Question

The moment of intertia of a system of four rods, each of length $l$ and mass $m$, about the axis shown is

• A. $\frac{2}{3}m{l}^2$
• B. $2m{l}^2$
• C. $3m{l}^2$
• D. $\frac{16m{l}^2}{3}$

Answer 1

The correct option is D $\frac{16m{l}^2}{3}$

We know that Moment of Inertia of a rod of length l and mass m about its end is $\frac{m{l}^2}{3}$
Also, by parallel axis theorem, when moment of inertia axis is shifted parallel x distance, an additional moment of $m{x}^2$ is added.
Keeping these in point, we can see that 2 rods have their ends at the required axis, so for them the moment of inertia will be $\frac{m{l}^2}{3}$. For the remaining two, we can see that the common end of those two is at distance of $l\sqrt{2}$ from the required axis. So their moment of inertia will be $\frac{m{l}^2}{3}+m{\left(l\sqrt{2}\right)}^2$
Adding all the moments we get, $I=2\times \frac{m{l}^2}{3}+2\times (\frac{m{l}^2}{3}+m{\left(l\sqrt{2}\right)}^2)=\frac{4m{l}^2}{3}+4m{l}^2=\frac{16m{l}^2}{3}$