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If we have a structure that rotates to create artificial gravity, then if the mass isn't perfectly distributed along the circumference the CM will be offset from the geometric center so there will be a wobble.

I made this illustration, to show the wobbles that happens becuse a person goes to visit his/her friend on the opposite side of the structure.

wobble

Some artist conceptions include water in the forms of lakes, and are often contiguous bodies of water that circle the entire circumference (Bernal sphere example). If we assume there is a free-flowing body of water around the entire circumference like this (and is sufficiently deep), then it's clear that water will move as a result of a mass movement since the previous water surface is no longer equipotential. This will also move the CM.

How would the presence of water affect the wobble?

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Hmmm ... you are obviously aware that this is generally neglected in fictional depictions. I think you need something on order of 10 meters water-equivalent (MWE) shielding for adequate solar storm protection which means that you are generally safe neglecting the motion of the inhabitants, but large bodies of water may be another mater entirely. –  dmckee Mar 3 '13 at 23:59
    
@dmckee Fictional depictions are one thing, but there have been serious attempts to study what would work for space habitats. The hydrodynamics would be of interest for that. I mean, in the worse case it could be unstable, and in the best case it could keep its axis from moving. Regarding radiation, I don't think anyone calls for the water to be under the people in the first place. My intent is that the smiley-face people are on solid ground and the water is next to them. It wouldn't be the same problem if they were in boats. –  AlanSE Mar 4 '13 at 0:27
    
Meters water equivalent are a common notion in neutrino physics for how much cosmic ray shielding you have that is independent of the local geology. If you had lots of water you could use it that way, but I assume you'll fuse a bunch of rock on the outside and stabilize it with structural steel. But what that does is put a lower limit on the areal density of the shell (around 10 tons per square meter) unless you are going to shield the thing magnetically. –  dmckee Mar 4 '13 at 0:52
    
@dmckee I can't tell if this is what you meant or not, but I would imagine this rock would be put around the station, and not a part of the rotating mass. The material strength needed to hold in one atmosphere of pressure grows linearly with volume and would be substantial for any such spacious habitat. One atmosphere is about 10 meters of water equivalent (which is no coincidence) so shielding would double the material requirements. Nonetheless, I've never read anyone else propose that idea. –  AlanSE Mar 4 '13 at 1:15

2 Answers 2

The system CM is the only thing we can identify that is truly inertial. Make that the origin.

It should also be noted that the CM is a combination of the mass-weighted CM of all the components. Those components are:

  • The man that moves
  • The rigid structure
  • The body of the water, which I will break down into geometric parts:

    1. The outer edge, or the "lake floor"
    2. The surface, which is always a circle about the system CM

The last point is absolutely crucial, but let's sum up the entire course of events. The man moves in the negative y-axis direction (down). Since the system CM remains in place, the CM of the (structure + water) moves in the positive y-axis direction (up). We can guess that the rigid structure moves "up" as in the OP illustration. As we imagine that, the lake floor moves but the surface doesn't. That means the lake's mass moves "up" - the same direction as the structure. From this information alone we conclude that the wobble is lessened by the presence of the water. Let's go deeper into the dynamics of the water.

The water is an annular region. As the rigid structure moves up, the outer edge moves up, making the annulus into a shape like a crescent moon.

cerscent

When the structure moves, it moves the outer circle but the inner circle remains in the exact same place relative to the CM of the system. So the inner circle can't factor into the calculation for the revised location of the water's CM. The center of mass of the water, is thus weighted as if the inner circle just isn't there. That will create an extremely large "artificial" mass of the water that will decrease the wobble of the system. The water decreases the amount of wobble as much as you would get if the entire thing was fully filled with water! To the extent that you're not drying out any part of it this will be true, and it will be irrelevant of the level of the water.

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Ce n'est pas une réponse. (Where are the hydrodynamicists?)

Have you considered the possibility of "tsunamis"?

Gravity waves have an approximate wave velocity $v_w = \sqrt{gh}$, with $h$ the water depth and $g=v_s^2/R$ the radial acceleration (of a cylindrical shell with radius $R$ and wall velocity $v_s$).

So $v_w = \left(\sqrt{h/R}\right) v_s $, and the wave velocity is less than the shell's.

The question is whether a particular standing wave pattern could become resonant and extract energy from the rotating shell to increase the wave's (and the wobble's) amplitude.

I have no idea how to figure that out.

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