Determine the time period of small oscillations for a uniform disc...

The time period of a physical pendulum is given by $T = 2\pi \sqrt{\dfrac{I}{mgd}}$, where $I$ is the moment of inertia about the point of suspension and $d$ is the distance of the centre of mass from the point of suspension.

Given $d = \dfrac{r}{2}$.

The moment of inertia of the uniform disc about its centre of mass is $I_{cm} = \dfrac{1}{2}mr^2$.

Using the parallel axis theorem, the moment of inertia about the point of suspension is:

$I = I_{cm} + md^2 = \dfrac{1}{2}mr^2 + m\left(\dfrac{r}{2}\right)^2 = \dfrac{1}{2}mr^2 + \dfrac{1}{4}mr^2 = \dfrac{3}{4}mr^2$

Substituting $I$ and $d$ into the formula for the time period:

$T = 2\pi \sqrt{\dfrac{\dfrac{3}{4}mr^2}{mg\left(\dfrac{r}{2}\right)}}$

$T = 2\pi \sqrt{\dfrac{3r}{2g}}$