Step 1: Derivation of Moment of Inertia about the Base
Consider a uniform triangular lamina of mass $M$, base $l$, and height $h$. We aim to find the moment of inertia about its base.
Let us select a thin strip parallel to the base at a distance $y$ from the base with thickness $dy$.
- By similar triangles, the width of the strip $w$ decreases linearly from $l$ (at $y=0$) to $0$ (at $y=h$). Thus, $w = l \left( 1 – \frac{y}{h} \right)$.
- The area of the strip is $dA = w \cdot dy = l \left( 1 – \frac{y}{h} \right) dy$.
- The surface mass density $\sigma = \frac{M}{\frac{1}{2}lh} = \frac{2M}{lh}$.
- The mass of the strip is $dm = \sigma dA = \frac{2M}{lh} \cdot l \left( 1 – \frac{y}{h} \right) dy = \frac{2M}{h} \left( 1 – \frac{y}{h} \right) dy$.
The moment of inertia of this strip about the base is $dI = dm \cdot y^2$. Integrating from $y=0$ to $y=h$:
$$ I_{\text{base}} = \int_0^h y^2 \left[ \frac{2M}{h} \left( 1 – \frac{y}{h} \right) \right] dy $$ $$ I_{\text{base}} = \frac{2M}{h} \int_0^h \left( y^2 – \frac{y^3}{h} \right) dy $$ $$ I_{\text{base}} = \frac{2M}{h} \left[ \frac{y^3}{3} – \frac{y^4}{4h} \right]_0^h = \frac{2M}{h} \left( \frac{h^3}{3} – \frac{h^3}{4} \right) $$ $$ I_{\text{base}} = \frac{2M}{h} \left( \frac{h^3}{12} \right) = \frac{Mh^2}{6} $$Step 2: Moment of Inertia about Centroidal Axes
Isotropy of the Plane Moment of Inertia:
We need to find the moment of inertia about the centroid. Let’s define two axes in the plane passing through the centroid $C$:
- Axis $x’$: Parallel to the base.
- Axis $y’$: Perpendicular to the base (along the altitude).
1. Calculating $I_{x’}$ (Parallel to Base):
Using the Parallel Axis Theorem, we shift the axis from the base to the centroid. The distance between the base and the centroid is $d = h/3$.
$$ I_{\text{base}} = I_{x’} + M\left(\frac{h}{3}\right)^2 $$ $$ \frac{Mh^2}{6} = I_{x’} + \frac{Mh^2}{9} $$ $$ I_{x’} = \frac{Mh^2}{6} – \frac{Mh^2}{9} = \frac{3Mh^2 – 2Mh^2}{18} = \frac{Mh^2}{18} $$2. Calculating $I_{y’}$ (Along Altitude) to prove symmetry:
Consider the triangle composed of two right-angled triangles of base $l/2$ and height $h$. We integrate horizontal strips of length $x$ at distance $y$ from the top vertex.
- For an equilateral triangle, side $l$ and height $h$ are related by $l^2 = l^2/4 + h^2 \Rightarrow l^2 = \frac{4h^2}{3}$.
- We can show via integration (integrating $x^2 dm$ for strips perpendicular to the altitude) that: $$ I_{y’} = \frac{Ml^2}{24} $$
- Substituting $l^2 = \frac{4h^2}{3}$: $$ I_{y’} = \frac{M}{24} \left( \frac{4h^2}{3} \right) = \frac{Mh^2}{18} $$
Conclusion on Planar Symmetry:
Since $I_{x’} = \frac{Mh^2}{18}$ and $I_{y’} = \frac{Mh^2}{18}$, the moment of inertia is the same for these two perpendicular axes. For any regular polygon (like an equilateral triangle), the moment of inertia about any axis passing through the centroid and lying in the plane is constant.
$$ I_{\text{plane}} = \frac{Mh^2}{18} $$Step 3: Moment of Inertia about Perpendicular Axis ($I_z$)
We apply the Perpendicular Axis Theorem. The axis $z$ is perpendicular to the plane and passes through the centroid.
$$ I_z = I_{x’} + I_{y’} $$Since $I_{x’} = I_{y’} = \frac{Mh^2}{18}$:
$$ I_z = \frac{Mh^2}{18} + \frac{Mh^2}{18} = \frac{Mh^2}{9} $$Step 4: Final Substitution
We express the result in terms of side length $l$. For an equilateral triangle, $h = \frac{\sqrt{3}}{2}l$.
$$ I_z = \frac{M}{9} \left( \frac{\sqrt{3}}{2}l \right)^2 $$ $$ I_z = \frac{M}{9} \left( \frac{3l^2}{4} \right) = \frac{3Ml^2}{36} = \frac{1}{12}Ml^2 $$