Question

The radius of gyration of a uniform rod of length $l$ about an axis passing through a point $\frac{l}{4}$ away from the center of the rod, and perpendicular to it, is

• A. $\sqrt{\frac{7}{48}}l$
• B. $\sqrt{\frac{3}{8}}l$
• C. $\frac{1}{4}l$
• D. $\frac{1}{8}l$

Answer 1

The correct option is A

$\sqrt{\frac{7}{48}}l$

Step 1: Given data

Lenth of rod=$l$

The point through which radius passages=$\frac{l}{4}$

Step 2: Formula used

According to the parallel axis theorem,

$I=I_c+M{d}^2$

$whereI=momentofinertiaofrod,I_c=Momentofinertiaaboutthecenter,M=massofrod,d=dis\tan cebetweentwoaxes$

But we know that $I=mass\left(m\right)\times radius{\left(R\right)}^2$. So,

$$\begin{gathered} m{R}^2=I_c+M{d}^2 \\ R=\sqrt{\frac{I_c+M{d}^2}{m}}----\left(1\right) \end{gathered}$$

Step 3: Compute the moment of inertia about the center

Suppose the mass of the rod is $m$ so the moment of inertia about the center is,

$I_c=\frac{1}{12}m{l}^2$

Step 3: Compute the radius of gyration

Consider the following figure:

Substitute the known values in the equation $\left(1\right)$ to compute radius,

$$\begin{gathered} R=\sqrt{\frac{\left({\displaystyle \frac{1}{12}}m{l}^2\right)+\left({\displaystyle \frac{m{l}^2}{16}}\right)}{m}} \\ R=\sqrt{\frac{1}{12}+\frac{1}{16}}l \\ R=\sqrt{\frac{4+3}{48}}l \\ R=\sqrt{\frac{7}{48}l} \end{gathered}$$

Thus, the radius of gyration of a uniform rod is $\sqrt{\frac{7}{48}}l$.

Hence, option A is the correct answer.