question_answer

Two identical smooth surfaced solid cylinders of radius r are placed touching along their lengths on a horizontal surface. A third cylinder of same material but twice the radius of that of the cylinders is placed lengthwise on them so that the system remains at rest. If all three cylinders have the same length, then minimum value of the coefficient of friction between smaller cylinders and the surface is:          

A) \[\frac{1}{\sqrt{2}}\]  

B) \[\frac{1}{3}\]

C) \[\frac{1}{3\sqrt{2}}\]  

D) None of these

Correct Answer: C

Solution :

[c] \[\sin  \theta = \frac{r}{3r}=\frac{1}{3},cos\,\,\theta =\sqrt{8/9}\] If mass of smaller cylinder is m, then mass of bigger one will be 4m. For the equilibrium of upper cylinder, \[2\,{{N}_{1}} sin \theta  = 4 mg\] \[\therefore {{N}_{1}} = \frac{2 mg}{sin \theta }\] Now for the equilibrium of lower cylinder,             \[{{\operatorname{N}}_{2}} =~{{N}_{1}}\,\,sin\,\theta  + mg~~~...\left( ii \right)\] and      \[f={{N}_{1}} \,\,cos\,\,\theta \] or         \[\mu {{N}_{2}}={{N}_{1}}\cos \,\,\theta \]          ?.(iii) After solving above equations, we get             \[\mu =\frac{1}{3\sqrt{2}}\].