The period of a physical pendulum of length $L$ and pivoted about its end is:

$$T = 2 \pi \sqrt{\frac{I}{mgd}}$$

where $I =$ moment of inertia about the pivot and $d =$ distance from the pivot to the center of mass. Then for the bar, $I = \frac{1}{3}mL^2$ and $d = \frac{L}{2}$. Therefore,

$$T = 2 \pi \sqrt{\frac{2L}{3g}}$$

Thus, a simple pendulum of length $\frac{2}{3}L$ will have the same period. A surprising fact is that there is a conjugate pivot point that yields the same period. The pair of conjugate support distances, $l_1$ and $l_2$ (distances from center of mass to respective pivot), are related by the formula:

$$l_1 l_2 = \frac{I_c}{m}$$

where $I_c$ is the moment of inertia about the center of mass.

For the bar, the conjugate support distances with equal period are $L/2$ and $L/6$. Exploitation of the above relation can be used to measure g with Kater’s pendulum.