JEE Main & Advanced Physics Rotational Motion Question Bank Self Evaluation Test - System of Particles Rotational Motion

  • question_answer
    A child with mass m is standing at the edge of a playground merry-go-round (A large uniform disc which rotates in horizontal plane about a fixed vertical axis in parks) with moment of inertia Vs I, radius R, and initial angular velocity w as shown in the figure. The child jumps off the edge of the merry-go-round with a velocity v with respect to the ground in direction tangent to periphery of the disc as shown. The new angular veloity of the merry-go-round is :

    A) \[\sqrt{\frac{I{{\omega }^{2}}-m{{v}^{2}}}{I}}\]

    B) \[\sqrt{\frac{(I+m{{R}^{2}}){{\omega }^{2}}-m{{v}^{2}}}{I}}\]

    C) \[\frac{I\omega -mvR}{I}\]

    D) \[\frac{(I\omega -m{{R}^{2}})\omega -mvR}{I}\]

    Correct Answer: D

    Solution :

    [d] Let the angular velocity of disc after child jumps off. be \[\omega '\] \[\therefore \] From conservation of angular momentum \[(I+m{{R}^{2}})\omega =mvR+I\omega \]. \[\therefore \,\,\,\omega '=\frac{(I+m{{R}^{2}})\omega +mvR}{I}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner