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

Kinetics: Angular Momentum

We previously showed that \(\begin{align*} \bs{v}_t(\bs{x})&=\bs{\Omega}(t)\bs{x}+\bs{c}(t)\\ \label{eq:SpatialVel} \tag{SpatialVel} \end{align*}\)

where

\[\begin{align} \bs{\Omega}_t&=\dot{\bs{R}}_t\bs{R}_t^{\text{T}}\\ \bs{c}_t&=-\dot{\bs{R}}_t\bs{R}_t^{\text{T}}\bs{t}_t+\dot{\bs{t}}_t\\ \bs{\varphi}_{t}(\bs{X})&=\bs{R}_t\bs{X}+\bs{t}_t \end{align}\]

We also previously showed (see: Angular Velocity Tensor) that $\bs{\Omega}_t$ is a skew-symmetric tensor. According to our knowledge of axial vectors, we know that corresponding to every skew-symmetric tensor there exits a vector called its axial vector. We termed the axial vector corresponding to $\bs{\Omega}_t$ the ‘angular velocity vector,’ $\bs{\omega}_t$. Using $\bs{\omega}_t$, $\eqref{eq:SpatialVel}$ can be written as

\[\begin{align} \bs{v}_t(\bs{x})&=\bs{\omega}_t \times \bs{x}+\bs{c}_t \end{align}\]

It can be shown that

\(\begin{align} \bs{v}_t(\bs{x})&=\bs{\omega}_t \times \left(\bs{x} -\bar{\bs{x}}_t\right)+\bar{\bs{v}}_t \end{align}\) where $\bar{\bs{v}}_t$ is the velocity of the center of mass, i.e.,

\[\begin{align} \bar{\bs{v}}_t&=\bs{v}_t(\bar{\bs{x}}_t) \end{align}\]

In terms of the relative position vector $\bs{\pi}_t(\bs{x})=\bs{x}-\bar{\bs{x}}_t$, the spatial velocity field can be written as, \(\begin{align} \bs{v}_t(\bs{x})&=\bs{\omega}_t \times \bs{\pi}_t+\bar{\bs{v}}_t \end{align}\)

Preliminary results

Result 1

Recall that $[\bs{a}\otimes \bs{b}]_{ij}=a_i b_j$, the trace of $\bs{a}\otimes \bs{b}$ is \(\begin{align} \mathsf{Tr}\left(\bs{a}\otimes \bs{b}\right)&=a_i b_i, \\ &=\bs{a}\cdot\bs{b} \end{align}\)

Specifically,

\[\begin{align} \mathsf{Tr}\left(\bs{\pi}_t\otimes \bs{\pi}_t\right)&=\bs{\pi}_t\cdot\bs{\pi}_t \end{align}\]

Result 2

\[\begin{align} \mathsf{Tr}\left(\int_{\Omega_t}\bs{A}(\bs{x})\, d\Omega_t\right)&= \int_{\Omega_t}\mathsf{Tr}\left(\bs{A}(\bs{x})\right)\, d\Omega_t \end{align}\]

Inertia tensor in terms of the relative position vector

We defined the intertia tensor in terms of the euler tensor as,

\[\begin{align} \bs{J}_t &= \mathsf{Tr}\left(\bs{E}_t\right)\bs{I}-\bs{E}_t \\ &=\mathsf{Tr}\left(\bs{E}_t\right)\bs{I}-\int_{\Omega_t}\bs{\pi}_t\otimes \bs{\pi}_t\rho_t\, d\Omega_t\\ &=\mathsf{Tr}\left(\int_{\Omega_t}\bs{\pi}\otimes \bs{\pi}\rho_t\, d\Omega_t\right) \bs{I}-\int_{\Omega_t}\bs{\pi}_t\otimes \bs{\pi}_t\rho_t\, d\Omega_t\\ &=\int_{\Omega_t}\mathsf{Tr}\left(\bs{\pi}\otimes \bs{\pi}\right)\rho_t\, d\Omega_t \bs{I}-\int_{\Omega_t}\bs{\pi}_t\otimes \bs{\pi}_t\rho_t\, d\Omega_t\\ &=\int_{\Omega_t}\left(\bs{\pi}\cdot \bs{\pi}\right)\bs{I}\rho_t\, d\Omega_t -\int_{\Omega_t}\bs{\pi}_t\otimes \bs{\pi}_t\rho_t\, d\Omega_t\\ &=\int_{\Omega_t}\left(\left(\bs{\pi}\cdot \bs{\pi}\right)\bs{I}-\bs{\pi}_t\otimes \bs{\pi}_t\right)\rho_t\, d\Omega_t\\ \end{align}\]

Proof that $\int_{\Omega_t}\bs{\pi}_t\rho_t\, d\Omega_t=0$

\[\begin{align} \int_{\Omega_t}\bs{\pi}_t\rho_t\, d\Omega_t&= \int_{\Omega_t}\bs{x}\rho_t\, d\Omega_t-\int_{\Omega_t}\bar{\bs{x}}_t\rho_t\, d\Omega_t,\\ &=M \bar{\bs{x}}_t-\bar{\bs{x}}_t \int_{\Omega_t}\rho_t\, d\Omega_t,\\ &=M \bar{\bs{x}}_t-\bar{\bs{x}}_t M,\\ &=0 \end{align}\]

Angular momentum

The angular momentum of a solid is defined as

\[\begin{align} \bs{H}_t&=\int_{\Omega_t}(\bs{x}-\bar{\bs{x}}_t) \times \bs{v}_t(\bs{x})\rho_t(\bs{x})\, d\Omega_t \end{align}\] \[\begin{align*} \bs{H}_t&=\int_{\Omega_t}\bs{\pi}_t(\bs{x})\times \bs{v}_t(\bs{x})\rho_t(\bs{x})\, d\Omega_t\\ &=\int_{\Omega_t}\bs{\pi}_t\times \left(\bar{\bs{v}}_t+\bs{\omega}_t \times \bs{\pi}_t\right)\rho_t\, d\Omega_t,\\ &=\int_{\Omega_t}\bs{\pi}_t\times \bar{\bs{v}}_t\rho_t\, d\Omega_t+\int_{\Omega_t}\bs{\pi}_t\times\bs{\omega}_t \times \bs{\pi}_t \rho_t\, d\Omega_t,\\ &=\left(\int_{\Omega_t}\bs{\pi}_t\rho_t\, d\Omega_t \right)\times \bar{\bs{v}}_t+\int_{\Omega_t}\bs{\pi}_t\times\bs{\omega}_t \times \bs{\pi}_t \rho_t\, d\Omega_t,\\ &=\int_{\Omega_t}\bs{\pi}_t\times\bs{\omega}_t \times \bs{\pi}_t \rho_t\, d\Omega_t,\\ &=\int_{\Omega_t}\left(\left(\bs{\pi}_t\cdot\bs{\pi}_t\right) \bs{\omega}_t - \left(\bs{\pi}_t\cdot\bs{\omega}_t\right)\bs{\pi}_t \right)\rho_t\, d\Omega_t,\\ &=\left(\int_{\Omega_t}\left(\left(\bs{\pi}_t\cdot\bs{\pi}_t\right) \bs{I} - \bs{\pi}_t\otimes \bs{\pi}_t \right)\rho_t\, d\Omega_t\right) \bs{\omega}_t,\\ &=\bs{J}_t\bs{\omega}_t \end{align*}\]

Angular momentum about an arbitrary point

The angular momemtum about a point $\mathcal{A}$ is defined as,

\[\begin{align} \bs{H}_t^{\mathcal{A}}&=\int_{\Omega_t}(\bs{x}-\bs{x}^{\mathcal{A}})\times \rho_t(\bs{x})\bs{v}_t(\bs{x})\, d\Omega_t \end{align}\]

How is $\bs{H}_t^{\mathcal{A}}$ related to $\bs{H}$?

\(\begin{align} \bs{H}_t^{\mathcal{A}}&=\int_{\Omega_t}(\bs{x}-\bs{x}^{\mathcal{A}})\times \rho_t(\bs{x})\bs{v}_t(\bs{x})\, d\Omega_t\\ &=\int_{\Omega_t}(\bs{x}-\bar{\bs{x}}+\bar{\bs{x}}-\bs{x}^{\mathcal{A}})\times \rho_t(\bs{x})\bs{v}_t(\bs{x})\, d\Omega_t\\ &=\int_{\Omega_t}(\bs{x}-\bar{\bs{x}})\times \rho_t(\bs{x})\bs{v}_t(\bs{x})\, d\Omega_t+\int_{\Omega_t}(\bar{\bs{x}}-\bs{x}^{\mathcal{A}})\times \rho_t(\bs{x})\bs{v}_t(\bs{x})\, d\Omega_t\\ &=\bs{H}+(\bar{\bs{x}}-\bs{x}^{\mathcal{A}})\times\int_{\Omega_t} \rho_t(\bs{x})\bs{v}_t(\bs{x})\, d\Omega_t\\ \end{align},\) Noting that $\int_{\Omega_t} \rho_t(\bs{x})\bs{v}_t(\bs{x})\, d\Omega_t$ is the linear momentum of the solid $\bs{G}_t$, we get

\[\begin{align} \bs{H}_t^{\mathcal{A}}&=\bs{H}+(\bar{\bs{x}}-\bs{x}^{\mathcal{A}})\times \bs{G}_t\\ \end{align}\]

Angular momentum about the origin

For the special case of $\mathcal{A}=\mathcal{O}$ (the origin), we have that,

\[\begin{align} \bs{H}_t^{\mathcal{O}}&=\bs{H}+\bar{\bs{x}}\times \bs{G}_t\\ \end{align}\]

From the last two equations, we get that,

\[\begin{align} \bs{H}_t^{\mathcal{A}}-\bs{H}_t^{\mathcal{O}}&=-\bs{x}^{\mathcal{A}}\times \bs{G}_t\\ \bs{H}_t^{\mathcal{A}}+\bs{x}^{\mathcal{A}}\times \bs{G}_t&= \bs{H}_t^{\mathcal{O}}\\ \end{align}\]

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