Bell-clapper loss of contact

Let a bell have axial moment of inertia $M{k}^2$ and centre of mass h from its axle. A clapper of point mass m at radius l touches the bell at fixed relative angle $\beta$. While they remain in contact, the clapper angle $\theta$ from the upward vertical obeys $(M{k}^2+m{l}^2)\ddot{\theta }=Mgh\sin (\theta -\beta )+mgl\sin \theta$. The contact force opposing increasing angle is $N=mg\sin \theta -ml\ddot{\theta }$. Eliminating the acceleration givesThe first zero on the physical trajectory marks detachment, because a nonadhesive contact cannot support negative normal force. Selecting the correct trigonometric branch is part of the contact problem.