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 through its center 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.

Options

A. `sqrt((2pi sigma A)/(mR))`
B. `sqrt((2pi sigma A)/(mR))`
C. `(3pi sigma A)/(mR)`
D. `(pi sigma A)/(2mR)`

Answer

A

Explanation

Everything element of the ring rotates in a circular path of radius `R` about the axis of rotation. The radical componenrt of tensionin the wire provides the centripental forces. The figures shows a free body diagram of a small element of a mass fo the ring.
`dm = (m//2pi) d theta`

The radical componenet of tension is `F_(r) = 2F sin .(d theta)/(2) = 2F ((d theta )/(2)) approx Fd theta`
Applying Newton's secound law, we get `Fd theta = (dm) omega^(2) R`
`Fd theta = ((md theta)/(2pi)) omega^(2) R` or `F = (m omega^(2)R)/(2pi)`
If the breaking stress is `sigma_(b)` then the maximum value of `F` can be `F_("max") = sigma_(b)A`
`(m omega_("max")^(2R))/(2pi) = sigma_(b) A, omega_("max") = sqrt((2pi sigma_(0) A)/(mR))`