Question

A metal ring of mass, $m$, and radius, $r$, is placed on a smooth horizontal table and is set rotating about its axis so that each part of the ring moves with a speed, $v$. The tension in the ring is

• A. $\frac{m{v}^2}{2\pi r}$
• B. $\frac{m{v}^2}{r}$
• C. $\frac{m{v}^2}{\pi r^2}$
• D. $\frac{m{v}^2}{2r}$

Answer 1

The correct option is A $\frac{m{v}^2}{2\pi r}$

Let us consider a small element of mass $\delta m$ and length $\delta l$ of the ring. We have
$2T\sin \theta =\frac{\delta m{v}^2}{r}$

$\delta m=\frac{m}{2\pi r}\delta l$ and $\sin \theta \approx \theta$; since $\theta$ is small.

$\Rightarrow 2T\theta =\frac{m\delta l}{2\pi r}\times \frac{v^2}{r}$

$\Rightarrow 2T\frac{\left(\delta l/2\right)}{r}=\frac{m\delta l}{2\pi r}\times \frac{v^2}{r}$$\because \theta =\frac{\left(\delta l/2\right)}{r}$

$\Rightarrow T=\frac{m{v}^2}{2\pi r}$