Physical meaning of $a + w\times v $? ($a=$ acceleration, $w=$ angular velocity, $v=$ linear velocity) Given that a transformation matrix, $T_{ab} \in \mathbb{SE}(3)$, allows the transformation of coordinates, e.g. of a point $p \in \mathbb{R}^3$, expressed in a frame $b$ to be expressed in a frame $a$ by:
$$\hat{p}_a = T_{ab}\,\,\hat{p}_b = \begin{bmatrix}
R_{ab} & t_{ab}\\
0 & 1
\end{bmatrix}\hat{p}_b$$
Where $R_{ab}$ is the rotation matrix that expresses the orientation of frame $b$ w.r.t. frame $a$, $t_{ab}$ is the translation vector that expresses (in frame $a$) the origin of frame $b$ and $\hat{p}$ is the point $p$ expressed in homogeneous coordinates.
Concretely, I'm interested in a situation where $a$ is a stationary frame and $b$ is also a stationary frame that is instantaneously coincident with a frame attached to a moving rigid body whose instantaneous pose is expressed by $T_{ab}$.
Given this, I'm struggling to see if there is any physical meaning in the "translation" component of the following expression (where dot notation expresses differentiation w.r.t. time):
$$ \begin{align}
\dot{T_{ab}^{-1}}\,\,\dot{T_{ab}} + T_{ab}^{-1}\,\,\ddot{T_{ab}} \quad\equiv\quad \dot{T_{ba}}\,\,\dot{T_{ab}} + T_{ba}\,\,\ddot{T_{ab}} && \text{Given that } T_{ab}^{-1}=T_{ba} 
\end{align}$$
By developing the expression, we end up with the following "translation" term:
$$ R_{ba}\,\,\ddot{t_{ab}} + \dot{R_{ba}}\,\,\dot{t_{ab}}$$
Thereby we get as first term: $R_{ba}\,\,\ddot{t_{ab}}$ which can be seen as $a_b$ i.e. the acceleration of the origin of frame $b$ expressed in frame $b$.
Regarding the second term: $\dot{R_{ba}}\,\,\dot{t_{ab}}$, it can be seen as the cross product between the angular velocity experimented by frame $b$ (and expressed also in frame $b$) with the velocity of the origin of frame $b$ expressed in frame $b$, i.e. $ w_b \times v_b$. This is because:
$$  \dot{R_{ba}}\,\,\dot{t_{ab}} = [w_b]_\times R_{ba}\,\,\dot{t_{ab}} = [w_b]_\times\,\,\dot{t_{ba}} =[w_b]_\times v_b = w_b \times v_b $$
Where $[w_b]_\times$ is the skew-symmetric matrix representation of the angular velocity $w_b$.
But what I am having difficulties to see is if there is any physical meaning of the sum of these terms: $w_b \times v_b + a_b$.
Can please someone shed some light on this?
Thanks in advance!
 A: There are two ways to look at linear acceleration of a rigid body.

*

*Material Acceleration You can look at the acceleration of a specific particle on the body in terms of how the linear velocity components change over time. We call this the material acceleration. So we are are tracking a point A that is riding on the rotating reference frame

$$\overline{a}_A^m = \tfrac{\rm d}{{\rm d}t} \overline{v}_A \tag{1}$$

*

*Spatial Acceleration You can also look at the change in speed of whatever particle is passing under a fixed reference point
$$\overline{a}_A^s = \tfrac{\partial}{\partial t} \overline{v}_A \tag{2}$$
 See spatial acceleration entry on wikipedia. 
The two are related with a familiar relationship
$$ \boxed{ \overline{a}_A^m = \overline{a}_A^s + \overline{\omega} \times \overline{v}_A } \tag{3}$$
This is a direct result of the formula for differentiating a vector $\overline{A}$ on rotating frame
$$ \tfrac{\rm d}{{\rm d}t} \overline{A} = \tfrac{\partial }{\partial t} \overline{A} + \overline{\omega} \times \overline{A} \tag{4}$$
See rotating frames on MIT lecture series
Looking at your notation $a_b$ is a spatial acceleration and $a_b + \omega \times v_b$ the material acceleration.

Take the simple example of a spinning disk with non-zero rotational acceleration.
Look at point A on the disc located at $\vec{r}_A$ from the center at some point the velocity vector is
$$ \overline{v}_A = \overline{\omega} \times \overline{r} \tag{5}$$
Looking at the material acceleration the point is moving with time
$$ \overline{a}_A^m = (\tfrac{\rm d}{{\rm d}t} \overline{\omega} ) \times \overline{r}_A + \overline{\omega} \times \tfrac{\rm d}{{\rm d}t} \overline{r}_A = \overline{\alpha} \times \overline{r}_A + \overline{\omega} \times \overline{v}_A \tag{6}$$
But the spatial acceleration where the point is fixed in space
$$\require{cancel} \overline{a}_A^s = (\tfrac{\rm d}{{\rm d}t} \overline{\omega} ) \times \overline{r}_A + \overline{\omega} \times \cancel{ \tfrac{\rm d}{{\rm d}t} \overline{r}_A }= \overline{\alpha} \times \overline{r}_A  \tag{7}$$
As you can see the spatial acceleration (which is sometimes called Euler acceleration) contains only the result of the change in rotational speed of the disk. This is why it is useful because it describes the acceleration state of the entire body and not only a specific point in space.
You can transfer spatial acceleration from one point to another (say B) just as you do velocities
$$ \begin{aligned} 
 \overline{v}_B & = \overline{v}_A + \overline{\omega} \times ( \overline{r}_B - \overline{r}_A) \\
 \overline{a}_B^s & = \overline{a}_A^s + \overline{\alpha} \times ( \overline{r}_B - \overline{r}_A) \end{aligned} \tag{8} $$
and then convert to material acceleration by (3)
$$ \overline{a}_B^m = \overline{a}_B^s + \overline{\omega} \times \overline{v}_B \tag{9}$$
