Question

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

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

Answer 1

The Correct Option is B

Total moment of inertia $=l_{1}+l_{2}+l_{3}+l_{4}=2 l_{1}+2 l_{2}$ $=2\left(l_{1}+l_{2}\right)\left[l_{3}=l_{1}, l_{1} l_{1}=l_{4}\right]$ Now $l_{2}=l_{3}=\frac{ml^{2}}{3}$ Using parallel axes theorem, we have $l=l_{cm}+m x^{2}$ and $x=\sqrt{l^{2}+\frac{l^{2}}{4}}$ $l_{1}=l_{4}=\frac{ml^{2}}{12}+M\left[\sqrt{l^{2}+\left(\frac{1}{2}\right)^{2}}\right]^{2}$ Putting all values we get Moment of inertia, $l=10\left(\frac{ml^{2}}{3}\right)$