# Meaning of angular velocity in a rotating system

When you study the motion of a rigid body you have $\vec\omega$, the vector associated to angular velocity. In the case you are using Euler angles and want a quick formula for the rotational kinetic energy you switch to a system that rotates with the body and express the components of $\vec\omega_{[e]}$ in terms of a basis $[e]$ attached to the principal axes of the body.

However what's the meaning of $\vec\omega_{[e]}$? If you are in a rotating system the body should seem still, so there should be no angular velocity. Moreover if we consider that $\vec v = \vec \omega \times \vec r$, which should remain valid in any basis, $\vec v_{[e]}$ should be zero in the rotating system, so $\vec \omega_{[e]}$ should be also zero...

I know I am confusing many things but could you clarify to me this point?

## 2 Answers

It is simple. $\vec{\omega}_{[e]}$ are not the components of angular velocity seen in the reference frame attached to the rigid body itself. As you point out, that angular velocity is zero.

It is the result of mathematical manipulation. You have a set of relations between the basis vectors of the inertial frame and the rotating frame, and you use that to write $\vec{\omega}$ in terms of the basis vectors of the rotating frame to simplify calculation. The physical meaning of $\vec{\omega}$ is still the angular velocity seen in the inertial frame of reference.

Why is the mathematical formalism for change of basis not sufficient here? Because both the change of basis matrix and the definition for (angular) velocity involves external parameter--time. In general relativity, time and space are merged together, and every vector in the 4-dimensional spacetime has both temporal and spatial parts. In that case, all vectors transform nicely as mathematics dictates.

Back in classical mechanics, because of the special status of time, there is no general formula transforming physical quantities from one frame to another with relative rotational motion. Angular velocity is a special case, however. The transformation is simple as $$\boldsymbol{\omega}=\boldsymbol{\omega}'+\boldsymbol{\Omega},$$

where $\boldsymbol{\Omega}$ is the relative angular velocity of the primed frame with respect to the unprimed one.

• I now understand. Is there a general theory about how vectors transform under a change of frame of reference? It seems that the mathematical formalism for the "change of basis" is not enough to express what happens to velocity, position, angular velocity, etc.. when you change the frame of reference – Ralph Jul 4 '12 at 11:26
• @Ralph: Answered in my new edits. – Siyuan Ren Jul 4 '12 at 12:53

What we do when we have a rigid body in motion is to attach some coordinates to the body, so an origin $O'$ and three base vectors. Now, the position of a point relative to the origin of a fixed reference frame of origin $O$ is

$$\vec R_i= \vec{OO'}+ \vec r_i$$

where $\vec r_i$ is the position vector respect to the origin of the body reference frame $O'$.

You can show, by means of the Euler rotation theorem and simple geometric considerations, that the velocity of a point respect to the fixed reference frame is
$$\vec V_i = \vec V_{O'} + \Omega \vec r_i$$ where $\Omega$ is a 3x3 matrix, called rotation tensor. It can be shown that this matrix is antisymmetric, and so we always are allowed to write $$\Omega\vec b=\vec\omega\times \vec b$$ where $\omega$ is a vector associated to $\Omega$ and its components are the indipendent components of $\Omega$. The strength of this formalism is that $\omega$ is unique for all the points (!) and only depends on the time. So, it is a instantaneous rotation angular velocity.

So, in the reference frame attached to the rigid body, we don't see rotation of any point, and that is intuitive because of the constraint of rigidity. In fact, if we wanted to write $V_i$ in the reference attached to the body we would have simply $\vec r_i=\vec 0 \Rightarrow \vec V_i = \vec 0$. Therefore, all the points are at rest in this reference frame.

When you say "we want a quick formula for the rotational kinetic energy" I guess you mean $$E_R = \frac {1}{2} I\omega^2$$. Now, for any mechanical system we know from the Koenig Theorem that the total kinetic energy of a system is the traslation energy of the center of mass + the energy respect to the center of mass. In the case of a rigid body, being forbidden the traslation of a point respect to another on the body volume, the only possible "internal" kinetic energy is the rotational one: $E_R$ where $I$ is the momentum of inertia. If we put ourselves in the reference frame attached to the rigid body, and we take as origin the center of mass, we just have the "internal" kinetic energy term, without the translating energy of the center of mass. Resumning: $$Koenig: \ \ \ K_{TOT}=Mv_C^2/2+\int_V dm v_i^2/2$$ $$Rigid body:\ \ \ \int_V dm V_i^2/2 = \int dm |\vec v_C + \vec\omega\times\vec r^{(C)} |^2/2$$

$$RF\ with\ origin\ in\ C:\ \ \ \int dm |\vec v_C + \vec\omega\times\vec r^{(C)} |^2/2= \int dm |\vec r^{(C)}|^2 \omega^2/2 = I_{C}\omega^2/2$$ where $r^{(C)}$ is the position of the element of mass $dm$ respect to the center of mass. At last, finally: $$K_{TOT}=M v_C^2+I_C\omega^2/2$$ where $\omega$ is well defined for a given $t$ and is the same for all the points. Moral: you don't need a rotating reference frame to have a "simple" form for the kinetic energy of a rigid body. I hope this helps.