Q. A body moves along a circular path of radius $10m$ and the coefficient of friction is 0.5. What should be its angular speed in rad/s if it is not to slip from the surface $(g = 9.8\ m/s^{2})$
A) 5
B) 10
C) 0.1
D) 0.7
Explanation
(d) For moving on circular path without slipping, centripetal force must equal frictional force
That is,
$\frac{mv^{2}}{r} = \mu\ m\ g$
$\Rightarrow mr\omega^{2} = \mu\ m\ g$ $\lbrack\therefore v = r\ \omega\rbrack$
$\Rightarrow r\ \omega^{2} = \ \mu\ g$
$\therefore\ \omega = \sqrt{\frac{\mu\ g}{r}} = \sqrt{\frac{0.5 \times 9.8}{10}} = 0.7\ rad/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.