Past exam of the physics 1 paper of the University of Oxford / 2016 / 8 / Solution

The parallel axis theorem gives $I=I_{\mathrm{CM}}+M{d}^2$ for parallel axes separated by $d$, with the reference axis through the centre of mass. For a plane lamina, the perpendicular axis theorem gives $I_z=I_x+I_y$ for mutually perpendicular axes meeting at one point, with $x,y$ in the plane.

Treat the removed material as negative mass. With the hole's centre at $b\hat{\mathbf{x}}$, its moment of inertia about $O$ is $m(a^2/2+b^2)$. The remaining moment of inertia and centre of mass areFor a uniform original disk, $m/M=a^2/R^2$. Write $d_1=mb/(M-m)$ and measure $\varphi$ from the downward equilibrium direction of its centre of mass. The physical pendulum then has LagrangianThe Euler-Lagrange equation is $I_1\ddot{\varphi }+mbg\sin \varphi =0$. For small oscillations, $\sin \varphi \simeq \varphi$, givingNow add the removed small disk on the opposite side, with its centre a distance $c=R-a$ from $O$ so that it is internally tangent to the large disk. The total mass returns to $M$, while its first moment about $O$ is $-m(b+c)\hat{\mathbf{x}}$. ThusReplacing both the inertia and gravitational restoring moment givesThe same first-moment method applies to a general perforated-disc pendulum. The added disk is attached to the lamina and shares its rotation; treating it as a massless patch or updating only the inertia would omit its restoring torque.