= Bell-clapper loss of contact
Let a bell have axial <moment of inertia> $Mk^2$ 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 $(Mk^2+ml^2)\ddot\theta=Mgh\sin(\theta-\beta)+mgl\sin\theta$. The contact <force> opposing increasing angle is $N=mg\sin\theta-ml\ddot\theta$. Eliminating the <acceleration> gives
$$
N=\frac{mMg}{Mk^2+ml^2}[k^2\sin\theta-lh\sin(\theta-\beta)].
$$
The 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.
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