A metal rign of mass m and radius R is placed on a smooth horizontal table and is set rotating abut 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

Consider a small prt ACB of the ring that subtends an angle `/_\theta` aty te centre as shown in ure. Let the tension in te ring be T

The forces on this small part ACB are
a. tension T by the part of the ring left to A.
b. tension T by the part of the ring right to B,
c. weight `(/_\m)g ` and
`d. normal force N by the table
The tension at A acts along the tangent at A and teh tension at B acts along the tangent at B. As the small part ACB moveds in a circle of radius R at a constant speed v, its acceleration is twoards the centre (along CO) and has a magnitude `(/_\m)^2/R`.
Resolving the forces along the radius CO
`Tcos(90^0-/_\theta)/2+T cos (90^0-/_\theta)/2)=(v^2/R)`
or` 2Tsin(/_\m)(v^2/R)` ..........i
The length of the part ACB is `R/_\theta`. As the total mass of the rignn ism the mass of the part ACB will be `/_\m=m(2piR)R/_\theta=(m/_\theta)/(2pi)`
Putting `/_\m` in i.
`2Tsin(/_\theta)/2=m/(2pi) /_\theta(v^2/R)`
or `T=(mv^2)/(2piR) (/_\theta/2)/(sin(/_\theta/2))`
`As /_\theta` is very small, `(/_\theta/2)/(sin(/_\theta/2))~~1 and T=(mv^2)/(2piR^2)`.