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Manipal Medical Manipal Medical Solved Paper-2014

  • question_answer
    A body rolls without slipping. The radius of gyration of the body above an axis passing through its centre of mass is K. The radius of the body is R The ratio of rotational kinetic energy to translational kinetic energy is

    A) \[\frac{{{v}^{2}}}{{{R}^{2}}}\]                                        

    B) \[\frac{{{R}^{2}}}{{{v}^{2}}+{{R}^{2}}}\]

    C) \[\frac{{{v}^{2}}}{{{v}^{2}}+{{R}^{2}}}\]                    

    D) \[{{v}^{2}}+{{R}^{2}}\]

    Correct Answer: A

    Solution :

    \[\frac{Rotational\,\,KE}{Translational\,\,KE}=\frac{\frac{1}{2}l{{\omega }^{2}}}{\frac{1}{2}m{{v}^{2}}}\]                                     \[=\frac{m{{v}^{2}}\frac{{{v}^{2}}}{{{R}^{2}}}}{m{{v}^{2}}}=\frac{{{v}^{2}}}{{{R}^{2}}}\]


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