$$\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}$$

Kinetics: Euler and Inertia tensors

Relative position vectors

The relative position vectors are defined as follows: $\begin{align} \bs{\Pi}(\bs{X})&=\bs{X}-\bar{\bs{X}},\\ \bs{\pi}_{t}(\bs{x})&=\bs{x}-\bar{\bs{x}}_t, \end{align}$

Euler tensors

The Euler tensors $\bs{E}_0$ and $\bs{E}$ are defined as follows:

$$\begin{align} \bs{E}_0&=\int_{\Omega_0}\bs{\Pi}\otimes \bs{\Pi} \rho_0 \, d\Omega_0\\ \bs{E}_t&=\int_{\Omega_t}\bs{\pi}\otimes \bs{\pi} \rho_t \, d\Omega_t \end{align}$$

Inertia tensors

The inertia tensors $\bs{J}_0$ and $\bs{J}_t$ are defined as follows:

$$\begin{align} \bs{J}_0&=\textsf{tr}\left(\bs{E}_0\right)\bs{I}-\bs{E}_0,\\ \bs{J}_t&=\textsf{tr}\left(\bs{E}_t\right)\bs{I}-\bs{E}_t,\\ \end{align}$$

Computation of material Euler and inertia tensors

Ellipsoid example: Calculation in Mathematica

Computation of spatial Euler and inertia tensors

The spatial tensors $\bs{E}_t$ and $\bs{J}_t$ can be computed from their material counterparts. Specifically, it can be shown (part of your HW 5 prblem set) that,

$$\begin{align} \bs{E}_t&=\bs{R}_t\bs{E}_0\bs{R}_t^{\textsf{T}}\\ \bs{J}_t&=\bs{R}_t\bs{J}_0\bs{R}_t^{\textsf{T}} \end{align}$$

Let $\uv{E}_i$ be Cartesian basis vectors. Say that $(J_0)_{ij}$ (resp. $(E_0)_{ij}$) are the components of $\bs{J}$ (resp. $\bs{E}_0$) w.r.t. $\uv{E}_i$, i.e.,

$$\begin{align} (J_{0})_{ij}&= \uv{E}_i\cdot(\bs{J}_0\uv{E}_j)\\ (E_{0})_{ij}&= \uv{E}_i\cdot(\bs{E}_0\uv{E}_j) \end{align}$$

Following the dyadic representation of tensors, we can also write $\begin{align} \bs{J}_0&=(J_{0})_{ij}\uv{E}_i\otimes \uv{E}_j\\ \bs{E}_0&=(E_{0})_{ij}\uv{E}_i\otimes \uv{E}_j\\ \end{align}$

Say $\uv{e}_i$ are corresponding co-rotational basis vectors corresponding to $\uv{E}_i$. It can be shown that $\begin{align} \bs{J}_t &=(J_{0})_{ij}\uv{e}_i\otimes \uv{e}_j\\ \bs{E}_t &=(E_{0})_{ij}\uv{e}_i\otimes \uv{e}_j \end{align}$

Proof. Let $\bs{F}_t$ stand for either $\bs{E}_t$ or $\bs{J}_t$, and Let $\bs{F}_0$ stand for either $\bs{E}_0$ or $\bs{J}_0$.

$$\begin{align} \bs{F}_t&=\bs{R}_t\bs{F}_0\bs{R}_t^{\textsf{T}}\\ &=\bs{R}_t\left((F_0)_{ij}\uv{E}_i\otimes \uv{E}_j \right)\bs{R}_t^{\textsf{T}}\\ &=(F_0)_{ij}\bs{R}_t\left(\uv{E}_i\otimes \uv{E}_j\right) \bs{R}_t^{\textsf{T}}\\ &=(F_0)_{ij}\bs{R}_t\left(\uv{E}_i\otimes \left(\left( \bs{R}_t^{\textsf{T}} \right)^{\textsf{T}}\uv{E}_j\right)\right)\\ &=(F_0)_{ij}\bs{R}_t\left(\uv{E}_i\otimes \left(\bs{R}_t \uv{E}_j\right)\right)\\ &=(F_0)_{ij}\left(\left(\bs{R}_t\uv{E}_i\right)\otimes \left(\bs{R}_t \uv{E}_j\right)\right)\\ &=(F_0)_{ij}\left(\bs{R}_t\uv{E}_i\right)\otimes \left(\bs{R}_t \uv{E}_j\right)\\ &=(F_0)_{ij}\left(\uv{e}_i\right)\otimes \left(\uv{e}_j\right)\\ &=(F_0)_{ij}\uv{e}_i\otimes \uv{e}_j \end{align}$$

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