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 and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the particle with respect to the table as its distance from the centre becomes L.

• A. $v=\omega \sqrt{L^2-a^2}$
• B. $v=\omega \sqrt{L^2-2a^2}$
• C. $v=\omega \sqrt{2L^2-a^2}$
• D. $v=\omega \sqrt{L^2}$

Answer 1

The correct option is A $v=\omega \sqrt{L^2-a^2}$

Let $v$ be the velocity of the particle w.r.t 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.

Using $a_r=\frac{vdv}{dx}$ where $a_r=x{w}^2$

$\therefore$ $x{w}^2=\frac{vdv}{dx}$

OR ${\int }_0^{V}vdv=w^2{\int }_a^{L}xdx$

OR $\frac{v^2{|}_0^V}{2}=w^2\times \frac{x^2{|}_a^L}{2}$

OR $V^2-0=w^2(L^2-a^2)$ $⟹V=w\sqrt{L^2-a^2}$