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 \[\sqrt{1+{{\mu }^{2}}}\]\[\therefore \,\,\,\,\,\,\,g\sqrt{1+{{\mu }^{2}}}\le \mu {{\omega }^{2}}R\] \[\therefore \,\,\,\,\,\,\,\omega \ge \sqrt{\frac{g\sqrt{1+{{\mu }^{2}}}}{R\mu }}\]You need to login to perform this action.
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