A thin circular wire of radius `R` rotatites about its vertical diameter with an angular frequency `omega` . Show that a small bead on the wire remain at its lowermost point for `omegalesqrt(g//R)` . What is angle made by the radius vector joining the centre to the bead with the vertical downward direction for `omega=sqrt(2g//R)` ? Neglect friction.

Explanation

`N cos theta = mg` ….(i)
`m rw^(2) = N sin theta` …..(ii)
`m (R sin theta) omega^(2) = N sin theta`
`m R omega^(2) = N`
From equation (i) `mg = m R omega^(2) cos theta`
`cos theta = (q)/(R omega^(2))`…..(iii)
As cos `le` 1 Therefore, bead will remain at its lowermost point for `(g)/(R omega^(2)) ge 1` or `omega le sqrt((g)/(R))`