A small ring P is threaded on a smooth wire bent in the form of a circle of radius a and centre O. The wire is rotating with constant angular speed `omega` about a vertical diameter xy, while the ring remains at rest relative to the wire at a distance `(a)/(2)` from xy. Then `omega^2` is equal to:

Options

A. `(2g)/(a)`
B. `(g)/(2a)`
C. `(2g)/(asqrt(3))`
D. `(g sqrt(3))/(2a)`

Answer

C

Explanation

`N cos theta = mg` …(i) and `N sin theta = (m omega^(2) a)/(2)` ….(ii)
Equation (ii) divided by equation (i)
`tan theta = (omega^(2) a)/(2g)`, `omega^(2) = (2g tan theta)/(a)`, `sin theta = (a)/(2) xx (1)/(a) = (1)/(2)` , `theta = 30^(@)`
`omega^(2) = (2 g tan 30^(@))/(a)` , `omega^(2) = (2g)/(sqrt(3) a)`