distance ri from the axis, we find

I=

i

mir_i². (10.1)

Note that I is a quantity that depends on how the mass is distributed rela-tive to the axis about which it is rotating. It plays the role that mass did in translational motion. If the body is continuous, the sum is replaced by the corresponding integral. The angular momentum L

L= Iω (10.2)

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plays the role of momentum p= mv, and F = ma =^dp_dt is replaced by τ = Iα =dL

dt (10.3)

where the torque is defined by τ =

i

riFisinγi (10.4)

whereγiis the angle measured from ri, the vector from the point of rota-tion to the point of applicarota-tion of the force, and force vector Fi. We may rewrite the expression for torque in two other useful ways:

τ = F⊥r= Fr⊥ (10.5)

where F_⊥ is the component of the force perpendicular to r and r_⊥is the component of r perpendicular to F.

In our convention a torque was positive (negative) if it implied counterclockwise (clockwise) acceleration.

The kinetic energy is K=1

2Iω². (10.6)

All the rotational equations, say the one for K, are simply the results from linear motion as applied to the constituents of the rigid body.

Finally, at each point in the body, the tangential displacement, velocity, and acceleration are just r times the angular ones

s= rθ vT= rω aT= rα. (10.7)

We practiced computing moments of inertia of various objects (rod and disk) and concluded with the observation that for a rod

Iend= ICM+ M

L 2

2

. (10.8)

We will now see that this is an example of a more general result:

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The parallel axis theorem: The moment of inertia for a planar body about any axis is

I= ICM+ Md² (10.9)

where d is the distance from the CM to the new axis. (In the rod example d=^L₂.)

Here is the proof. As a warm-up, we will do it in d= 1 for a rod of a given linear mass density ρ(x), that is, where ρ(x)dx is the mass of an infinitesimal segment of width dx at the point x. Let us choose as our origin the CM, and measure x from it. Let the new axis pass through the point x= d, as shown in Figure 10.1. The moment of inertia about the new axis is

The first two terms are welcome: they are required in the theorem. The cross term is like a leftover part after you have assembled your bookshelf from Ikea. We need to get rid of it. Luckily it is zero, for the following reason. By definition

where X is the CM. What is X? Our origin is at the CM itself and with respect to that origin, the CM has zero coordinate, X= 0.

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Figure 10.1 (Top) The moment of inertia calculation about an axis a distance d from the CM, which is not necessarily at the midpoint of the rod if the density is not uniform. The rod is made up of segments of width dx, one of which is shown at a distance x from the CM and x− d from the new axis. (Bottom) The parallel axis now passes through a point with a vector separation d from the CM. The body is made up of tiny areas of size dx dy, one of which is shown, separated by r from the CM and r^from the new parallel axis.

Thus, the moment of inertia is the smallest with respect to the CM;

any other axis adds an Md².

Now let us do it in d= 2. I remind you that the length squared of A+ B is

|A + B|²= (A + B) · (A + B)

= A · A + B · B + 2A · B = A²+ B²+ 2A · B. (10.15)

Letρ(x, y) be the mass per unit area, let r be the position vector of a point with the origin at the CM, and let d be the location of the new axis. Clearly the position vector of a point relative to the new axis is

r^= r − d. (10.16)

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Now we repeat the earlier proof with vectors galore:

I=

where the last term vanishes for the same reason as in d= 1: R is the CM position in a coordinate system with the CM itself as the origin.

Here is an illustration of the power of this result. Suppose that instead of rotating a disk about its center you wanted to hold fixed a point at its circumference. We cannot view the disk as a union of concentric annuli centered around the new axis as we did for ICM. Only a part of every annulus would fit into the disk, and we would need to figure out how much, if we want to compute I directly. Of course, we will do no such thing: we will invoke the parallel axis theorem to say, for a disk of radius R

Iedge= ICM+ MR²=3MR²

2 . (10.21)

Consider for example a coin that is standing on its rim on some surface. It has just one point of contact with the surface. If we demand that there be no slipping, the point of contact cannot move relative to the surface as it rolls. So the coin will simply rotate around this point (at this instant) and the relevant I will be ^3MR2₂ .