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JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    A uniform rod \[AB\] of length \[l\] and mass \[m\] is free to rotate about point\[A\]. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about \[A\] is \[\frac{m{{l}^{2}}}{3}\] the initial angular acceleration of the rod will be

    A) \[\frac{2g}{3l}\]                               

    B) \[mg\frac{l}{2}\]

    C)  \[\frac{3}{2}gl\]                                              

    D)  \[\frac{3g}{2l}\]

    Correct Answer: D

    Solution :

    The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is                                 \[I=\frac{m{{l}^{2}}}{3}\] where \[m\] is mass of rod and \[l\] is length. Torque \[(\tau =I\alpha )\] acting on centre of gravity of rod is given by                 \[\tau =mg\frac{l}{2}\] or            \[I\alpha =mg\frac{l}{2}\] or            \[\frac{m{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\] or            \[\alpha =\frac{3g}{2l}\]


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