JEE Main & Advanced Physics Rotational Motion Question Bank Self Evaluation Test - System of Particles Rotational Motion

  • question_answer
    A solid cylinder of mass m & radius R rolls down inclined plane without slipping. The speed of C.M. when it reaches the bottom is   

    A) \[\sqrt{2gh}\]

    B) \[\sqrt{4gh/3}\]

    C) \[\sqrt{3/4gh}\]

    D) \[\sqrt{4gh}\]

    Correct Answer: B

    Solution :

    [b] By energy conservation \[{{(K.E)}_{i}}+{{(P.E)}_{i}}={{(K.E)}_{f}}+{{(P.E)}_{f}}\] \[{{(K.E)}_{i}}=0,{{(P.E)}_{i}}=mgh,{{(P.E)}_{f}}=0\] \[{{(K.E)}_{f}}={\scriptstyle{}^{1}/{}_{2}}I{{\omega }^{2}}+{\scriptstyle{}^{1}/{}_{2}}m{{v}^{2}}_{cm}\] Where I (moment of inertia) \[={\scriptstyle{}^{1}/{}_{2}}m{{R}^{2}}\] (for solid cylinder) so \[mgh={\scriptstyle{}^{1}/{}_{2}}({\scriptstyle{}^{1}/{}_{2}}m{{R}^{2}})\left( \frac{{{v}^{2}}cm}{{{R}^{2}}} \right)+{\scriptstyle{}^{1}/{}_{2}}m{{v}^{2}}_{cm}\] \[\Rightarrow \,\,\,\,{{v}_{cm}}=\sqrt{4gh/3}\]


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