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

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 $x = 2$.

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 $x = 2$.