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A smooth wire of length $2 π r$ is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed $ω$ about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of $ω^{2}$ is equal to

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$\frac{(g \sqrt{3})}{r}$

$\frac{2 g}{r}$

$\frac{2 g}{(r \sqrt{3})}$

$\frac{\sqrt{3} g}{2 r}$

answer is C.

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Detailed Solution

Let $θ$ be the angle with the vertical, then $\sin θ = \frac{r / 2}{r} = \frac{1}{2}$

$\Rightarrow θ = \sin^{- 1} (\frac{1}{2}) = 30^{o} N \cos θ = m g N \sin θ = F_{c} = \frac{m ω^{2} r}{2} \Rightarrow \tan θ = \frac{ω^{2} r}{2 g} \Rightarrow ω^{2} = \frac{2 g \tan θ}{r} = \frac{2 g}{r \sqrt{3}}$

Hence, the correct answer is (C)

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