Past exam of the physics 1 paper of the University of Oxford / 2013 resit / 8

Choose parallel axes along $z$, one through the centre of mass and one displaced transversely by $\mathbf{a}$. Write the perpendicular position as ${\mathbf{r}}_{\perp }={\mathbf{r}}_{\perp }^{\prime }+\mathbf{a}$. Direct expansion of the moment of inertia givesThe integral in the cross term is zero by the definition of the centre of mass. Therefore the parallel axis theorem isThe axes must be parallel and the reference axis must pass through the centre of mass; the formula is not generally valid between arbitrary parallel axes using either axis as the reference.

Let $\theta =0$ place the centre of mass vertically below a horizontal pivot axis. The restoring torque and rotational kinetic energy are $-mgh\sin \theta$ and $I{\dot{\theta }}^2/2$, soFor small angles, $\sin \theta =\theta +O({\theta }^3)$, and the equation reduces to simple harmonic motion. Thus the physical pendulum hasThis assumes negligible friction and a horizontal axis, with $h>0$. For a tilted axis the projected gravitational restoring torque changes; an axis through the centre of mass has no gravitational restoring torque.

For the shorter uniform rod, the parallel axis theorem gives $I=m(a^2/12+x^2)$. Its equivalent simple pendulum length is $I/(mx)=x+a^2/(12x)$. The rod of length $2a$ pivoted at an end has equivalent length $2(2a)/3=4a/3$. Matching the periods therefore requiresA pivot on the shorter rod requires $0<x\le a/2$, leavingEither side of the rod's centre of mass gives this distance. The other algebraic root is beyond the rod and would require a support extension of negligible mass.