Park and Lynch $F = ma$ derivation for a single rigid body

Question

I'm having some conceptual difficulties with a derivation presented in the Modern Robotics text by Park and Lynch. Here are the relevant parts:

At the beginning of the book, they insist that ${\it all}$ frames used in the book are stationary, ${\it even}$ body-fixed frames. This has been extremely confusing for me, as we are constantly talking about body frames evolving in time, and there are many animations on the YouTube videos that depict embedded body-fixed frames that are ${\it not}$ stationary.

I assume they are trying to say that all body-fixed frames follow a sequence of stationary frames? Anyway...

My main confusion comes from the $[\omega_b] v_b$ term in their $F = m a$ statement. What is this term? Suppose a planar rigid body is simultaneously translating and rotating with, say, $v_b = [1,0,0]$ and $\omega_b = [0,0,1]$. The center of mass does not accelerate for this motion. Clearly $\dot{v}_b = 0$, but $[\omega_b]v_b = [0,1,0]$, so the equations tell me that a force must be present in the body y-direction to support this motion.

What am I missing?

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EDIT:

Here is a possibility that makes sense to me. Suppose that the frame $\{{\bf b}_1, {\bf b}_2, {\bf b}_3\}$ is affixed to the body at its CG. Then, we can use this basis to describe the velocity of the CG of the body: $\bar{\bf v} = \bar{v}_i {\bf b}_i$. Now, the acceleration of the CG using this basis becomes $\bar{\bf a} = \dot{\bar{v}}_i {\bf b}_i + \boldsymbol{\omega} \times \bar{\bf v}$. Now if everything is represented on the $\{{\bf b}_i \}$ basis including the force ${\bf F} = F_i {\bf b}_i$ and angular velocity $\boldsymbol{\omega} = \omega_i {\bf b}_i$, then the matrix form of the three equations you get from ${\bf F} = m \bar{\bf a}$ becomes that obtained by the book.

This derivation is clean, but it assumes a $\textbf{non-inertial}$, $\textbf{non-stationary}$ frame $\{ {\bf b}_i \}$.

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EDIT 2: Here are the confusing remarks in question about stationary versus non-stationary frames:

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EDIT 3: Interpretation of (8.22)

Consider a wheel rolling without slip to the right. Mark two neighboring points, $A$ and $B$.

These two material points follow their own flows through space, as in the perspective of system dynamics or fluid mechanics. Their paths look something like this:

Velocity profile of disk at any time looks something like this:

The interpretation of (8.22) is similar to the Eulerian description of fluid mechanics. If we choose the inertial point coincident with material point $A$, then at the next time instance, $B$ will occupy that point in space. Therefore, $\dot{v}_b$ is upwards, and clearly $[\omega_b] v_b$ is downwards. The two terms cancel to tell us at the current time instance that the acceleration of material point $A$ is zero, as expected.

Answer 1

The key thing to remember is that none of the equations used in the Modern Robotics textbook use "body-fixed frames". The {b} frame is defined as a "body frame" which is instantaneously coincident (same position & orientation) with a body-fixed frame attached to the body (as stated in the grey box in your second edit).

This means that any body-fixed frame attached to any given body in a system can move as much as you want, but for every instant of time in which you need to formulate an equation, you will be using a fixed, inertial, instantaneous copy of the frame which the authors have (somewhat confusingly) labelled as the "body frame".

The second point that should be made is that the calculations you did in Edit 1 were for a stationary frame, not a rotating frame. Otherwise, you would have to also include one additional term in the velocity calculation and two additional terms in the acceleration calculation to deal with the rotating frame (see https://en.wikipedia.org/wiki/Centrifugal_force).