# Moment of Inertia of Triangular loop made of rods only about an axis passing through one of the vertex and in the plane of loop parallel to the base

This question is from electrodynamics although I am stuck in the mechanics part here.

$$AB$$,$$BC$$,$$CA$$ are wires/rods, we needed to calculate Moment of Inertia(MOI) about an axis $$xx'$$ parallel to BC and in the plane of loop, each wire/rod is of mass $$m$$ and length $$l$$.

I tried calculating MOI about the centroid i.e center of mass(COM), and then using parallel axes theorem, the distance between side and centroid is d = $$\frac{l}{2 \sqrt3}$$ so calculating MOI perpendicular to the plane about COM would be 3($$\frac{ml^2}{12}$$ + $$md^2$$). Now using perpendicular axes theoram $$I_x$$ + $$I_y$$ = $$I_z$$, here I know $$I_y$$ would be equals $$\frac {ml^2}{12}$$ but how do I calculate $$I_x$$?

Should I use this approach or is there any other approach for this?

The answer given is $$\frac{5ml^2}{4}$$.

Any help would be appreciated

• It would be easier to figure out the moment of inertia of one of the slanted sides straight from the definition $\int r^2 dm$, and recognize the total as twice that plus a rod about an axis parallel to it's axis. Commented Mar 22, 2022 at 12:38
• @Triatticus thanks for pointing out the rod BC, I was considering MOI from that rod to be 0, but I was wrong, MOI of that rod is 0 about its own axis, but can be considered as a normal point mass for the axis $xx'$ about which I have to calculate and adding that gave me the final wanted result.
– Melo
Commented Mar 23, 2022 at 9:51

Look at the Fig.

The moment of Inertia towards the y' axes (green arrow) must be transformed to (x,y,z) this can obtain with this equation

$$\mathbf I_{xyz}=\mathbf R\,\mathbf I_{x'y'z'}\,\mathbf R^T$$

where $$~\mathbf R~$$ isthe transformation matrix between $$~x'y'z'~$$ and $$~xyz~$$ system and $$~\mathbf I~$$ is the inertia tensor

$$\mathbf R= \left[ \begin {array}{ccc} \cos \left( \alpha \right) &\sin \left( \alpha \right) &0\\-\sin \left( \alpha \right) & \cos \left( \alpha \right) &0\\ 0&0&1\end {array} \right] \quad,\alpha=\frac{\pi}{3}\\ \mathbf I_{x'y'z'}=\left[ \begin {array}{ccc} 0&0&0\\ 0&{\it I_{y'}}&0 \\ 0&0&0\end {array} \right]\quad\Rightarrow\\ \mathbf I_{xyz}=\left[ \begin {array}{ccc} \left( \sin \left( \alpha \right) \right) ^{2}{\it I_{y'}}&\sin \left( \alpha \right) {\it I_{y'}}\,\cos \left( \alpha \right) &0\\ \sin \left( \alpha \right) {\it I_{y'}}\,\cos \left( \alpha \right) & \left( \cos \left( \alpha \right) \right) ^{2}{\it I_{y'}}&0\\ 0&0&0 \end {array} \right]$$

thus $$~I_x=\sin^2(\alpha)\,I_{y'}=\frac{3}{4}\,\underbrace{\frac{m\,L^2}{12}}_{I_{y'}}~$$

for the 3 rods you obtain

$$I_\color{red}X=2\,\left(I_{x_1}+m\,H{_1}^2\right)+m\,H_3^2\\ H_1=\frac L2\,\sin(\pi/3)\\ H_3=L\,\sin(\pi/3)$$

$$\Rightarrow~I_\color{red}X=\frac 54 m\,L^2$$

• Thanks for the contribution, I don't know any math related to rotation matrix and Tensors(only heard of them), can I consider the Rotation matrix as analogous to complex $e^{\iota \theta}$ which rotates the complex number by $\theta$ angle?
– Melo
Commented Mar 23, 2022 at 9:58
UPDATE(Final solution): We can use the basic $$\int r^2 dm$$ to calculate MOI of a rod through its end point about an axis at an angle $$\theta$$ , which comes out to be $$\frac{1}{3} ML^2 \sin^2(\theta)$$, The angle $$x'AC$$ and $$x'AB$$ equals $$\frac{\pi}{3}$$ and $$\frac{2\pi}{3}$$ respectively so we can calculate MOI of $$AB$$ and $$AC$$, now we can consider BC to be a normal point mass concentrated at the COM i.e mid point of BC and calculate its MOI using $$\lambda^2 M$$ where $$\lambda$$ is the distance of A from COM of BC and now adding MOI due to all three rods we get the final result.