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 remains 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:

l = Rsinθ … (iii)

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

m(Rsinθ) ω² = Nsinθ

mR ω² = N ... (iv)

Substituting equation (iv) in equation (i), we get:

mg = mR ω² cosθ

... (v)

Since cosθ ≤ 1, the bead will remain at its lowermost point for, i.e., for

For or

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