A circular ring of radius `R` and mass `m` made of a uniform wire of cross sectional area `A` is rotated about a stationary vertical axis passing throgh its centre and perpendicular to the plane of the ring. If the breaking stress of the material of the ring is `sigma_(b)`, then determine the maximum angular speed `omega_("max")` at which the ring may be rotated without failure.

Explanation

Every element of the ring rotates in a circular path of radius `R` about the axis of rotation.The radial component of tension in the wire provides the centrifugal forces.
Figure shows a free body diagram of a small element of mass `dm=(m//2pi)d theta` of the ring.

The radial component of tension is
`F_(r)=2Fsin(d theta)/2~~2F((d theta)/2)~~Fd theta`
Applying Newton's second law ,we get `Fd theta=(dm)omega^(2)R`
`Fd theta=((md theta)/(2pi))omega^(2)R` or `F=(momega^(2)R)/(2pi)`
If the breaking stress is `sigma_(b)` then the maximum value of `F` can be
`F_("max")=sigma_(b)A`
`(momega_("max")^(2R))=sigma_(b)A`
`omega_("max")=sqrt((2pisigma_(0)A)/(mR))`