Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    A body rolls without slipping. The radius of gyration of the body about 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)  \[\frac{{{K}^{2}}}{{{R}^{2}}}\]

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

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

    D)  \[{{K}^{2}}+{{R}^{2}}\]

    Correct Answer: A

    Solution :

     \[\frac{Rotational\text{ }KE}{Translational\text{ }KE}=\frac{\frac{1}{2}I{{\omega }^{2}}}{\frac{1}{2}m{{v}^{2}}}\] \[=\frac{m{{K}^{2}}\frac{{{v}^{2}}}{{{R}^{2}}}}{m{{v}^{2}}}=\frac{{{K}^{2}}}{{{R}^{2}}}\]


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