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

We have shown that radius joining the bead to the center of the wire mkes an angle `theta` with the verticle downward direction If N is reaction then as is clear from the figuree
`mg = N cos theta`
`m r omega^(2) = N sin theta`
or `m (R sin theta) omega^(2) = N sin theta` or `m R omega^(2) = N`
From `mg = m R omega^(2) cos theta` or `cos theta = (g)/(R omega^(2)`
As `cos theta le 1` therefore bead will remain at its lowermost point for
`(g)/(R omega^(2)) le 1` , or `omega le sqrt((g)/(R)) `
When `omega = sqrt((2 g )/(R))`
From `cos theta = (g)/(R) ((R)/(2g)) = (1)/(2) :. theta = 60^(@)`
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