A thin circular loop of radius R rotates about its vertical diameter with an angular
Question.
A thin circular loop of radius $R$ rotates about its vertical diameter with an angular frequency $\omega$. Show that a small bead on the wire loop remains at its lowermost point for $\omega \leq \sqrt{\mathrm{g} / R}$. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega=\sqrt{2 \mathrm{~g} / R} ? \mathrm{Neglect}$ friction.

solution:

Let the radius vector joining the bead with the centre make an angle $\theta$, 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:

$m g=N \cos \theta$

$m / \omega^{2}=N \sin \theta \ldots$ (ii)

In $\triangle \mathrm{OPQ}$, we have:

$\sin \theta=\frac{l}{R}$

$I=R \sin \theta \ldots(i i i)$

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

$m g=m R \omega^{2} \cos \theta$

$\cos \theta=\frac{\mathrm{g}}{R \omega^{2}}$

Since $\cos \theta \leq 1$, the bead will remain at its lowermost point for $\frac{\mathrm{g}}{R \omega^{2}} \leq 1$, i.e., for $\omega \leq \sqrt{\frac{\mathrm{g}}{R}}$

For $\omega=\sqrt{\frac{2 g}{R}}$ or $\omega^{2}=\frac{2 g}{R}$ $\ldots(v i)$

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

$\frac{2 \mathrm{~g}}{R}=\frac{\mathrm{g}}{R \cos \theta}$

$\cos \theta=\frac{1}{2}$

$\therefore \theta=\cos ^{-1}(0.5)=60^{\circ}$
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