A ring of mass `m` and radius `R` is being rotated about its axis with constant angular velocity `omega` in the gravity free space. Find tension in the ring.

Explanation

Consider a small length of ring that subtends an angle `d theta` at the center.
`dm=(m)/(2piR)dl=(m)/(2piR)Rd theta=(m)/(2pi)d theta`
`dm` is moving in the circle and the necessary centripetal force is provide by component of tension.
`2T sin ((d theta)/(2))=dm omega^(2) R=(momega^(2)R)/(2pi)dtheta`
As `d theta` is small
`sin( (d theta)/(2))`
`2T(d theta)/(2)=Txxd theta=(momega^(2)R)/(2pi)d theta`
`T=(momega^(2)R)/(2pi)`