 Solution

For the edge-hinged cube with a central spherical cavity, use the parallel-axis theorem: for a parallel axis at perpendicular distance $d$ from the centre-of-mass axis, $I=I_C+M{d}^2$. Let $M=\rho a^3$. The centre lies $d=a/\sqrt{2}$ from the edge hinge, soAt equilibrium the centre is vertically below the hinge. Its gravitational torque at angular displacement $\theta$ is $-Mgd\sin \theta$. Therefore the small-angle equation is $I_h\ddot{\theta }+Mgd\theta =0$, withThe removed sphere has mass $M_s=\pi M/6$ and the same centre. Its hinge inertia isSuperposition leaves the centre at the same location but changes both inertia and weight:The frequency decreases: the removed central material contributes less hinge inertia per unit mass than the original cube average.