A circular table with smooth horizontal surface is rotating at an angular speed `omega` about its axis. A groove is made on the surface along a radius and a small particle is gently placed inside the groove at a distance l from the centre. Find the speed of the particle with respect to the table as its distance from the centre becomes L.

Options

A. `v=omegal`
B. `v=omega(l-a)`
C. `v=(omega(l+a))/(2)`
D. `v=omegasqrt(l^(2)-a^(2))`

Explanation

`a=omega^(2)xor(vdv)/(dx)=omega^(2)x`
`int_(0)^(v) vdv =int_(a)^(t)omega^(2)xdx`
`v=omegasqrt(l^(2)-a^(2))`