Q. A coin, placed on a rotating turn-table slips, when it is placed at a distance of $9\ cm$ from the centre. If the angular velocity of the turn-table is tripled, it will just slip, if its distance from the centre is
A) $27\ cm$
B) $9\ cm$
C) $3\ cm$
D) $1\ cm$
Explanation
(d) In the given condition friction provides the required centripetal force and that is constant. $i.e.$ $m\omega^{2}r =$ constant
$\Rightarrow r \propto \frac{1}{\omega^{2}}\therefore r_{2} = r_{1}\left( \frac{\omega_{1}}{\omega_{2}} \right)^{2} = 9\left( \frac{1}{3} \right)^{2} = 1\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.