Question

A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop *** at its lowermost point for ≤ √/. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for = √/? Neglect friction.

Answer 1

Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

OP = R = Radius of the circle

N = Normal reaction

The respective vertical and horizontal equations of forces can be written as:

Mg = Ncosθ ………….. (i)

mlω2 = Nsinθ ………... (ii)

In ΔOPQ, we have:

sinθ = l/R

l = Rsinθ ……………………… (iii)

Substituting equation (iii) in equation (ii), we get:

mg = mR ω2 cosθ

Since cosθ ≤ 1, the bead will remain at its lowermost point for /2 ≤ 1, i.e., for ≤ √g/R

On equating equations (v) and (vi), we get: