JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    A circular ring of mass m and radius r is rolling on a smooth horizontal surface with speed \[v\]. Its kinetic energy is:

    A) \[\frac{1}{8}m{{v}^{2}}\]                             

    B) \[\frac{1}{4}m{{v}^{2}}\]

    C) \[\frac{1}{3}m{{v}^{2}}\]                             

    D) \[m{{v}^{2}}\]

    Correct Answer: B

    Solution :

    Kinetic energy of rotation of a body having moment of inertia \[I\]I and angular velocity \[\omega \] is given by                 \[K=\frac{1}{2}I{{\omega }^{2}}\] For a circular ring of radius \[r\], and mass \[m\] moment of inertia \[(I)\] about its diameter is given by                 \[I=\frac{M{{r}^{2}}}{2}\] \[\therefore \]  \[K=\frac{1}{2}\left( \frac{M{{r}^{2}}}{2} \right){{\omega }^{2}}\] Also,\[v=r\omega ,\] therefore\[\omega =\frac{v}{r}\] \[\therefore \]  \[K=\frac{1}{2}\left( \frac{m{{r}^{2}}}{2} \right){{\left( \frac{v}{r} \right)}^{2}}=\frac{1}{4}m{{v}^{2}}\]


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