# Q3 : A thin circular loop of radius R  rotates about its vertical diameter with an angular frequency $\omega \le \sqrt{g/R.}$ What is the angle made by radius vector joining the centre to the bead with the vertical download direction for $\omega =\sqrt{2g/R?}$ Neglect friction.

Dear student

Let O be the centre of the circular loop and $\theta$ is the angle made by radius vector joining the centre to bead with vertical.

At P the normal reaction of the loop is along PO and its resolved components are

where $mr{\omega }^{2}$ is the centrifugal force on the bead and r=PR=$R\mathrm{sin}\theta$ where R is the radius of the circle so

at lowest point $\theta$=0o so $\omega =\sqrt{\frac{g}{R}}$ and thus if $\omega$ becomes higher this value bead rises. so bead remains at lowest position if $\omega \le \sqrt{\frac{g}{R}}$.
angle when $\omega =\sqrt{\frac{2g}{R}}$
$\mathrm{cos}\theta =\frac{g}{R}·\frac{R}{2g}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\theta ={60}^{°}$

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