Question

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 $

Answer 1

The Correct Option is C

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} $