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 point for `omegalesqrt(g//R)` . What is the angle made by the radius vector joining the centre to bead with the vertical downward direction for `omega=sqrt(2g//R)` ? Neglect friction .

Explanation

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 theta omega^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^@`