Question

A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end $A$ of the rod. The rod starts from rest in angular motion about $A$ with a constant angular acceleration $\alpha$ .If the coefficient of friction between the rod and the bead is $\mu$ and gravity is neglected then what is the time after which the bead starts slipping

• A. $\sqrt{\mu /\alpha }$
• B. $\frac{\mu }{\sqrt{\alpha }}$
• C. $\frac{1}{\sqrt{\mu \alpha }}$
• D. infinitesimal

Answer 1

Answer: (A)

Tangential force $\left(F_t\right)$ of the bead will be given by the normal reaction $(N)$, while centripetal force $\left(F_c\right)$ is provided by friction $\left(f_r\right)$. The bead starts sliding when the centripetal force is just equal to the limiting friction.

Therefore,
$F_t=ma=m\alpha L=N$
$\therefore$ Limiting value of friction
${\left(f_r\right)}_{\max }=\mu N=\mu m\alpha L....$ (i)
Angular velocity at time $t$ is $\omega =\alpha t$
$\therefore$ Centripetal force at time $t$ will be
$F_c=mL{\omega }^2=mL{\alpha }^2{t}^2.....$(ii)
Equating Eqs. (i) and (ii), we get
$t=\sqrt{\frac{\mu }{\alpha }}$
For $t>\sqrt{\frac{\mu }{\alpha }},F_c>{\left(f_r\right)}_{\max }$ i.e. the bead starts sliding.
In the figure, $F_t$ is perpendicular to the paper inwards.