= Solution
Let $\theta$ measure the rim <point mass>'s displacement from vertically below the fixed axis. The disk has <moment of inertia> $I_D=\tfrac12(2M)R^2=MR^2$, and the rim <mass> contributes another $MR^2$. Its varying <gravitational potential energy> is $-MgR\cos\theta$. Thus
$$
\boxed{L=MR^2\dot\theta^2+MgR\cos\theta.}
$$
The disk's <centre of mass> remains fixed, so its own gravitational contribution is a constant and has been omitted. The total fixed-axis <moment of inertia> is $2MR^2$.
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