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

Question

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 m on the groove. All the surfaces are smooth. If the radius of the table is 3 m , the radial velocity of the ball with respect to the table at the time ball leaves the table is $x \sqrt{2} \omega \mathrm{~m} / \mathrm{s}$, where the value of $x$ is ______.

Professor PC Thomas Classes · Answer

(2) : Let the table rotate with angular velocity, $\omega$ acceleration $a$ and velocity $v$. Centrifugal acceleration of the ball, $a=\omega^{2} r$\ \ or $v \frac{d v}{d r}=\omega^{2} r \Rightarrow v d v=\omega^{2} r d r$\ $\int_{0}^{v} v d v=\omega^{2} \int_{r_{0}}^{r} r d r ; \frac{v^{2}}{2}=\frac{\omega^{2}}{2}\left(r^{2}-r_{0}^{2}ight)$\ $
u=\omega \sqrt{r^{2}-r_{0}^{2}}$\ $v=\omega \sqrt{9-1} \Rightarrow v=2 \sqrt{2} \omega \mathrm{~m} / \mathrm{sec}$\ So, the value of $x$ is 2 .\