Question

A smooth wire of length 2πr is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed w about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of ω²
is equal to:
(A) √3 g/2r
(B) g√3/r (C) 2g/r (D) 2g/(r √3)

Answer 1

correct option 4 2g/(r √3)

Answer 2

Thus the angular speed of the wire is ω^2 = 2g / √3r

Option (D) is correct.

Explanation:

Given data:

• Length of wire = 2πr
• Angular speed = ω

Nsinθ = mr^2ω^2   --------(1)

Ncosθ = mg     ------(2)

tanθ = rω^2 / 2g

r / 2√3r / 2 = rω^2 / 2g

ω^2 = 2g / √3r

Thus the angular speed of the wire is ω^2 = 2g / √3r