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 $\omega \leq \sqrt{g / R} .$ What is the angle made by the radius vector jotning the centre to the bead with the vertical downward direction for $\omega=\sqrt{2 g / R} ?$ Neglect friction.
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 $\omega \leq \sqrt{g / R} .$ What is the angle made by the radius vector jotning the centre to the bead with the vertical downward direction for $\omega=\sqrt{2 g / R} ?$ Neglect friction.
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:
$M g=N \cos \theta$
$m l \omega^{2}=N$
In $\Delta$ $OPQ$, we have:
$\sin \theta=\frac{l}{R}$
$l=R \sin \theta$
$m(R \sin \theta) \omega^{2}=N \sin \theta$
$m R \omega^{2}=N$
$m g=m R \omega^{2} \cos \theta$
$\cos \theta=\frac{ g }{R \omega^{2}}$
since $\cos \theta \leq 1,$ the bead will remain at its lowermost point for $\frac{g}{R \omega^{2}} \leq 1,$ i.e., for $\omega \leq \sqrt{\frac{g}{R}}$
For $\omega=\sqrt{\frac{2 g }{R}} {\text { or }} \omega^{2}=\frac{2 g }{R}$
On equatingabove equations
$\frac{2 g}{R}=\frac{g}{R \cos \theta}$
$\cos \theta=\frac{1}{2}$
$\therefore \theta=\cos ^{-1}(0.5)=60^{\circ}$
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