Past exam of the physics 1 paper of the University of Oxford / 2014 / 7 / Solution

Let $\theta$ measure the spoke's displacement from downward equilibrium. The total moment of inertia about the axle is $I=M{R}_2^2+m{R}_1^2$: the rim is a thin thin hoop, and the attached body is a point mass. The rim's weight has no axle torque, while the attached mass gives restoring torque $-mg{R}_1\sin \theta$. HenceEquivalently conservation of energy gives $\frac{1}{2}I{\dot{\theta }}^2=mg{R}_1(\cos \theta -\cos {\theta }_0)$. For small angles, $\sin \theta \simeq \theta$, so the motion is approximately $\theta ={\theta }_0\cos (\omega t)$ with ${\omega }^2=mg{R}_1/I$. The small-amplitude period isAt finite amplitude the nonlinear pendulum equation remains valid but this period approximation does not. The $M\to 0$ limit recovers a simple pendulum of length $R_1$.