Question

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

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

Answer 1

Answer: (B)

Let's suppose points are $A,B,C,D$
Total moment of inertia about an axis passing through point $O$ and perpendicular to the plane of paper is
$I_{net}=I_{AB}+I_{BO}+I_{OC}+I_{CD}\dots (i)$
Let's suppose points $P,Q,R,S$ are the middle points of all rods respectively.
Now using parallel axis theorem
$I_{BO}=I_{OC}=\frac{m{l}^2}{12}+(m){\left(\frac{l}{2}\right)}^2$
$I_{BO}=I_{OC}=\frac{m{l}^2}{3}$
Now,
Distance $PO=?$


Using Pythagoras theorem,
$x=PO=\sqrt{l^2+\frac{l^2}{4}}=\frac{\sqrt{5}l}{2}$
Now,$I_{AB}=I_{CD}=\frac{m{l}^2}{12}+m{\left(\frac{\sqrt{5}l}{2}\right)}^2$
(Using parallel axis theorem)
So, $I_{AB}=I_{CD}=\frac{4}{3}m{l}^2$
Now putting values in eq. (i)
$I_{net}=\frac{m{l}^2}{3}+\frac{m{l}^2}{3}+\frac{4}{3}m{l}^2+\frac{4}{3}m{l}^2$
$I_{net}=\frac{10}{3}m{l}^2$
So option (2) is correct