A table with smooth horizontal surface is turning at an angular speed `omega` about 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

Here the motion of the partical is constrained with in the groowe radial direction . During rotation of the table . When particle is at a distance a from the center of the table a show in fig 87.57. its acceleration is given as

`a = omega^(2) r`
`v(dv)/(dx) = omega^(2) pi`
`v dv = omega^(2) x dx dx`
Integrating `int_(0)^(v) v dv = int_(l)^(L) omega^(2) x dx`
`[v^(2)/(2)]_(0)^(v) = omega^(2) [x^(2)/(2)]_(l)^(l)rArr v = omega (L^(2) - t^(2)) ^(1//2)`