A thin circular wire of radius $R$ rotates about its vertical diameter with an angular frequency $ω$ . Show that a small bead on the wire remains at its lowermost point for $ω \leq \sqrt{g / R}$ . What is angle made by the radius vector joining the centre to the bead with the vertical downward direction for $ω = \sqrt{2 g / R}$ ? Neglect friction.

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$60 °$

$40 °$

$80 °$

$20 °$

answer is D.

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Detailed Solution

$Ncosθ = mg$

$Nsinθ = {mrω}^{2} = m (Rsinθ) ω^{2}$

Dividing these two equations, we get

$cosθ = \frac{g}{{Rω}^{2}}$

$ω = \sqrt{\frac{g}{Rcosθ}}$

At lowermost point, $θ = 0 °$

$ω = \sqrt{\frac{g}{R}}$

Substituting $ω = \sqrt{2 g / R}$ in Eq. (i), we have

$cosθ = \frac{1}{2}$

$θ = 60 °$

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