Question

A circular table is rotating with an angular velocity of \( \omega \, \text{rad/s} \) 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 \( 1 \, \text{m} \) on the groove. All the surfaces are smooth. If the radius of the table is \( 3 \, \text{m} \), the radial velocity of the ball with respect to the table at the time the ball leaves the table is \( x\sqrt{2}\omega \, \text{m/s} \), where the value of \( x \) is \(\dots\).

Answer 1

Correct Answer: 2

The centripetal acceleration acting on the ball is:
\[a_c = \omega^2 x.\]
Using the relationship between velocity and position:
\[v \frac{dv}{dx} = \omega^2 x.\]
Integrate both sides:
\[\int_0^v v \, dv = \int_1^3 \omega^2 x \, dx.\]
Solve the integrals:
\[\frac{v^2}{2} = \omega^2 \int_1^3 x \, dx = \omega^2 \left[\frac{x^2}{2}\right]_1^3.\]
Substitute the limits:
\[\frac{v^2}{2} = \omega^2 \times \frac{1}{2} \left[3^2 - 1^2\right].\]
Simplify:
\[\frac{v^2}{2} = \omega^2 \times \frac{1}{2} \times (9 - 1) = \omega^2 \times \frac{1}{2} \times 8 = 4\omega^2.\]
Solve for $v$:
\[v^2 = 8\omega^2 \implies v = \sqrt{8}\omega = 2\sqrt{2}\omega.\]
Thus, the radial velocity of the ball as it leaves the table is:
\[v = 2\sqrt{2}\omega,\]
where $x = 2$.