Problem

Problem

A thin circular wire of radius RR rotates about its vertical diameter with an angular frequency ω\omega. Show that a small bead on the wire remains at its lowermost point for ω≤g/R\omega \le \sqrt{g/R}. What is angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω=2g/R\omega = \sqrt{2g/R} ? Neglect friction.

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