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MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • question_answer
    Two discs of moment of inertia \[{{I}_{1}}\] and \[{{I}_{2}}\]and angular speeds \[{{\omega }_{1}}\] and \[{{\omega }_{2}}\] are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate combindly along the same axis the rotational KE of system will be

    A) \[\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{2({{I}_{1}}+{{I}_{2}})}\]

    B) \[\frac{({{I}_{1}}+{{I}_{2}})\,{{({{\omega }_{1}}+{{\omega }_{2}})}^{2}}}{2}\]

    C) \[\frac{({{I}_{1}}\omega {{I}_{2}}+\,{{({{I}_{2}}{{\omega }_{2}})}^{2}}}{2({{I}_{1}}+{{I}_{2}})}\]  

    D) None of these  

    Correct Answer: C

    Solution :

    Conservation of angular momentum \[{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}=({{I}_{1}}+{{I}_{2}})\omega \] Angular velocity of system\[\omega =\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{{{I}_{1}}+{{I}_{2}}}\] \[\therefore \]Rotational kinetic energy\[=\frac{1}{2}({{I}_{1}}+{{I}_{2}}){{\omega }^{2}}\] \[=\frac{1}{2}({{I}_{1}}+{{I}_{2}}){{\left( \frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{{{I}_{1}}+{{I}_{2}}} \right)}^{2}}\] \[=\frac{{{({{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}})}^{2}}}{2({{I}_{1}}+{{I}_{2}})}\]


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