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A thin circular wire of radius R rotates...

A thin circular wire of radius `R` rotates 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.

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The correct Answer is:
`theta = 60^(@)`
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