Q. A long horizontal rod has a bead, which can slide along its length and initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular acceleration $\alpha$. If the coefficient of friction, between the rod and the bead is $\mu$ and gravity is neglected, then the time after which the bead starts slipping is
A) $\sqrt{\mu/\alpha}$
B) $\mu/\sqrt{\alpha}$
C) $1/\sqrt{\mu\alpha}$
D) Infinitesimal
Explanation
(a) Tangential acceleration $a = L\alpha$
$\therefore\ \ \ $Normal relation $N = Ma = ML\alpha$
$\therefore$ Frictional force $F = mN = \mu ML\alpha$
For no sliding along the length frictional force $\geq$ centripetal force
$ie,\mu ML\alpha \geq ML\omega^{2}\ $
As $\omega = \omega_{0} + \alpha t = \alpha t$
$\therefore\ \mu ML\alpha \geq ML(\alpha t)^{2} \Rightarrow t = \sqrt{\frac{\mu}{\alpha}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.