Question

A thin uniform rod, of length $4l$ and mass $4m$, is bent at the points as shown in the fig. What is the moment of inertia of the rod about the axis passing through point $O$ & perpendicular to the plane of the paper?

• A. $\frac{m{l}^2}{3}$
• B. $\frac{10m{l}^2}{3}$
• C. $\frac{m{l}^2}{12}$
• D. $\frac{m{l}^2}{24}$

Answer 1

The correct option is B $\frac{10m{l}^2}{3}$


Let us apply perpendicular axis theorem.
$I_z=I_x+I_y$

Now, $I_X=0+\frac{m{l}^2}{3}+m{l}^2+\frac{m{l}^2}{3}$
$=\frac{5}{3}m{l}^2$

From the symmetry of mass distribution and axes of rotation, we can say $I_x=I_y=\frac{5}{3}m{l}^2$
$\Rightarrow I_z=I_x+I_y=\frac{10}{3}m{l}^2$

Hence, $\left(B\right)$ is the correct answer.