An equilateral-triangle rigid-material model: inertia matrix, principal axes, rotational term and the threefold-symmetry check

Finite-atomic integration formula: let $f$ be $\nu$-integrable. If $s={\sum }_j{a}_j{\chi }_{E_j}$ is a nonnegative simple function, then $\int sd\nu ={\sum }_j{a}_j\nu (E_j)={\sum }_j{a}_j{m}_0{\sum }_k{\delta }_{x_k}(E_j)=m_0{\sum }_{k}s(x_k)$ by [L1]; taking the supremum over simple $0\le s\le f$ and using the minorant $s_0:={\sum }_{k}f(x_k){\chi }_{\{x_k\}}\le f$ together with the bound $\int sd\nu =m_0{\sum }_{k}s(x_k)\le m_0{\sum }_{k}f(x_k)$ valid for every simple $0\le s\le f$ gives $\int fd\nu =m_0{\sum }_{k}f(x_k)$ for nonnegative $f$; the real case follows from $f=f^{+}-f^{-}$ and linearity [L2], and the complex case by real/imaginary parts.

Trigonometric values and sums: put $\theta =2\pi \mathrm{/}3$ and $c=\cos \theta$. The period in [L4] gives $\cos 3\theta =1$. Repeated addition and the Pythagorean identity give $\cos 3\theta =4c^3-3c$, hence $(c-1)(2c+1{)}^2=0$. Since $0<\theta <\pi$ and cosine strictly decreases on $[0,\pi ]$, $c<\cos 0=1$, so $c=-1\mathrm{/}2$. Positivity of sine and its squared value $1-c^2=3\mathrm{/}4$ give $\sin \theta =\sqrt{3}\mathrm{/}2$. The addition formulas now give $(\cos 2\theta ,\sin 2\theta )=(-1\mathrm{/}2,-\sqrt{3}\mathrm{/}2)$. Together with $(\cos 0,\sin 0)=(1,0)$ these three explicit pairs yield all five finite sums in [L4] by addition.

The measure is a mass distribution: by [L1] it is finite and bounded with support $\{x_0,x_1,x_2\}$, and $\nu ({\mathbb{R}}^3)=3m_0\in (0,\mathrm{\infty })$. Its first moment is $\int \rho d\nu =m_0{\sum }_k{x}_k=m_{0}a({\sum }_k\cos (2\pi k\mathrm{/}3),{\sum }_k\sin (2\pi k\mathrm{/}3),0)=0$ by step 1.1 and [L4]: the reference coordinates are centred.

Inertia matrix: by step 1.1 and [L3], $I=\int (\parallel \rho {\parallel }_2^2{I}_3-\rho {\rho }^{\mathsf{T}})d\nu (\rho )=m_0{\sum }_k(\parallel x_k{\parallel }_2^2{I}_3-x_k{x}_k^{\mathsf{T}})=m_0(3a^2{I}_3-{\sum }_k{x}_k{x}_k^{\mathsf{T}}),$ because $\parallel x_k{\parallel }_2=a$ for every $k$. The matrix ${\sum }_k{x}_k{x}_k^{\mathsf{T}}$ has entries $a^2{\sum }_k{\cos }^2(2\pi k\mathrm{/}3)$, $a^2{\sum }_k\cos (2\pi k\mathrm{/}3)\sin (2\pi k\mathrm{/}3)$ and $a^2{\sum }_k{\sin }^2(2\pi k\mathrm{/}3)$ in the plane and zero in the third row and column, so by [L4] it equals $\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\frac{3a^2}{2},\frac{3a^2}{2},0)$. Hence $I=m_0\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(3a^2-\frac{3a^2}{2},3a^2-\frac{3a^2}{2},3a^2)$, which is the displayed matrix of clause 2.

Principal data: the displayed matrix is diagonal, so $I{e}_1=\frac{3m_0{a}^2}{2}{e}_1$, $I{e}_2=\frac{3m_0{a}^2}{2}{e}_2$ and $I{e}_3=3m_0{a}^2{e}_3$: $e_3$ is a principal axis with moment $3m_0{a}^2$, and every $g$-orthonormal pair in the plane $\{\rho \in {\mathbb{R}}^3:{\rho }_3=0\}$ is a pair of principal axes with the common moment $\frac{3m_0{a}^2}{2}$. The integral identities are checked directly: $\int \parallel e_3\times \rho {\parallel }_2^{2}d\nu =m_0{\sum }_k((x_k{)}_1^2+(x_k{)}_2^2)=m_0\cdot 3a^2=3m_0{a}^2$ and $\int \parallel e_1\times \rho {\parallel }_2^{2}d\nu =m_0{\sum }_k((x_k{)}_2^2+(x_k{)}_3^2)=m_0\cdot \frac{3a^2}{2}$ by [L4] and [L6], in agreement with the diagonal entries by clause 2.

Nondegeneracy and invertibility: no line through the origin contains two of the points $x_k$, because their arguments modulo $\pi$ are $0,\frac{\pi }{3},\frac{2\pi }{3}$ (using [L4]), which are pairwise distinct; hence for every line $L$ through the origin at least two of the three points lie outside $L$ and $\nu ({\mathbb{R}}^3\setminus L)\ge 2m_0>0$. The model is therefore nondegenerate, and by clause (d) of [L3] all principal moments are positive (as they visibly are) and $I$ is invertible.

Rotational term: by clause (b) of [L3] and clause 2, for a motion with body angular velocity $\omega ={\sum }_j{\omega }_j{e}_j$ the quadratic form is ${\omega }^{\mathsf{T}}I\omega ={\sum }_j{I}_j{\omega }_j^2=\frac{3m_0{a}^2}{2}({\omega }_1^2+{\omega }_2^2)+3m_0{a}^2{\omega }_3^2$; multiplying by $\frac{1}{2}$ and citing clause 3 of Total momentum, kinetic-energy splitting and centre-of-mass angular momentum of a rigid-material motion for the reading of the term as the rotational kinetic energy gives clause 4. In particular a rotation about the symmetry axis $e_3$ with rate $\varpi$ has rotational term $\frac{3}{2}{m}_0{a}^2{\varpi }^2$, and a rotation about an in-plane principal axis has term $\frac{3}{4}{m}_0{a}^2{\varpi }^2$.

Symmetry check: $S=R_z(2\pi \mathrm{/}3)$ is the rotation of the reference plane that carries $x_k$ to $x_{k+1}$ for $k=0,1,2$ (indices modulo $3$): its matrix acts on $a(\cos \theta ,\sin \theta ,0)$ as $a(\cos (\theta +\frac{2\pi }{3}),\sin (\theta +\frac{2\pi }{3}),0)$ by the addition formulas and [L5]. Since $S$ is orthogonal and preserves norms, $SI{S}^{\mathsf{T}}=m_0(3a^2{I}_3-{\sum }_k(S{x}_k)(S{x}_k{)}^{\mathsf{T}})=m_0(3a^2{I}_3-{\sum }_k{x}_{k+1}{x}_{k+1}^{\mathsf{T}})=I$ by step 2.2, the preservation of norms and the relabelling of the finite sum. Thus $SI=IS$, and the invariance of the inertia data under the threefold symmetry is exhibited; on the plane $\{\rho \in {\mathbb{R}}^3:{\rho }_3=0\}$, where $S$ acts as the rotation by $\frac{2\pi }{3}$, the matrix $I$ is scalar, which is exactly the equality of the two transverse principal moments in step 3.1.

Interpretation: the model is a rigid-material model with three equal point masses whose reference shape is symmetric under the threefold rotation; its inertia matrix, principal axes, principal moments and rotational term have been computed, and the symmetry invariance of the inertia matrix checked. The numbers are proportions of the model parameters $m_0$ and $a$; no measured moment of inertia and no concrete physical body is asserted, and the statements are conditional on the model of Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope.