= Solution
For the <rolling disk with a rim point mass>, impose <rolling without slipping> and take the clockwise angular displacement $\theta$ from the configuration with the extra <mass> at the bottom. The disk centre has $x_C=R\theta$ and $y_C=R$. The extra <mass> has
$$
x_M=R(\theta-\sin\theta),\qquad y_M=R(1-\cos\theta),\qquad v_M^2=2R^2(1-\cos\theta)\dot\theta^2.
$$
The disk's translational and <rotational kinetic energies> sum to $3MR^2\dot\theta^2/2$. Including the rim <mass> gives
$$
T=\frac12MR^2(5-2\cos\theta)\dot\theta^2,\qquad U=MgR(1-\cos\theta).
$$
The exact <Euler-Lagrange equation> is
$$
MR^2(5-2\cos\theta)\ddot\theta+MR^2\sin\theta\,\dot\theta^2+MgR\sin\theta=0.
$$
To first order, $3MR^2\ddot\theta+MgR\theta=0$. Therefore
$$
\boxed{\omega_{\rm roll}=\sqrt{\frac{g}{3R}},\qquad \frac{T_{\rm roll}}{T_{\rm fixed}}=\sqrt{\frac32}.}
$$
At the bottom the extra <mass> is instantaneously at the contact point and has no first-order <velocity>. The increased effective inertia comes from the moving disk. Static <friction> enforces the rolling constraint without doing <work> at the stationary contact.
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