Question

A thin rod of length $4l$ and $4m$ is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passes through point $O$ and perpendicular to the plane of 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 $\frac{10M{l}^2}{3}$

Total moment of inertia
$=I_1+I_2+I_3+I_4=2I_1+2I_2$
$=2(I_1+I_2)[I_3+I_1,I_1=I_4]$
Now, $I_2=I_3=\frac{M{I}^2}{3}$
Using parallel axes theorem, we have
$I=I_{CM}+M{x}^2$ and $x=\sqrt{l^2+\frac{l^2}{4}}$
$I_1=I_4=\frac{M{l}^2}{12}+M{\left[\sqrt{l^2+{\left(\frac{1}{2}\right)}^2}\right]}^2$
Putting all values we get
Moment of inertia, $I=10\left(\frac{M{I}^2}{3}\right)$.