﻿ 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 ω = ? Neglect friction. from Physics Laws of Motion Class 11 Manipur Board

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A thin circular loop of radius rotates about its vertical diameter with an angular frequency ωShow 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 ω ? Neglect friction.

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

OP = R = Radius of the circle

N = Normal reaction

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

mg = N Cosθ                         ... (i)

mlω2 = Sinθ                       … (ii)

In ΔOPQ, we have

Sin θ = l / R

R Sinθ                              … (iii)

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

m(R Sinθω2 = N Sinθ

mR ω2 = N                            ... (iv)

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

mg = mR ω2 Cosθ

Cosθ = g / Rω2                        ...(v)

Since cosθ ≤ 1, the bead will remain at its lowermost point for g / Rω2 ≤ 1.

i.e., for                      ω ≤ (g / R)1/2

For ω = (2g / R)1/2 ω2 = 2g / R                  ...(vi)

On equating equations (v) and (vi), we get = g / RCos θ

Cos θ = 1 / 2

∴ θ = Cos-1(0.5)  =  600 , is the angle made by the radius vector joining the centre to the bead with the vertical downward direction.
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