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I am confused about the important concept expressed by these three, I think equivalent, theorems. Let me use an example: a rigid body is initially in a certain position and orientation in space (configuration A). After 2 seconds, the same body occupies a different position and has a different orientation (let's call it configuration B). The body arrived to the new configuration B through different intermediate configurations that we don't know unless we took a video of the entire motion and have all the instantaneous configurations between A and B. However, the mentioned theorem state that it is possible, mathematically, to bring the rigid body from configuration A to configuration B with just a specific translation followed by a specific rotation (or vice versa). Is that correct? That may be mathematically/geometrically true but it does not consider the actual, physical motion and sequence of actual configurations that the rigid body passed through as it went from configuration A to configuration B, right? Why are these theorems so useful?

thank you!

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"Why are these theorems so useful?"

I find the use in the kinematics of a rigid body when dealing with infinitesimal motion and/or velocities. Fundamentally, at any instant, the motion of a rigid body can be decomposed into a rotation about an arbitrary axis, coupled with a parallel translation along that axis.

First of you recognize that given the rotational velocity $\boldsymbol{\omega}$ of a rigid body and the velocity $\boldsymbol{v}_A$ of a specific point A you can find the velocity of any other point B by $$\boldsymbol{v}_B = \boldsymbol{v}_A + \boldsymbol{\omega} \times ( \boldsymbol{r}_B - \boldsymbol{r}_A )$$ where $\boldsymbol{r}_A$ and $\boldsymbol{r}_B$ are the location vectors. And the same for any other point riding on the rigid body.

So the entire motion of a rigid body is specified by the three rotation components $\boldsymbol{\omega}$ and the three velocity components $\boldsymbol{v}_A$ of any arbitrary point A. Hence it is common to say that a rigid body has 6 degrees of freedom.

Now consider the case where the body is rotating about the origin. Additionally, it moves parallel to the rotation axis by $\boldsymbol{v}_\parallel = h\, \boldsymbol{\omega}$. The scalar quantity $h$ (called the screw pitch) is the ratio of the translational velocity to the rotational velocity. Again, with that we can specify the velocity of any other point

$$ \boldsymbol{v}_A = h\,\boldsymbol{\omega} + \boldsymbol{\omega} \times \boldsymbol{r}_A \tag{1}$$

The above states that the three components of $\boldsymbol{v}_A$ can be composed by the scalar pitch $h$ and the two components of $\boldsymbol{r}_A$ that are not parallel to the rotation axis. So knowing where the rotation axis is and what the pitch is we know the motion of the body.

The interesting part is working the problem in reverse. Given the velocity of the body at a point, where is the axis of rotation? In the system above, given $\boldsymbol{v}_A$ we need to recover $\boldsymbol{r}_A$ as well as $h$.

This is done as follows:

$$\begin{aligned} h & = \frac{ \boldsymbol{v}_A \cdot \boldsymbol{\omega} }{ \| \boldsymbol{\omega} \|^2} \\ \boldsymbol{r}_A & = \frac{ \boldsymbol{v}_A \times \boldsymbol{\omega} }{ \| \boldsymbol{\omega} \|^2} \end{aligned} \ \tag{2}$$

In summary, Chasle's Theorem allows us to extract the geometry of motion at any instance but extracting the location of the rotation axis, as well as the screw pitch. The 6 degrees of freedom of a rigid body can be specified by either $(\boldsymbol{\omega},\,\boldsymbol{v}_A)$ or by

  • Rotation axis direction $\boldsymbol{\hat{z}}$ (two components)
  • Rotation speed $\omega$ (one component), such that $\boldsymbol{\omega} = \omega \boldsymbol{\hat{z}}$
  • Location of screw axis $\boldsymbol{r}_A$ (two components)
  • Screw pitch $h$ (one component)

For a total of 6 degrees of freedom. The decomposition of the motion is summarized as

$$ \begin{aligned} \boldsymbol{\omega} & = \omega \, (\boldsymbol{\hat{z}}) \\ \boldsymbol{v}_A & = \omega \left( h \boldsymbol{\hat{z}} + \boldsymbol{\hat{z}} \times \boldsymbol{r}_A \right) \end{aligned} \tag{3}$$

The above pair $(\boldsymbol{\hat{z}}, \, h\,\boldsymbol{\hat{z}} + \boldsymbol{\hat{z}}\times \boldsymbol{r}_A)$ designate the screw axis of the motion, in what is called Plucker line coordinates.


Proof

Use (2) in (1) to get

$$ \begin{aligned} \boldsymbol{v}_A & = \frac{ (\boldsymbol{v}_A \cdot \boldsymbol{\omega}) }{ \| \boldsymbol{\omega} \|^2} \boldsymbol{\omega} + \boldsymbol{\omega} \times \frac{ (\boldsymbol{v}_A \times \boldsymbol{\omega}) }{ \| \boldsymbol{\omega} \|^2} \\ & = \frac{ (\boldsymbol{v}_A \cdot \boldsymbol{\omega}) \boldsymbol{\omega}}{ \| \boldsymbol{\omega} \|^2} + \frac{ \boldsymbol{v}_A ( \boldsymbol{\omega} \cdot \boldsymbol{\omega}) - \boldsymbol{\omega} ( \boldsymbol{\omega} \cdot \boldsymbol{v}_A) }{ \| \boldsymbol{\omega} \|^2} \\ & = \frac{ \boldsymbol{v}_A \| \boldsymbol{\omega} \|^2}{\| \boldsymbol{\omega} \|^2} = \boldsymbol{v}_A \; \checkmark \end{aligned} $$

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  • $\begingroup$ The instantaneous axis of rotation IAR is defined by points with velocity parallel to $\omega$ at time t. The instantaneous motion of the entire rigid body is a helical displacement (rotation about IAR and translation along IAR). When a free body moves unconstrained in 3D space, its motion can be viewed as a sequence of helical displacements. The IAR is very useful in kinematics, less in dynamics where the CM is preferred. The body is seen to rotate about the CM and translate at the speed of the CM. When using the CM, the instantaneous displacement is not helical in this case. $\endgroup$ – user34203 Dec 30 '19 at 1:25
  • $\begingroup$ I would not say that a rigid body has a unique axis of rotation about which the body physically rotates (unless we have rotation about a fixed axis?). The fact that we can choose the IAR or an axis parallel to $\omega$ and passing through the CM as axes of rotation is a proof of that. What is unique and pertains to all points of the rigid body at time $t$ is the angular velocity vector $\omega$. That said, is it possible to physically track the instantaneous center of rotation (2D motion) using a high speed camera? How? $\endgroup$ – user34203 Dec 30 '19 at 1:30
  • $\begingroup$ In general, the CM has some velocity so the body is not rotating about the axis through the CM. At any instant, the motion of a rigid body is consistent with a rotation about IAR. I camera system that detects motion can identify the locations where there is no motion (in 2D). Those points are the centers of rotation. $\endgroup$ – John Alexiou Dec 30 '19 at 2:05
  • $\begingroup$ Thanks, I did not know that cameras could detect points on zero instantaneous speed, for 2D motion. For 3D motion, the IAR has a nonzero but minimal speed relative to all the other points. I guess a camera can detect those minimal speed points too. In 3D, both the IAR or the CM have some nonzero velocity. The thing is that the CM's velocity is not parallel to $\omega$ as it is for the points composing the IAR. $\endgroup$ – user34203 Dec 30 '19 at 2:58
  • $\begingroup$ still thinking about the screw axis...with a nonzero $\omega(t)$, if we sat on any point on a freely moving rigid body, we would see all other points appear to follow circular paths, correct? Also, the screw axis is unique...but there is really no true axis of rotation; the screw axis is just a convenient choice. The velocity distribution the body's points is as if translation with velocity v_A and rotation with $\omega$ about an axis through A were superimposed. This is true for every point A. But it causes confusion if the body has a fixed stationary point and if A is not this fixed point $\endgroup$ – user34203 Jan 3 '20 at 0:14

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