Proof for Moment of Inertia of Triangle (with respect to the axis out of the page at a origin) [duplicate]

Question

I found a formula for a "Triangle with vertices at the origin and at $P$ and $Q$, with mass $m$, rotating about an axis perpendicular to the plane and passing through the origin" given on this Wikipedia page. The given formula was $I=\frac{1}{6}m(P \cdot P + P \cdot Q+ Q \cdot Q)$. I assume they're talking about the following case, where $P$ and $Q$ are vectors representing the triangle legs and the triangle pivots around the origin.

I can't find a proof for the given formula. Could someone provide me with a link to one?

Answer 1

1. Parametrization - The first step is to parametrically describe the position of each point interior to the triangle.

   $$ {\rm \boldsymbol{pos}}(s,t)=t\,\left(\left(1-s\right)\boldsymbol{P}+s\,\boldsymbol{Q}\right) \tag{1} $$

   where $s$ and $t$ are independent parameters that range from 0..1

2. Volume Integral - Assume the triangle has mass, and thus has some infinitesimal width $w$ and derive the volume element used for integration

   $$ {\rm d}V=w\left\Vert \left(\tfrac{\partial}{\partial t}{\rm \boldsymbol{pos}}\right)\times\left(\tfrac{\partial}{\partial s}{\rm \boldsymbol{pos}}\right)\right\Vert {\rm d}s\,{\rm d}t \tag{2} $$

   _{The expansion of the above is $$\begin{gathered}{\rm d}V=w\left\Vert \left(\tfrac{\partial}{\partial t}{\rm pos}\right)\times\left(\tfrac{\partial}{\partial s}{\rm pos}\right)\right\Vert {\rm d}s\,{\rm d}t\\ =w\left\Vert \left(\left(1-s\right)\boldsymbol{P}+s\,\boldsymbol{Q}\right)\times\left(t\,\boldsymbol{Q}-t\,\boldsymbol{P}\right)\right\Vert {\rm d}s\,{\rm d}t\\ =w\left\Vert t\,\left(\left(\left(1-s\right)\boldsymbol{P}+s\,\boldsymbol{Q}\right)\times\boldsymbol{Q}-\left(\left(1-s\right)\boldsymbol{P}+s\,\boldsymbol{Q}\right)\times\boldsymbol{P}\right)\right\Vert {\rm d}s\,{\rm d}t\\ =w\left\Vert t\,\left(\left(1-s\right)\left(\boldsymbol{P}\times\boldsymbol{Q}\right)-s\,\left(\boldsymbol{Q}\times\boldsymbol{P}\right)\right)\right\Vert {\rm d}s\,{\rm d}t\\ =w\,t\,\left\Vert \boldsymbol{P}\times\boldsymbol{Q}\right\Vert {\rm d}s\,{\rm d}t \end{gathered}$$}

   where $\| \boldsymbol{P} \times \boldsymbol{Q} \| = 2({\rm area})$ is the magnitude of the cross product, which equals to two times the area of the triangle

   The volume integral is

   $$ V=w\,\int_{0}^{1}\int_{0}^{1}t\,\left\Vert \boldsymbol{P}\times\boldsymbol{Q}\right\Vert {\rm d}t\,{\rm d}s=\tfrac{w}{2}\|\boldsymbol{P}\times\boldsymbol{Q}\|=w\,({\rm area}) \tag{3} $$

3. Mass Properties - The density of the material is $$\rho = \frac{m}{V} = \frac{m}{w\,({\rm area})} \tag{4} $$

   or you can re-write the volume integral as a mass integral in terms of the density and area of the triangle

   $$m=w\int_{0}^{1}\int_{0}^{1}2t\,\rho({\rm area}){\rm d}s\,{\rm d}t = \rho\,w\,({\rm area}) \tag{5}$$

   This means that $\rho\,({\rm area})=\frac{m}{w}$ which is found inside the volume/mass integrals.

   _{The center of mass for example is found with $$\begin{gathered}\boldsymbol{C}=\tfrac{1}{m}w\int_{0}^{1}\int_{0}^{1}2t\,{\rm pos}(s,t)\,\tfrac{m}{w}{\rm d}s\,{\rm d}t\\ =\int_{0}^{1}\int_{0}^{1}2t^{2}\left(\left(1-s\right)\boldsymbol{P}+s\,\boldsymbol{Q}\right){\rm d}s\,{\rm d}t\\ =\int_{0}^{1}t^{2}\left(\boldsymbol{P}+\boldsymbol{Q}\right){\rm d}t\,\\ =\tfrac{1}{3}\left(\boldsymbol{P}+\boldsymbol{Q}\right) \end{gathered}$$}

4. Mass Moment of Inertia Tensor - The MMOI tensor of a single particle ${\rm d}m$ located at ${\rm pos}$ is defined as

   $${\rm d}{\bf I}=\left(\left({\rm \boldsymbol{pos}}\cdot{\rm \boldsymbol{pos}}\right){\bf 1}+{\rm \boldsymbol{pos}}\otimes{\rm \boldsymbol{pos}}\right){\rm d}m \tag{6}$$

   where ${\bf 1}$ is the identity matrix, $\cdot$ is the dot/inner product and $\otimes$ is the outer product.

   The full MMOI integral is

   $${\bf I}=m\int_{0}^{1}\int_{0}^{1}2t\left(\left({\rm \boldsymbol{pos}}\cdot{\rm \boldsymbol{pos}}\right){\bf 1}+{\rm \boldsymbol{pos}}\otimes{\rm \boldsymbol{pos}}\right)\,{\rm d}s\,{\rm d}t \tag{7}$$

   _{Now consider how each vector product expands to $$\begin{gathered}{\rm \boldsymbol{pos}}\cdot{\rm \boldsymbol{pos}}=t^{2}\left(\left(1-s\right)^{2}\left(\boldsymbol{P}\cdot\boldsymbol{P}\right)+2s\left(1-s\right)\left(\boldsymbol{P}\cdot\boldsymbol{Q}\right)+s^{2}\left(\boldsymbol{Q}\cdot\boldsymbol{Q}\right)\right)\\ {\rm \boldsymbol{pos}}\otimes{\rm \boldsymbol{pos}}=t^{2}\left(\left(1-s\right)^{2}\left(\boldsymbol{P}\otimes\boldsymbol{P}\right)+s\left(1-s\right)\left(\boldsymbol{P}\otimes\boldsymbol{Q}+\boldsymbol{Q}\otimes\boldsymbol{P}\right)+s^{2}\left(\boldsymbol{Q}\otimes\boldsymbol{Q}\right)\right) \end{gathered}$$}

   If I do the math for you I get

   $$\begin{gathered}{\bf I}=m\frac{\left(\boldsymbol{P}\cdot\boldsymbol{P}\right)+\left(\boldsymbol{P}\cdot\boldsymbol{Q}\right)+\left(\boldsymbol{Q}\cdot\boldsymbol{Q}\right)}{6}+m\frac{\left[\boldsymbol{P}\otimes\boldsymbol{P}\right]+\left[\boldsymbol{Q}\otimes\boldsymbol{Q}\right]+\left[\left(\boldsymbol{P}+\boldsymbol{Q}\right)\otimes\left(\boldsymbol{P}+\boldsymbol{Q}\right)\right]}{12}\end{gathered} \tag{8}$$

   which is close to the desired expression. Only if there is a way to simplify all those outer products, but at this point, this is deep into the vector algebra weeds and I do not see a clear path forward.

   _{I used the following identity $\left(\boldsymbol{P}+\boldsymbol{Q}\right)\otimes\left(\boldsymbol{P}+\boldsymbol{Q}\right)\equiv\left[\boldsymbol{P}\otimes\boldsymbol{P}\right]+\left[\boldsymbol{Q}\otimes\boldsymbol{P}\right]+\left[\boldsymbol{P}\otimes\boldsymbol{Q}\right]+\left[\boldsymbol{Q}\otimes\boldsymbol{Q}\right]$ to simplify the result to the form above.}