 Solution

The uniform rod has linear mass density $m/\ell$. Integrating squared distance from the pivot gives its moment of inertia about the axis perpendicular to the plane of motion:Measure $\theta$ from the downward vertical. The center of mass is at distance $\ell /2$ from the pivot, so gravity produces the restoring torque $-mg(\ell /2)\sin \theta$. The fixed-axis rotational equation of motion is $I_O\ddot{\theta }={\tau }_O$. Therefore the physical pendulum obeysThe pivot force has zero torque about the pivot.