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

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?

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 $$\{ {\bf b}_i \}$$.

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

• From the book, All frames in this book are stationary, inertial, frames. When we refer to a body frame {b}, we mean a motionless frame that is instantaneously coincident with a frame that is fixed to a (possibly moving) body. This is important to keep in mind, since you may have had a dynamics course that used non-inertial moving frames attached to rotating bodies. Do not confuse these with the stationary, inertial, body frames of this book. Dec 23, 2021 at 16:56
• cont., I'm reading this book now and it is confusing when they say all frames are stationary as I quoted above. This remark is not totally clear to me what they infer. Dec 23, 2021 at 16:58
• My understanding to their remark when they say stationary to the body frame, they mean it is aligned with the inertial frame but translated to a fixed point in the rigid body so that they can track the orientation; technically, there are two frames attached to the body frame. Not sure though. Dec 23, 2021 at 17:05
• Hi CroCo. Thanks for the reply. I too am not sure about the insistence about body frames being stationary. I am trying to understand their perspective and why they want to stress that; it seems to make things unnecessarily difficult. I imagine there is some point they are trying to get across that is getting lost in translation. I think they might want us to imagine at an instant that the body is passing through a stationary frame because when you get to the part about twists, you need to imagine how the body is moving at that instant, and maybe they don't want us rotating the body frame in Dec 23, 2021 at 17:24
• our heads while we are doing that. Check out my edit though, I think I figured out another perspective for how they are deriving their formula for a single rigid body. Feel free to message me if you have any other ideas. Dec 23, 2021 at 17:24

• So in the context of the wheel moving to the right at constant axle speed, if we look at the center point, then at the next time instance, a new material point will occupy that same point in space with a velocity that is to the right and up. Hence, $\dot{v_b}$ is up, while $[\omega_b] v_b$ is down, and the two terms cancel to tell us that the acceleration of the center of the wheel is zero. Jan 29 at 17:45