 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 down by $R\theta$. Both speeds have magnitude $R|\dot{\theta }|$.

Thus the hoop with a suspended counterweight hasand $L=T-V$. The Euler-Lagrange equation givesAt equilibrium,The positive curvature $V^{\prime \prime }({\theta }_0)=MgR\cos {\theta }_0$ establishes stability. Put $\theta ={\theta }_0+\delta$ and linearize. The equilibrium force cancels, leavingHenceFor $M\gg m$, ${\theta }_0\simeq m/M$ 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.