= Solution
{c}
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:
$$
\boxed{I_O=\int_0^\ell s^2\frac m\ell\,ds=\frac13m\ell^2.}
$$
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> obeys
$$
\boxed{\ddot\theta=-\frac{3g}{2\ell}\sin\theta.}
$$
The pivot <force> has zero <torque> about the pivot.
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