RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    Mass of a ring is M and its radius is R. It is rotating about its own axis with angular velocity \[\omega \]. Two particles each of mass m sticks to the rim at diameter opposite point. Now angular velocity of ring is:

    A) \[\frac{M\omega }{2m}\]                                            

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

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

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

    Correct Answer: D

    Solution :

    Angular moment of ring will be conserved. \[{{I}_{1}}\omega ={{I}_{2}}\omega \] \[\frac{M{{R}^{2}}}{2}\omega =\left( \frac{M+2m}{2} \right){{R}^{2}}\omega \] \[\omega =\frac{M\omega }{(M+2m)}\]


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