Question

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

• A. $2\sqrt{2gC}$
• B. $2\sqrt{gC}$
• C. $\sqrt{\frac{2gC}{ab}}$
• D. $\sqrt{\frac{2g}{C}}$

Answer 1

The correct option is A $2\sqrt{2gC}$

Given that, $y=4C{x}^2$

$\Rightarrow \frac{dy}{dx}=\tan \theta =8Cx$

At $P,\tan \theta =8Ca$


For the bead to stay at point $P$,

$m{\omega }^{2}a\cos \theta =mg\sin \theta$

$\Rightarrow \omega =\sqrt{\frac{g\tan \theta }{a}}$

$\therefore \omega =\sqrt{\frac{g\times 8aC}{a}}=2\sqrt{2gC}$.

Hence, option $\left(A\right)$ is correct.