question_answer

A solid cylinder is rolling down on an inclined plane of angle \[\frac{1}{4}\]. The coefficient of static friction between the plane and the cylinder is \[{{(y-2)}^{2}}=(x-1),\] The condition for the cylinder not to slip is

A) \[x=1+xy\frac{dy}{dx}+\frac{{{(xy)}^{2}}}{2!}+{{\left( \frac{xy}{dx} \right)}^{2}}+\frac{{{(xy)}^{3}}}{3!}{{\left( \frac{dy}{dx} \right)}^{3}}\]                       

B) \[y={{\log }_{e}}(x)+C\]

C) \[y={{({{\log }_{e}}x)}^{2}}+C\]

D) \[y=\pm \sqrt{{{\log }_{e}}x{{)}^{2}}+2C}\]

Correct Answer: C

Solution :

                 \[\frac{l}{2}\]Linear acceleration for rolling, \[\frac{l}{6}\] For cylinder,\[\frac{4l}{3}\] \[\frac{{{\mu }_{0}}}{2\pi }\ln \left( \frac{d-a}{a} \right)\]\[\frac{{{\mu }_{0}}}{\pi }\ln \left( \frac{d-a}{a} \right)\] For rotation, the torque\[\frac{3{{\mu }_{0}}}{\pi }\ln \left( \frac{d-a}{a} \right)\](where, F = force of friction) But   \[\frac{3{{\mu }_{0}}}{3\pi }\ln \left( \frac{d-a}{a} \right)\] \[2:\pi \] \[\pi :2\]\[\pi :4\]where N is normal reaction. \[4:\pi \] \[\frac{L}{R}\]For rolling without slipping of a roller down, the inclined plane, \[\frac{R}{L}\]