Question

A 1.4 m long thin uniform rod is pivoted at one point so that it oscillates like a physical pendulum. Compare the angular frequency of this physical pendulum with a simple pendulum of the same length, both oscillating near the earth’s surface.

Answer 1

1) Moment of inertia of the rod rotating round either end:

$$I=\dfrac13mL^2.$$

2) Period of such a physical pendulum:

$$T_p=2\pi\sqrt\dfrac{I}{mgL}=2\pi\sqrt\dfrac{L}{3g}=1.37\text{ s}.$$

3) Period of a simple pendulum:

$$T_s=2\pi\sqrt\dfrac{l}{g}=2.37\text{ s}.$$