How can I prove the existence of rest points in rigid motion with all speeds orthogonal to $\vec{\omega}$?

If in a rigid motion there is a point at rest, then all speeds are normal to $$\vec{\omega}$$ (assumed $$\neq \vec{0}$$), to prove choose the rest point as origin and explot fundamental equation of rigid motion: you find $$\vec{v}_P = \vec{\omega} \times \vec{r}_P$$. But how can i prove the vice versa: if all speeds are $${\perp}$$ to $$\vec{\omega}$$ then it exists a rest point (and so infinite rest points)?

1. Show that all rigid motion is a combination of rotation plus translation
2. Rotation is rotation around an axis. If there is no translation, the axis is fixed.
3. If there is translation, the axis is moved to another place. Find the place.

I think you need to add at least one extra assumption and/or to more precisely state your hypothesis and thesis.

Here is the (silly, but still...) counterexample: consider a translation at a uniform velocity perpendicular to your $$\vec{\omega}$$... it is rigid and it has no rest point.

Sure, you assumed that $$\vec{\omega}\neq\vec{0}$$ but you have not declared WHAT $$\vec{\omega}$$ is. In the absence of further info, to me this is just a generic vector. Based on your comment I guess you also want to assume that $$\vec{\omega}$$ is some sort of "rotation direction"? Then my next question is: what do you mean by that, exactly?

I continue with guessing here... then I guess that indeed your initial assumption is that the velocity field has the form

$$\vec{v}(\vec{r}) = \vec{v}_0 + \vec{r}\times\vec{\omega}$$

for some unknown $$\vec{v}_0$$ and what you are wondering here is if/how this can be written as $$\vec{v}(\vec{r})=(\vec{r}-\vec{r}_0)\times\omega$$, with a fixed point $$\vec{r}=\vec{r}_0$$? If I am guessing right and I am interpreting correctly your initial hypothesis, then the proof is not difficult. Requiring $$\vec{v}(\vec{r})$$ to be always perpendicular to $$\vec{\omega}$$ is equivalent to state that

$$\vec{v}(\vec{r})\cdot\vec{\omega}=0$$

for any $$\vec{r}$$, which in turn implies

$$(\vec{v}_0+\vec{r}\times\vec{\omega})\cdot\vec{\omega}=\vec{v}_0\cdot\vec{\omega}=0$$

so the hypothesis implies that our unknown $$\vec{v}_0$$ has to be perpendicular to $$\vec{\omega}$$ too. You can then exploit the expansion of the triple product

$$(\vec{v}_0\times\vec{\omega})\times\vec{\omega} = -\vec{v}_0(\vec{\omega}\cdot\vec{\omega})+\vec{\omega}(\vec{v}_0\cdot\vec{\omega}) = -\omega^2\vec{v}_0$$

where $$\omega$$ is obviously the norm $$|\vec{\omega}|$$ (note here we used $$\vec{v}_0\cdot\vec{\omega}=0$$ to drop one term, otherwise we end up with a screw transformation). If $$\omega$$ is not zero (and here we exclude the silly counterexample above), then you can rewrite your velocity field as follows

$$\vec{v}(\vec{r}) = -(\vec{v}_0\times\vec{\omega})\times\vec{\omega}/\omega^2+\vec{r}\times\vec{\omega} = [\vec{r}-(\vec{v}_0\times\vec{\omega})/\omega^2]\times\vec{\omega}$$.

So here is your fixed point:

$$\vec{r}_0 = (\vec{v}_0\times\vec{\omega})/\omega^2$$

and the whole axis $$\vec{r}_0+\lambda\vec{\omega}$$ for $$\lambda\in\mathbb{R}$$ is fixed... the rotation axis.

• I assumed $\vec{\omega}\neq\vec{0}$ so translatory motion Is excluded.
– user291161
Commented May 31, 2021 at 15:05
• Ok but you have not said what $\vec{\omega}$ is for you! In the absence of info to me this is just a vector ¯_(ツ)_/¯. But I guess that to you this is not a generic vector, so next question is ... what is ur exact initial hypothesis on the velocity field? Possible guessed answers in the updated text above!
– Ste
Commented May 31, 2021 at 20:13
• $\vec{\omega}$ is nothing but the vector that we know exists by Poisson's theorem. Anyway thank you I will reflect on what you and others have written.
– user291161
Commented May 31, 2021 at 21:28
• Welcome. Can you formulate Poisson's theorem or provide a link? I know about Poisson brackets, about the Poisson distribution... I cannot recall any "Poisson's theorem" in connection to rigid motion.
– Ste
Commented Jun 1, 2021 at 10:09
• I found this theorem in the beginning of this very interesting page : dma.unifi.it/~frosali/meclab/labanc/motrigid/motrigid.htm I can't understadnd the proof after "rimarrebbero da da definire" but anyway I developed a different proof by myself
– user291161
Commented Jun 1, 2021 at 11:32

Theorize there exists such a point C such that the velocity of any other arbitrary point P is found with

$$\vec{v}_P = \vec{\omega} \times (\vec{r}_P -\vec{r}_C) \tag{1}$$

Now find where $$\vec{r}_C$$ by using the vector triple product $$a\times(b \times c) = b ( a \cdot c) - c (a \cdot b)$$

\require{cancel} \begin{aligned} \vec{\omega} \times \vec{v}_P & = \vec{\omega} \times \left( \vec{\omega} \times (\vec{r}_P -\vec{r}_C) \right) \\ & = \vec{\omega}( \cancel{\vec{\omega}\cdot(\vec{r}_P -\vec{r}_C)}) - (\vec{r}_P -\vec{r}_C) ( \vec{\omega}\cdot \vec{\omega}) \\ & = (\vec{r}_C - \vec{r}_P) \omega^2 \\ \vec{r}_C &= \vec{r}_P + \frac{\vec{\omega}\times \vec{v}_P}{\omega^2} \end{aligned}

The above assumes we are considering only points on the plane of rotation about P which is why $$\vec{\omega}\cdot ( \vec{r}_P - \vec{r}_C) = 0$$. Also above I use the shortcut $$\omega = \| \vec{\omega} \|$$, or $$\omega^2 = \sqrt{ \vec{\omega}\cdot \vec{\omega}}$$.

So we can always find a point C located at $$\vec{r}_C = \vec{r}_P + \frac{\vec{\omega}\times \vec{v}_P}{\omega^2} \tag{2}$$

which describes the location of the rotation axis. Now, this point isn't unique, as any point parallel to $$\vec{\omega}$$ would also be valid for (1).

Edit 1

The general motion of a rigid body is described as the translational velocity of a point P, and a rotation about P at the same time. This motion has 6 independent quantities that can be specified.

There exist an axis in space (rotation axis), parallel to $$\vec{\omega}$$ and through a point C such that the translational velocity of the body on this locus in space is purely parallel to the rotation, $$\vec{v}_C = h\,\vec{\omega}$$. The ratio of magnitudes between the translational and rotational velocity at C is a scalar value called pitch $$h$$.

The velocity of P is found by

$$\vec{v}_P = h\,\vec{\omega} + \vec{\omega} \times (\vec{r}_P - \vec{r}_C) \tag{3}$$

So given the rotation $$\vec{\omega}$$ (3 components) the location long the rotation axis $$\vec{r}_C$$ (2 components, since C can slide along the axis) and the scalar pitch value $$h$$ (1 component) , equation (3) gives the translational velocity of any other point. There is a 1:1 relationship between the general motion $$\{\vec{v}_P,\vec{\omega}\}$$, and describing the motion by the rotation axis, rotational velocity and pitch $$\{\vec{r}_C, \vec{\omega},h \}$$.

So how do we go in reverse? Use $$\vec{\omega} \cdot \vec{v}_P$$ and $$\vec{\omega} \times \vec{v}_P$$ to come up with the rotation decomposition parameters

$$\vec{r}_C = \vec{r}_P + \frac{\vec{\omega} \times \vec{v}_P}{\omega^2} \tag{4}$$

$$h = \frac{\vec{\omega} \cdot \vec{v}_P}{\omega^2} \tag{5}$$

You can prove the 1:1 relationship by plugging in (4) and (5) into(3). The 1:1 relationship also proves the existence since each motion possible can be described by the rotation axis location and pitch.

There is a special case of pure translation with $$\omega=0$$, and in that case, the rotation axis is at infinity, and pitch is infinite also. When this is the case then $$\vec{v}_P = \vec{v}_C$$.

Appendix

• I thought what follow. You generalize by using (3) instead of (1). This doesn't trouble me, it is Mozzi's theorem. And it is a good idea because going on the previous proof is the same because of cross product with $\vec{\omega}$. But in this case the point $\vec{r}_C$ we found is in motion at speed $h\vec{\omega}$. It is at rest iff $h=0$. But looking to (3) this condiction is the same as saying that $\vec{v}_P \perp \vec{\omega} \quad \forall P$. This would answer my question. This looks working.