Shielding off an external gravitational field by using a centrifuge 
Imagine a colonizer of Mars needing his/her daily dose of a full Earth-like gravitational pull, $g$. Since the pull of Mars, $g_M$, is smaller than $g$, let's place the astronaut in a centrifuge rotating in a vertical plane. If we simply set it to have a constant rotational velocity, $\dot\theta$, the astronaut may become dizzy from the changing normal force s/he is feeling due to the combined effect of $g_M$ and the centrifuge (e.g., s/he is lighter while being at the top of the loop and heavier when at the bottom). 
The idea is to find $\theta(t)$ such that the astronaut in the centrifuge experiences a constant radial acceleration of magnitude $g$ (i.e. if s/he were to stand on a weight, it would show the same as on Earth) as well as no tangential acceleration. I.e., if the chamber of the centrifuge s/he is in has no windows, s/he should not be able to tell whether s/he was in a centrifuge or on Earth. 

From this Wikipedia-article, I found that the acceleration of an object undergoing non-uniform circular motion is $$\vec a=r\pmatrix{-\dot \theta^2 \\ \ddot \theta}.$$ In this case, I would think that the forces constituting this acceleration would be force of gravity from Mars, as well as the normal force from the centrifuge-chamber. My understanding is that the normal force is what we feel, so therefore we demand that (in polar coordinates) 
\begin{align} 
\vec a &=\vec g_M+\vec a_N \implies \\
r\pmatrix{-\dot \theta^2 \\ \ddot \theta} &= -g_M \pmatrix{\sin \theta \\ \cos \theta}-\pmatrix{g \\ 0} \implies \\
r \dot \theta^2&=g_M \sin \theta +g \\-r\ddot \theta&=g_M \cos \theta.
\end{align}
However, if I differentiate the first equation and insert the result into the second, I get $\dot \theta=2$, which is obviously wrong. 
Where is my mistake(s)?
Thanks. 
 A: I'd start with a horizontal setup, since that's easier to imagine. You'd have acceleration $g_M$ towards the ground. You want acceleration $g$ in any direction. Accelerations are additive (just like forces), so come up with a rectangle which has $g_M$ as one edge (vertical downward) and $g$ as its diagonal (down and outwards). The other edge (horizontal outwards) will tell you how much acceleration your centrifuge has to contribute. And the direction of the diagonal tells you how you should align the floor of your chamber so that the resulting force at target speed looks like “down” from the inside.
I think you can do the same for your vertical setup as well, but it's far harder. Essentially you have to do the above considerations for every point along the path, which means you'd have to compute the required acceleration and the direction of the chamber floor anew for every point. Furthermore, due to the constant changes in speed, your centrifuge-induced acceleration would not only be centrifugal, but contain a significant tangential component as well, further complicating things. Rotating the chamber will introduce yet another frame of reference where accelerations occur, and I doubt these can be controlled by some clever motion of the chamber itself. On the engineering side, having a fixed cabin endure such centrifugal forces is one thing, but having the motor precisely adjust speed and having some mechanism to constantly adjust the orientation of the chamber, is far more problematic.
So if you really really really want a vertical centrifuge, and think it has to be possible, and have make some initial computations but hit a roadblock somewhere, then I think you can ask a more specific question about this toy setup. But if practicality comes into this, simply abandon the vertical layout and be happy with the easy and reliable description of a horizontal centrifuge.
