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 given problem involves the analysis of motion on a rotating table. We need to determine the value of \( x \) in the expression for the radial velocity of a steel ball leaving the table.

Consider the forces acting on the ball. As the table rotates with angular velocity \( \omega \), the centrifugal force acts outward on the ball. This force is given by \( F = m \omega^2 r \), where \( r \) is the radial distance from the center of the table.

The ball is initially placed at a distance \( r_0 = 1 \, \text{m} \) from the center. As it moves outward, it experiences no tangential force (due to the smooth groove) and gains only radial kinetic energy due to centrifugal force. As the ball reaches the edge of the table at \( r = 3 \, \text{m} \), its radial kinetic energy can be expressed as:

\[ \frac{1}{2} m v_r^2 = \int_{r_0}^{3} m \omega^2 r \, dr \]

Solving the integral, we find:

\[ \int_{1}^{3} \omega^2 r \, dr = \left. \frac{\omega^2 r^2}{2} \right|_{1}^{3} = \frac{\omega^2 (3^2 - 1^2)}{2} = \frac{\omega^2 (9 - 1)}{2} = 4\omega^2 \]

Equating the kinetic energy:

\[ \frac{1}{2} m v_r^2 = m \cdot 4\omega^2 \]

Simplifying,

\[ v_r^2 = 8\omega^2 \]

\[ v_r = \sqrt{8}\omega = 2\sqrt{2}\omega \]

Thus, the final radial velocity of the ball is \( 2\sqrt{2}\omega \, \text{m/s} \), and comparing with \( x\sqrt{2}\omega \, \text{m/s} \), we determine that \( x = 2 \).

This value fits perfectly within the specified range of \( 2,2 \).

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 $x = 2$.