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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    A solid cylinder of mass \[\text{2}0\text{ kg}\] rotates about its axis with angular speed\[\text{100 rad }{{\text{s}}^{-1}}\]. The radius of the cylinder is\[0.\text{25 m}\]. What is the . kinetic energy associated with the rotation of cylinder? What is the magnitude of angular momentum of cylinder about its axis?

    A) \[62.5T-s\]         

    B) \[70.4T-s\]

    C) \[79.6T-s\]                         

    D) \[60.5T-s\]

    Correct Answer: A

    Solution :

    If \[I\] be the moment of inertia of the cylinder about its axis. then \[I=\frac{1}{2}M{{R}^{2}}=\frac{1}{2}\times 20\times {{(0.25)}^{2}}=0.625\text{kg-}{{\text{m}}^{\text{2}}}\] \[\therefore \]KE associated with the rotating cylinder is given by, \[KE=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\times 0.625\times {{(100)}^{2}}=3125J\] Also using the relation, \[L=\sqrt{2I(KE)}\] We have    \[L=\sqrt{2\times 0.625\times 3125}\]\[L=62.5J-s\]                


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