The system consists of a cylindrical shell ($M$) and a particle ($m$) attached to its rim. The cylinder rolls without slipping. This means the point of contact with the ground is the Instantaneous Axis of Rotation (IAR).

We calculate the Moment of Inertia about the IAR (contact point).
1. For the Shell ($M$): Using the Parallel Axis Theorem ($I = I_{cm} + Md^2$): $$ I_{shell} = MR^2 + MR^2 = 2MR^2 $$ (Assuming a thin shell where $I_{cm} = MR^2$).

2. For the particle ($m$): Since $M \gg m$ and we are considering small oscillations where the particle stays near the bottom (close to the IAR), the contribution of $m$ to the total inertia is negligible ($mr^2 \approx 0$).

Total Inertia: $$ I_{sys} \approx 2MR^2 $$

Let the cylinder be displaced by a small angle $\theta$ from the vertical.
The mass $m$ is displaced from the bottom. The horizontal distance (lever arm) of the mass $m$ from the contact point (IAR) is $d = R \sin \theta$.
The gravitational force $mg$ creates a restoring torque about the IAR: $$ \tau = -mg \times (R \sin \theta) $$ For small angles, $\sin \theta \approx \theta$: $$ \tau \approx -mgR\theta $$ (Note: The weight of the shell $Mg$ acts through the vertical geometric center, which passes directly above the contact point, so it produces zero torque).

Using the rotational analog of Newton’s Second Law ($\tau = I \alpha$): $$ I_{sys} \frac{d^2\theta}{dt^2} = \tau $$ $$ (2MR^2) \ddot{\theta} = -mgR\theta $$ $$ \ddot{\theta} = – \frac{mgR}{2MR^2} \theta $$ $$ \ddot{\theta} = – \left( \frac{mg}{2MR} \right) \theta $$

The equation represents Simple Harmonic Motion with angular frequency squared: $$ \omega^2 = \frac{mg}{2MR} \implies \omega = \sqrt{\frac{mg}{2MR}} $$ The period of oscillation $T = \frac{2\pi}{\omega}$ is: $$ T = 2\pi \sqrt{\frac{2MR}{mg}} $$