Q. A body moves along a circular path of radius 10 m and the coefficient of friction is 0.5. What should be its angular speed in ${rad\ s}^{- 1},$ if it is not to slip from the surface? $(g = 9.8\ {ms}^{- 2})$
A) 5
B) 10
C) 0.1
D) 0.7
Explanation
(d) For body to move on circular path. Frictional force provides the necessary centripetal force,
$ie,\ $frictional force = centripetal force
$or\ \ \mu mg = \frac{{mv}_{0}^{2}}{r} = {mr\omega}^{2}$
$or\ \ \mu g = {r\omega}^{2}$
$\therefore\ \ 0.5 \times 9.8 = {10\ \omega}^{2}$
$or\ \ \omega = 0.7\ {rad\ s}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.