I am dealing with the motion of rigid bodies and in studying it I came up with some questions about the nature of rotation.

(1) Does a rotation always have to happen about one fixed axis?

(2) Is there always a center of istantaneous rotation (point with zero velocity at a given instant) for a moving rigid body?

Actually, I already know the answer to the first one, since a rigid body accelerating along the axis of rotation can not have a point with zero velocity; nevertheless I was curious of what would happen in other cases.

As always, any answer or comment is highly appreciated!


1 Answer 1


Essentially, your questions boil down to the study of velocity fields arising from rigid motion. The short answer is that this is a well studied subject, the mathematical concept is that of the Lie algebra of the $3$-D special euclidean group. In general, you don't have a center of rotation due to possible translations, but if your motion is not purely translational, you always have an instantaneous axis of rotation.

Mathematically, writing $\vec v(\vec r)$ the velocity field at a given time, at position $\vec r$ with an arbitrary origin, you can find at all time, spatially constant $\vec v_0,\vec \omega$ such as $$ \vec v = \vec v_0+\vec \omega \times r $$ The motion is purely translational when $\vec omega = 0$. When it is not, there is always an instantaneous axis of ration is where $\vec v ||\vec \omega$, i.e. the line $\frac{\vec \omega\times\vec v_0}{\omega^2}+\mathbb R \vec \omega$.

To prove the above fact, you need to study the flow under rigid motions: $\vec y(\vec r,t)$ with $\vec v = \partial_t \vec y$, $y(\vec r,0)= \vec r$. Rigid motions are affine, so you can write $\vec y = \vec u(t) +R(t)\vec r$, with $R$ a time dependent rotation and the initial conditions give $\vec u(0)=0,R(0)=id$.

You immediately recoginze $\vec v_0 = \frac{d}{dt}\vec u$. The tricky part is convincing yourself that you can find $\vec \omega$ such that: $$ \frac{dR}{dt}_{|t=0} \vec r= \vec \omega \times \vec r $$

By taking the derivative of the dot product preserving property, you easily find that $\frac{dR}{dt}_{|t=0}$ is an antisymmetric operator, and you need to invoke the fact that all antisymmetric operators can be written as a cross product by a vector (specific to 3D).

Hope this helps.


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