= Inertia tensor of a displaced uniform ball
{title2=$I(P)=M[(2R^2/5+|X|^2)I-XX^T],\quad X=P-G$}
For a uniform ball of radius $R>0$ and positive <mass> $M$, let $G$ be its <centre of mass>. Spherical symmetry gives the second-moment matrix $\int uu^Tdm=(MR^2/5)I$, so the central <inertia tensor> is $(2MR^2/5)I$. The <tensor parallel-axis theorem> gives the displayed shifted tensor. For $X\ne0$ its <eigenvector> along $X$ has <eigenvalue> $2MR^2/5$, while the two-dimensional <orthogonal complement> has repeated eigenvalue $M(2R^2/5+|X|^2)$. Thus it is an <isotropic tensor> exactly at $P=G$. Displacement along a chosen axis does not change that axis's moment; the transverse moments increase by $M|X|^2$.
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