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

`(mv^(2))/(2 pi R)`

Explanation

Take a small part of the ring which subtends an angle ` Delta 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 Fig.7.118 . The mass of the small part of ring `Delta m = (m)/(2 pi) Delta theta `
The centripetal force on this part is 2T sin `(Delta 0)/(2)`.

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