Answer to Question #152910 in Mechanics | Relativity for Sudarshan

Question

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 of and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the particle as its distance from the centre becomes L.
Plz solve it from inertial frame or without using centrifugal force

Answer 1

Answer

Now suppose v is the velocity of the particle with respect to the table at the instant the particle is at a distance x.When the particle is at a distance a from the centre, at that instant its velocity is zero.

Diagram can be shown as

Therefore acceleration

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

To taking this integration

$\int^v_0 vdv=\omega^2\int_a^Lxdx$

$v^2/2=\frac{\omega^2[x^2]_a^L}{2}$

So speed of particle is

$v=\omega\sqrt{L^2-a^2}$