Inertia matrix

The inertia matrix $\overline{I}$ is a 3X3 matrice that describes how a rigid body’s mass is distributed relative to a chosen origin.

It relates the body’s angular velocity $\vec{\omega}$ to its angular momentum $\vec{L}$:

$$ \vec{L} = \overline{I}\vec{\omega} $$

Definition in orthonormal basis

In a orthonormal basis (Cartesian coordinate system) $B$ with origin at point $O$, and position vector $\vec{r}_m=(x,y,z)$ of each mass element $dm$, the inertia matrix of an object or system $(S)$ is

$$\overline{I}(O,S)_B = \begin{pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} & I_{yz}\\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix}$$

where $\mathbf{I}(O,S)_B$ is the inertia tensor of object or system (S) according to cartesian coordinate system $B$ of origin $O$

with

$$ I_{xx}=\int_S (y^2+z^2)\,dm,\qquad I_{yy}=\int_S (x^2+z^2)\,dm,\qquad I_{zz}=\int_S (x^2+y^2)\,dm $$

and

$$ I_{xy}=I_{yx}=-\int_S xy\,dm,\qquad I_{xz}=I_{zx}=-\int_S xz\,dm,\qquad I_{yz}=I_{zy}=-\int_S yz\,dm $$

It can also be written as:

$$ \overline{I}(O,S)_B = \int_S \left[(\vec{r}_m\cdot\vec{r}_m)\,\mathbf{1}-\vec{r}_m\otimes\vec{r}_m\right]dm $$

where $\vec{r}_m=(x,y,z)$ is always the position vector of $dm$, the $\mathbf{1}$ is the 3x3 identity matrix, and $\otimes$ denotes the outer product.

Derivation

If the mass element $dm \in S$ at point $M$ rotates about $O$ with angular velocity $\vec{w}$, then its linear velocity

$$ \vec{v}(M)=\vec{w}\wedge\vec{r}. $$

Its angular momentum about $O$ $\vec{L}_O$ satisfies

$$ \vec{L}_O = \int_S \vec{r}\wedge\vec{v}(M)\,dm = \int_S \vec{r}\wedge(\vec{w}\wedge\vec{r})\,dm. $$

Since

$$ \vec{r}\wedge(\vec{w}\wedge\vec{r}) = (\vec{r}\cdot\vec{r})\vec{w} - (\vec{r}\cdot\vec{w})\vec{r} $$

Writing $\vec{r}=(x,y,z)$ and $\vec{w}=(w_x,w_y,w_z)$,

$$ \vec{r}\wedge(\vec{w}\wedge\vec{r}) = \begin{pmatrix} (y^2+z^2)u_x-xyu_y-xzu_z\\ -xyu_x+(x^2+z^2)u_y-yzu_z\\ -xzu_x-yzu_y+(x^2+y^2)u_z \end{pmatrix} $$

Integrating over the solid gives

$$ \int_S \vec{r}\wedge(\vec{w}\wedge\vec{r})\,dm = \begin{bmatrix} \int_S(y^2+z^2)\,dm & -\int_S xy\,dm & -\int_S xz\,dm\\ -\int_S xy\,dm & \int_S(x^2+z^2)\,dm & -\int_S yz\,dm\\ -\int_S xz\,dm & -\int_S yz\,dm & \int_S(x^2+y^2)\,dm \end{bmatrix} \begin{pmatrix} w_x\\ w_y\\ w_z \end{pmatrix} $$

Hence

$$ \overline{I}(O,S)_B\cdot\vec{w} = \begin{bmatrix} \int_S(y^2+z^2)\,dm & -\int_S xy\,dm & -\int_S xz\,dm\\ -\int_S xy\,dm & \int_S(x^2+z^2)\,dm & -\int_S yz\,dm\\ -\int_S xz\,dm & -\int_S yz\,dm & \int_S(x^2+y^2)\,dm \end{bmatrix} \vec{w} $$

By definition, the inertia matrix $\overline{I}(O,S)_B$ satisfies that

$$ \vec{L}_O=\overline{I}(O,S)_B\vec{w}. $$

Therefore,

$$ \overline{I}(O,S)_B = \begin{bmatrix} \int_S(y^2+z^2)\,dm & -\int_S xy\,dm & -\int_S xz\,dm\\ -\int_S xy\,dm & \int_S(x^2+z^2)\,dm & -\int_S yz\,dm\\ -\int_S xz\,dm & -\int_S yz\,dm & \int_S(x^2+y^2)\,dm \end{bmatrix} $$

Principal basis of inertia

$\forall O, \exists$ orthonormal basis $B \Rightarrow$

$$ \overline{I}_O(S) = \begin{pmatrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0 & 0 & I_{zz} \end{pmatrix}$ $$

$I_{xx},I_{yy},I_{zz}$ are then known as the principal moments of inertia for $(S)$ according to $O$.

Change of origin

$\forall$ orthonormal basis $B$ defined by orthonormal vectors $(e_1,e_2,e_3)$, if the centre of inertia of $(S)$ is at $C$, and $\overrightarrow{OC}=(a,b,c)$, then

$$ \overline{I}(O,S)_B = \overline{I}(C,S)_B + \begin{bmatrix} m(b^2+c^2) & -mab & -mac \\ -mab & m(a^2+c^2) & -mbc \\ -mac & -mbc & m(a^2+b^2) \end{bmatrix}_{B} $$

Derivation

For any angular velocity $\vec{\omega}$,

$$\begin{gathered} \overline{I}(O,S)_B\vec{\omega} = \vec L(O,S)_B \\ = \int_S \overrightarrow{OM}\wedge(\vec{\omega}\wedge\overrightarrow{OM})\,dm \end{gathered}$$

Since

$$ \overrightarrow{OM}=\overrightarrow{OC}+\overrightarrow{CM}, $$

$$ \begin{aligned} \overline{I}(O,S)_B\vec{\omega} ={}& \int_S \overrightarrow{OC}\wedge(\vec{\omega}\wedge\overrightarrow{OC})\,dm +\int_S \overrightarrow{OC}\wedge(\vec{\omega}\wedge\overrightarrow{CM})\,dm \\ &+\int_S \overrightarrow{CM}\wedge(\vec{\omega}\wedge\overrightarrow{OC})\,dm +\int_S \overrightarrow{CM}\wedge(\vec{\omega}\wedge\overrightarrow{CM})\,dm ={}& \overrightarrow{OC}\wedge(\vec{\omega}\wedge\overrightarrow{OC})\int_S dm +\overrightarrow{OC}\wedge\left(\vec{\omega}\wedge\int_S \overrightarrow{CM}\,dm\right) \\ &+\left(\int_S \overrightarrow{CM}\,dm\right)\wedge(\vec{\omega}\wedge\overrightarrow{OC}) +\int_S \overrightarrow{CM}\wedge(\vec{\omega}\wedge\overrightarrow{CM})\,dm \end{aligned} $$

Note $\int_S \overrightarrow{CM}\wedge(\vec{\omega}\wedge\overrightarrow{CM})\,dm=\vec{L}(C,S)_B = \overline{I}(C,S)_B\vec{\omega}$,

Because $C$ is the centre of inertia,

$$ \int_S \overrightarrow{CM}\,dm=0. $$

Hence

$$ \overline{I}(O,S)_B\vec{\omega} = \overline{I}(C,S)_B\vec{\omega} + m\,\overrightarrow{OC}\wedge(\vec{\omega}\wedge\overrightarrow{OC}). $$

To calculate $\overrightarrow{OC}\wedge(\vec{\omega}\wedge\overrightarrow{OC})$, use the vector triple product identity:

$$ \vec{a}\wedge(\vec{b}\wedge\vec{c}) = \vec{b}(\vec{a}\cdot\vec{c}) -\vec{c}(\vec{a}\cdot\vec{b}). $$

Taking

$$\begin{gathered} \vec{a}=\overrightarrow{OC},\qquad \vec{b}=\vec{\omega},\qquad \vec{c}=\overrightarrow{OC}, \\ \overrightarrow{OC}\wedge(\vec{\omega}\wedge\overrightarrow{OC}) = \vec{\omega}(\overrightarrow{OC}\cdot\overrightarrow{OC}) -\overrightarrow{OC}(\overrightarrow{OC}\cdot\vec{\omega}) \\ \because \overrightarrow{OC}=a e_1+b e_2+c e_3,\qquad \vec{\omega}=\omega_x e_1+\omega_y e_2+\omega_z e_3, \\ \therefore \overrightarrow{OC}\cdot\overrightarrow{OC}=a^2+b^2+c^2 \end{gathered}$$

Hence

$$ \vec{\omega}(\overrightarrow{OC}\cdot\overrightarrow{OC}) = \begin{bmatrix} a^2+b^2+c^2 & 0 & 0 \\ 0 & a^2+b^2+c^2 & 0 \\ 0 & 0 & a^2+b^2+c^2 \end{bmatrix}_{B} \begin{bmatrix} \omega_x\\ \omega_y\\ \omega_z \end{bmatrix}_{B} $$

Also,

$$ \overrightarrow{OC}\cdot\vec{\omega} = (a e_1+b e_2+c e_3)\cdot(\omega_x e_1+\omega_y e_2+\omega_z e_3) = a\omega_x+b\omega_y+c\omega_z. $$

Thus

$$ \begin{aligned} \overrightarrow{OC}(\overrightarrow{OC}\cdot\vec{\omega}) &= (a e_1+b e_2+c e_3)(a\omega_x+b\omega_y+c\omega_z) \\ &= a(a\omega_x+b\omega_y+c\omega_z)e_1 +b(a\omega_x+b\omega_y+c\omega_z)e_2 +c(a\omega_x+b\omega_y+c\omega_z)e_3 \\ &= \begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{bmatrix}_{B} \begin{bmatrix} \omega_x\\ \omega_y\\ \omega_z \end{bmatrix}_{B}. \end{aligned} $$

Hence

$$ \begin{aligned} \overrightarrow{OC}\wedge(\vec{\omega}\wedge\overrightarrow{OC}) &= \begin{bmatrix} a^2+b^2+c^2 & 0 & 0 \\ 0 & a^2+b^2+c^2 & 0 \\ 0 & 0 & a^2+b^2+c^2 \end{bmatrix}_{B} \begin{bmatrix} \omega_x\\ \omega_y\\ \omega_z \end{bmatrix}_{B} - \begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{bmatrix}_{B} \begin{bmatrix} \omega_x\\ \omega_y\\ \omega_z \end{bmatrix}_{B} \\ &= \begin{bmatrix} b^2+c^2 & -ab & -ac \\ -ab & a^2+c^2 & -bc \\ -ac & -bc & a^2+b^2 \end{bmatrix}_{B} \begin{bmatrix} \omega_x\\ \omega_y\\ \omega_z \end{bmatrix}_{B}. \end{aligned} $$

Thus

$$ \overline{I}(O,S)_B\vec{\omega} = \overline{I}(C,S)_B\vec{\omega} + m \begin{bmatrix} b^2+c^2 & -ab & -ac \\ -ab & a^2+c^2 & -bc \\ -ac & -bc & a^2+b^2 \end{bmatrix}_{B} \vec{\omega} $$

Divide both sides by $\vec{\omega}$ and multiply $m$ into the matrix,

$$ \overline{I}(O,S)_B = \overline{I}(C,S)_B + \begin{bmatrix} m(b^2+c^2) & -mab & -mac \\ -mab & m(a^2+c^2) & -mbc \\ -mac & -mbc & m(a^2+b^2) \end{bmatrix}_{B}. $$