Astronaut at centre of tetrahedron

[voko]

Maye I misguessed where you were going, but it seemed like you were heading towards concluding that the result would not apply in a gravitational field.

No, I never meant that. I was in fact hinting at the opposite.

If I interpreted correctly, the flaw in your argument (treating a string as a very stiff spring) is that a spring of arbitrarily great stiffness and zero relaxed length would always have zero length.

You analyzed that too deeply. I only mentioned strings ( = very stiff springs) so that it was especially obvious that purely rotary motion was possible with springs. Rotary motion is possible with soft springs just the same, but that is not usually emphasized in intro-level physics.

Where I did indeed go wrong was in assuming that there are particular arrangements of springs that could prevent rotary motion. That is impossible. The equations obtained in the vector analysis developed in the earlier posts are those of the radial harmonic oscillator no matter how many springs and how they are arranged, so rotary motion is always possible if the initial conditions are just right.

So the presence of the tetrahedron and ISS ( = absence of gravity) in the formulation of the problem is unnecessary. The same result would follow for any other arrangement of springs with or without gravity. The only condition required for that is the zero natural length of springs.

 

[haruspex]

No, I never meant that. I was in fact hinting at the opposite.



You analyzed that too deeply. I only mentioned strings ( = very stiff springs) so that it was especially obvious that purely rotary motion was possible with springs. Rotary motion is possible with soft springs just the same, but that is not usually emphasized in intro-level physics.

Where I did indeed go wrong was in assuming that there are particular arrangements of springs that could prevent rotary motion. That is impossible. The equations obtained in the vector analysis developed in the earlier posts are those of the radial harmonic oscillator no matter how many springs and how they are arranged, so rotary motion is always possible if the initial conditions are just right.

So the presence of the tetrahedron and ISS ( = absence of gravity) in the formulation of the problem is unnecessary. The same result would follow for any other arrangement of springs with or without gravity. The only condition required for that is the zero natural length of springs.

OK, thanks for clarifying.