= Solution
In the <rim-mass pulley coupled to a hanging mass>, the printed figure shows the smaller pulley fixed overhead, and the <mechanical string> leaves the right-hand tangent of the larger pulley. Let $\theta$ increase counterclockwise from the downward vertical to the radius carrying $M$. Then the hanging <mass> $m$ moves downward by $R\theta$. The string is taken to grip the rim without slipping. Its <tension> must be solved dynamically, rather than set equal to $mg$ during motion.
The <kinetic energy> and <potential energy> are
$$
T=\frac12(M+m)R^2\dot\theta^2,\qquad U=-MgR\cos\theta-mgR\theta.
$$
The <Euler-Lagrange equation> and the condition for <equilibrium> are
$$
\boxed{(M+m)R\ddot\theta+Mg\sin\theta-mg=0,\qquad \sin\theta_e=\frac mM.}
$$
One can also eliminate the <tension> from $MR^2\ddot\theta=TR-MgR\sin\theta$ and $mR\ddot\theta=mg-T$. For $0<m<M$, the solution with $\cos\theta_e>0$ is stable; the other solution in a full turn is unstable.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-rim-mass-pulley.png]
{title=Coordinates and string motion in the coupled pulley system}
{description=The original schematic records the printed figure's fixed upper pulley, right-hand tangent string, rim mass, and angle from downward vertical. Positive rotation raises the tangent string and lowers the hanging mass.}
Back to article page
