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


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