Question

Three particles of same mass $m$ are fixed to a uniform circular hoop of mass $m$ and radius $R$ at the corner of an equilateral triangle. The hoop is free to rotate in a vertical plane about a point on the circumference opposite to one of the masses. Find the equivalent length of the simple pendulum.

• A. $4R$
• B. $2R$
• C. $3R$
• D. $5R$

Answer 1

The correct option is B $2R$

Moment of inertia of all parts$,$

$I=m{R}^2+m{R}^2+m{R}^2+m{R}^2$

$=4m{R}^2$

Total mass $\left(M\right)=4m$

Time period$,T=2\pi \surd \left(l/g\right)$

Here$,T=2\pi \surd (Icm+M\alpha ²)/Mgd$

$=2\pi \surd (4mR²+4mR²)/4mgd$

$=2\pi \surd \left(2R/g\right)$

Hence equivalent length $l=2R$

Hence,

option $\left(B\right)$ is correct answer.