Question

A thin rod of length $4l$, mass $4m$ is bent at the points shown in the figure. What is the moment of inertia of the rod about the axis passing point $\text{O}$ and 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 $\frac{10M{l}^2}{3}$


Each rod is of mass $M$ and length $l$. As we know, moment of inertia of a rod of mass $M$ and length $l$ about its center is
$I=\frac{M{l}^2}{12}$
For rod $\text{OB}$ and $\text{OC}$, moment of inertia about point $\text{O}$, by parallel axis theorem.
$I_{OB}=I_{OC}=\frac{M{l}^2}{12}+M{\left(\frac{l}{2}\right)}^2=\frac{M{l}^2}{3}$

Distance between point $\text{O}$ and center of rod $\text{AB}$, or point $\text{O}$ and center of rod $\text{CD}$.
$=\sqrt{l^2+{\left(\frac{l}{2}\right)}^2}=\sqrt{\frac{5l^2}{4}}$
Hence, Moment of inertia of rod $\text{AB}$ and $\text{CD}$ about point $\text{O}$ by parallel axis theorem:
$I_{AB}=I_{CD}=\frac{M{l}^2}{12}+M\left(\frac{5l^2}{4}\right)$
$I_{AB}=I_{CD}=\frac{4M{l}^2}{3}$

Therefore, MOI of bent rod of length $4l$ and mass $4M$ about point $\text{O}$ will be,
$I=I_{AB}+I_{OB}+I_{OC}+I_{CD}$
$I=2(I_{AB}+I_{OB})$
$I=2(\frac{M{l}^2}{3}+\frac{4M{l}^2}{3})$
$I=\frac{10M{l}^2}{3}$