= Solution
Use the angle orientation in the PDF: $M$ has coordinates $(R\sin\theta,-R\cos\theta)$ relative to the hoop centre. The wrapped string segment has length $R(\pi/2-\theta)$. As $\theta$ increases, it releases length $R\theta$, so the suspended mass moves \i[down] by $R\theta$. Both speeds have magnitude $R|\dot\theta|$.
Thus the <hoop with a suspended counterweight> has
$$
T=\frac12(M+m)R^2\dot\theta^2,\qquad V=-MgR\cos\theta-mgR\theta+\text{constant},
$$
and $L=T-V$. The <Euler-Lagrange equation> gives
$$
\boxed{(M+m)R\ddot\theta=g(m-M\sin\theta).}
$$
At equilibrium,
$$
\boxed{\theta_0=\arcsin(m/M),\qquad0<\theta_0<\pi/2.}
$$
The positive curvature $V''(\theta_0)=MgR\cos\theta_0$ establishes stability. Put $\theta=\theta_0+\delta$ and linearize. The equilibrium force cancels, leaving
$$
\ddot\delta+\omega^2\delta=0,\qquad\omega^2=\frac{Mg\cos\theta_0}{(M+m)R}=\frac gR\sqrt{\frac{M-m}{M+m}}.
$$
Hence
$$
\boxed{\omega=\left(\frac{M-m}{M+m}\right)^{1/4}\sqrt{\frac gR}.}
$$
For $M\gg m$, $\theta_0\simeq m/M$ and $\omega\to\sqrt{g/R}$, the small-oscillation frequency of a <simple pendulum> of length $R$. The <linearization> assumes an amplitude small enough to remain inside the stated angular range and a taut string throughout.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-q9-hoop-counterweight.png]
{title=Independent hoop-and-counterweight geometry and the stable minimum of its potential energy}
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