 Solution

Let $\theta =0$ place the centre of mass vertically below a horizontal pivot axis. The restoring torque and rotational kinetic energy are $-mgh\sin \theta$ and $I{\dot{\theta }}^2/2$, soFor small angles, $\sin \theta =\theta +O({\theta }^3)$, and the equation reduces to simple harmonic motion. Thus the physical pendulum hasThis assumes negligible friction and a horizontal axis, with $h>0$. For a tilted axis the projected gravitational restoring torque changes; an axis through the centre of mass has no gravitational restoring torque.