Manipal Medical Manipal Medical Solved Paper-2007

  • question_answer
    A round disc of moment of inertia\[{{I}_{2}}\]about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia\[{{I}_{1}}\]rotating with an angular velocity co about the same axis. The final angular velocity of the combination of discs is

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

    B)  \[\omega \]

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

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

    Correct Answer: C

    Solution :

     Key Idea: Wien no external torque acts on a system of particles, then the total angular momentum of the system remains always a constant.   The angular momentum of a disc of moment of inertia\[{{I}_{1}}\]and rotating about its axis with angular velocity\[\omega \]is \[{{L}_{1}}={{I}_{1}}\omega \] When a round disc of moment of inertia\[{{I}_{2}}\]is placed on first disc, then angular momentum of the combination is \[{{L}_{2}}=({{I}_{1}}+{{I}_{2}})\omega \] In the absence of any external torque, angular momentum remains conserved i.e., \[{{L}_{1}}={{L}_{2}}\] \[{{I}_{1}}\omega =({{I}_{1}}+{{I}_{2}})\omega \] \[\Rightarrow \] \[\omega =\frac{{{I}_{1}}\omega }{{{I}_{1}}+{{I}_{2}}}\]


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