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

Options

A. 60
B. 30
C. 90
D. 0

Explanation

To solve the problem step by step, we will analyze the forces acting on the bead and derive the necessary conditions for it to remain at the lowermost point of the rotating wire.

### Step 1: Analyze the Forces Acting on the Bead
When the bead is on the wire, it experiences two main forces:
1. The gravitational force acting downward, which is \( mg \).
2. The normal force \( N \) exerted by the wire, which acts perpendicular to the wire at the point of contact.

### Step 2: Set Up the Coordinate System
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