JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2006

  • question_answer
        A particle performs uniform circular motion with an angular momentum L. If the frequency of particles motion is doubled and its KE is halved. the angular momentum becomes

    A)  \[\frac{L}{2}\]                                  

    B)  \[2L\]

    C)  \[4L\]                                  

    D)  \[\frac{L}{4}\]

    Correct Answer: D

    Solution :

                    The angular momentum is measure for the amount of torque that has been applied over time to the object. For a particle with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as \[L=I\omega \]                        ....(i) where I is moment of inertia of particle and\[\omega \]the angular velocity. Also,      \[K=\frac{1}{2}I{{\omega }^{2}}\]                                     ...(ii) where K is kinetic energy of rotation. From Eqs. (i) and (ii), we get                 \[L=\frac{2K}{{{\omega }^{2}}}\omega =\frac{2K}{\omega }\]                 \[L=\frac{2(K/2)}{2\omega }=\frac{1}{4}\left( \frac{2K}{\omega } \right)=\frac{L}{4}\] Note: In a closed system angular momentum is constant.


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