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. $\sqrt {\dfrac{\mu }{\alpha }}$
B. $\dfrac{\mu }{{\sqrt \alpha }}$
C. $\dfrac{1}{{\sqrt {\mu \alpha } }}$
D. infinitesimal

Answer 1

The bead given in the question can slide along its length as shown. The bead is moving along the length so it will be acted by different forces. The force of friction due to the surface of the rod. The tangential force on the bead will be found out using the normal reaction. The condition for it to start sliding is the centripetal force to be just equal to limiting friction.

Complete step by step answer:
Given, The friction coefficient between the bead and the rod is $\mu$. The angular acceleration with which the bead is moving is $\alpha$.

Consider the different forces as;
-Tangential force ${F_t}$ which will be given by the Normal reaction which Is (N).
-The bead is moving along the length which actually means It is moving along the circular path. It will have a centripetal force acting towards the center ${F_c}$.
-It is provided by the friction (f).

The bead starts sliding when the limiting friction matches the value of centripetal force as stated earlier. So we can draw the equations as
{F_t} - ma = m\alpha L \\\ and \\\ m\alpha L = N \\\
So the limiting value of friction will be
{f_{\max }} = \mu N \\\ \Rightarrow {f_{\max }} = \mu m\alpha L\,\,\,\,...(i) \\\
The angular velocity can be found out from angular acceleration by multiplying with time
$\omega = \alpha t$
The centripetal force at time t
${F_c} = mL{\omega ^2} \\\ substitute\,\omega \\\ \Rightarrow {F_c} = mL{(\alpha t)^2} \\\$
After squaring the term in bracket
${F_c} = mL{\alpha ^2}{t^2}...(ii)$
From (i) and (ii) we get
$\mu m\alpha L = mL{\alpha ^2}{t^2} \\\ \Rightarrow {t^2} = \dfrac{\mu }{\alpha } \\\$
Taking the square root on the both sides we get
$t = \sqrt {\dfrac{\mu }{\alpha }}$
As we found out the time by equating centripetal force and limiting friction
For
$t > \sqrt {\dfrac{\mu }{\alpha }} ,F > {f_{\max }}$ , the bead will start sliding.

Hence, option A is correct.

Note: If we are going on a curved path in the car.The static frictional force can point towards the centre of the circle but the kinetic frictional force will be directed opposite to the direction of motion. It will be difficult to regain control of the car due to this kinetic friction if we travel along the curve.