Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 6

The uniform rod has linear mass density $m/\ell$. Integrating squared distance from the pivot gives its moment of inertia about the axis perpendicular to the plane of motion:Measure $\theta$ from the downward vertical. The center of mass is at distance $\ell /2$ from the pivot, so gravity produces the restoring torque $-mg(\ell /2)\sin \theta$. The fixed-axis rotational equation of motion is $I_O\ddot{\theta }={\tau }_O$. Therefore the physical pendulum obeysThe pivot force has zero torque about the pivot.

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

The rod and disc contribute restoring torques $-mg(\ell /2)\sin \theta$ and $-Mgd\sin \theta$, respectively. Using i in the rotational equation of motion givesOnly the disc's center of mass enters its gravitational torque; its finite radius enters its moment of inertia. The formula is also regular at $M=m$, although the source happens to exclude that value.

The small-angle approximation $\sin \theta =\theta +O({\theta }^3)$ reduces ii to $\ddot{\theta }+{\omega }^2\theta =0$, whereThe coefficient is positive, so the solution is simple harmonic motion, $\theta (t)=C\cos \omega t+D\sin \omega t$. Here $\omega$ is the angular frequency; the ordinary frequency and period are

For a fixed attachment ratio $\delta >0$, making the disc much heavier than the rod givesIf the disc is also small relative to the attachment distance, $\alpha /\delta =a/d\to 0$, its moment of inertia about its own center is negligible compared with $M{d}^2$. ThusThis is a simple pendulum with a point mass at the disc's center and effective length $d$; it is length $\ell$ only if $d=\ell$. Even before the small-angle approximation, its equation tends to $\ddot{\theta }=-(g/d)\sin \theta$.

If $\delta$ is allowed to tend to zero together with the other parameters, the bare conditions $\alpha \ll 1$ and $\mu \gg 1$ do not alone justify that reduction. The needed dominance conditions are $\mu \delta \gg 1$, $\mu {\delta }^2\gg 1$, and $\alpha /\delta \ll 1$, so that the rod contributes negligible gravitational torque and moment of inertia, and the disc behaves as a point mass. They follow automatically in the combined limit with fixed $\delta >0$.