question_answer

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. If the coefficient of friction between the rod and the bead is, and gravity is neglected, then the time after, which the bead starts slipping is

A) \[\sqrt{\frac{\mu }{\alpha }}\]                                   

B) \[\frac{\mu }{\alpha }\]

C) \[\frac{1}{\sqrt{\mu \alpha }}\]                     

D)  None of these  

Correct Answer: A

Solution :

As shown in the figure, the bead will start slipping, when centripetal force exceeds the force of friction. i.e., \[m{{\omega }^{2}}L=\mu R=\mu m({{a}_{t}})\] Where \[{{a}_{t}}=\]tangential acceleration\[=L\,\alpha \] Also, \[\omega =\alpha t\]\[\therefore \,\,m{{(\alpha t)}^{2}}L=\mu m(L\alpha )\] \[t=\sqrt{\frac{\mu }{\alpha }}\] Hence, the correction option is [a],