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 lower most point for $\omega \le \sqrt{g/R}$. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega =\sqrt{2g/R}$ ? Neglect friction.

Answer 1

$Nsin\theta =mr{\omega }^2$
here, $r=Rsin\theta$
So,

$Nsin\theta =mRsin\theta {\omega }^2$
$\Rightarrow N=mR{\omega }^2---\left(i\right)$
and,

$Ncos\theta =mg-----\left(ii\right)$

Dividing these two equations (ii)/(i), we get:
$cos\theta =\frac{mg}{mR{\omega }^2}$
$\Rightarrow cos\theta =\frac{g}{R{\omega }^2}$

$\therefore \omega =\sqrt{\frac{g}{Rcos\theta }}$

At lowermost point, $\theta =0^{\circ }$

$\omega =\sqrt{\frac{g}{R}}$

Substituting $\omega =\sqrt{\frac{2g}{R}}$ in Eqs, $\left(i\right)$ we have

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

$\therefore \theta ={60}^{\circ }$