Question

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.

Answer 1

Mass per unit length of the ring $=\frac{m}{2\pi R}$

$\therefore$ Mass of $dx$ element $=\frac{mdx}{2\pi R}=dm$

For the $dx$ element the two $Tcos\theta$ componenets cancel and $Tsin\theta$ components add up to provide required centripetal force.

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

Since $\theta$ is small $\therefore sin\frac{\theta }{2}\approx \frac{\theta }{2}$

$\therefore 2T\frac{\theta }{2}=\frac{dm{v}^2}{R}\Rightarrow T\left(\frac{dx}{R}\right)=\frac{mdx}{2\pi R}\times \frac{v^2}{R}\Rightarrow T=\frac{m{v}^2}{2\pi R}$