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.

Explanation

Take a small part of the ring which subtends an angle `Delta theta ` at the conter of the ring .Let `T` be the tension in the ring . The force act on the this part in the small part of ring `Delta m = (m)/(2 pi) Delta theta `

Thus, by Newton 's second law we have
`2T sin (Delta theta)/(2) = (Delta m v^(2))/(R )`
As `Delta theta` small, `sin (Delta theta)/(2) = (Delta theta )/(2)`
The above equation reduces in
`2T ((Delta theta)/(2)) = ((m)/(2 pi) Delta theta )(v^(2))/(R )`
or `T = (mv^(2))/(2 pi R)`