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KVPY Sample Paper KVPY Stream-SX Model Paper-2

  • question_answer
    A dumbbell consists of two masses connected by a rigid rod of negligible mass and length d. A physics student takes the dumbbell and rotates it about its center of mass with an angular velocity\[\omega \], giving it an angular momentum\[{{L}_{1}}\]. The student then takes a second dumbbell, with masses 2m and length 2d, and rotates them with the same angular velocity\[\omega \]. What is the angular momentum \[{{L}_{2}}\] of this second dumbbell?

    A) \[2{{L}_{1}}\]                         

    B) \[4{{L}_{1}}\]

    C) \[6{{L}_{1}}\]                         

    D) \[8{{L}_{1}}\]

    Correct Answer: D

    Solution :

    [D]
    Although the problem asks us to compare angular momenta of the two dumbbells, \[L=I\omega ,\] and \[\omega \]is identical for both dumbbells. Therefore, this is really a problem asking us to compare the moments of inertia I
    \[{{I}_{1}}=\sum{{{r}^{2}}m}\]
    \[{{I}_{1}}={{\left( \frac{1}{2}d \right)}^{2}}m+{{\left( \frac{1}{2}d \right)}^{2}}m=\frac{{{d}^{2}}m}{2}\]
    Now let?s calculate \[{{I}_{2}}\]is 8 times \[{{I}_{1}}\]
    \[{{I}_{2}}=\sum{{{r}^{2}}m}\]
    \[{{I}_{2}}={{(d)}^{2}}2m+{{(d)}^{2}}2m\]
    \[{{I}_{2}}=2{{d}^{2}}2m=4{{d}^{2}}m\]
    \[{{I}_{2}}=8\left( \frac{{{d}^{2}}m}{2} \right)=8{{I}_{1}}\]
    Because the moment of inertia \[{{I}_{2}}\] is 8 times \[{{I}_{1}},\]\[{{L}_{2}}\]the angular momentum is\[8{{L}_{1}}\] as well.


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