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 $$ \rho_G\times(\omega_B\times\omega_B\times\rho_G - \rho_G\times\dot{\omega}_B) = 0 $$ ?$$ \rho_G\times(\omega_B\times\omega_B\times\rho_G - \rho_G\times\dot{\omega}_B) = 0~? $$
