## conventions and setup

As far as possible, I am going to stay with my notation. r and $\rho$ are the old and new (fixed and rotating) components of the position vector; v and $\nu$ are derivatives wrt time of r and $\rho$ respectively; a and $\alpha$ are derivatives wrt time of v and $\nu$ respectively. (But R is a convenient scalar value, and will no longer denote the position vector whose components are r and $\rho\$.)

$v = \dot{r}$

$\nu = \dot{\rho}$

$a = \dot{v}$

$\alpha = \dot{\nu}$

The rotating frame is the same in all these problems, so get its matrices early (hence not often). The z-axis is our axis of rotation.

The attitude matrix for a CCW rotation of the axes (about the z-axis) is…

$A = \left(\begin{array}{lll} \cos (t \omega ) & \sin (t \omega ) & 0 \\ -\sin (t \omega ) & \cos (t \omega ) & 0 \\ 0 & 0 & 1\end{array}\right)$

The transition matrix is…

$T = \left(\begin{array}{lll} \cos (t \omega ) & -\sin (t \omega ) & 0 \\ \sin (t \omega ) & \cos (t \omega ) & 0 \\ 0 & 0 & 1\end{array}\right)$

And N is the derivative of A, at zero, divided by $\omega\$ (to “get it” from a unit vector):

$N = \frac{1}{\omega}\ \frac{\partial A}{\partial t}\ \left(0\right) = \left(\begin{array}{lll} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$

I like to calculate N from first principles, in order to be certain I have the right signs. Of course, my notation for those has not changed: T is the transition matrix, A is the attitude matrix: $T = A^T\$, and because A is orthogonal, $A^{-1} = A^T \left(= T\right)\$.

## fixed in the rotating frame

We imagine something held fixed in a rotating coordinate system. We might think of an object on a merry-go-round. If it’s at a distance R from the center, we put it on the new (rotating) x axis (or, equivalently, we put the new x-axis thru it!). The position vector has new components $\rho$, and we differentiate to get $\nu$ and again to get $\alpha\$:…

$\rho = \{R,\ 0,\ 0\}$

$\nu = \{0,\ 0,\ 0\}$

$\alpha = \{0,\ 0,\ 0\}$

Now recall

$v = T\ \nu - \omega\ N\ T\ \rho\$.

(I note that I have an appalling tendency to write the utterly false –
false$v = T\ \nu - \omega\ N\ T\ \nu\$. false)

but since $\nu$ is zero, $T\ \nu$ is zero…

$T\ \nu = \{0,\ 0,\ 0\}$

so the other term is v…

$v = \{-R \omega \sin (t \omega ),\ R \omega \cos (t \omega ),\ 0\}$

Next, we compute a

$a = T\ \alpha - 2\ \omega\ N\ T\ \nu + \omega^2\ N^2\ T\ \rho$

Since $\nu$ and $\alpha$ are zero, $T\ \nu$ and $T\ \alpha$ are zero, so the only nonzero term is the last(and i’ve marked it “a”)… but we could compute them all, and add them up:

$T\ \alpha = \{0,\ 0,\ 0\}$

$2\ \omega\ N\ T\ \nu = \{0,\ 0,\ 0\}$

$a = \left\{-R \omega ^2 \cos (t \omega ),-R \omega ^2 \sin (t \omega ),0\right\}$

Let me write out the pieces of that nonzero term. We have

$N = \left(\begin{array}{lll} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$

Then

$\omega^2\ N^2\ T\ \rho$

=
$\left(\begin{array}{lll} -\omega ^2 & 0 & 0 \\ 0 & -\omega ^2 & 0 \\ 0 & 0 & 0\end{array}\right)$ $\left(\begin{array}{lll} \cos (t \omega ) & -\sin (t \omega ) & 0 \\ \sin (t \omega ) & \cos (t \omega ) & 0 \\ 0 & 0 & 1\end{array}\right)$ $\left(\begin{array}{c} R \\ 0 \\ 0\end{array}\right)$

So, we have the acceleration a. Of course, those are old components.

What are the new components of a? (Strictly speaking, what are the new components of the vector whose old components are a?) They are

$T^{-1}\ a = T^T\ a = A\ a$

=

$\left(\begin{array}{lll} \cos (t \omega ) & \sin (t \omega ) & 0 \\ -\sin (t \omega ) & \cos (t \omega ) & 0 \\ 0 & 0 & 1\end{array}\right)$ $\left(\begin{array}{l} -R \omega ^2 \cos (t \omega ) \\ -R \omega ^2 \sin (t \omega ) \\ 0\end{array}\right)$

$= \left(\begin{array}{l} -R \omega ^2 \\ 0 \\ 0\end{array}\right)$

The acceleration is along the negative x-axis, i.e. radial. Good. In fact, perfect.

If we assume that our fixed frame is inertial, then a radial physical force is required to hold that mass in circular motion at constant speed.

Did it confuse you that I found the new components of a? Well, a in an inertial frame tells me what force is required, but it’s in the rotating frame that the answer takes a simple form.

Using polar coordinates, instead, is a different way of coming at that. I think I prefer my way: compute the new, rotating, components of a.

It’s so nice to be clear about a and $\alpha\$, and the new components of a.

We could check the cross product form – and I have – but I’ll leave it to you. For the $\omega$ vector, just take scalar $\omega$ times the unit vector (0, 0, 1) along the z-axis.