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 center to the bead with the vertical downward direction for `omega=sqrt(2g//R)` ? Neglect friction.

Explanation

To solve the problem, we will break it down into two parts as stated in the question.

### Part 1: Proving the Condition for the Bead to Remain at the Lowermost Point

1. **Understanding the Forces Acting on the Bead**:
- The bead is subjected to two forces: its weight \( mg \) acting downward and the normal force \( N \) exerted by the wire.
- The bead is also in circular motion due to the rotation of the wire.

...