I'm following an example on finding the principal axes of a $45^\circ$ right triangle lamina with constant mass density. The example first shows how to find the principal moments of inertia about the center of mass, which gave the following result for the inertia tensor:
\begin{align}
\mathbf{I}
=\dfrac{ma^2}{3}
\begin{pmatrix}
\dfrac{1}{2} & 0 & 0
\\
0 & \dfrac{1}{2} & 0
\\
0 & 0 & 1
\end{pmatrix}
\end{align}
Here, $m$ is the mass and $a$ is the length of one side of the triangle.
Now, the example uses the parallel axis theorem to find the principal axes of the system. I attempted to find the first principal axis on my own before referencing the example using the parallel axis theorem as a starting point. My approach was to find the center of mass (COM) of the $x$ and $y$ components, which gave the position of the COM for the triangle. Then, using the parallel axis theorem, calculate the principal axes. But, my result differed by a small amount due to the COM calculation.
I'll first show my work for one principal axis, then state the result of the example.
\begin{align}
x_{cm} &= \dfrac{1}{m} \int_A x dA
\\[.5em]
&= \dfrac{2}{a^2} \int_0^a \int_0^{a-y} x dx dy
\\[.5em]
&= \dfrac{a}{3}
\end{align}
By symmetry, $y_{cm} = x_{cm} = \dfrac{a}{3}$. So the position of the COM of the triangle is $\big ( \dfrac{a}{3}, \dfrac{a}{3}, 0 \big )$.
Calculating the principal axis for the $x$-component, we have
\begin{align}
I_1 & = I_{xx} + mr_0^2
\end{align}
Let $r_0^2 = x_{cm}^2 + y_{cm}^2 = \dfrac{2a^2}{9}$. So,
\begin{align}
I_1 & = \dfrac{ma^2}{6} + \dfrac{2ma^2}{9} = \dfrac{7ma^2}{18}
\end{align}
According to the example, this is incorrect. The example says the center of mass is $y_{cm} = \dfrac{a}{3}$, with coordinates $(0,\dfrac{a}{3},0)$. Which then changes $r_0$ to $r_0^2 = y_{cm}^2 = \dfrac{a^2}{9}$. This changes the result slightly, giving $I_1 = \dfrac{ma^2}{18}$.
Why is the $x$-component of the COM disregarded here?
Best Answer
When you are computing a moment of inertia around the $x$ axis, what matters is the distance from the mass to the $x$ axis, not the distance from the mass to the origin.
For example, the moment of inertia of a unit point mass at $(0,1,0)$ around the $x$ axis is the same as for a unit point mass at $(1,1,0)$ or a unit point masss at $(-17,1,0).$ In short, the $x$ coordinate doesn't matter in this particular calculation, because it does not affect the moment arm around the $x$ axis.
Changing the $x$ coordinate has just as little impact on the moment of inertia around the $x$ axis when we move a lot of points at once as when we move just one point, so you can move your entire triangle in a direction parallel to the $x$ axis without affecting its moment of inertia around the $x$ axis. The $x$ coordinate of the center of mass is therefore irrelevant for this calculation.
Of course as we change the $x$ coordinate of the point mass, the moment of inertia around the $y$ axis changes. When you are calculating the moment of inertia around the $y$ axis, the $x$ coordinates matter and the $y$ coordinates don't.