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 is set in angular motion about A with a constant angular acceleration $\alpha$. If the coefficient of friction between the rod and bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is

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

Answer 1

Answer: (A)

$\sqrt{\mu/\alpha}$

Answer 2

Tangential force $(F_t)$ of the bead will be given by the normal reaction (N), while centripetal force $(F_c)$ is provided by friction $(f_r)$. 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
\hspace15mm (f_r)_{max}=\mu N=\mu m \alpha L\hspace15mm ...(i)
Angular velocity at time t is $\omega=\alpha t$
$\therefore$Centripetal force at time t will be
\hspace15mm F_c=mL \omega^2=mL\alpha^2 t^2\hspace15mm ...(ii)
Equating Eqs. (i) and (ii), we get
\hspace25mm t=\sqrt{\frac{\mu}{\alpha}}
For $t > \sqrt{\frac{\mu}{\alpha}}, F > (f_r)_{max} i.e.$ the bead starts sliding.
In the figure, $F_t$ is perpendicular to the paper inwards.