Question

The radius of gyration of rod of length ${}^{\prime }{L}^{\prime }$ and mass ${}^{\prime }{M}^{\prime }$ 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 $\frac{L}{3}$

Moment of inertia of a rod about its centre of mass is

$I_{cm}=\frac{M{L}^2}{12}$

To calculate the moment of inertia about a point at a distance of $\frac{L}{3}$ from the end we need to find the value of $x$ which is the distance between the COM and this point.

As the COM for a rod lies at the centre of the rod at $\frac{L}{2}$ we have,


We have from parallel axis theorem,

$I=I_{\text{cm}}+M{x}^2$

$I=\frac{M{L}^2}{12}+M{\left(\frac{L}{6}\right)}^2$

$I=M{L}^2[\frac{1}{12}+\frac{1}{36}]$

$I=M{L}^2\left[\frac{4}{36}\right]$

If $K$ is the radius of gyration,

$M{K}^2=I=M{L}^2\left[\frac{4}{36}\right]$

$K^2=L^2\left[\frac{4}{36}\right]$

$\Rightarrow K=L\times \frac{2}{6}=\frac{L}{3}$