Inertia Tensors

Tensor inertia

There's a Wikipedia page that explains the Moment of inertia very well. For a more precise definition, go there. Here we get only some of the concepts using tensorial notation

Definition

If we have a point in the space, with mass m at position p = (x, y, z) the inertia tensor II is given by

$\mathbb{II} = m \cdot \begin{bmatrix} y^2 + z^2 & -xy & -xz \\ -xy & x^2 + z^2 & -yz \\ -xz & -yz & x^2 + y^2 \end{bmatrix}$

Now we call the kinematic inertia tensor IIk when we divide the inertia tensor by the mass

$\mathbb{II}_{k} = \begin{bmatrix} y^2 + z^2 & -xy & -xz \\ -xy & x^2 + z^2 & -yz \\ -xz & -yz & x^2 + y^2 \end{bmatrix}$

$\mathbb{II}_{k} = (x^2+y^2+z^2) \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} x^2 & xy & xz \\ xy & y^2 & yz \\ xz & yz & z^2 \end{bmatrix}$

A better way to write them is using the tensorial notation and get:

$\mathbb{II}_{k} = \vec{p} \circ \vec{p} \ \cdot \bold{I} - \vec{p} \otimes \vec{p}$

Tensor for any object

If we have not only one point but a complete domain we have the inertia tensor II like

$\mathbb{II} = \int \left( \vec{p} \circ \vec{p} \ \cdot \bold{I} - \vec{p} \otimes \vec{p} \right) dm = \int_{\Omega} \rho \left( \vec{p} \circ \vec{p} \ \cdot \bold{I} - \vec{p} \otimes \vec{p} \right) dV$

And as consequence, we have the IIk defined by

$\mathbb{II}_{k} = \dfrac{1}{\int_{\Omega} \rho \ dV} \int_{\Omega} \rho \left( \vec{p} \circ \vec{p} \ \cdot \bold{I} - \vec{p} \otimes \vec{p} \right) dV$

If the density of mass rho is constant we have only

$\mathbb{II}_{k} = \dfrac{1}{V} \int_{\Omega} \left( \vec{p} \circ \vec{p} \ \cdot \bold{I} - \vec{p} \otimes \vec{p} \right) dV$

Exemples of inertia tensor with constant density

Bar

For a bar with length L, centered in the middle and its axis in the x direction

$\mathbb{II}_{k} = \dfrac{L^2}{12} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Disk

For a disk of radius R and its axis in the z direction

$\mathbb{II}_{k} = \dfrac{R^2}{4} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Ring

For a ring of radius R and its axis in the z direction

$\mathbb{II}_{k} = \dfrac{R^2}{2} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Solid cylinder

For a cylinder of radius R and height h centered in the origin, with axis in z direction

$\mathbb{II}_{k} = \dfrac{1}{12} \begin{bmatrix} 3R^2 +h^2 & 0 & 0 \\ 0 & 3R^2+h^2 & 0 \\ 0 & 0 & 6R^2 \end{bmatrix}$

hollowed cylinder

For a hollowed cylinder with internal radius Ri, external radius Re, height h with axis in z direction

$\mathbb{II}_{k} = \dfrac{1}{12} \begin{bmatrix} 3(R_i^2+R_{e}^2) +h^2 & 0 & 0 \\ 0 & 3(R_i^2+R_{e}^2)+h^2 & 0 \\ 0 & 0 & 6(R_i^2+R_{e}^2) \end{bmatrix}$

The bar, disk, ring and solid cylinder are specific cases of the hollowed cylinder with different values of Ri, Re and h

Solid sphere

A solid sphere of radius R centered in the origin

$\mathbb{II}_{k} = \dfrac{2}{5} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Hollowed sphere

A hollowed sphere of internal radius Ri and external radius Re centered in the origin

$\mathbb{II}_{k} = \dfrac{2}{5} \cdot \dfrac{R_e^5-R_i^5}{R_e^3 - R_e^3} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Retangular plate

A planar plate of size a in the direction x, size b in the direction y and size c in direction z, with its center in the origin

$\mathbb{II}_{k} = \dfrac{1}{12} \begin{bmatrix} b^2 + c^2 & 0 & 0 \\ 0 & a^2 + c^2 & 0 \\ 0 & 0 & a^2 + b^2 \end{bmatrix}$

Parallel axis theorem

If we have a kinematic inertia tensor IIk in the referential frame 1 translated by the vector c relation to the referential 0, using the parallel axis theorem the kinematic inertia tensor in the referential frame 0 will is

$^{0}\mathbb{II}_{k} = \ ^{1}\mathbb{II}_{k} + \vec{c} \circ \vec{c} \cdot \bold{I} - \vec{c} \otimes \vec{c}$

And if we have a referential rotation of an angle theta in the direction of the unit vector u we will have

$^{0}\mathbb{II}_{k} = \bold{R}_{01}\circ \ ^{1}\mathbb{II}_{k} \circ \bold{R}_{01}^T$

Where R01 is the rotation matrix and it's defined for any u

$\bold{R}_{01} = \bold{I} \cdot \cos \theta + \left[\bold{u}\right]_{\times} \cdot \sin \theta + (1-\cos \theta) \cdot \vec{u} \otimes \vec{u}$

Once the tensor II is a positive semi-definite matrix, all of its eigenvalues are not negative.

TODO

Put the equations to get the center of mass using the inertia tensor
Show the principal directions of a non-diagonal tensor
Show that the minimal tensor inertia eigenvalues are gotten when the tensor is calculated in the center of mass

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