Question

A disc of radius a and 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

Answer: (B)

$\frac{3}{2}a$

Answer 2

We know that period of a physical pendulum $T=2\pi \sqrt{\frac{{{l}_{0}}}{mgd}}=2\pi \sqrt{\frac{\frac{1}{2}m{{a}^{2}}+m{{a}^{2}}}{mga}}=2\pi \sqrt{\frac{3a}{2g}}$ T for simple pendulum $=2\pi \sqrt{\frac{l}{g}}$ ?(ii) From Eqs. (i) and (ii), we get $l=\frac{3}{2}a$