With your finger you are changing the direction of force to always be at the COM. So it's very similar to the LM situation.
I don't see how. If you presume an RCS solution, the RCS doesn't have any more heinous a job to do if thrust is always directed through the center of mass. If you presume a thrust-vectored solution, then rotating the ball by adding a tangential force is the same regardless of where on the ball you apply it. Now if you add gravity, the object will
fall faster in some thrust orientations than in others, but that's a different problem. Not only can you decouple them, you
must do so in order to understand what's happening.
I'll create the physics demo in short time, and capture the result in video capture.
How do you plan to assure us that you won't simply be illustrating your misconception? Why not just write down the physics?
And I do accept your potential correction about the "location of the center of thrust" not making a difference on angular stability. In the simple model, there is sense to this, in that neither force is exerting an angular acceleration on the rigid body.
I believe you are correct, and I thank you for this correction. I consider this a "small miss" that you easily corrected.
Funny how every time you're caught making a glaring elementary mistake, it's somehow inconsequential.
This a bit mind-bending. Because there's a stability advantage for a helicopter to have its rotors above it.
Aerodynamic factors for stability do not apply. You can unbend your mind by not trying to argue by analogy and not interpolating irrelevant factors. Just do the math.
The simplified concept logic would indicate that maintaining attitude/balance for a 300' rocket is the same as doing it for flat disk. Would you affirm or refute this derived conclusion?
The problem has never stopped being
only about moment arms, thrust vectors, and moment of inertia. Stick to those tools and you'll be able to solve all the problems.