two pendulums, which pendulum has shorter period [duplicate]

Question

Possible Duplicate:
A pendulum clock problem

There are two pendulums.

First pendulum consists of a rod of length L and flat heavy disk of radius R (R < L), disk is connected rigidly to the rod such that the plane of the disk is vertical. Mass of the rod is negligible compared to mass of the disk. Center of the disk is at distance L from the upper point of the rod.

Second pendulum has same rod of length L and same disk, but the disk is not connected rigidly to the rod. Disk can rotate freely around its center at which it is attached to the rod at distance L from the upper end of the rod. Disk remains in vertical plane, as if there is ball bearing atached to the rod.

Which pendulum has smaller period at small deviations, if their periods are different.

Answer 1

In the case of the first pendulum, disk has to rotate and this is classical physical pendulum. In the case of the second pendulum, you don't expect disk to rotate (the only force is in the center of mass, so there is no torque), so therefore you simply have only transitional movement - mathematical pendulum.

Using equation for periods of physical pendulum

$$T = 2 \pi \sqrt{\frac{I}{m g l}}$$

and noting, that

$$I = I_\textrm{cm} + m l^2,$$

where $I_\textrm{cm} > 0$ is center-of-mass inertia of the disk, you get

$$T = 2 \pi \sqrt{\frac{l}{g} + \frac{I_\text{cm}}{m g l}}.$$

On the other hand equation for periods of mathematical pendulum

$$T = 2 \pi \sqrt{\frac{l}{g}}.$$

Obviously, period is smaller for the second pendulum.