# Instantaneous axis of rotation of a rigid body

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 $\Omega$ and of the velocity of $O$, independently of the choice of $O$.

$$\dot{P} = \dot{ O} + \Omega∧(P −O)\tag{1}$$

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 $\gamma$.

In which cases it is correct to say that $\gamma$ is an instantaneous axis of rotation of the rigid body?

In order to do that must the point $O$ (on the axis $\gamma$) have zero velocity (i.e. $\dot{O}=0$)? Or can I define $\gamma$ as an instantaneous axis of rotation in any case when I write down $(1)$?

• – ja72 Apr 10 '16 at 17:23

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 $\vec{v}_O = h \vec{\omega}$

Consider a point O on the IAR and a point P outside of it. You have

$$\vec{v}_P = \vec{v}_O + (\vec{r}_P-\vec{r}_O) \times \vec{\omega}$$

Now I can prove that given the motion $\vec{v}_P$ of an arbitrary point P and the rotational velocity $\vec{\omega}$. The motion can always be decomposed into an axis of rotation point $\vec{r}_O$ and a parallel velocity $\vec{v}_O$.

Lemma Given $\vec{v}_P$ and $\vec{\omega}$ there exists a unique (relative) location $\vec{r}$ such that $$\vec{v}_O = \vec{v}_P + \vec{\omega}\times \vec{r} = h \vec{\omega}$$ so the velocity $\vec{v}_O = h \vec{\omega}$ is parallel to $\vec{\omega}$ only. This location is on the instantaneous axis of rotation closest to P such that $\vec{r}_O = \vec{r}_P + \vec{r}$.

Proof

Take $\vec{r}= \frac{\vec{\omega} \times \vec{v}_P}{\| \vec{\omega}\|^2}$ 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?