= Solution
{c}
The disc is rigidly attached, so its rotation axis is perpendicular to its face. Its <moment of inertia> about its center is $Ma^2/2$: with <surface mass density> $M/(\pi a^2)$, an annulus of <radius> $s$ contributes $s^2[M/(\pi a^2)]2\pi s\,ds$, whose integral is $Ma^2/2$. The <parallel axis theorem> adds $Md^2$ at the pivot. Adding the rod's <moment of inertia> gives
$$
\boxed{I_O=\frac13m\ell^2+M\left(d^2+\frac{a^2}{2}\right)
=m\ell^2\left[\frac13+\mu\left(\delta^2+\frac{\alpha^2}{2}\right)\right].}
$$
Back to article page
