= Inertia tensor of paired point masses
An exact planar example makes the off-diagonal entries tangible. Join four equal <point masses> $m$ with negligible-mass rigid connectors at $a(2,1,0)$, $a(-2,-1,0)$, $a(1,-1,0)$ and $a(-1,1,0)$, where $a>0$ is a length. Their <centre of mass> is $O=(0,0,0)$. Summing $x^2$, $y^2$ and $xy$ gives $10a^2$, $4a^2$ and $2a^2$, respectively, so
$$
\mathbf I_{\mathrm{planar}}=ma^2\begin{pmatrix}4&-2&0\\-2&10&0\\0&0&14\end{pmatrix}.
$$
The $z$ axis is a <principal axis>, and $I_{zz}=I_{xx}+I_{yy}$ illustrates the <perpendicular axis theorem>. The $x$ axis is not a <principal axis>: <angular velocity> $\boldsymbol\omega=(\Omega,0,0)$ gives <angular momentum> $ma^2\Omega(4,-2,0)$, which points in a different direction. A bearing can enforce fixed-axis rotation despite this difference, supplying the necessary transverse <torque>.
\Image[/inertia-tensor-paired-masses-rotational-energy.png]
{title=Paired <point masses> and direction-dependent <rotational kinetic energy>}
{description=Original exact model. Four equal <point masses> lie in the $xy$ plane with negligible-mass connectors. The curves use <angular velocity> $\boldsymbol\omega=\Omega(\cos\theta,\sin\theta,0)$ and plot $2K/(ma^2\Omega^2)$. Adding two more equal <point masses> at $a(0,0,\pm1)$ raises this curve by two. Markers depict ideal <point masses>, not finite-sized weights; no experimental data are plotted.}
For an in-plane <unit vector> $\mathbf n=(\cos\theta,\sin\theta,0)$, the planar model has
$$
\frac{I_{O,\mathbf n}}{ma^2}=4\cos^2\theta+10\sin^2\theta-4\sin\theta\cos\theta.
$$
Its extrema are the <principal moments of inertia> $(7-\sqrt{13})ma^2$ and $(7+\sqrt{13})ma^2$; the remaining <principal moment of inertia> is $14ma^2$. These values follow from the two-by-two <characteristic polynomial> $(4-\lambda)(10-\lambda)-4$, so they can be checked without trusting a numerical <eigenvalue> routine. At a stipulated <angular speed> $\Omega=1\,\mathrm{rad/s}$, with $m=1\,\mathrm{kg}$ and $a=1\,\mathrm m$, rotation about $x$, $y$ or $z$ has <rotational kinetic energy> $2$, $5$ or $7\,\mathrm J$, respectively.
A truly three-dimensional example adds two more equal <point masses> at $a(0,0,1)$ and $a(0,0,-1)$. Each contributes $ma^2\operatorname{diag}(1,1,0)$ to the <inertia tensor>, giving
$$
\mathbf I_{\mathrm{3D}}=ma^2\begin{pmatrix}6&-2&0\\-2&12&0\\0&0&14\end{pmatrix},
\qquad
(I_1,I_2,I_3)=ma^2(9-\sqrt{13},9+\sqrt{13},14).
$$
The in-plane <principal axes> stay the same while their <principal moments of inertia> each increase by $2ma^2$. The $z$ <moment of inertia> stays unchanged because the added <point masses> lie on that axis. This comparison changes the total <mass> from $4m$ to $6m$; it holds the original four <point masses> fixed rather than redistributing a fixed total <mass>.
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