Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    The moment of inertia of a disc of mass m and radius\[R\]about an axis, which is tangential to the circumference of the disc and parallel to its diameter is

    A) \[\frac{3}{2}m{{R}^{2}}\]                             

    B) \[\frac{2}{3}m{{R}^{2}}\]

    C) \[\frac{5}{4}m{{R}^{2}}\]             

    D)        \[\frac{4}{5}m{{R}^{2}}\]

    Correct Answer: C

    Solution :

    Use parallel axis theorem, from the theorem of parallel axis, the moment of inertia \[(I)\] of a body about given axis is equal to its moment of inertia \[I\] about its diameter, plus the product of the mass \[M\] of the body and square of perpendicular distance between the two axes. That is \[I={{I}_{d}}+M{{R}^{2}}=\frac{1}{4}M{{R}^{2}}+M{{R}^{2}}=\frac{5}{4}M{{R}^{2}}\]


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