Question

A disc is suspended at a point $\frac{R}{2}$ above its center, Find its period of oscillation.

• A. $2\pi \sqrt{\frac{2g}{3R}}$
• B. $2\pi \sqrt{\frac{2R}{3g}}$
• C. $2\pi \sqrt{\frac{3R}{2g}}$
• D. $2\pi \sqrt{\frac{R}{2g}}$

Answer 1

The correct option is C

$2\pi \sqrt{\frac{3R}{2g}}$

When a disc is moved through on angle $\theta$ about 0,the restoring torque is

$\tau =-,g\frac{R}{2}sin\theta$

Taking $\theta$ to be small ,sin$\theta \approx \left(\theta \right)$

$\tau =-mg\frac{R}{2}(\theta$

Also $\tau =I_0\frac{d^2\theta }{d{t}^2}$

$\Rightarrow I_0\frac{d^2\theta }{d{t}^2}=-mg\left(\frac{R}{2}\right)\theta$

$\frac{d^2\theta }{d{t}^2}$+$\frac{mgR}{2I_0}\theta =0$

Comparing it with step SHM equation ,$\frac{d^2\theta }{d{t}^2}+{\omega }^2\theta =0$