Find the moments of inertia of a homogeneous isosceles
Find the moments of inertia of a homogeneous isosceles triangle with base $a$ and height $h$ ($\varrho=1$) with respect to each of its vertices.
Explanation of Method
With the triangle centered symmetrically about the $y$-axis and the base on the $x$-axis, the two axial moments $I_{Ox}=\iint y^2\,dA$ and $I_{Oy}=\iint x^2\,dA$ about the origin (base midpoint) are computed first, since the moment about any vertex can then be built from them by expanding $(x-x_0)^2+(y-y_0)^2$ and using that the odd first moment $\iint x\,dA$ vanishes by symmetry.
- Compute the two axial moments $I_{Ox},I_{Oy}$ about the base midpoint.
- Compute the moment $I_C$ about the apex.
- Compute the moment $I_A=I_B$ about each base vertex.
Step 1 · Axial moments about the base midpoint
With $A=\left(-\frac a2,0\right),\,B=\left(\frac a2,0\right),\,C=(0,h)$ and width $w(y)=a\left(1-\frac yh\right)$:
$$ I_{Ox} = \int_0^hy^2w(y)\,dy = \frac{ah^3}{12}, \qquad I_{Oy} = \frac{a^3}{12}\int_0^h\left(1-\frac yh\right)^3dy = \frac{a^3h}{48}. $$
Step 2 · Moment about the apex $C$
$I_C=I_{Oy}+\iint(y-h)^2\,dA$. Using $(y-h)=-h\left(1-\frac yh\right)$ and $t=1-\frac yh$:
$$ \iint_D(y-h)^2\,dA = ah^3\int_0^1t^3\,dt = \frac{ah^3}4, \qquad I_C = \frac{a^3h}{48}+\frac{ah^3}4. $$
Step 3 · Moment about a base vertex $A$ (and by symmetry $B$)
$I_A=\iint\left[\left(x+\frac a2\right)^2+y^2\right]dA=I_{Oy}+a\underbrace{\iint x\,dA}_{=0}+\frac{a^2}4m+I_{Ox}$, with $m=\frac{ah}2$:
$$ I_A = \frac{a^3h}{48} + \frac{a^3h}8 + \frac{ah^3}{12} = \frac{7a^3h}{48}+\frac{ah^3}{12} = I_B. $$
Additional Notes
- This derivation matches the source exactly. In terms of the mass $m=\frac{ah}2$: $I_C=m\left(\frac{a^2}{24}+\frac{h^2}2\right)$, $I_A=I_B=m\left(\frac{7a^2}{24}+\frac{h^2}6\right)$.
- Verified numerically (e.g. $a=2.2,\,h=1.7$): $I_C\approx3.0793$, $I_A=I_B\approx3.5405$, both matching the boxed closed forms exactly.
Result
The moments of inertia about the apex and about each base vertex are: