question_answer

A cylinder with radius R spins about its horizontal axis with angular speed \[\omega \]. There is a small block lying on the inner surface of the cylinder. The coefficient of friction between the block and the cylinder is \[\mu \]. Find the value of co for which the block does not slip, i.e., stays at rest with respect to the cylinder.

A) \[\sqrt{\frac{g\sqrt{1+{{\mu }^{2}}}}{R}}\]     

B)        \[\sqrt{\frac{g}{R\sqrt{1+{{\mu }^{2}}}}}\]

C) \[\sqrt{\frac{g\sqrt{1+{{\mu }^{2}}}}{R\mu }}\]          

D)        \[\sqrt{\frac{g\mu }{R\sqrt{1+{{\mu }^{2}}}}}\]

Correct Answer: C

Solution :

[c] Consider the block at any position \[\theta \] shown in the figure. [In reference frame of cylinder] \[N=m{{\omega }^{2}}R-mg\,\sin \theta \]          .....(i) \[f=mg\,\cos \theta \]                   .....(ii) \[\because \,\,\,\,\,f\le \mu N\] \[\therefore \,\,\,\,\,\,\,\,mg\cos \theta \le \mu m{{\omega }^{2}}R-\mu mg\sin \theta \] \[\therefore \,\,\,\,\,\,\,g[\cos \theta +\mu \sin \theta ]\le \mu m{{\omega }^{2}}R\] [for all value of \['\theta '\]] \[\because \] Maximum value of \[\cos \theta +\mu \sin \theta \] will be                                                 \[\sqrt{1+{{\mu }^{2}}}\]\[\therefore \,\,\,\,\,\,\,g\sqrt{1+{{\mu }^{2}}}\le \mu {{\omega }^{2}}R\] \[\therefore \,\,\,\,\,\,\,\omega \ge \sqrt{\frac{g\sqrt{1+{{\mu }^{2}}}}{R\mu }}\]