This question is an extension of question Understanding terms Twist and Wrench.
Assuming there is a rigid body with body twist denoted as $\mathcal{V}_{b}=\left(\boldsymbol \omega_{b}, \boldsymbol v_{b}\right)$.
In eq.8.40 of "Modern Robotics" by K. Lynch, the author presents the 6×6 equations of motion of this rigid body based on wrench and twist:
$\begin{aligned} \mathcal{F}_{b} & =\mathcal{G}_{b} \dot{\mathcal{V}}_{b}-\operatorname{ad}_{\mathcal{V}_{b}}^{\mathrm{T}}\left(\mathcal{P}_{b}\right) \\ & =\mathcal{G}_{b} \dot{\mathcal{V}}_{b}-\left[\operatorname{ad}_{\mathcal{V}_{b}}\right]^{\mathrm{T}} \mathcal{G}_{b} \mathcal{V}_{b}\end{aligned}$
where $\mathcal V_b$ denotes the body twist of this rigid body, and $\mathcal G_b$ is the spatial inertia matrix defined as $\mathcal{G}_{b}=\left[\begin{array}{cc}\mathcal{I}_{b} & 0 \\ 0 & \mathfrak{m} I\end{array}\right]$
In many literature within the field of robotics, including the previously mentioned question, we already know that the derivative of the body twist is
$\dot {\mathcal{V}}_{b}=\left(\begin{array}{c} \boldsymbol{\alpha}_{b}\\ \mathbf{a}_{b}-\boldsymbol{\omega}_{b} \times \boldsymbol{v}_{A}\end{array}\right)$
where $\boldsymbol{\alpha}_{b}$ represents the angular acceleration of the rigid body, $\mathbf{a}_{b}$ represents the traditional acceleration, and the term $\mathbf{a}_{b}-\boldsymbol{\omega}_{b} \times \boldsymbol{v}_{A}$ is called the spatial acceleration.
This dual part of the derivative of the body twist has the physical meaning of the acceleration of the rigid body without its centripetal acceleration component.
Now, I am really confused about:
Since the derivative of the body twist has already lost the content of centrifugal acceleration, does this make the 6×6 equations of motion above incomplete?
In other words, what I would like to determine in the equations of motion is the specific form of the dual part of $\mathcal {\dot V}_b$. Is it the traditional acceleration $\mathbf a_b$or spatial acceleration$\mathbf{a}_{b}-\boldsymbol{\omega}_{b} \times \boldsymbol{v}_{A}$???
