RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy and translational kinetic energy is:

    A)  \[\frac{2}{9}\]                                  

    B)  \[\frac{2}{5}\]

    C)  \[\frac{2}{7}\]                  

    D)         \[\frac{7}{2}\]

    Correct Answer: B

    Solution :

    Moment of inertia of sphere                 \[I=\frac{2}{5}M{{R}^{2}}\]                                          ... (1) and for pure rolling \[\upsilon -R\,\omega \]                                       ... (2) \[\frac{Rotationl\,KE}{Translationl\,KE}=\frac{\frac{1}{2}I{{\omega }^{2}}}{\frac{1}{2}m{{\upsilon }^{2}}}\] from (1) and (2) \[=\frac{\frac{1}{2}\times \frac{2}{5}M{{R}^{2}}{{\omega }^{2}}}{\frac{1}{2}M{{R}^{2}}{{\omega }^{2}}}=\frac{2}{5}\]


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