Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    Four thin rods of same mass M and same length I, form a square as shown in figure. Moment of inertia of this system about an axis through centre O and perpendicular to its plane is

    A) \[\frac{{{x}^{2}}}{4}+\frac{4{{y}^{2}}}{7}=1\]                      

    B) \[\left( \frac{1}{2},2 \right)\]

    C) \[\sqrt{5}\]                                        

    D) \[\sqrt{12}\]

    Correct Answer: A

    Solution :

    Moment of inertia of rod AB about point\[2as{{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)}^{1/2}}\]Ml    of    rod    AB    about    point\[2a\frac{{{R}^{2}}}{s}\] [from the theorem of parallel axis] and the system consists of 4 rods of similar type so by the symmetry \[{{T}_{1}}=\]


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