Question

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. Then the speed of the particle with respect to the table as its distance from the centre becomes $L$ is given by $v=\omega \sqrt{L^2-n{l}^2}$. The value of $n$ is

Answer 1

The motion of the particle is constrained within the groove along radial direction.

During the rotation of the table, when particle is at a distance $x$ from the centre of the table as shown in figure, its acceleration is given as,
$a={\omega }^{2}x\Rightarrow v\frac{dv}{dx}={\omega }^{2}x\Rightarrow vdv={\omega }^{2}xdx\text{integrating,}{\int }_0^{v}vdv={\int }_l^L{\omega }^{2}xdx\Rightarrow {\left[\frac{v^2}{2}\right]}_0^v={\omega }^2{\left[\frac{x^2}{2}\right]}_l^L\Rightarrow v=\omega \sqrt{L^2-l^2}$