= 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 $I\dot\theta^2/2$, so
$$
I\ddot\theta+mgh\sin\theta=0.
$$
For small angles, $\sin\theta=\theta+O(\theta^3)$, and the equation reduces to <simple harmonic motion>. Thus the <physical pendulum> has
$$
\boxed{\omega_0=\sqrt{\frac{mgh}{I}},\qquad T=2\pi\sqrt{\frac{I}{mgh}}.}
$$
This 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>.
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