RAJASTHAN PMT Rajasthan - PMT Solved Paper-2007

  • question_answer
    A solid sphere is rotating about a diameter at an angular velocity \[\omega .\] If it coots so that its radius reduces to \[\frac{1}{n}\] of its original value, its angular velocity becomes

    A) \[\frac{\omega }{n}\]                                    

    B)  \[\frac{\omega }{{{n}^{2}}}\]

    C)  \[n\omega \]                   

    D)         \[{{n}^{2}}\omega \]

    Correct Answer: D

    Solution :

    From law of conservation of angular momentum, if no external torque is acting upon a body rotating about an axis, then the angular momentum of the body remains constant that is                 \[J=I\omega \] Also,\[I=\frac{2}{5}M{{R}^{2}}\]for a solid sphere. Given,\[{{R}_{1}}=R,\,\,{{R}_{2}}=\frac{R}{n}\] \[\therefore \]  \[\frac{2}{5}M{{R}^{2}}{{\omega }_{1}}=\frac{2}{5}M{{\left( \frac{R}{n} \right)}^{2}}\times {{\omega }_{2}}\] \[\Rightarrow \]               \[{{\omega }_{2}}={{n}^{2}}{{\omega }_{1}}\]                      \[={{m}^{2}}\omega \]


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