A bead of mass m stays at point P(a,b) on a wire bent in the shape of a parabola $y = 4 c x^{2}$ and rotating with angular speed $ω$ (see figure). The value of $ω$ is (neglect friction):

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$\sqrt{\frac{2 g}{C}}$

$2 \sqrt{g C}$

$\sqrt{\frac{2 g C}{a b}}$

$2 \sqrt{2 g C}$

answer is C.

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Detailed Solution

$\begin{array}{l} F o r p a r t i c l e t o b e i n e q u i l i b r i u m, \\ m g \sin θ = m x ω^{2} \cos θ \\ \Rightarrow \tan θ = \frac{ω^{2} x}{g} \\ A l s o, y = 4 c x^{2} \\ \Rightarrow \frac{d y}{d x} = 8 c x = s l o p e a t p o i n t P = \tan θ \\ E q u a t i n g b o t h v a l u e s o f \tan θ w e g e t, \\ \frac{ω^{2} x}{g} = 8 c x \\ \Rightarrow ω^{2} = 8 c g \\ \Rightarrow ω = \sqrt{8 c g} = 2 \sqrt{2 c g} \end{array}$