Question

A ring of diameter 2m oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is

• A. $2m$
• B. $4m$
• C. $1.5m$
• D. $3m$

Answer 1

Answer: (C)

$1.5m$

Time period of a physical pendulum is given by
$T=2\pi \sqrt{\frac{I_{support}}{mg{l}_{cm}}}$

In case of the given ring
$I_{support}=(\frac{m{R}^2}{2}+m{R}^2)=\frac{3}{2}m{R}^2{l}_{cm}=R$

So we have,
$T_{ring}=2\pi \sqrt{\frac{\left(\frac{3}{2}m{R}^2\right)}{mgR}}=2\pi \sqrt{\frac{3R}{2g}}$

and we know that the time period of a simple pendulum is given by
$T_p=2\pi \sqrt{\frac{l}{g}}$

for equal time period
$T_{ring}=T_p\Rightarrow \frac{3R}{2g}=\frac{l}{g}\Rightarrow l=\frac{3R}{2}=1.5m$
hence the equivalent length of the simple pendulum is $1.5m$