question_answer

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)  \[\frac{\sqrt{7}}{6}L\]                  

B)  \[\frac{{{L}^{2}}}{9}\]

C)  \[\frac{L}{3}\]                                  

D)  \[\frac{\sqrt{5}}{2}L\]

Correct Answer: C

Solution :

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