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

Answer: (C)

Moment of inertia of the rod about a perpendicular axis $P Q$ passing through centre of mass $C$

$I_{ CM }=\frac{M L^{2}}{12}$
Let $N$ be the point which divides the length of rod $A B$ 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 $P Q$ 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
$K=\sqrt{\frac{I^{'}}{M}}=\sqrt{\frac{L^{2}}{9}}=\frac{L}{3}$