Time Period of Oscillations of Uniform Disc About Axis

To find the time period of small oscillations of the uniform disc about the given axis $A$:

1. Identify the axis of rotation and Moment of Inertia ($I_A$):

   • The axis $A$ is a tangent to the circular disc, lying in the plane of the disc at its top point.
   • The moment of inertia of a uniform disc of mass $M$ and radius $R$ about its diameter (an axis passing through its center of mass and lying in its plane) is: $I_{\text{cm}} = \frac{1}{4} M R^2$
   • The distance from the center of mass to the axis $A$ is $d = R$.
   • By the parallel axis theorem, the moment of inertia about axis $A$ is: $I_A = I_{\text{cm}} + M d^2 = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$

2. Equation of Motion for Small Oscillations:

   • When the disc is displaced by a small angle $\theta$ from its equilibrium position, the restoring torque $\tau$ about axis $A$ due to gravity acting at the center of mass is: $\tau = - M g d \sin\theta$
   • For small angles ($\sin\theta \approx \theta$): $\tau \approx - M g R \theta$
   • Using Newton's second law for rotation ($\tau = I_A \alpha$): $I_A \frac{d^2\theta}{dt^2} = - M g R \theta$ $\left(\frac{5}{4} M R^2\right) \frac{d^2\theta}{dt^2} + M g R \theta = 0$ $\frac{d^2\theta}{dt^2} + \left(\frac{4g}{5R}\right) \theta = 0$

3. Time Period Calculation:

   • The angular frequency of small oscillations $\omega$ is: $\omega = \sqrt{\frac{4g}{5R}}$
   • Therefore, the time period of oscillations $T$ is given by: $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{5R}{4g}}$

This corresponds to Option A.