BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    A rigid body of mass m rotates with angular velocity \[\omega \]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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