Question

A uniform thin ring of radius $R$ and mass $m$ suspended in a vertical plane from a point in its circumference its time period of oscillation is

• A. $2\pi \sqrt{\frac{2R}{g}}$
• B. $2\pi \sqrt{\frac{3R}{2g}}$
• C. $\frac{\pi }{2}\sqrt{\frac{R}{g}}$
• D. $\pi \sqrt{\frac{R}{2g}}$

Answer 1

Answer: (B)

$2\pi \sqrt{\frac{3R}{2g}}$

The time period of the disc is $2\pi \sqrt{\left(3r/2g\right)}$

We know that the time period of an object$,$

$T=2\pi \sqrt{\left(1/mgL\right)}$

where$,$

$I=$ moment of inertia from the suspended point

$L=$ distance of its centre from suspended point $=r$

we know that$,$ the moment of inertia of disc about its centre $=m{r}^2/2$

using parallel axis theoram the moment of inertia from a point in its periphery$,$

$I=m{r}^2+m{r}^2/2=3m{r}^2/2$

putting the values in the above equation we get$,$

$2\pi \sqrt{\left(3m{r}^2/2mgr\right)}$

$=2\pi \sqrt{\left(3r/2g\right)}$

therefore$,$ the time period of the disc is $2\pi \sqrt{\left(3r/2g\right)}$

Hence,

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