Astronaut at centre of tetrahedron

[Saitama]

So what would need to be done to prevent circular motion?

I would displace the mass slightly along the string to make it perform SHM, but I am not sure why you ask me to "prevent" circular, I simply won't give it enough velocity perpendicular to string for performing circular motion.

Would you not need to consider the pull of gravity?

That comes in the force expression but that won't affect the angular frequency. :)

 

[haruspex]

In principle, any string is also a spring (of great stiffness). What kind of motion is possible for a mass on a string?

Need to be a bit careful there. One attribute that was definitely critical to the neat result is the zero relaxed length of the springs. A single such of great stiffness would therefore be of very short length. [STRIKE] I can see where you are going, and your argument is sound, but keep the stiffness moderate.[/STRIKE]
Edit:
The zero relaxed length does invalidate the argument. Back to the equations:
net force = $\Sigma D_i(\vec{v_i} - \vec{x} - \vec{\Delta x}) + m \vec{g}$
At equilibrium:
0 = $\Sigma D_i(\vec{v_i} - \vec{x}) + m \vec{g}$
whence
net force = $-\vec{\Delta x}\Sigma D_i$

 

[voko]

I would displace the mass slightly along the string to make it perform SHM, but I am not sure why you ask me to "prevent" circular

In certain spring arrangements you have rotary motion in addition to oscillations. The original problem used the tetrahedron arrangement to prevent rotary motion and my question is what is the general condition on the arrangement to prevent rotary motion.

 

[voko]

Need to be a bit careful there. One attribute that was definitely critical to the neat result is the zero relaxed length of the springs. A single such of great stiffness would therefore be of very short length. [STRIKE] I can see where you are going, and your argument is sound, but keep the stiffness moderate.[/STRIKE]
Edit:
The zero relaxed length does invalidate the argument.

I have lost you here. What argument is invalidated?

 

[haruspex]

I have lost you here. What argument is invalidated?

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. The equations show that it does.
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.

 

[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.