0
$\begingroup$

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!

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.