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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 joining the center to the bead with the vertical downward direction for $\omega=\sqrt{2} g / R$ ? Neglect friction.
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Verified Answer
$\theta$ - angle between radius vector joining the bead to the center and vertical downward direction.
$N$ - normal reaction.
$$
m g=N \cos \theta \text {...(i) }
$$
and $m r \omega^2=\mathrm{N} \sin \theta$
or, $m(R \sin \theta) \omega^2=N \sin \theta$
$\Rightarrow R \omega^2=$ N...(ii)
From (i) and (ii),
$m g=m R \omega^2 \cos \theta$
$\Rightarrow \cos \theta=g / R \omega^2$
Since $|\cos \theta| \leq 1$, bead will emain its lower most point. $\Rightarrow g / R \omega^2 \leq 1$ or, $\omega \leq \sqrt{ }(g / R)$
If, $\omega=\sqrt{(} 2 g / R), \cos \theta=g / R(2 g / R)=\frac{1}{2}$
i.e. $\theta=60^{\circ}$
$N$ - normal reaction.
$$
m g=N \cos \theta \text {...(i) }
$$
and $m r \omega^2=\mathrm{N} \sin \theta$
or, $m(R \sin \theta) \omega^2=N \sin \theta$
$\Rightarrow R \omega^2=$ N...(ii)
From (i) and (ii),
$m g=m R \omega^2 \cos \theta$
$\Rightarrow \cos \theta=g / R \omega^2$
Since $|\cos \theta| \leq 1$, bead will emain its lower most point. $\Rightarrow g / R \omega^2 \leq 1$ or, $\omega \leq \sqrt{ }(g / R)$
If, $\omega=\sqrt{(} 2 g / R), \cos \theta=g / R(2 g / R)=\frac{1}{2}$
i.e. $\theta=60^{\circ}$
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