# inertia tensor of rigid body in generalized coordinate frame?

Assuming we know the inertial tensor of a homogeneous rigid body about a coodinate frame at its COM and aligned to it principal axes, how do we find the inertial tensor for the body in some other general coordinate frame which has a linear transformation (4x4) (which accounts for both rotation and translation) T from the principal C.F at the COM ?

The general 4×4 transformation matrix has a structure where the top left 3×3 submatrix is the rotation + scaling factors. If there is no scaling, then extract the this matrix $$\mathrm{R}$$ from $$\text{transform}= \left| \matrix{ \mathrm{R} & \vec{t} \\ \vec{0}^\intercal & 1 } \right|$$

Then you do the standard

$$\mathrm{I}_{\rm world} = \mathrm{R}\,\mathrm{I}_{\rm body} \mathrm{R}^\intercal$$

This assumes the transformation is defined as local -> world sense.

Now if you want to include the parallel axis theorem to move the MMMOI definition to a new point them you use the following rule

$$\mathrm{I}_{\rm world} = \mathrm{R}\,\mathrm{I}_{\rm body} \mathrm{R}^\intercal - m [\vec{t}\times] [\vec{t}\times]$$

where $$[\vec{t}\times]$$ is the 3×3 skew-symmetric cross product operator:

$$\pmatrix{x\\y\\z} \times = \left[ \matrix{0 & -z & y\\ z & 0 & -x \\ -y & x & 0} \right]$$

such that $$\vec{a} \times \vec{b}$$ becomes the vector/matrix product $$[\vec{a} \times] \vec{b}$$.

• Thank you for the answer . I have a follow up question ! The now resulting inertia tensor matrix in the general C.F which is not at the COM necessarily, is also a symmetric matrix. Therefore, as with all symmetric real 3x3 matrices, this will also have real orthogonal eigen vectors. What is their significance ? I know had this general C.F positioned at the COM, the eigen vector basis would be the principal frame for the inertia matrix. And in that case you can set the rigid body in perpetual rotation (assume no external Moment) about an axes through the COM along any of the eigen vectors. – programmer Mar 28 '19 at 16:54
• The eigenvectors should give the direction of principal inertia about the new pivot point. Think of a mass attached at the end of a massless rod that is fixed on the other end. The inertia matrix is $$\mathrm{I} = -m [\vec{r}\times] [\vec{r}\times] = m \pmatrix{y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+z^2}$$ where $\vec{r}=\pmatrix{x&y&z}$ is the location of the mass (on the end of the rod). The eigenvectors are along the rod, and perpendicular to the rod. – ja72 Mar 29 '19 at 13:41