A table with smooth horiztonal surface is truning at an angular speed `omega` abouyt its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the speed of the particle as its distance from teh centre becomes L.

Explanation

The situation is hwon in ure.

Let us work from the frame of reference of the table. Let us take the origin at the centre of rottion O an the X-axis along the groove ure. The Y-axis is along the lineperpendicular to OX coplaN/Ar with the surface of the table and the Z-axis is along the vertical.
Suppose at time t particle in the groove is at a distance x fromn the origin and is moving along The X-axis witha speed v. The forces acting on the particle (including the pseudeo forces that we must assume because we have taken our frame on the table which is rotating and is nonintertial) are
a. weight mg vetically downward
b. normal contact force `N_1` verticall upward by te bottom surfce of the groove,
c. normal contact force `N_2` parallel to the Y-axis by the side walls of the groove,
d. centrifugal force `momega^2x` alongthe 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 teh X-axis is the centrifugal force `momega^2x`. all the other forces are perpendicular to the X-axis adn have no components along the X-axis.
Thus, the acceleration along the X-axis is
`a=F/m =(momega^2x)/m = omega^2x`
or, `(dv)/(dt)=omega^2x`
or ` (dv)/(dx).(dx)/(dt)=omega^2x`
or `(dv)/(dx).v=omega^2x`
or, ` v dv = omega^2 x dx `
or, `int^v_0 v dv = int^L_aomega^2 x dx `
or, ` [1/2 v^2]_0^v=[1/2 omega^2x^2]^L_a`
or, `v^2/2=1/2 omega^2(L^2-a^2)`
or, v=omega sqrt(L^2-a^2)`