Question

A ring is suspended at a point on its rim and it behaves as a second's pendulum when it oscillates such that its centre move in its own plane. The radius of the ring would be $(g={\pi }^2)$

• A. 0.5 m
• B. 1.0 m
• C. 0.67 m
• D. 1.5 m

Answer 1

The correct option is A 0.5 m

We know that the time period of a physical pendulum is given by
$T=2\pi \sqrt{\frac{I_{support}}{mg{l}_{cm}}}{I}_{support}=m{R}^2+m{R}^2=2m{R}^2{l}_{cm}=RT=2\pi \sqrt{\frac{\left(2m{R}^2\right)}{mgR}}=2\pi \sqrt{\frac{2R}{g}}\Rightarrow R=\frac{T^{2}g}{8{\pi }^2}$
A seconds pendulum is a pendulum whose period is precisely two seconds; one second for a swing in one direction and one second for the return swing. i.e. $T=2s$, so we have
$R=\frac{2^2\times {\pi }^2}{8{\pi }^2}=0.5m$