Q 11-04-185JEE MainJEE Main 2020 (2 Sep, Shift 1)Medium
A bead of mass $m$ stays at point $P(a, b)$ on a wire bent in the shape of a parabola $y = 4Cx^{2}$ and rotating with angular speed $\omega$ about the $y$-axis (see figure). The value of $\omega$ is (neglect friction)
Answer: (A) $2\sqrt{2gC}$
In the rotating frame the bead is in equilibrium under gravity $mg$, the centrifugal force $m\omega^{2}a$ and the normal reaction (perpendicular to the wire).
The resultant of $mg$ and $m\omega^2 a$ must be along the normal, so the slope of the wire at $P$ is
$$\tan\theta = \frac{dy}{dx} = \frac{\omega^{2}a}{g}$$
For $y = 4Cx^{2}$, $\dfrac{dy}{dx} = 8Ca$. Hence $8Ca = \dfrac{\omega^{2}a}{g}$, so $\omega^{2} = 8gC$ and
$$\omega = 2\sqrt{2gC}$$
Solution by Sreeraj P, M.Sc Physics