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

Kinetic energy

In this section we will derive the expression for the kinetic energy of a rigid solid. As you will see, the kinetic energy can be written as the two terms: one depending purely on the speed of the solid’s center of mass, and the second depending only on the angular velocity vector and inertia tensor (which is defined w.r.t to the center of mass).

Recall that the kinetic energy of a system of particles is defined as $m_i v_i/2$ where $m_i$ is the mass of the $i^\text{th}$ particle and $v_i$ is the speed of that particle. From this definition, it follows that the kinetic of a solid

\[\begin{align} \mathcal{T} &= \frac{1}{2} \int_{\Omega_t}\bs{v}_t(\bs{x})\cdot \bs{v}_t(\bs{x})\, \rho_t(\bs{x})d\Omega_t \label{eq:KE} \end{align}\]

It was shown on this page that

\[\begin{align} \bs{v}_t(\bs{x})&=\bs{\omega}_t \times \bs{\pi}_t(\bs{x})+\bar{\bs{v}}_t, \label{eq:vrel} \end{align}\]

and recall that $\bs{\pi}_t(\bs{x}):=\bs{x}-\bar{\bs{x}}_t$ is the relative position vector field. After substituting $\bs{v}_t(\bs{x})$ in $\eqref{eq:KE}$ with the right hand side of $\eqref{eq:vrel}$ and simplyfying, we get \(\begin{eqnarray} \mathcal{T}& =& \frac{1}{2} \int_{\Omega_t}\bar{\bs{v}}_t\cdot \bar{\bs{v}}_t\, \rho_t(\bs{x})d\Omega_t+\\ & &\frac{1}{2}\int_{\Omega_t}\bar{\bs{v}}_t\cdot (\bs{\omega}_t\times \bs{\pi}_t(\bs{x}))\, \rho_t(\bs{x})d\Omega_t+ \frac{1}{2}\int_{\Omega_t} (\bs{\omega}_t\times \bs{\pi}_t(\bs{x}))\cdot \bar{\bs{v}}_t\, \rho_t(\bs{x})d\Omega_t+\\ & &\frac{1}{2}\int_{\Omega_t} (\bs{\omega}_t\times \bs{\pi}_t(\bs{x}))\cdot (\bs{\omega}_t\times \bs{\pi}_t(\bs{x}))\, \rho_t(\bs{x})d\Omega_t\\ \end{eqnarray}\)

\[\begin{align} \mathcal{T} &= \frac{1}{2} M \bar{\bs{v}}_t\cdot \bar{\bs{v}}_t + \bar{\bs{v}}_t\cdot\bs{\omega}_t \times \int_{\Omega_t} \bs{\pi}_t(\bs{x})\, \rho_t(\bs{x})d\Omega_t + \frac{1}{2}\int_{\Omega_t} (\bs{\omega}_t \times \bs{\pi}_t(\bs{x}))\cdot (\bs{\omega}_t \times \bs{\pi}_t(\bs{x}))\, \rho_t(\bs{x})d\Omega_t\\ \end{align}\]

It can be shown that the second term on the RHS of the last equation vanishes. Therefore, we get that

\[\begin{align} \mathcal{T}&=\frac{1}{2} M \bar{\bs{v}}_t\cdot \bar{\bs{v}}_t + \frac{1}{2}\int_{\Omega_t} (\bs{\omega}_t \times \bs{\pi}_t(\bs{x}))\cdot (\bs{\omega}_t \times \bs{\pi}_t(\bs{x}))\, \rho_t(\bs{x})d\Omega_t \label{eq:KE2} \end{align}\]

The integrand of the last term in $\eqref{eq:KE2}$ can be written using ESC as

\[\begin{align} (\bs{\omega}_t \times \bs{\pi}_t(\bs{x}))\cdot (\bs{\omega}_t \times \bs{\pi}_t(\bs{x})) &= \epsilon_{ipq} (\omega_t)_p (\pi_t(x_\mathsf{i}))_q \epsilon_{imn} (\omega_t)_m (\pi_t(x_\mathsf{i}))_n \notag \\ &= (\delta_{pm}\delta_{qn}-\delta_{pn}\delta_{qm}) (\omega_t)_p (\pi_t(x_\mathsf{i}))_q (\omega_t)_m (\pi_t(x_\mathsf{i}))_n \notag \\ &= \delta_{pm}\delta_{qn}(\omega_t)_p (\pi_t(x_\mathsf{i}))_q (\omega_t)_m (\pi_t(x_\mathsf{i}))_n- \delta_{pn}\delta_{qm}(\omega_t)_p (\pi_t(x_\mathsf{i}))_q (\omega_t)_m (\pi_t(x_\mathsf{i}))_n \notag \\ &= (\omega_t)_m (\pi_t(x_\mathsf{i}))_n (\omega_t)_m (\pi_t(x_\mathsf{i}))_n- (\omega_t)_n (\pi_t(x_\mathsf{i}))_m (\omega_t)_m (\pi_t(x_\mathsf{i}))_n \notag \\ &= (\bs{\omega}_t\cdot \bs{\omega}_t) (\bs{\pi}(\bs{x})_t\cdot \bs{\pi}(\bs{x})_t) - (\bs{\omega}_t\cdot \bs{\pi}_t(\bs{x}))^2 \\ &= \bs{\omega}_t \cdot \left\{ (\bs{\pi}(\bs{x})_t\cdot \bs{\pi}(\bs{x})_t)\bs{I}-(\bs{\pi}(\bs{x})_t\otimes \bs{\pi}(\bs{x})_t) \right\}\bs{\omega}_t \end{align}\] \[\begin{align} \mathcal{T} &= \frac{1}{2} M \bar{\bs{v}}_t\cdot \bar{\bs{v}}_t + \frac{1}{2} \bs{\omega}_t \cdot \left[ \int_{\Omega_t} \left\{ (\bs{\pi}(\bs{x})_t\cdot \bs{\pi}(\bs{x})_t)\bs{I}-(\bs{\pi}(\bs{x})_t\otimes \bs{\pi}(\bs{x})_t) \right\}\, \rho_{t}(\bs{x})\, d\Omega_t \right]\bs{\omega}_t\\ &= \frac{1}{2} M \bar{\bs{v}}_t\cdot \bar{\bs{v}}_t + \frac{1}{2} \bs{\omega}_t \cdot \bs{J}_t\bs{\omega}_t \end{align}\]

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