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