A circular table is rotating with an angular velocity of $\omega \mathrm{rad} / \mathrm{s}$ about its axis…

A circular table is rotating with an angular velocity of $\omega \mathrm{rad} / \mathrm{s}$ about its axis (see figure). There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of $1 \mathrm{~m}$ on the groove. All the surfaces are smooth. If the radius of the table is $3 \mathrm{~m}$, the radial velocity of the ball w.r.t. the table at the time ball leaves the table is $x \sqrt{2} \omega \mathrm{m} / \mathrm{s}$, where the value of $x$ is _____.

Solution

$\begin{aligned} & a_c=\omega^2 x \\ & \frac{v d v}{d x}=\omega^2 x \\ & \int_0^v v d v=\int_1^3 \omega^2 x d x \\ & \frac{v^2}{2}=\omega^2\left[\frac{x^2}{2}\right] \\ & \frac{v^2}{2}=\frac{\omega^2}{2}\left[3^2-1^2\right] \\ & v=2 \sqrt{2} \omega \\ & x=2\end{aligned}$

Asked in: JEE Main 2024 (08 Apr Shift 2)

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