19-3 Finding the moment of inertia Now let us discuss the problem of finding the moments of inertia of various objects. The formula for the moment of inertia about the z-axis of an object is =∑m(x2+y?) 1=(2+y)mn=/(x2+y)d (19.4) That is, we must sum the masses, each one multiplied by the square of its distance (xi+ yfrom the axis. Note that it is not the three-dimensional distance, only the two-dimensional distance squared, even for a three-dimensional object. For the most part, we shall restrict ourselves to two-dimensional objects, but the ormula for rotation about the z-axis is just the same in three dimensions A ople example, consider a rod rotating about a perpendicular axis through one end( Fig. 19-3). Now we must sum all the masses times the x-distances quare(the y's being all zero in this case). What we mean by"the sum, "of course, is the integral of x2 times the little elements of mass. If we divide the rod into small elements of length dx, the corresponding elements of mass are propor Fig. 19-3. A straight rod of length tional to dx, and if dx were the length of the whole rod the mass would be M. L rotating about an axis through one end. Therefore dm= M dx/L Tmax m/x dx- mia (195) The dimensions of moment of inertia are always mass times length squared, so all we really had to work out was the factor 1/3 Now what is I if the rotation axis is at the center of the rod? We could just do the integral over again, letting x range from -iL to +2L. but let us notice a few things about the moment of inertia. We can imagine the rod as two rods, each of mass M/2 and length L/2; the moments of inertia of the two small rods are equal, and are both given by the formula(19.5). Therefore the moment of Inertia Is r=2(M2)(L/2)2ML2 Thus it is much easier to turn a rod about its center, than to swing it around an end Of course, we could go on to compute the moments of inertia of various other bodies of interest. However, while such computations provide a certain amount of important exercise in the calculus, they are not basically of interest to us as such. There is, however, an interesting theorem which is very useful. Suppose we have an object, and we want to find its moment of inertia around some axis. That means we want the inertia needed to carry it by rotation about that axis low if we support the object on pivots at the center of mass, so that the object does not turn as it rotates about the axis(because there is no torque on it from inertial effects, and therefore it will not turn when we start moving it), then the forces needed to swing it around are the same as though all the mass were concen- trated at the center of mass, and the moment of inertia would be simply I1 MRCM, where Rcm is the distance from the axis to the center of mass. But of course that is not the right formula for the moment of inertia of an object which is really being rotated as it revolves, because not only is the center of it moving in a circle, which would contribute an amount I, to the moment of inertia, but also we must turn it about its center of mass. So it is not unreasonable that we must that the total moment of inertia about any axis will be s. So it is a good guess add to I 1 the moment of inertia I about the center of ma Ie t M