= Rim-mass pulley coupled to a hanging mass
{title2=$(M+m)R\ddot\theta+Mg\sin\theta-mg=0$}
A massless fixed-axis <pulley> carries a rim <point mass> $M$ at radius $R$. A <mechanical string> gripping its rim without slipping passes over a fixed guide <pulley> to a hanging <mass> $m$. Choose the angle $\theta$ from downward vertical so the hanging <mass> descends by $R\theta$. The <Lagrangian> is $(M+m)R^2\dot\theta^2/2+MgR\cos\theta+mgR\theta$. The stable <equilibrium> satisfies $\sin\theta_e=m/M$ with $\cos\theta_e>0$, and exists for $0<m<M$. Its small-amplitude <angular frequency> is $[Mg\cos\theta_e/((M+m)R)]^{1/2}$.
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