We balance the torques about the axle. Torque from hanging mass \(m_0\): \( \tau_1 = T \cdot r = m_0 g r \).
Torque from rod mass \(m\) at distance \(l\): \( \tau_2 = m g l \cos \theta \) (where \(\theta\) is angle with horizontal). In equilibrium: \[ m_0 g r = m g l \cos \theta \] \[ \cos \theta = \frac{m_0 r}{m l} \implies \theta = \cos^{-1}\left( \frac{m_0 r}{m l} \right) \] The other case of above horizontal would be unstable.

For a small angular displacement \(\alpha\) from equilibrium, the restoring torque is: \[ \tau_{net} = – (m g l \sin \theta) \alpha \] (The tension torque remains constant \(m_0gr\) for small angles if we ignore higher order terms, but strictly speaking, we balance the change in gravitational torque). The Total Moment of Inertia \(I\) of the system about the axle: \[ I = I_{rod} + I_{drum\_mass} = m l^2 + m_0 r^2 \] Equation of Motion: \(\tau = I \ddot{\alpha}\) \[ – (m g l \sin \theta) \alpha = (m l^2 + m_0 r^2) \ddot{\alpha} \] \[ \omega^2 = \frac{m g l \sin \theta}{m l^2 + m_0 r^2} \] Using \(\sin \theta = \sqrt{1 – \cos^2 \theta} = \sqrt{1 – (\frac{m_0 r}{ml})^2} = \frac{1}{ml}\sqrt{m^2l^2 – m_0^2r^2}\). \[ T = 2\pi \sqrt{\frac{m l^2 + m_0 r^2}{g \sqrt{m^2 l^2 – m_0^2 r^2}}} \]