A metal ring of mass $m$ and radius $R$ is placed on a smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed $v$. Find the tension in the ring.

Take a small part of the ring which subtends an angle $∆\theta$ at the centre of the ring. Let $T$ be the tension in the ring. The forces act on this part in the plane of rotation are shown in figure. The mass of the small part of ring

$∆m=\frac{m}{2\pi }∆\theta$

The centripetal force on this part is $2T \sin \frac{∆\theta }{2}$.

By Newton’s second law, we have

$2T \sin \frac{∆\theta }{2}=\frac{∆m{v}^2}{R}$

As $∆\theta$ is small,

$\therefore \sin \frac{∆\theta }{2}\approx \frac{∆\theta }{2}$

The above equation reduces to, 

$$\begin{gathered} 2T \left(\frac{∆\theta }{2}\right)=\left(\frac{m}{2\pi }.∆\theta \right) \frac{v^2}{R} \\ or T=\frac{m{v}^2}{2\pi R} \end{gathered}$$