A ring of radius `r` is rotating about a vertical axis along its diameter with constant angular velocity `omega.A` read of mass `m` remains at rest `w.r.t.` ring at the position shown in figure. Then `w^(2)` is:

Options

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

Answer

c

Explanation

`N = cos theta = mg `…(i)
`N sin theta = (m omega^(2)a)/(2)`…(ii)
from eqs(ii)(i),`tan theta = (omega^(2)a)/(2g) rArr omega^(2) = (2g tan theta)/(a)`
Now` sin theta = (a)/(2) xx(1)/(a) or theta = 30^(@)`
`or omega^(2) = (2g)/(sqrt(3a)`