BCECE Medical BCECE Medical Solved Papers-2012

  • question_answer
    A rigid body of mass m rotates with angular velocity g) about an axis at a distance a from the centre of mass C. The radius of gyration about C is K. Then, kinetic energy of rotation of the body about new parallel axis is

    A)  \[\frac{1}{2}m{{K}^{2}}{{\omega }^{2}}\]

    B)  \[\frac{1}{2}m{{a}^{2}}{{\omega }^{2}}\]

    C)  \[\frac{1}{2}m(a+{{K}^{2}}){{\omega }^{2}}\]

    D)  \[\frac{1}{2}m({{a}^{2}}+{{K}^{2}}){{\omega }^{2}}\]

    Correct Answer: D

    Solution :

    MI of body about centre of mass                 \[{{I}_{cm}}=m{{k}^{2}}\] MI of body about new parallel axis                 \[{{I}_{new}}{{I}_{cm}}+m{{a}^{2}}\]                 \[=m{{k}^{2}}+m{{a}^{2}}\]                 \[=m\,({{K}^{2}}+{{a}^{2}})\] \[\therefore \] Kinetic energy, \[K=\frac{1}{2}{{I}_{new}}{{\omega }^{2}}\] \[=\frac{1}{2}\,m({{K}^{2}}+{{a}^{2}}){{\omega }^{2}}\]


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