Question

A uniform semicircular ring having mass $m$ and radius $r$ is hanging at one of its ends freely as shown in the figure. The ring is slightly disturbed so that it oscillates in its own plane. The time period of oscillation of the ring is

• A. $2\pi \sqrt{\frac{r}{g(1+\frac{1}{{\pi }^2})}}$
• B. $2\pi \sqrt{\frac{r}{g{(1-\frac{4}{{\pi }^2})}^{\frac{1}{2}}}}$
• C. $2\pi \sqrt{\frac{r}{g{(1-\frac{2}{{\pi }^2})}^{\frac{1}{2}}}}$
• D. $2\pi \sqrt{\frac{2r}{g{(1+\frac{4}{{\pi }^2})}^{\frac{1}{2}}}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{2r}{g{(1+\frac{4}{{\pi }^2})}^{\frac{1}{2}}}}$


COM of a semicircular ring is $\frac{2r}{\pi }$ height above the centre
$r^{\prime }=\frac{2r}{\pi }\Rightarrow d=\sqrt{r^2+(r^{\prime }{)}^2}$

$d=\sqrt{r^2+\frac{4r^2}{{\pi }^2}}=r\sqrt{1+\frac{4}{{\pi }^2}}$
Time period of physical pendulum is
$T=2\pi \sqrt{\frac{I_o}{mgd}}$

Here, moment of inertia about $\left(O^{\prime }\right)$ centre of semicircular ring is
$\Rightarrow I_{o^{\prime }}=I_c+m(r^{\prime }{)}^2$
$\Rightarrow m{r}^2=I_c+m(r^{\prime }{)}^2\Rightarrow I_c=m(r^2-\left(r^{\prime }{)}^2\right)$
Moment of inertia about point of suspension is
$\Rightarrow I_o=I_c+m{d}^2\Rightarrow I_o=m{r}^2-m(r^{\prime }{)}^2+m(r^2+\left(r^{\prime }{)}^2\right)$
$\Rightarrow I_0=2m{r}^2$

Time period is
$T=2\pi \sqrt{\frac{2m{r}^2}{mgr{(1+\frac{4}{{\pi }^2})}^{\frac{1}{2}}}}=2\pi \sqrt{\frac{2r}{g{(1+\frac{4}{{\pi }^2})}^{\frac{1}{2}}}}$