Newton-Euler equations in frame not centered at center of mass I'm having trouble with a step in the derivation of the Newton-Euler equations for rigid body translation and rotation when the body frame is not centered at the center of gravity.
The Newton-Euler equations in a frame $G$ attached to the center of gravity are
$$
\begin{bmatrix}
mI & 0 \\
0 & J
\end{bmatrix}
\begin{bmatrix}
\dot{v}_G \\
\dot{\omega}_G
\end{bmatrix} + 
\begin{bmatrix}
m\omega_G\times v_G \\
\omega_G\times J\omega_G
\end{bmatrix} = 
\begin{bmatrix}
f \\
\tau_G
\end{bmatrix},
$$
where $v_G$ is the body velocity, $\omega_G$ is the body angular velocity about the center of mass, $m$ is the mass of the rigid body and $J$ is the inertia tensor, $f$ is the applied force and $\tau_G$ is an applied torque about the center of mass.
Assume now that we are interested in describing the motion with respect to some other frame $B$ such that the center of mass is at a point $\rho_G$ in this frame (the orientation of the frame $B$ being the same as that of $G$, that is, the frames are related by pure translation). Then
$$
\omega_G = \omega_B \\
v_G = v_B - \rho_G\times\omega_B \\
\tau_G = \tau_B - \rho_G\times f
$$
from which it follows that the equations become
$$
m\dot{v}_B - (m\rho_G)\times\dot{\omega}_B + m(\omega_B\times v_B + \omega_B\times\omega_B\times\rho_G) = f \\
J\dot{\omega}_B + (m\rho_G)\times\dot{v}_B + \omega_B\times J\omega_B + (m\rho_G)\times(\omega_B\times v_B + \omega_B\times\omega_B\times\rho_G - \rho_G\times\dot{\omega}_B) = \tau_B.
$$
The derivation I'm following at this point simplifies the second equation to
$$
J\dot{\omega}_B + (m\rho_G)\times\dot{v}_B + \omega_B\times J\omega_B + (m\rho_G)\times(\omega_B\times v_B) = \tau_B.
$$
I can't understand why this is allowed. Why is it true that
$$
\rho_G\times(\omega_B\times\omega_B\times\rho_G - \rho_G\times\dot{\omega}_B) = 0~?
$$
 A: It seems that you are using spatial accelerations for $\dot{v}_G$. So your equations of motion at the center of mass G are correct
$$ \pmatrix{f \\ \tau_G} = \left[ \matrix{ m & \\ & J_G} \right] \pmatrix{\dot{v}_G \\ \dot{\omega}} + \pmatrix{m \omega \times v_G \\ \omega \times J_G \omega}  $$ 
Notice I added a subscript on the MMOI at it is expressed about G and removed it from $\omega$ since it is shared by the entire rigid body.
The first parts of the equation are easily transformed to a different point B with 
  $$\pmatrix{f \\ \tau_B} = \left[ \matrix{ 1 & 0 \\ \rho_G \times & 1} \right] \pmatrix{f \\ \tau_G}$$
and
  $$\pmatrix{\dot{v}_G \\ \dot{\omega}} = \left[ \matrix{ 1 & -\rho_G \times \\ 0 & 1} \right] \pmatrix{\dot{v}_G \\ \dot{\omega}}$$
Your question is, how do you transform the bias forces (the last part of the equations)?
Note that $$\begin{pmatrix}m\omega\times v_{G}\\
\omega\times J_{G}\omega
\end{pmatrix} = \begin{bmatrix}\omega\times\\
v_{G}\times & \omega\times
\end{bmatrix}\begin{bmatrix}m\\
 & J_{G}
\end{bmatrix}\begin{pmatrix}v_{G}\\
\omega
\end{pmatrix}$$
So the equations at B are
$$\begin{align}  
\pmatrix{f \\ \tau_B}  & = \underbrace{
   \left[ \matrix{ 1 & 0 \\ \rho_G \times & 1} \right]\left[ \matrix{ m & \\ & J_G} \right] 
\left[ \matrix{ 1 & -\rho_G \times \\ 0 & 1} \right] }_\mbox{6×6 spatial inertia at B}
\pmatrix{\dot{v}_G \\ \dot{\omega}} +\\ &
+ \left[ \matrix{ 1 & 0 \\ \rho_G \times & 1} \right]
\begin{bmatrix}\omega\times\\v_{G}\times & \omega\times\end{bmatrix}
\begin{bmatrix}m\\ & J_{G}\end{bmatrix}
\begin{bmatrix}1 & -\rho_{G}\times\\0 & 1\end{bmatrix}
\begin{pmatrix}v_{B}\\\omega\end{pmatrix} 
\end{align} $$
In the end you are trying to show that the last part is 
$$ 
+\begin{bmatrix}
\omega\times\\v_{B}\times & \omega\times\end{bmatrix}
\begin{bmatrix}1 & 0\\\rho_{G}\times & 1\end{bmatrix}
\begin{bmatrix}m\\ & J_{G}\end{bmatrix}
\begin{bmatrix}1 & -\rho_{G}\times\\0 & 1\end{bmatrix}
\begin{pmatrix}v_{B}\\\omega\end{pmatrix}$$
I can show this with
$$\begin{aligned}
\begin{bmatrix}\omega\times\\v_{B}\times & \omega\times\end{bmatrix}\begin{bmatrix}1 & 0\\
\rho_{G}\times & 1
\end{bmatrix}&=\\\begin{bmatrix}\omega\times\\
\left(v_{G}+\rho_{G}\times\omega\right)\times & \omega\times
\end{bmatrix}\begin{bmatrix}1 & 0\\
\rho_{G}\times & 1
\end{bmatrix}&=\begin{bmatrix}\omega\times\\
\left(v_{G}+\rho_{G}\times\omega\right)+\omega\times\rho_{G}\times & \omega\times
\end{bmatrix}\\\begin{bmatrix}\omega\times\\
v_{G}\times+\left(\rho_{G}\times\omega\right)\times+\omega\times\left(\rho_{G}\times\right) & \omega\times
\end{bmatrix}&=\begin{bmatrix}\omega\times\\
v_{G}\times+\left(\rho_{G}\times\omega\times-\omega\times\rho_{G}\times\right)+\omega\times\rho_{G}\times & \omega\times
\end{bmatrix}\\\begin{bmatrix}\omega\times\\
v_{G}\times+\rho_{G}\times\omega\times+ & \omega\times
\end{bmatrix}&=\begin{bmatrix}1 & 0\\
\rho_{G}\times & 1
\end{bmatrix}\begin{bmatrix}\omega\times\\
v_{G}\times & \omega\times
\end{bmatrix}
\end{aligned}$$
Use the vector identity $\left(\rho_{G}\times\omega\right)\times = \rho_{G}\times\omega\times-\omega\times\rho_{G}\times$
So, in the end, the equations of motion are
$$ \begin{aligned} 
& \mbox{at G} & & \mbox{at B}  \\
{\bf f}_G & = {\rm J}_G \dot{{\bf v}}_G + {\bf v}_G \times {\rm J}_G {\bf v}_G
&
{\bf f}_B & = {\rm J}_B \dot{{\bf v}}_B + {\bf v}_B \times {\rm J}_B {\bf v}_B
\\
{\bf f}_G & = \pmatrix{f \\ \tau_G} & {\bf f}_B & = \pmatrix{f \\ \tau_B} \\
{\bf v}_G & = \pmatrix{v_G \\ \omega} & {\bf v}_B & = \pmatrix{v_B \\ \omega} \\
\dot{{\bf v}}_G & = \pmatrix{\dot{v}_G \\ \dot{\omega}} & \dot{{\bf v}}_B & = \pmatrix{\dot{v}_B \\ \dot{\omega}} \\
{\rm J}_G & = \begin{bmatrix}m\\  & J_{G}\end{bmatrix} & {\rm J}_B & = \begin{bmatrix}m & -m \rho_G \times\\m \rho_G \times  & J_{G}-m \rho_G \times \rho_G \times\end{bmatrix} \\
{\bf v}_G \times & = \begin{bmatrix}\omega\times\\v_{G}\times & \omega\times\end{bmatrix} &{\bf v}_B \times & = \begin{bmatrix}\omega\times\\v_{B}\times & \omega\times\end{bmatrix}
\end{aligned}$$
NOTE:
It is worth in to use material accelerations to derive the equation of motion like with my answer here and then transform back to spatial accelerations (with $a_B = \dot{v}_B + \omega \times v_B$) to simplify the terms.
Also, The term $J_{B}=J_{G}-m \rho_G \times \rho_G \times$ is the vector form of the parallel axis theorem. All throughout I am using $\times$ as the 3×3 cross product operator $$r \times = \pmatrix{x\\y\\z} \times = \left[ \matrix{0&-z&y\\z&0&-x\\-y&x&0} \right]$$
