A thin circular loop of radius R rotates about its vertical diameter with an angular frequency omega .Show that a small bead on the wire loop remains at its lowermost pointfor omega lt sqrt( g// R)Whatis theangle made by the radius vector joining the centre to the bead with the vertical downward direction for omega = sqrt(2g// R) ?neglect friction.

A thin circular loop of radius R rotates about its vertical diameter w

1.	A thin circular loop of radius R rotates about its vertical diameter with an angular frequency omega .Show that a small bead on the wire loop remains at its lowermost pointfor omega lt sqrt( g// R)Whatis theangle made by the radius vector joining the centre to the bead with the vertical downward direction for omega = sqrt(2g// R) ?neglect friction.
Answer» Solution :Consider the free-body diagram of the bead when the RADIUS vector joining the centre of the wire MAKES an angle `theta` with the vertical downward direction.We have mg `= N cos theta` and `m R sin thetaomega^2 = N sin theta` .These equations GIVE` cos theta = g//Romega^2`.Since `cos theta lt= 1,` the bead remains at its lowemost point `omega lt= sqrt(g/R )` For `omega = sqrt((2g)/(R ), cos theta = 1/2 i.e theta = 60^@`