Q. The moment of inertia of a rod (length I mass m) about an axis perpendicular to the length of the rod and passing through a J point equidistant from its mid-point and one end, is
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Solution:
Moment of inertia of thin rod $ I=\frac{m{{l}^{2}}}{12} $
Distance of axis CD from AB $ =\frac{L}{4} $ From theorem of parallel axis, we have $ {{I}_{CD}}={{I}_{AB}}+M{{\left( \frac{l}{4} \right)}^{2}} $ $ =\frac{m{{l}^{2}}}{12}+\frac{m{{l}^{2}}}{16} $
