\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}}
\renewcommand{\b}[1]{\boldsymbol{#1}}
\newcommand{\t}[1]{\textsf{#1}}
\newcommand{\m}[1]{\mathbb{#1}}
\def\RR{\bf R}
\def\bold#1{\bf #1}
\def\mbf#1{\mathbf #1}
\def\uv#1{\hat{\usf {#1}}}
\def\dl#1{\underline{\underline{#1}}}
\newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}}
\def\bs#1{\usf #1}\]
Kinetics: Linear momentum
The liner momentum of the solid is defined as
\[\begin{align}
\bs{G}_t&=\int_{\Omega_t}\bs{v}_t(\bs{x})\rho_t(\bs{x})\, d\Omega_t
\end{align}\]
It can be shown that
\[\begin{align}
\bs{G}_t&=M \bs{v}_t^{\textsf{c.o.m}},
\end{align}\]
where
\(\begin{align}
\bs{v}_t^{\textsf{c.o.m}}=\bs{v}_t(\bs{x}_t^{\textsf{c.o.m}})
\end{align}\)
and $\bs{v}_t$ is the spatial velocity field.
Proof:
Starting from
$$
\begin{align}
\bs{\varphi}_t(\bs{X})&=\bs{R}_t\bs{X}+\bs{t}_t\\
\end{align}
$$
We previously showed that the velocity field of the rigid body can be computed as,
$$
\begin{align*}
\bs{v}_t(\bs{x})&=\bs{\Omega}_t\bs{x}+\bs{c}_t\\
\label{eq:vel}
\tag{SpatialVel}
\end{align*}
$$
where
$$
\begin{align*}
\bs{\Omega}_t&=\dot{\bs{R}_t}\bs{R}_t^{\textsf{T}}\\
\bs{c}_t&=-\dot{\bs{R}_t}\bs{R}_t^{\textsf{T}}\bs{t}_t+\dot{\bs{t}_t}
\end{align*}
$$
Integrating both sides of equation $\eqref{eq:vel}$, we get that
$$
\begin{align}
\int_{\Omega_t}\bs{v}_t(\bs{x})\rho_t(\bs{x})\, d\Omega_t
&=\int_{\Omega_t}(\bs{\Omega}_t\bs{x}+\bs{c}_t)\rho_t(\bs{x})\, d\Omega_t\\
&=\int_{\Omega_t}\bs{\Omega}_t\bs{x}\rho_t(\bs{x})\, d\Omega_t+\int_{\Omega_t}\bs{c}_t\rho_t(\bs{x})\, d\Omega_t\\
&=\bs{\Omega}_t\int_{\Omega_t}\bs{x}\rho_t(\bs{x})\, d\Omega_t+\bs{c}_t\int_{\Omega_t}\rho_t(\bs{x})\, d\Omega_t\\
&=\bs{\Omega}_t M \bs{x}_t^{\textsf{c.o.m}}+\bs{c}_t M\\
&=M\left(\bs{\Omega}_t \bs{x}_t^{\textsf{c.o.m}}+\bs{c}_t \right)\\
&=M\bs{v}_t(\bs{x}_t^{\textsf{c.o.m}})\\
\end{align}
$$
[edit]