A circular table is rotating with an angular velocity of $\omega\ \text{rad/s}$ about its axis. There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of $1\ \text{m}$ on the groove. All the surfaces are smooth. If the radius of the table is $3\ \text{m}$, the radial velocity of the ball w.r.t. the table at the time the ball leaves the table is $x\sqrt2\,\omega\ \text{m/s}$, where the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 2
In the rotating frame, the only radial force is the centrifugal force $m\omega^2r$ (the groove's normal force is perpendicular to the groove):
$$v\frac{dv}{dr} = \omega^2r \;\Rightarrow\; v^2 = \omega^2(r^2 - r_0^2)$$
At the edge, $r = 3\ \text{m}$ with $r_0 = 1\ \text{m}$:
$$v = \omega\sqrt{9 - 1} = 2\sqrt2\,\omega\ \text{m/s} \;\Rightarrow\; x = 2$$
Solution by Sreeraj P, M.Sc Physics