Table Dance

Let us work from the frame of reference of the table. Let us take the origin at the centre of rotation O and the X-axis along the groove (as shown). The Y-axis is along the line perpendicular to OX coplanar with the surface of the table and the Z-axis is along the vertical.

Suppose at time t the particle of mass m in the groove is at a distance x from the origin and is moving along the X-axis with a speed v. The forces acting on the particle (including the pseudo forces that we must assume because we have taken our frame on the table, which is rotating and is non-inertial) are:

(a) weight mg vertically downward

(b) normal contact force $N_{1}$ , vertically upward by the bottom surface of the groove

(c) normal contact force $N_{2}$ , parallel to the Y-axis by the side walls of the groove

(d) centrifugal force m $\omega^{2}x$ along the X-axis, and

(e) coriolis force along Y-axis (coriolis force is perpendicular to the velocity of the particle and the axis of rotation)

As the particle can only move in the groove, its acceleration is along the X-axis. The only force along the X-axis is the centrifugal force m $\omega^{2}x$ . All the other forces are perpendicular to the X-axis and have no components along the X-axis.

Thus, the acceleration along the X-axis is:

a = $\frac{F}{m}$ = $\omega^{2}x$

$\implies$ $\frac{dv}{dt}$ = $\omega^{2}x$

$\implies$ $\frac{dv}{dx}$ . $\frac{dx}{dt}$ = $\omega^{2}x$

$\implies$ $\frac{v.dv}{dx}$ = $\omega^{2}x$

Now integrating both sides and applying appropriate limits, we get:

$\int^v_0 v.dv$ = $\int^{25}_{15} \omega^{2}x.dx$

Given $\omega$ = 10 rad/s, solving the above integral, we get:

v = 200 m/s