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BHU PMT BHU PMT (Screening) Solved Paper-2009

  • question_answer
    A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity\[\omega \]. Two objects, each of mass m, are connected gently to the ring. The ring now rotates with an angular velocity

    A)  \[\frac{\omega M}{M+m}\]

    B)  \[\frac{\omega (M-2M)}{(M+2m)}\]

    C)  \[\frac{\omega (M+2M)}{M}\]

    D)  \[\frac{\omega M}{M+2m}\]

    Correct Answer: D

    Solution :

                     Initial angular momentum of ring \[L=I\omega =M{{R}^{2}}\omega \] Final angular momentum of ring and two particle system\[L=(M{{R}^{2}}+2m{{R}^{2}})\omega '\] As there is no torque on the system, therefore angular momentum remains constant \[M{{R}^{2}}\omega =(M{{R}^{2}}+2m{{R}^{2}})\omega '\] \[\Rightarrow \]               \[\omega '=\frac{m\omega }{M+2m}\]


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