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Physics Question on Moment Of Inertia

The radius of gyration of a rod of length L and mass M about an axis perpendicular to its length and passing through a point at a distance L/3 from one of its ends is

A

76L\frac{\sqrt{7}}{6}L

B

L29\frac{{{L}^{2}}}{9}

C

L3\frac{L}{3}

D

52L\frac{\sqrt{5}}{2}L

Answer

L3\frac{L}{3}

Explanation

Solution

Moment of inertia of the rod about a perpendicular axis PQ passing through centre of mass C ICM=ML212{{I}_{CM}}=\frac{M{{L}^{2}}}{12} Let N be the point which divides the length of rod AB in ratio 1:3. This point will be at a distance L6\frac{L}{6} from C. Thus, the moment of inertia I about an axis parallel to PQ and passing through the point N. I=ICM+M(L6)2I={{I}_{CM}}+M{{\left( \frac{L}{6} \right)}^{2}} =ML212+ML236=ML29=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{36}=\frac{M{{L}^{2}}}{9} If K be the radius of gyration, then K=IM=L29=L3K=\sqrt{\frac{I}{M}}=\sqrt{\frac{{{L}^{2}}}{9}}=\frac{L}{3}