With a good sketch derive the equation of motion of a compound pendulum

The gravitational force $mg$ creates a restoring torque about the pivot O. The perpendicular distance from the pivot to the line of action of the force is $L \sin \theta$.

$$\tau = -mgL \sin \theta$$

The negative sign indicates that the torque acts to reduce the angular displacement $\theta$.

Newton's second law for rotation states that the net torque $\tau$ is equal to the moment of inertia $I_O$ about the pivot multiplied by the angular acceleration $\alpha$. Here, $\alpha = \ddot{\theta}$.

$$\tau = I_O \ddot{\theta}$$

Equating the two expressions for torque:

$$I_O \ddot{\theta} = -mgL \sin \theta$$

Move all terms to one side to get the differential equation:

$$I_O \ddot{\theta} + mgL \sin \theta = 0$$

Divide by $I_O$ to isolate $\ddot{\theta}$:

$$\ddot{\theta} + \frac{mgL}{I_O} \sin \theta = 0$$

For small oscillations (i.e., when $\theta$ is small), we can approximate $\sin \theta \approx \theta$.
Substituting this into the equation:

$$\ddot{\theta} + \frac{mgL}{I_O} \theta = 0$$

This is the equation of motion for a compound pendulum under small oscillations.

The equation of motion for a compound pendulum is:

$$\boxed{\ddot{\theta} + \frac{mgL}{I_O} \sin \theta = 0}$$

For small oscillations, it simplifies to:

$$\boxed{\ddot{\theta} + \frac{mgL}{I_O} \theta = 0}$$