Why can we use just one angular velocity vector to describe the rotation of a whole non-inertial reference frame?

Question

The other day in class the professor was explaining non-inertial reference frames. We were working out how to find the acceleration of a point as measured from the non-inertial reference frame, and one problem was finding the time derivatives of the unit vectors as measured from the non-inertial reference frame.

To explain how to do it, he drew a nice diagram that showed that, if $\mathbf{\Omega}$ is the vector representing angular velocity of the rotation of $\hat{\imath}$ (that is, its direction is the axis of rotation), then $\frac{\mathrm{d}\hat{\imath}}{\mathrm{d}t} = \mathbf{\Omega} \times \hat{\imath}$. This makes sense. Next, he said that we can use the same argument for the other unit vectors: $\frac{\mathrm{d}\hat{\jmath}}{\mathrm{d}t} = \mathbf{\Omega} \times \hat{\jmath}$ and $\frac{\mathrm{d}\hat{k}}{\mathrm{d}t} = \mathbf{\Omega} \times \hat{k}$. I don't understand why we use the same $\mathbf{\Omega}$ for all of them. Since their axes of rotation are different, shouldn't we use different angular velocity vectors?

Answer 1

As Ron pointed out in the comment, it is important to know that there is only one axis of rotation. In a rotating frame, every vector rotates around the same axis, in the same direction, with the same speed.

Now consider this: what makes $\hat{i}$ special in the first place? The answer, of course, is "nothing" - when you set up your coordinate system, you're picking $\hat{i}$ to lie in an arbitrary direction. But you could just as well have set up a different coordinate system in which $\hat{j}$ (or $\hat{k}$) points in that direction. So whatever formula you come up with for the rate of change of $\hat{i}$ is going to work just as well for any other vector. In a sense, the formula doesn't "know" what vector it's being applied to.

Alternatively, of course, you can just do the math to compute the rates of change from scratch, and you'll find out that $\Omega\times\vec{r}$ works for any vector $\vec{r}$.