# N Particle Rigid Body System

Rigid bodies can be defined as a system of mass points, whose positions relative to each other are fixed. We can also regard the body as a continuous and define the mass of a particle as $\rho dV$ the density times volume element. To specify the motion of a rigid body we introduce two coordinate systems. An inertial system $XYZ$ and a moving system $xyz$ that is fixed to the rigid body. The origin of the moving coordinate frame shall be at the bodies center of mass. The radial vector $\mathbf{R}$ points to the origin of the moving coordinate system $xyz$. A rigid body can be fully described by the three coordinates of $\mathbf{R}$ and three angles between the axis of the $xyz$ coordinate frame and $\mathbf{R}$.

A rigid body is fully characterized by 6 degrees of freedom

In the $xyz$ system a point has the coordinates $\mathbf{r}$, while in the system $XYZ$ the same point has coordinates $\bar{\mathbf{r}}$. An infinitesimal movement $d\bar{\mathbf{r}}$ of the rigid body consists of the movement of the center of mass $d\mathbf{R}$ and the rotation of the $xyz$ system $\vec{\varphi} \times \mathbf{r}$ \begin{align} d\bar{\mathbf{r}}=d\mathbf{R}+d\vec{\varphi} \times \mathbf{r} \end{align} If this movement happens in the time $dt$ we can define the velocities \begin{align} \frac{d\bar{\mathbf{r}}}{dt}=\mathbf{v} \quad \frac{d\mathbf{R}}{dt}=\mathbf{V} \quad \frac{d\vec{\varphi}}{dt}=\vec{\omega} \end{align} The rigid body's velocity thus is \begin{align} \mathbf{v}=\mathbf{V} +\vec{\omega} \times \mathbf{r}\tag{1} \end{align} $\mathbf{V}$ is called translatory velocity and $\vec{\omega}$ is known as angular velocity.

## Moment of Inertia Tensor

To calculate the kinetic energy of the rigid body system, we assume the system consists of mass points and write the kinetic energy as sum over all point masses \begin{align} T=\sum \frac{m \mathbf{v}^2}{2} \end{align} With the relation (1) the above equation becomes \begin{align} T=\sum \frac{m }{2} \left( \mathbf{V}+\vec{\omega} \times \mathbf{r} \right)^2=\sum \frac{m }{2} \mathbf{V}^2 +\sum m \left( \mathbf{V} \cdot\vec{\omega} \times \mathbf{r}\right)+\sum \frac{m }{2} \left( \vec{\omega} \times \mathbf{r} \right)^2\\ \end{align} In the first term we can pull out $\frac{\mathbf{V}^2}{2}$ since it's the same for every point mass and call $\sum m=\mu$ \begin{align} \sum \frac{m \mathbf{V}^2}{2} =\frac{\mathbf{V}^2}{2} \sum m = \frac{\mu \mathbf{V}^2}{2} \end{align} The second term can be rewritten according to the rule, that even permutations of $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ are equal $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}),$ we get \begin{align} \sum m \left( \mathbf{V} \cdot\vec{\omega} \times \mathbf{r}\right)=\sum m \left( \mathbf{r} \cdot \mathbf{V}\times \vec{\omega}\right) \end{align} The cross product $\mathbf{V}\times \vec{\omega}$ is the same for all mass points, we can pull this factor out \begin{align} \mathbf{V}\times \vec{\omega} \sum m \mathbf{r} \end{align} The sum $\sum m \mathbf{r}$ is zero, thus the second term cancels. Ultimately the square of the third term by Lagrange's identity is \begin{align} \left( \vec{\omega} \times \mathbf{r} \right)^2=\vec{\omega}^2 \cdot \mathbf{r}^2 - (\vec{\omega} \cdot \mathbf{r})^2 \end{align}

The kinetic energy can now be written as \begin{align} T=\frac{\mu \mathbf{V}^2}{2}+\frac{1}{2} \sum m \{ \omega^2 r^2 - (\vec{\omega}\mathbf{r})^2 \} \end{align}

Let us introduce the tensor notation for the angular kinetic energy \begin{align} T_{rot}&=\frac{1}{2} \sum m \{ \omega_i^2 x_l^2 - \omega_i \, x_i \, \omega_k \, x_k \}\\ &=\frac{1}{2} \sum m \{ \omega_i \, \omega_k \, \delta_{ik} x_l^2 - \omega_i \, \omega_k \, x_i \, x_k \}\\ &=\frac{1}{2} \omega_i \, \omega_k \sum m \{ \, \delta_{ik} x_l^2 - \, x_i \, x_k \} \end{align} where we used $\omega_i=\omega_k \, \delta_{ik}$. Now we define the sum as moment of inertia tensor \begin{align} I_{ik}= \sum m \{ \, \delta_{ik} x_l^2 - \, x_i \, x_k \} \end{align} and write the kinetic energy \begin{align} T=\frac{\mu \mathbf{V}^2}{2}+\frac{1}{2} I_{ik} \, \omega_i \, \omega_k \end{align}

The Lagrangian for the rigid body system is \begin{align} \mathcal{L}=\frac{\mu \mathbf{V}^2}{2}+\frac{1}{2} I_{ik} \, \omega_i \, \omega_k-U \end{align}

The moment of inertia for each particle is given by $I = \sum_{i} m_{i} \begin{pmatrix} y_i^2+z_i^2 & -x_i y_i & -x_i z_i \\ -y_i x_i & x_i^2+z_i^2 & -y_i z_i \\ -z_i x_i & - z_i y_i & x_i^2+y_i^2 \\ \end{pmatrix}$ The moment of inertia is additive, this means we get the total moment of inertia as sum over inertia of each particle. For a continuous system the sum becomes an integral $I_{ik}=\int \limits_{V} \rho(x_l^2 \delta_{ik} - x_i x_k) \,\mathrm{d}V$

## Steiner's Theorem

Steiner's theorem states that the moment of inertia of a rigid body about an arbitrary axis, is given by the moment of inertia about the bodies center of mass plus the moment of inertia of the bodies point mass around the chosen axis. \begin{align} I'_{ik}=I_{ik} + \mu (a^2 \, \delta_{ik} - a_i \, a_k) \end{align}

## Angular Momentum

The most convenient description of angular momentum is the one with the center of mass as origin for the moving coordinate frame. In this case we can write the angular momentum which is generally defined as \begin{align} \mathbf{L}=\sum m \cdot \mathbf{r} \cdot \mathbf{v} \end{align} with the equation $\mathbf{v}=\vec{\omega} \times \mathbf{r}$ as \begin{align} \mathbf{L}=\sum m \cdot \mathbf{r} \times (\vec{\omega} \times \mathbf{r}) \end{align} which is by the "BAC-CAB" fromula $\mathbf{A} \times (\mathbf{B} \times \mathbf{C})=\mathbf{B}(\mathbf{A}\cdot \mathbf{C})-\mathbf{C}(\mathbf{A}\cdot \mathbf{B})$ \begin{align} \mathbf{L}=\sum m \cdot \left(\vec{\omega} \cdot (\mathbf{r} \cdot \mathbf{r})-\mathbf{r} \cdot (\vec{\omega} \cdot \mathbf{r}) \right) \end{align} In tensor notation this is \begin{align} L_i=\sum m \{x_l^2 \, \omega_i - x_i \, x_k \, \omega_k \}=\omega_k \, \sum m \{x_l^2 \, \delta_{ik} - x_i \, x_k \} \end{align} By the definition of the moment of inertia this is

the angular momentum in tensor notation \begin{align} L_i= I_{ik} \omega_k \end{align} or in vector notation \begin{align} \mathbf{L}= \mathbf{I} \cdot \vec{\omega} \end{align}