 Solution

The disc is rigidly attached, so its rotation axis is perpendicular to its face. Its moment of inertia about its center is $M{a}^2/2$: with surface mass density $M/(\pi a^2)$, an annulus of radius $s$ contributes $s^2[M/(\pi a^2)]2\pi sds$, whose integral is $M{a}^2/2$. The parallel axis theorem adds $M{d}^2$ at the pivot. Adding the rod's moment of inertia gives