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BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    The moment of inertia of a body about a given axis is \[1.2\,\,kg\,{{m}^{2}}\]. Initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of \[25\,rad/{{s}^{2}}\] must be applied about that axis for a duration of

    A)  4 s                                         

    B)  2 s

    C)  8 s                         

    D)         10 s

    Correct Answer: B

    Solution :

    KE of rotation = 1500 \[\frac{1}{2}I{{\omega }^{2}}=1500\] \[\frac{1}{2}\times 1.2{{\omega }^{2}}=1500\] \[{{\omega }^{2}}=\frac{1500\times 2}{1.2}=2500\] \[\omega =\sqrt{2500}\] \[=50\,rad/s\] From equation of rotational motion \[\omega ={{\omega }_{0}}+\alpha t\] \[50=0+25\times t\]                 \[\therefore \]  \[t=\frac{50}{25}=2s\]


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