# The period of a physical pendulum

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 $$T = T_0 \sqrt{\frac{2}{3}}$$. The period gets shorter as the mass of the rod increases