Does the axis of rotation of a rigid body depend on the frame of reference?

Question

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?

Answer 1

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.

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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.

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