RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

  • question_answer
    The moment of inertia of a body about a given axis is \[2.4\,\,kg-{{m}^{2}}\]. To produce a rotational kinetic energy of 750 J, an angular acceleration of \[5\,rad/{{s}^{2}}\] must be applied about that axis for

    A)  6 s                                         

    B)  5 s

    C)  4 s                         

    D)         3 s

    Correct Answer: B

    Solution :

     The rotational energy                 \[\frac{1}{2}I{{\omega }^{2}}=750\,\,J\] \[\Rightarrow \]               \[{{\omega }^{2}}=\frac{750\times 2}{2.4}=625\] \[\Rightarrow \]               \[\omega =25\,\,rad/s\]                 \[\alpha =\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}\] \[\Rightarrow \]               \[5=\frac{25-0}{t}\] \[\Rightarrow \]               \[t=5s\]


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