Question

A ring of the radius $r$ is suspended from a point on its circumference. If the ring is made to oscillate in the plane of the figure, then the angular frequency of these small oscillations is

• A. $\sqrt{\frac{\text{2} \textit{g}}{\text{3} \textit{r}}}$
• B. $\sqrt{\frac{\textit{g}}{\text{2} \textit{r}}}$
• C. $\sqrt{\frac{\text{5} \textit{g}}{\text{4} \textit{r}}}$
• D. $\sqrt{\frac{\text{4} \textit{g}}{\text{3} \textit{r}}}$

Answer 1

Answer: (B)

It is a physical pendulum, the time period of which is,
$ T =2 \pi \sqrt{\frac{ I }{ mgl }} $
Here, $I=$ moment of inertia of the ring about the point of suspension
$ \begin{array}{l} =m r^{2}+m r^{2} \\ =2 m r^{2} \end{array} $
and $I=$ distance of the point of suspension from the centre of gravity
$=r$
$ \therefore T =2 \pi \sqrt{\frac{2 mr ^{2}}{ mgr }}=2 \pi \sqrt{\frac{2 r }{ g }} $
$\therefore$ Angular frequency $ \omega=\frac{2 \pi}{T}$
or $\omega=\sqrt{\frac{g}{2 r}}$