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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    A uniform rod of mass m and length\[L\]is rotated about an axis passing through the point \[P\] as shown in figure. The magnitude of angular momentum of the rod about the rotational axis\[yy'\] passing through the point \[P\]is

    A) \[\frac{M{{L}^{2}}}{18}\]                                             

    B) \[\frac{M{{(x+y)}^{2}}}{9}\]

    C) \[\frac{M({{L}^{2}}+{{x}^{2}})}{9}\]                        

    D)  \[\frac{M({{L}^{2}}+2{{y}^{2}})}{18}\]

    Correct Answer: B

    Solution :

    We have\[\frac{x}{y}=\frac{2}{1}\]or\[\frac{x+y}{y}=\frac{3}{1}\] or            \[y=\frac{x+y}{3}=\frac{L}{3}\] \[\Rightarrow \]               \[x=2y=\frac{2L}{3}\] Now,     \[{{I}_{P}}={{I}_{com}}+M{{(d)}^{2}}\]                 \[=\frac{M{{(L)}^{2}}}{12}+M{{\left( \frac{L}{2}-\frac{L}{3} \right)}^{2}}\]                 \[=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{36}\]                 \[=\frac{4M{{L}^{2}}}{36}=\frac{M{{L}^{2}}}{9}=\frac{M{{(x+y)}^{2}}}{9}\]


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