CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity cd. Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

    A)  \[4\omega \]                   

    B)         \[\frac{M}{4m}\omega \]

    C)  \[\left( \frac{M+4m}{M} \right)\omega \]

    D)        \[\left( \frac{M}{M-4m} \right)\omega \]

    E)  \[\left( \frac{M}{M+4m} \right)\omega \]

    Correct Answer: E

    Solution :

    According to conservation of angular momentum, \[I\omega =constant\] ie, we can write                 \[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\] or            \[M{{R}^{2}}\omega =(M+4m){{R}^{2}}{{\omega }_{2}}\] or            \[{{\omega }_{2}}=\left( \frac{M}{M+4m} \right)\omega \]


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