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 $\omega \le \sqrt{\frac{g}{r}}$. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega =\sqrt{\frac{g}{r}}$? Neglect friction.

Answer 1

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:
$Mg=Ncos\theta \dots \dots \dots \left(i\right)ml{\omega }^2=Nsin\theta \dots \dots \dots \left(ii\right)$
In $\Delta OPQ$, we have:
$sin\theta =\frac{l}{R}l=Rsin\theta \dots \dots \dots \left(iii\right)$
Substituting equation (iii) in equation (ii), we get:
$m\left(Rsin\theta \right){\omega }^2=Nsin\theta mR{\omega }^2=N\dots \dots \dots \left(iv\right)$
Substituting equation (iv) in equation (i), we get:
$mg=mR{\omega }^{2}cos\theta cos\theta =\frac{g}{R{\omega }^2}\dots \dots \dots \left(v\right)$
Since $cos\theta \le 1$, the bead will remain at its lowermost point for $\frac{g}{R{\omega }^2}\le 1,i.e.,for\omega \le \sqrt{\frac{g}{R}}$
For $\omega =\sqrt{\frac{2g}{R}}or{\omega }^2=\frac{2g}{R}\dots \dots \dots \left(vi\right)$
On equating equations (v) and (vi), we get:
$\frac{2g}{R}=\frac{g}{Rcos\theta }cos\theta =\frac{1}{2}\therefore \theta =co{s}^{-1}\left(0.5\right)={60}^{\circ }$