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When studying the kinematic motion of a rigid body, angular velocity $\omega$ is a vector that not seem to specify a unique axis of rotation... When looking at the free rigid body motion of a wheel rolling without sliding, we can talk about the wheel's rotation from the point of view of a fixed frame of reference and in that case talk about rotation about the instantaneous center of rotation (which is the contact point) or we can talk about rotation from the center of mass frame of reference and in that acase the center of rotation is the center of mass itself. From a frame of reference that is fixed with the wheel (body axes), the wheel does not rotate at all because the frame of reference rotates and every point looks stationary. Chasles theorem states that a rigid body can pass from one configuration to the next one via one of the infinite combinations of translation/rotation about any arbitrary point which becomes the point of rotation for that transformation. All transformations share the same $\omega$... Does that mean that rotation is a relative concept and there is no unique, physical, axis of rotation for a rigid body? I have read about the instantaneous screw axis where the points of the rigid body with the same velocity parallel to the axis reside... Certainly, when a free rigid body rotates while translating, maybe tumbling in some random fashion, the initial conditions (how the body starts, the forces in action) should uniquely determine the rigid body's configuration at every instant $t$ and how it kinematically moves from one configuration to the next: even if Chasles theorem states that there are infinite possible combinations (translation+rotation), the body certainly moves from its current configuration to the next configuration in a very specific way...Can anyone shine some clarity on this topic?

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    $\begingroup$ Would you mind refining your question into categories such as what you understand, what you are confused about etc.. it makes it much easier to address it $\endgroup$ – InertialObserver Jan 6 '19 at 0:46
  • $\begingroup$ I am having a hard time understanding what exactly you are asking, and I consider myself an expert in rigid body kinematics and screw theory. $\endgroup$ – John Alexiou Jan 6 '19 at 1:59
  • $\begingroup$ the apparent axis of rotation will depend on your frame of reference, same as your velocity, kinetic energy, and many other variables $\endgroup$ – Wolphram jonny Jan 6 '19 at 4:00
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I don't understand the question.

At any instance, in any one coordinate frame if the position $\boldsymbol{r}_A$, velocity $\boldsymbol{v}_A$ of a point A in a rigid body rotating with $\boldsymbol{\omega}$ is known or measured then the location of the instantaneous rotation axis is given by the following calculation

  • Direction (Vector) - The direction of rotation is $$ \boldsymbol{e} = \frac{ \boldsymbol{\omega} }{ \| \boldsymbol{\omega} \| }$$

  • Position (Vector) - The point on the rotation axis closest to the coordinate frame is $$\boldsymbol{r}_{\rm COR} = \boldsymbol{r}_A + \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_A} { \| \boldsymbol{\omega} \|^2 }$$

  • Magnitude (Scalar) - The magnitude of rotation $$ \omega = \| \boldsymbol{\omega} \| $$

  • Pitch (Scalar) - The ratio of the parallel motion (translation) along the rotation axis to the rotation magnitude $$ h = \frac{ \boldsymbol{\omega} \cdot \boldsymbol{v}_A }{ \| \boldsymbol{\omega} \|^2 } $$

    Here $\cdot$ is the vector inner product and $\times$ the vector cross product. Vector quantities are shown in boldface.

I can provide proof of the above if needed. So from the basis of the statements above can you comment below and rephrase./summarise your question.


Equations of Motion

Equations of motion for a rigid body are derived from the time derivative of momentum. When expressed at the center of mass (Point C) the momentum equations are

$$ \begin{aligned} \boldsymbol{p} & = m \boldsymbol{v}_C \\ \boldsymbol{L}_C & = \mathrm{I}_C \,\boldsymbol{\omega} \end{aligned} $$

Where $\mathbf{I}_C$ is the mass moment of inertia 3×3 tensor about the center of mass, and $\boldsymbol{v}_C$ the velocity of the center of mass.

As you can see, as expressed at the center of mass the equations are rather simple.

Now transform the above equations to an arbitrary point A and the above become

$$ \begin{aligned} \boldsymbol{p} & = m ( \boldsymbol{v}_A - \boldsymbol{c} \times \boldsymbol{\omega}) \\ \boldsymbol{L}_A & = \mathrm{I}_A \boldsymbol{\omega} + \boldsymbol{c} \times m \boldsymbol{v}_A \end{aligned} $$

Where $\boldsymbol{c}$ is the position vector of the center of mass from the reference point A. Also $\mathbf{I}_A$ is the mass moment of inertia tensor at A.

As you can see, the level of complexity has increased significantly since now there are cross terms (linear momentum depends on $\boldsymbol{\omega}$ and angular momentum on $\boldsymbol{v}_A$

Now ask the question if resolving the equations of motion onto the center of rotation makes things any simpler. The answer is just marginally yes. If A is the center of rotation, then $\boldsymbol{v}_A = h\, \boldsymbol{\omega}$ where the scalar $h$ is the pitch. That is it. That is all the simplifies at the center of rotation.

$$ \begin{aligned} \boldsymbol{p} & = m ( h\,\boldsymbol{\omega} - \boldsymbol{c} \times \boldsymbol{\omega}) \\ \boldsymbol{L}_{\rm COR} & = \mathrm{I}_{\rm COR} \boldsymbol{\omega} + m\,h\,(\boldsymbol{c} \times \boldsymbol{\omega}) \end{aligned} $$

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  • $\begingroup$ Sorry for being unclear. Most physical quantities are relative (velocity, position, etc.). I am confused on which axis of rotation a free rigid body spinning/tumbling in the air in some arbitrary fashion is rotating about. Is there a "unique" axis of rotation? From the fixed reference frame (earth), is the unique, instantaneous rotational axis the screw axis (Mozzi axis)? The particular kinematic motion of a body spinning is determined by the initial conditions. Mozzi axis may not pass from the center of mass but we can describe rotation about the CM. There are choices for the rotation axis. $\endgroup$ – Brett Cooper Jan 6 '19 at 15:27
  • $\begingroup$ The axis is like a physical object. You can define its position relative to the coordinate system used as described above. At any instant, it is unique and finite or at infinity (pure translation). The initial conditions play no specific role here. At any instant, it is the motion (velocity) vectors that matter only. For a body constrained by a joint (like a hinge) the axis is a physical object. $\endgroup$ – John Alexiou Jan 6 '19 at 22:39
  • $\begingroup$ Thanks. @Ja72, you mention that the initial conditions play no role. But one body launched in certain way tumbles differently that an object launched in a different manner. The Mozzi rotation axis is unique at any instant... so how should I view the fact that, geometrically, it is possible to move the body from one configuration to the next via a translation+rotation about any arbitrary point? If the point is arbitrary, then the axis of rotation is arbitrary even if the angular velocity is the same for any point about which the rotation happens.... $\endgroup$ – Brett Cooper Jan 7 '19 at 0:51
  • $\begingroup$ A free floating body will always have the rotation axis either on the center of mass, or co-moving with the center of mass. I still don't understand the question. The rotation axis is an idealization that helps describe the motion and from one instant to the next it might change rapidly. The location of the axis doesn't have to be a smooth function. its like the contact point of a body that can jump around. $\endgroup$ – John Alexiou Jan 8 '19 at 0:13
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    $\begingroup$ Hello, just to summarize: the various points of a freely moving rigid body have their own velocity but all share the same $\omega$. Knowing the velocity of a certain point and $\omega$, we can find the velocity of any other point. If that "certain point" has a velocity instantaneously parallel to $\omega$, then the inst axis of rotation passes through that point. $\endgroup$ – user34203 Dec 15 '19 at 17:28

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