Kinetic energy and angular momentum of a rigid placement: $K=\frac{1}{2}m\parallel c^{\mathrm{\prime }}{\parallel }_2^2+\frac{1}{2}{\omega }^{\mathsf{T}}I\omega$ and $L_c=RI\omega$

By [L1], $R^{\mathrm{\prime }}=R[\omega {]}_{\times }$, and $[\omega {]}_{\times }\rho =\omega \times \rho$ by [L2]; hence for every $\rho$ the derivative of $s\mapsto x(s,\rho )$ at $t$ is ${\mathrm{\partial }}_{t}x(t,\rho )=c^{\mathrm{\prime }}(t)+R^{\mathrm{\prime }}(t)\rho =c^{\mathrm{\prime }}(t)+R(t)(\omega (t)\times \rho )$.

For every $\rho ,\omega$ one has the vector identity $\rho \times (\omega \times \rho )=(\parallel \rho {\parallel }_2^2{I}_3-\rho {\rho }^{\mathsf{T}})\omega$: for instance the first coordinate of the left side is ${\rho }_2({\omega }_1{\rho }_2-{\omega }_2{\rho }_1)-{\rho }_3({\omega }_3{\rho }_1-{\omega }_1{\rho }_3)={\omega }_1({\rho }_2^2+{\rho }_3^2)-{\rho }_1({\omega }_2{\rho }_2+{\omega }_3{\rho }_3)$, which equals the first coordinate $\parallel \rho {\parallel }_2^2{\omega }_1-{\rho }_1⟨\rho ,\omega ⟩$ of the right side, and the second and third coordinates are the cyclic analogues.

The first integrand satisfies $\parallel {\mathrm{\partial }}_{t}x(t,\rho ){\parallel }_2^2\le 4\parallel c^{\mathrm{\prime }}(t){\parallel }_2^2+4\parallel \omega (t){\parallel }_2^2\parallel \rho {\parallel }_2^2$ by step 1.1, [L3] and the two elementary bounds $\parallel a+b{\parallel }_2^2\le 2\parallel a{\parallel }_2^2+2\parallel b{\parallel }_2^2$ and $\parallel u\times v{\parallel }_2\le \parallel u{\parallel }_2\parallel v{\parallel }_2$; the second integrand satisfies $\parallel (x-c)\times ({\mathrm{\partial }}_{t}x-c^{\mathrm{\prime }}){\parallel }_2\le \parallel \rho {\parallel }_2\parallel \omega (t)\times \rho {\parallel }_2\le \parallel \omega (t){\parallel }_2\parallel \rho {\parallel }_2^2$ by step 1.1 and [L3]. Both are bounded on the support ball of $\nu$ by constants depending on $t$, so they are $\nu$-integrable by [L4]; this is the integrability clause of the statement.

Expanding the square with bilinearity and symmetry of the inner product and using [L3], $\parallel {\mathrm{\partial }}_{t}x{\parallel }_2^2=\parallel c^{\mathrm{\prime }}{\parallel }_2^2+2⟨c^{\mathrm{\prime }},R(\omega \times \rho )⟩+\parallel \omega \times \rho {\parallel }_2^2$.

By step 1.1 the relative velocity is $(x-c{)}^{\mathrm{\prime }}={\mathrm{\partial }}_{t}x-c^{\mathrm{\prime }}=R(\omega \times \rho )$, so the angular-momentum integrand is $(x-c)\times (x-c{)}^{\mathrm{\prime }}=R\rho \times R(\omega \times \rho )=R(\rho \times (\omega \times \rho ))$ by [L2].

Integrating step 2.2 with [L5] and pulling the constant $R(t)$ and the linear map $\rho \mapsto \omega \times \rho$ through the integral, $\int \parallel {\mathrm{\partial }}_{t}x{\parallel }_2^{2}d\nu =m\parallel c^{\mathrm{\prime }}{\parallel }_2^2+2⟨c^{\mathrm{\prime }},R(\omega \times \int \rho d\nu )⟩+\int \parallel \omega \times \rho {\parallel }_2^{2}d\nu \mathrm{.}$ The middle term vanishes because $\int \rho d\nu =0$ by [L4], and the last term equals ${\omega }^{\mathsf{T}}I\omega$ by [L6]; substituting into the definition of $K$ gives clause 2.

Integrating step 2.3 and substituting step 1.2, $L_c(t)=R(t)\int (\parallel \rho {\parallel }_2^2{I}_3-\rho {\rho }^{\mathsf{T}})d\nu \omega (t)=R(t)I\omega (t)$, the constant matrix $R(t)$ and the finite sum of integrals being interchanged by [L5] componentwise, and the last equality being the definition of $I$ in Mass distributions on ${\mathbb{R}}^3$: total mass, first moment and inertia matrix. This is clause 3.