For the description of rigid body motion, any point O of the rigid body could be taken as reference, since the velocity of a generic point P can be written in function of the angular velocity Ω and of the velocity of O, independently of the choice of O.
˙P=˙O+Ω∧(P−O)
That means that the motion of P is seen as the composition of the traslation of O, plus a rotational motion about an axis passing through O: let's call this axis γ.
In which cases it is correct to say that γ is an instantaneous axis of rotation of the rigid body?
In order to do that must the point O (on the axis γ) have zero velocity (i.e. ˙O=0)? Or can I define γ as an instantaneous axis of rotation in any case when I write down (1)?
Answer
A body can have parallel velocity on the instantaneous axis of rotation. This parallel velocity is sometimes designated with a scalar pitch value h, such that →vO=h→ω
Consider a point O on the IAR and a point P outside of it. You have
→vP=→vO+(→rP−→rO)×→ω
Now I can prove that given the motion →vP of an arbitrary point P and the rotational velocity →ω. The motion can always be decomposed into an axis of rotation point →rO and a parallel velocity →vO.
Lemma Given →vP and →ω there exists a unique (relative) location →r such that →vO=→vP+→ω×→r=h→ω so the velocity →vO=h→ω is parallel to →ω only. This location is on the instantaneous axis of rotation closest to P such that →rO=→rP+→r.
Proof
Take →r=→ω×→vP‖ in the transformation and expand out the terms
\vec{v}_O = \vec{v}_P + \vec{\omega}\times \left( \frac{\vec{\omega} \times \vec{v}_P}{\| \vec{\omega}\|^2} \right)
Now use the vector triple product identity a \times (b \times c) = b (a\cdot c) - c (a \cdot b)
\require{cancel} \vec{v}_O =\vec{v}_P +\frac{\vec{\omega}(\vec{\omega}\cdot \vec{v}_P) - \vec{v}_P (\vec{\omega}\cdot \vec{\omega})}{\| \vec{\omega}\|^2} = \cancel{\vec{v}_P} + \frac{\vec{\omega}(\vec{\omega}\cdot \vec{v}_P)}{\| \vec{\omega}\|^2}-\cancel{\vec{v}_P}
Now define the scalar pitch as h = \frac{\vec{\omega} \cdot \vec{v}_P}{\| \vec{\omega} \|^2} and the above becomes \vec{v}_O = h \vec{\omega}
So the velocity on O is parallel to the rotation only.
Example A body rotates by \vec{\omega} = (0,0,10) and a point P located at \vec{r}_P=(0.8,0.2,0) has velocity \vec{v}_P = (2,-3,1). Find the IAR point O and pitch value.h.
\vec{r}_O = \vec{r}_P + \frac{\vec{v}_P \times \vec{\omega}}{\| \vec{\omega} \|^2} = (0.8,0.2,0) + \frac{(-30,20,0)}{10^2} = (0.5,0,0)
h = \frac{\vec{\omega} \cdot \vec{v}_P}{\|\vec{\omega}\|^2} = \frac{10}{10^2} = 0.1
Confirmation
\vec{v}_P = h \vec{\omega} + (\vec{r}_P - \vec{r}_O) \times \vec{\omega} = (0,0,1) + (0.3,0.2,0) \times (0,0,10) = (2,3,-1) \; \checkmark
NOTES: See also related answer to question Why is angular velocity of any point about any other point of a rigid body always the same?
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