moment of inertia problem-- kleppner #6.18 Ok here's the problem: [from An Introduction to Mechanics, Kleppner&Kolenkow, problem 6.18, page 281] Find the period of a pendulum consisting of a disk of mass M and radius R fixed to the end of a rod of length l and mass m. How does the period change if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin? For both cases---the unclamped disk and the clamped disk---the torque is the same. Correct? The difference between these two cases is in the calculation of the moment of inertia. In the unclamped case, there is only translational motion of the disk. That is, if we were to draw an arrow on the disc at one point in time, that arrow would still point in the same direction at a later point in time. In the clamped case, however, the motion is both rotational and translational. That is, if we were to draw an arrow on the disk as before, the direction of the arrow would change in time. In the unclamped case, we were able to treat the disk as a point particle. But in the clamped case, we are not able to treat the disk as a point particle because for different points on the disk, the speeds are different. My big problem is in the calculation of the moment of inertia. I'm just... really confused as to what is truly going on here. I'd love some help ASAP ttt___ttt In the unclamped case, there are two pivot points---one at the end of the rod, and one in the middle of the disk, which is what is throwing me off. So how do I calculate the moment of inertia of both cases with the parallel axis theorem? For the clamped case, you could use Ip = Ic + ml^2 (general case, where Ip is moment of inertia about pivot, and Ic is moment of inertia about center of mass. But here, what is Ic, since there is the disc at the very end? and "m" here is really the mass of the rod plus the mass of the disk?) And how would the parallel axis theorem be used with the unclamped case???