The period \(T\) of a physical pendulum is given by

$$T = 2 \pi \sqrt{\frac{I}{m g R}} \mathrm{,}$$

where

- \(I\) is the moment of inertia of the pendulum about the pivot point.
- \(m\) is the mass of the pendulum.
- \(g\) is the accelleration of gravity.
- \(R\) is the distance between the pivot point and the center of mass.

The formula holds for all sizes and shapes; however, when someone says the word “pendulum”, we usually think of a bob attached to a rod, in which case the formula can be simplified as follows.

Let the length of the rod be \(l\) and the mass of the rod be \(m_r = \alpha m\) for \(\alpha \in [0, 1]\). Then

- The mass of the bob is \(m_b = (1 - \alpha) m\).
- The moment of inertia of the rod about the pivot point is \(I_r = \frac{\alpha}{3} m l ^ 2\).
- The moment of inertia of the bob about the pivot point is \(I_b = (1 - \alpha) m l^2\).
- The distance between the pivot point and the center of mass is \(R = \left(1 - \frac{\alpha}{2} \right) l\).

Therefore

$$T = T_0 \sqrt{\frac{1 - \frac{2 \alpha}{3}}{1 - \frac{\alpha}{2}}} ,$$

where \(T_0 = 2 \pi \sqrt{\frac{l}{g}}\) is the period of a simple pendulum, which we get by setting \(\alpha = 0\), i.e., by making the rod massless. On the other hand, for \(\alpha = 1\) the bob is massless and we have \(T = T_0 \sqrt{\frac{2}{3}}\).