Question

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 motion about A with constant 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)/(sqrtmualpha)`
D. `infinitesimal

Answer 1

Correct Answer - A
After time `t,omega=0+alphat=alphat`
`N=ma_(t)=mLalpha`
Friction provides centripetal force
`f=muN=mumLalpha=momega^(2)L`
`mualpha=(alphat)^(2)impliest=sqrt((mu)/(alpha))