Question

A table with smooth horizontal surface is rotating at a 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 center. Find the speed $v$ of the particle with respect to the table as its distance from the center becomes $l$.

• A. $v=\omega l$
• B. $v=\omega (l-a)$
• C. $v=\frac{\omega (l+a)}{2}$
• D. $v=\omega \sqrt{l^2-a^2}$

Answer 1

Answer: (D)

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

= $\frac{F}{m}$ = $\frac{{m\omega }^{2}x}{m}$

or $\frac{dv}{dt}={\omega }^{2}x$

$\frac{dv}{dx}.\frac{dx}{dt}={\omega }^{2}x$

$vdv={\omega }^{2}xdx$; ${\int }_0^{v}vdv={\int }_a^l{\omega }^{2}xdx$
$v$ = $\omega \sqrt{l^2-a^2}$