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 a .If the coefficientof friction between the rod and the bead is u and gravity is neglected then what is the time after which the bead starts slipping

Answer 1

Assuming the given motion takes place in Horizontal plane.

Steps:

1) The bead will start sliding when centripetal force is equal to limiting friction.

i. e.

$m{w}^{2}L=f\left(max\right)$

=> $m{w}^{2}L=uN$ ----(1)

2) We know that,

Here. Normal Reaction is provided by Tangential force.

F(t) = mLα =N ---(2)

3 ) Substituting value of N in eq. (1),

Now,

At time t,.

w = 0 + αt

w = αt

Now,

$m{w}^{2}L=umL\alpha$

=>$m(\alpha t{)}^{2}L=umL\alpha$

=>$\alpha t^2=u$

=> $t=\sqrt{\frac{u}{\alpha }}$ .

At time t , bead will start sliding.