# Translate inertia tensor, caculate distance matrix

I know that inertia tensor of a sphere is

$$2mr^2/5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

Now if I translate it into a coordinate (x,y,z). What will be the inertia tensor then ?

All I know is, after translation :

$$I_{new}=I_{old}-m*r^2$$ and here $$r$$ is a matrix, actually I want to know how can I calculate that matrix ?

$$\mathbf{I}_{\rm new} = \mathbf{I}_{\rm old} + m \left[ \matrix{y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & - y z \\ -x z & - y z & x^2+y^2 } \right]$$
where $$\pmatrix{x\\ y \\z}$$ is the location of $$({\rm old})$$ relative to $$({\rm new})$$.
• You mean how $$-\left[ \matrix{0 & -z & y \\ z & 0 & -x \\ -y & x & 0} \right] \left[ \matrix{0 & -z & y \\ z & 0 & -x \\ -y & x & 0} \right] = \left[ \matrix{y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & - y z \\ -x z & - y z & x^2+y^2 } \right]$$ ? In that post $r$ is a 3×1 vector and $\tilde{r}$ is a 3×3 matrix. The overbar denotes the cross product matrix operator. Dec 22 '19 at 7:09