A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is ______. ( g is acceleration due to gravity)

For small-angle oscillations about a fixed horizontal axis, any rigid body behaves as a physical pendulum. Its time period is

$$T = 2\pi\sqrt{\frac{I_A}{Mgd}} \quad -(1)$$
where
  • $$I_A$$ is the moment of inertia about the axis of suspension A,
  • $$M$$ is the mass of the body,
  • $$d$$ is the distance between the axis A and the centre of mass (C.O.M.),
  • $$g$$ is the acceleration due to gravity.

Moment of inertia about axis A

The disc is uniform, has radius $$R$$ and mass $$M$$. The axis A passes through a point on the rim and is perpendicular to the plane of the disc (a tangential axis). Using the parallel-axis theorem,

$$I_A = I_{\text{cm dia}} + Md^{2} \quad -(2)$$
Here,
  • $$I_{\text{cm dia}}$$ for a disc about its central axis (parallel to the plane) is $$\frac{1}{4}MR^{2}$$.
  • The distance from the rim to the centre is $$d = R$$.

Substitute in $$(2):$$

$$I_A = \frac{1}{4}MR^{2} \;+\; M R^{2} \;=\; \frac{5}{4} M R^{2} \quad -(3)$$

Distance $$d$$ between axis and C.O.M.

The C.O.M. of a uniform disc is at its centre, so $$d = R$$.

Time period

Insert $$I_A$$ from $$(3)$$ and $$d = R$$ into the physical-pendulum formula $$(1):$$

$$T = 2\pi\sqrt{\frac{\frac{5}{4}MR^{2}}{M g R}} = 2\pi\sqrt{\frac{5}{4}\,\frac{R}{g}} = 2\pi\sqrt{\frac{5R}{4g}}$$

Thus the time period of oscillation is

$$T = 2\pi\sqrt{\frac{5R}{4g}}$$

Option A which is: $$2\pi\sqrt{\frac{5R}{4g}}$$