Tensor of inertia of a triangular plate
I consider a right-angled triangular plate with height h and base h, placed in the xy-plane with the right angle at the origin. The plate has uniform density \rho and mass m.
The centroid (center of mass) is a key reference point. I calculate its coordinates (x_c, y_c) using integration:
\mathbf{C} = \frac{1}{A} \int_A \mathbf{r} \,dA
where A is the area of the triangle. I find:
x_c = \frac{2h}{3}, \quad y_c = \frac{h}{3}
I first compute the moments of inertia about the origin (I_{xx}^0, I_{yy}^0, I_{zz}^0) using integration. For example, for I_{xx}^0:
I_{xx}^0 = \iint_A y^2 \rho \,dA
After integration, I obtain:
I_{xx}^0 = \frac{mh^2}{6}, \quad I_{yy}^0 = \frac{mh^2}{2}, \quad I_{zz}^0 = \frac{2mh^2}{3}
I also calculate the product of inertia about the origin:
I_{xy}^0 = -\iint_A xy \rho \,dA = -\frac{mh^2}{4}
To find the moments of inertia about the center of mass (I_{xx}, I_{yy}, I_{zz}, I_{xy}), I use the parallel axis theorem. For example, for I_{xx}:
I_{xx} = I_{xx}^0 - m \bar{y}^2
where \bar{y} is the y-coordinate of the centroid.
I can now express the moment of inertia tensor both about the origin:
\mathbf{I}^0 = \frac{mh^2}{18} \begin{bmatrix} 3 & -2 & 0 \\ -2 & 9 & 0 \\ 0 & 0 & 12 \end{bmatrix}
and the center of mass:
\mathbf{I} = \frac{mh^2}{36} \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}
For more insights into this topic, you can find the details here.