## 2: released in rotating frame (linear motion, inertial frame)

(See the previous post, example 1, for notation.)

Suppose we are holding an object fixed on the merry-go-round at a distance R on the x-axis: it is stationary in the rotating frame. Now suppose that the surface is frictionless. We release the object at t= 0. What is it’s motion?

At t = 0 we have initial values of $\rho$ and $\nu$:

$\rho0 = \{R,\ 0,\ 0\}$

$\nu0 = \{0,\ 0,\ 0\}$

We apply the transition matrix (setting t = 0; note that all axes coincide) to get initial r and v from initial $\rho$ and $\nu$:

$r0 = T\ \rho0 = \{R,\ 0,\ 0\}$

$v0 = T\ \nu0 - \omega\ N\ T\ \rho0 = \{0,\ R \omega ,\ 0\}$

(Yes, wrt the fixed frame, the initial position r0 is on the x-axis, and the initial velocity v0 is tangential, i.e. having only a y-component.)

We know that the acceleration a is zero, so v is constant, and r is linear in time:

$a = 0$

$v = v0 = \{0,\ R \omega ,\ 0\}$

$r = r0 + v0\ t = \{R,\ R t \omega ,\ 0\}$

To see what it looks like in the rotating frame, I want $\rho\$, so I apply the attitude matrix to r (because $A = T^{-1}\$):

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

=

$\left(\begin{array}{l} R \cos (t \omega )+R t \omega \sin (t \omega ) \\ R t \omega \cos (t \omega )-R \sin (t \omega ) \\ 0\end{array}\right)$

Draw it! Choose numerical values: $R\to 1, \omega \to 1$.

Viewed from the rotating frame, it appears to spiral out.

## 3: linear motion in rotating frame

Suppose that a person starts (t=0) at the center of the merry-go-round and walks straight out along his (new, rotating) x-axis at constant speed $\nu0$. Then the new components of his position $\rho$ are…

$\{t\ \nu0,\ 0,\ 0\}$

and $\nu$ and $\alpha$ are…

$\nu = \{\nu0,\ 0,\ 0\}$

and

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

We compute the old components a.

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

The three terms are

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

$- 2\ \omega\ N\ T\ \nu = \{-2\ \nu0\ \omega\ \sin (t \omega ),\ 2\ \nu0\ \omega\ \cos (t \omega ),\ 0\}$

$\omega^2\ N^2\ T\ \rho = \left\{-t\ \nu0\ \omega ^2\ \cos (t \omega ),\ -t\ \nu0\ \omega ^2\ \sin (t \omega ),\ 0\right\}$

and their sum is

$a = \left(\begin{array}{l} -\nu0\ \omega\ (t\ \omega\ \cos (t \omega )+2 \sin (t \omega )) \\ \nu0\ \omega (2 \cos (t \omega )-t\ \omega\ \sin (t \omega )) \\ 0\end{array}\right)$

Now, what are the components of a in the rotating frame? it want to see the radial and tangential components of a, hence implicitly the radial and tangential components of force exerted by our walker as he moves outward in a straight line at constant speed on the merry-go-round.

The new components of a are found by applying A to a:

$A\ a = \left(\begin{array}{lll} \cos (t \omega ) & \sin (t \omega ) & 0 \\ -\sin (t \omega ) & \cos (t \omega ) & 0 \\ 0 & 0 & 1\end{array}\right)$ $\left(\begin{array}{l} -\nu0\ \omega\ (t\ \omega\ \cos (t \omega )+2 \sin (t \omega )) \\ \nu0\ \omega (2 \cos (t \omega )-t\ \omega\ \sin (t \omega )) \\ 0\end{array}\right)$

=

$\left(\begin{array}{l} -t\ \nu0\ \omega ^2 \\ 2\ \nu0\ \omega \\ 0\end{array}\right)$

We have a component $-t\ \nu0\ \omega ^2$ along the new x-axis (i.e. radial): the negative sign says it points inward. It grows with time, i.e. as he moves outward. (It reflects only his increasing distance from the center, because he is walking out at constant speed $\nu = \nu0\$.)

We also have a constant positive acceleration $2\ \nu0\ \omega$ in the tangential direction: as he moves out, he must increase his tangential speed wrt the fixed frame.

I say again that I really like having unambiguous definitions of, for example, a and $\alpha\$, and then being able to confidently ask for the new components of a.