= Solution
{c}
The rod and disc contribute restoring <torques> $-mg(\ell/2)\sin\theta$ and $-Mgd\sin\theta$, respectively. Using <6/b/i> in the <rotational equation of motion> gives
$$
\boxed{\ddot\theta=-\frac g\ell\,
\frac{\frac12+\mu\delta}{\frac13+\mu(\delta^2+\alpha^2/2)}\sin\theta.}
$$
Only the disc's <center of mass> enters its gravitational <torque>; its finite <radius> enters its <moment of inertia>. The formula is also regular at $M=m$, although the source happens to exclude that value.
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