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 A.

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Detailed Solution

$y = 4 C x^{2} \Rightarrow \frac{d y}{d x} = \tan θ = 8 C x$

At P, $\tan θ = 8 C a$
For steady circular motion $m ω^{2} a \cos θ = m g \sin θ \Rightarrow ω = \sqrt{\frac{g \tan θ}{a}}$ $∴ ω = \sqrt{\frac{g \times 8 a C}{a}} = 2 \sqrt{2 g C}$

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