Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section B / B9

For a fixed rotation axis with constant moment of inertia, the axial torque obeys $\tau =I\ddot{\theta }$, with $\tau =\sum (\mathbf{r}\times \mathbf{F})\cdot {\hat{\mathbf{e}}}_{\mathrm{axis}}$. For a point mass at radius $l$, the tangential equation is $F_t=ml\ddot{\theta }$; multiplying by $l$ gives $l{F}_t=m{l}^2\ddot{\theta }$, as required. Radial forces have zero moment about the axis.

Take increasing $\theta$ in the direction of the clapper's initial fall and let the contact force $N$ oppose that tangential motion. Initial equilibrium of the clapper givesWhile contact persists the bell and clapper rotate together. Their respective offsets from the upward vertical are $\theta -\beta$ and $\theta$. Summing external gravitational moments eliminates their internal contact force:The clapper alone satisfies $m{l}^2\ddot{\theta }=mgl\sin \theta -Nl$. Hence for $m\ll M$,A nonadhesive contact cannot exert tension. The first zero of this expression on the released trajectory is the loss-of-contact angle. Write $q=lh/k^2=0.9375$. ThenThe second-quadrant root is reached from $\theta =\beta =30^{\circ }$; selecting the principal negative arctangent would choose the wrong branch. Indeed the exact contact force before taking $m/M\to 0$ is $mMg[k^2\sin \theta -lh\sin (\theta -\beta )]/(M{k}^2+m{l}^2)$, so its zero is unchanged.