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 \[K=\sqrt{\frac{I}{M}}=\sqrt{\frac{{{L}^{2}}}{9}}=\frac{L}{3}\]