= Solution
For a fixed rotation axis with constant <moment of inertia>, the axial <torque> obeys $\tau=I\ddot\theta$, with $\tau=\sum(\mathbf r\times\mathbf F)\cdot\widehat{\mathbf e}_{\rm axis}$. For a <point mass> at radius $l$, the tangential equation is $F_t=ml\ddot\theta$; multiplying by $l$ gives $lF_t=ml^2\ddot\theta$, 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 gives
$$
Nl=mgl\sin\beta,\qquad\boxed{N_{\rm initial}=mg\sin\beta.}
$$
While 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>:
$$
\boxed{\ddot\theta=\frac{Mgh\sin(\theta-\beta)+mgl\sin\theta}{Mk^2+ml^2}.}
$$
The clapper alone satisfies $ml^2\ddot\theta=mgl\sin\theta-Nl$. Hence for $m\ll M$,
$$
\boxed{N\simeq mg\left[\sin\theta-\frac{lh}{k^2}\sin(\theta-\beta)\right].}
$$
A nonadhesive contact cannot exert tension. The first zero of this expression on the released trajectory is the loss-of-contact angle. Write $q=lh/k^2=0.9375$. Then
$$
(1-q\cos\beta)\sin\theta_0+q\sin\beta\cos\theta_0=0,
\qquad
\boxed{\theta_0=111.9^\circ.}
$$
The 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 $m/M\to0$ is $mMg[k^2\sin\theta-lh\sin(\theta-\beta)]/(Mk^2+ml^2)$, so its zero is unchanged.
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