This theorem is called the parallel-axis theorem, and may be easily proved The moment of inertia about any axis is the mass times the sum of the x,'s and the yi,s, each squared: I=2(x+ yi)mj. We shall concentrate on the x' s. but of course the y's work the same way. Now x is the distance of a particular point mass from the origin, but let us consider how it would look if we measured xfrom the CM, instead of x from the origin. To get ready for this analysis, we write x 'i+ Xo Then we just square this to find xi= xi+2XcMx':+XCM So, when this is multiplied by mi and summed over all i, what happens? Ta the constants outside the summation sign, we get ∑ ∑mx+Xx∑m The third sum is easy; it is just MYcy. In the second sum there are two pieces,one of them is 2 mixi, which is the total mass times the x -coordinate of the center of mass. But this contributes nothing, because x is measured from the center of mass and in these axes the average position of all the particles, weighted by the masses, is zero. The first sum, of course, is the x part of Ie. Thus we arrive at Eq. 19.7) Just as we guessed Let us check(19.7) for one example. Let us just see whether it works for the od. For an axis through one end, the moment of inertia should be mL /3, for ve calculated that. The center of mass of a rod, of course, is in the center of the rod, at a distance L/2. Therefore we should find that ML2/3=ML212+ M(L/2). Since one-quarter plus one-twelfth is one-third, we have made no fundamental error Incidentally, we did not really need to use an integral to find the moment of inertia(19.5). If we simply assume that it is ML times t, an unknown coefficient, and then use the argument about the two halves to get 1 r for (19.6), then from ou argument about transferring the axes we could prove that y=4r+4,so y must be 1/3. There is always another way to do it! In applying the parallel-axis theorem, it is of course important to remember that the axis for Ie must be parallel to the axis about which the moment of inertia is wanted One further property of the moment of inertia is worth mentioning because it is often helpful in finding the moment of inertia of certain kinds of objects This property is that if one has a plane figure and a set of coordinate axes with origin in the plane and z-axis perpendicular to the plane, then the moment of inertia of this figure about the z-axis is equal to the sum of the moments of inertia about the x- and y-axes. This is easily proved by noting that L=∑m(n2+21)=∑m2 L=∑ but ∑mx2+y2)=∑mx2+∑m2 As an example, the moment of inertia of a uniform rectangular plate of M, width w, and length L, about an axis perpendicular to the plate and through its center is simply I=M(w2+L2)/12, because its moment of inertia about in its plane and parallel to its length Mw /12, i. e, just as for a rod of length w, and the moment of inertia about the other axis in its plane is ML 2 /12, just as for a rod of length L