A circular table is rotating with an angular velocity of ω rad/s about its axis (see figure). There is a smooth groove along a radial direction

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 \(1 \mathrm{m}\) on the groove. All the surface are smooth. If the radius of the table is 3 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............ 

Answer 1

Correct answer is : 2 

\(\mathrm{a}_{\mathrm{c}}=\omega^{2} \mathrm{x}\)

\( \frac{v d v}{d x}=\omega^{2} x \)

\(\int_{0}^{v} v d v=\int_{1}^{3} \omega^{2} x d x\)

\(\frac{\mathrm{v}^{2}}{2}=\omega^{2}\left[\frac{\mathrm{x}^{2}}{2}\right]\)

\(\frac{\mathrm{v}^{2}}{2}=\frac{\omega^{2}}{2}\left[3^{2}-1^{2}\right]\)

\(\mathrm{v}=2 \sqrt{2} \omega\)

\(\mathrm{x}=2\)