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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 ?

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Use the parallel axis theorem in vector/tensor form

$$ \mathbf{I}_{\rm new} = \mathbf{I}_{\rm old} + m \left[ \matrix{y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & - y z \\ -x z & - y z & x^2+y^2 } \right] $$

where $\pmatrix{x\\ y \\z}$ is the location of $({\rm old})$ relative to $({\rm new})$.

For an example, see the end of this related answer (equation 4).

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  • $\begingroup$ can you tell me why here, how the value of r came here, it's a 3*3 matrix, where you have given me a 1*3 matrix $\endgroup$
    – Danial
    Dec 22 '19 at 5:53
  • $\begingroup$ 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. $\endgroup$ Dec 22 '19 at 7:09

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