# Cancelling internal forces/moments term when deriving inertial matrix

I am attempting to derive the inertial matrix for a general rigid body of mass $m$ as shown in the following diagram: The green vectors indicate the key position vectors:

Position of centroid relative to fixed coordinate system: $\pmb{R_G}$
Position of an element of mass $dm$ relative to fixed coordinate system: $\pmb{R}$
Position of element relative to centroid: $\pmb{r}$

The centroid moves with velocity $\pmb{\dot R_G}$ and acceleration $\pmb{\ddot R_G}$, and the body rotates with angular velocity $\pmb{\omega}$ and angular acceleration $\pmb{\dot \omega}$. Note that the body is not laminar, and that the direction of rotation is generalised.

The red quantities are the forces/moments acting on the element of mass $dm$:

Mass: $dm$
External force: $\pmb{dF}$
Internal force: $\pmb{dF_{int}}$
Internal moment: $\pmb{dM_{int}}$

THE ISSUE
I am attempting to derive the inertial matrix by determining the net moment about the centroid, $\pmb{M_G}$, and hoping that an equation of the form $\pmb{M_G} = I\pmb{\dot \omega}$ arises, where $I$ is the 3x3 inertial matrix.

First of all, I am assuming that $\pmb{M_G} = I \pmb{\dot \omega}$ is true, by extension of the 2D laminar rotation equation $M_G = J\dot \omega$, with $J$ being the scalar polar moment of inertia. Is that the case?

My first step is to take moments about the centroid by adding up all the contribution to moments for all elements of mass $dm$:

$$\pmb{M_G} = \int\limits_{m} \pmb{r} \times \pmb{dF}$$

Applying 2nd Law on the element:

$$\pmb{dF} + \pmb{dF_{int}} = dm \pmb{\ddot R}$$

$$\therefore \pmb{M_G} = \int\limits_{m} \pmb{r} \times \pmb{\ddot R} dm - \int\limits_{m} \pmb{r} \times \pmb{dF_{int}}$$

My main issue lies in my next step where I assume that the integral containing $\pmb{dF_{int}}$ is zero as I thought it was intuitive that the effects of the internal forces would cancel out when you add it up over the entire body. However, I have no way of confirming this, by reasoning or mathematics. Therefore, my question is: Is this assumption valid? If so, is there any way of proving why it is?

## 1 Answer

I think what you are trying to get to is

$${\rm d}{\bf I}_G = -[{\bf r}\times][{\bf r}\times]{\rm d}m$$

where ${\rm d}{\bf I}_G$ is the contribution to the mass moment of inertia of a mass ${\rm d}m$ located a distance ${\bf r}={\bf R}-{\bf R}_G$ from the center of mass. The 3×3 skew symmetric cross product operator $[{\bf r}\times]$ is defined as

$$\begin{pmatrix}x\\y\\z \end{pmatrix}\times = \begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}$$

(See example use at: https://physics.stackexchange.com/a/164413/392)

You arive at this using angular momentum. Each mass has linear momentum $${\rm d}p = {\bf v} {\rm d}m = ({\bf v}_G+{\bf \omega}\times{\bf r}){\rm d}m$$ where ${\bf v}_G$ is the velocity of the center of mass and ${\bf \omega}$ the rotational velocity of the rigid body. The angular momentum about the center of mass is defined as ${\rm d}{\bf h}_G = {\bf r}\times {\rm d}{\bf p}$ which is expanded as

$${\rm d}{\bf h}_G = {\bf r}\times ({\bf v}_G+{\bf \omega}\times{\bf r}){\rm d}m \\ = ({\bf r}\times {\bf v}_G-{\bf r}\times {\bf r}\times {\bf \omega}){\rm d}m$$

Together they form a 6×6 system:

$$\begin{pmatrix} {\rm d}{\bf p} \\{\rm d}{\bf h}_G \end{pmatrix} = \begin{bmatrix} {\rm d}m & -{\rm d}m [{\bf r}\times] \\ {\rm d}m [{\bf r}\times] & -{\rm d}m[{\bf r}\times][{\bf r}\times] \end{bmatrix} \begin{pmatrix} {\bf v}_G \\{\bf \omega} \end{pmatrix}$$

The element connecting angular momentum to rotational velocity we define as mass moment of inertia $${\rm d}{\bf h}_G = {\bf I}_G {\bf \omega} =(-[{\bf r}\times][{\bf r}\times] {\rm d}m) {\bf \omega}$$

(See also https://physics.stackexchange.com/a/88566/392 and links therein)