JEE Main & Advanced Physics NLM, Friction, Circular Motion Coefficient of Friction Between Body and Edge

Coefficient of Friction Between Body and Edge A body slides on a smooth wedge of angle \[\theta \] and its time of descent is t.             If the same wedge made rough then time taken by it to come down becomes n times more (i.e. nt) The length of path in both the cases are same.

For smooth wedge             \[S=u\,t+\frac{1}{2}a{{t}^{2}}\]             \[S=\frac{1}{2}(g\,\sin \theta )\,{{t}^{2}}\]   .....(i) \[[\text{As}\,u=0\,\text{and }\]\[a=g\sin \theta ]\]

For rough wedge \[S=u\,t+\frac{1}{2}a{{t}^{2}}\] \[S=\frac{1}{2}g\,(\sin \theta -\mu \cos \theta )\,{{(nt)}^{2}}\]           .....(ii)                                     \[[\text{As}\,u=0\,\text{and}\,a=g\,(\sin \theta -\mu \,\cos \theta )]\]

From equation (i) and (ii) \[\frac{1}{2}(g\,\sin \theta )\,{{t}^{2}}=\] \[\frac{1}{2}g\,(\sin \theta -\mu \cos \theta )\,{{(nt)}^{2}}\]                                     \[\Rightarrow \,\,\,\sin \theta =\,(\sin \theta -\mu \cos \theta )\,{{n}^{2}}\]                                      \[\Rightarrow \,\,\mu =\tan \theta \,\left[ 1-\frac{1}{{{n}^{2}}} \right]\] Problem 26. A body takes just twice the time as long to slide down a plane inclined at \[{{30}^{o}}\] to the horizontal as if the plane were frictionless.  The coefficient of friction between the body and the plane is [JIPMER 1999] (a) \[\frac{\sqrt{3}}{4}\]                         (b) \[\sqrt{3}\]                (c) \[\frac{4}{3}\]                       (d) \[\frac{3}{4}\] Solution: (a) \[\mu =\tan \theta \,\left( 1-\frac{1}{{{n}^{2}}} \right)=\tan 30\,\left( 1-\frac{1}{{{2}^{2}}} \right)\]\[=\frac{\sqrt{3}}{4}\].