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

(We are going to be working exclusively with non-dimensional quantities in this section.)

Kinetics: Time Derivative of Angular Momentum

Previously, in this section, we derived the angular momentum of the solid to be

\[\begin{align} \usf{H}(\tau) &= \usf{J}(\tau) \usf{w}(\tau) \end{align}\]

Differentiating the above equation w.r.t time, we get,

\[\begin{align} \usf{H}'(\tau) &= \usf{J}'(\tau) \usf{w}(\tau)+ \usf{J}(\tau) \usf{w}'(\tau) \label{eq:Hder} \end{align}\]

In a previous section we showed that \(\begin{align} \usf{J}'(\tau) &=\usf{W}(\tau)\usf{J}(\tau)+\usf{J}(\tau)\usf{W}^{\textsf{T}}(\tau),\\ \end{align}\)

From the above equation it follows that \(\begin{align} \usf{J}'(\tau) \usf{w}(\tau) &= \usf{W}(\tau)\usf{J}(\tau) \usf{w}(\tau)+\usf{J}(\tau)\usf{W}^{\textsf{T}}(\tau)\usf{w}(\tau), \label{eq:dotJ} \end{align}\)

In a previous section we showed that the angular velocity tensor is skew symmetric. Therefore, using the rule $\usf{W}^{\textsf{T}}\to - \usf{W}$ in $\eqref{eq:dotJ}$, we get

\[\begin{align} \usf{J}'(\tau) \usf{w}(\tau) &= \usf{W}(\tau)\usf{J}(\tau) \usf{w}(\tau)-\usf{J}(\tau)\usf{W}(\tau)\usf{w}(\tau),\\ &= \usf{W}(\tau)\usf{J}(\tau) \usf{w}(\tau)-\usf{J}(\tau) \usf{w}(\tau) \times \usf{w}(\tau),\\ &= \usf{W}(\tau)\usf{J}(\tau) \usf{w}(\tau),\\ \end{align}\]

or equivalently,

\[\begin{align} \usf{J}'(\tau) \usf{w}(\tau) &= \usf{w}(\tau) \times \usf{J}(\tau) \usf{w}(\tau),\\ \end{align}\]

Using the above result in $\eqref{eq:Hder}$ we get that, \(\begin{align} \usf{H}'(\tau) &= \usf{W}(\tau) \usf{J}(\tau) \usf{w}(\tau)+ \usf{J}(\tau) \usf{w}'(\tau) \label{eq:Hder2} \end{align}\)

When reviewing the time derivative of angular velocity tensors, we showed that

\(\begin{align} \usf{w}'(\tau) &=w_i'(\tau)\usf{e}_i(\tau). \end{align}\) Using the above result in $\eqref{eq:Hder2}$, we get

\[\begin{align} \usf{H}'(\tau) &= \usf{W}(\tau) \usf{J}(\tau) \usf{w}(\tau)+ \usf{J}(\tau) w_i'(\tau) \usf{e}_i(\tau). \end{align}\]

Classical form of the Euler equations

In this section we derived that the intertial tensor $\usf{J}(\tau)$ can be written as,

\[\begin{align} \usf{J}(\tau)&=(J_0)_{ij}\usf{e}_i(\tau)\otimes \usf{e}_j(\tau), \end{align}\]

where $(J_0)_{ij}$ are the componenets of the inertia tensor $\usf{J}_0$ w.r.t $\usf{E}_i$. That is,˛ \(\begin{align} \usf{J}_0&=(J_{0})_{ij} \usf{E}_i\otimes \usf{E}_j, \end{align}\)

In this subsection we assume that $\usf{E}_i$ have been chosen to be in the directions of the eigenvector of $\usf{J}_0$. This can always be accomplised because $\usf{J}_0$ is a symmetric tensor. It follows that

\(\begin{align} \usf{J}_0=\sum_{i}^{3}\lambda_i \usf{E}_i \otimes \usf{E}_i \end{align}\) and

\(\begin{align} \usf{J}(\tau)=\sum_{i=1}^{3}\lambda_i \usf{e}_i(\tau) \otimes \usf{e}_i(\tau) \end{align}\) The numbers $\lambda_i$ are called the principal moments of inertia.

For this special case, the rate of change of angular momentum simplifies to, \(\begin{align} \usf{H}'(\tau)=&\left(\lambda_1 w_1'(\tau)-(\lambda_2-\lambda_3)w_2(\tau) w_3(\tau)\right)\usf{e}_1+\notag \\ &\left(\lambda_2 w_2'(\tau)-(\lambda_3-\lambda_1)w_1(\tau) w_3(\tau)\right)\usf{e}_2+\notag\\ &\left(\lambda_3 w_3'(\tau)-(\lambda_1-\lambda_2)w_1(\tau) w_2(\tau)\right)\usf{e}_3 \label{eq:Hder3} \end{align}\)

Mathematica files

• Details of the simplification to arrive at $\eqref{eq:Hder3}$

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