Question

A disc of radius a mass $m$ is pivoted at the rim and is set in small oscillation. If a simple pendulum have the same period as that of the disc, then the length of the simple pendulum should be

• A. $\frac{5}{2}a$
• B. $\frac{3}{2}a$
• C. $\frac{7}{2}a$
• D. $\frac{1}{2}a$

Answer 1

The correct option is B $\frac{3}{2}a$

$T_{disc}=2\pi \sqrt{\frac{I_0}{mgd}}=2\pi \sqrt{\frac{3a}{2g}}$.......(1)

$T_{Pendulum}=2\pi \sqrt{\frac{l}{g}}$..............(2)

solving equation (1) and (2) we get

$l=\frac{3a}{2}$