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Centrifuge and Radial Bouyancy in space

  1. Oct 4, 2013 #1
    Hi all,

    I want to calculate how tell if a particle in a centrifuge in deep space will move towards the rotation axis or to the rim of the fluid. There is no container walls. This is not a HW problem. I watched the video



    and wanted to learn more about this. (it is under the you-tube search subject: rotating fluids microgravity)

    As the video shows, less dense particles move towards the rotation axis, more dense particles move towards the rim.

    Here's what I have so far.

    Suppose I have a centrifuge in a space station in orbit. The rotation axis points along the +Z axis. I have a particle of mass M, volume V, and density ρ[itex]_{p}[/itex]. To keep things simple, the particle remains at latitude 0 degrees (equator, no north and south motion, etc, but it can move radially in or out) The particle is immersed in a fluid of density ρ[itex]_{f}[/itex] and spins at constant angular velocity ω.

    Since the particle moves in a circle, by Newton's Laws, its net force is mrω[itex]^{2}[/itex]. mrω[itex]^{2}[/itex] points inwards radially. This term takes the place of "ma" in the usual Newton's Laws. There should be a bouyant force ρ[itex]_{f}[/itex]Vg? that points radially outwards (except that there's no "g" for gravity in this situation, so I don't know how to replace "g" in this context). Since the mrω[itex]^{2}[/itex] is a net force, there must be some force to cause this and overcome the bouyant force to keep the particle moving in a circle. In a centrifuge, the Normal force of the walls of the container on the object would provide the mrω[itex]^{2}[/itex] but I don't know what that force would be if the fluid has no walls.

    If I could just write down Newton's Laws with this I would be good to go. I could not find any old HW problems or websites that talked about "Centrifuge Mechanics".

    What force works against the bouyant force to move the particle in a circle?
    How do I replace "g" in ρ[itex]_{f}[/itex]Vg for this context?
    Between the bouyant force and the force that acts towards the center, is there an "ma" to categorize how the particle moves radially until it hits the rotation axis or the rim?

    Thanks for you help, as always.
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Oct 4, 2013 #2

    AlephZero

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    The basic idea here has nothing to do with microgravity as such. It only depends on the centripetal acceleration of the rotating material being much greater than the acceleration due to gravity. For example in a high speed centrifuge on earth, the centripetal accelerations may be of the order of 105g.

    If the fluid is all rotating with the same angular velocity, there must be a radial force on each particle of fluid that causes the centripetal acceleration of at particle. That radial force is caused by a radial pressure gradient in the fluid, low near the axis of rotation and high at the outside.

    There must be "something" to oppose the pressure at the outer radius. For a conventional centrifuge, it is opposed by the structure that contains the fluid. In your video link, I guess the surface tension at the outside of the sphere is enough to balance the relatively small forces involved in the slow rotation of a small object.

    So, inside the fluid, you have a pressure gradient that increases linearly with radius from the axis. if you have particles with different densities in the fluid, that has exactly the same effect as pressure varying with depth in a fluid on earth - i.e. Archimedes' principle. Particles with higher density "sink" to the outside, and particles with lower density "float" to the center.
     
  4. Oct 4, 2013 #3
    Exactly so, doesn't that answer your question?

    Be careful, "bouyant force" doesn't really exist, it is a shorthand for the effect of a true force which is proportional to mass (usually gravity, but here the centripetal force) on substances with different densities.

    g is not "gravity", it is the acceleration due to gravity. What acceleration do you think is analogous in the rotating sphere?

    What force holds drops of water together? What force is there at the surface of a fluid?

    This is where the false nature of the buoyant force is confusing you - there is nothing to do any work against, any more than the string does any work when you swing a bucket on it. Just like the tension in the string, the buoyant effects are a consequence of the rotational motion.
     
  5. Oct 4, 2013 #4
    How does the bouyant force not exist? It's the net force exerted by the surrounding liquid, and the only force acting on the particle, so it is also the force causing the particle to go round in a circle.
     
  6. Oct 7, 2013 #5
    Exactly so, doesn't that answer your question?

    Not exactly. I was looking for something with ƩF = ma. For example, I could easily do this problem:
    A ball of mass M, volume V, and density ρ[itex]_{p}[/itex] sinks in a fluid of density ρ[itex]_{f}[/itex]. If I wanted to know the acceleration, I would have, assuming down is negative

    -Mg + ρ[itex]_{f}[/itex]Vg = -Ma

    for a particle more dense than the fluid it is immersed in. A lighter particle less dense than the fluid would have the Ma be positive as the Bouyant Force would win over the weight of the particle in that case. Either way, I could calculate the acceleration. I'm trying to write down the analogous equation for the situation in this post. I am assuming that the Ma would be replaced by MRω[itex]^{2}[/itex].

    I'm not sure where the Mg goes. It does have weight towards the Earth, but I think that weight cancels out or at least am not sure hot to split it into components. For example, let the rotation axis of the rotating bubble be || and tangent to the orbit of the satellite (+ x axis) at a snapshot. Then, Mg, pointing towards the Earth, would be perpendicular to the rotation axis, pointing in the -y direction, and therefore not influencing any radial motion of the particle (x-direction).


    e careful, "bouyant force" doesn't really exist, it is a shorthand for the effect of a true force which is proportional to mass (usually gravity, but here the centripetal force) on substances with different densities.

    Do you mean the MRω[itex]^{2}[/itex] term? Yes, this is not an external force, like tension, friction, weight, etc but it a net force. It is the net force where, after all the other external forces add together or partially cancel out, the particle moves in a circle, or more generally, accelerates because of a change in direction. It is the Ma for a particle to change direction for constant circular motion. For example, swinging a ball around on a string. Assuming a bird's eye view, the tension of the string is the only real force acting on the ball, and therefore causes the net acceleration v[itex]^{2}[/itex]/R because the ball changes direction.

    The bouyant force (archimedes) is certainly real. If you drop an object in a fluid, the weight of the fluid displaced exerts the bouyant force on you. For example, if you try to dunk a beach ball, a rather light substance in air, in a swimming pool, you have a hard time. Why? Because the volume of the beach ball is large. If you try to hold a beach ball in the water, you push up the same beach-ball-volume worth of fluid. Water is heavy, so an object with a large volume will displace a lot of water, and you have to hold back all that weight of water to keep the beach ball submerged.

    g is not "gravity", it is the acceleration due to gravity. What acceleration do you think is analogous in the rotating sphere?

    I'm not sure here. I am certain that the force must be radial. It seems like it has to be since particles move either radially inwards (towards the axis is less dense) or radially outwards (to the rim if the particle is more dense) The pressure gradient in an earlier answer may have something to do with it, but I don't have the function verses distance to calculate a force: P(x) = F(x)/A. There may be two parts to this problem, now that I think of it: a net acceleration radially and one to keep the particle of mass M moving in a circle once it hits equilibrium. Either way, I'm hoping for such to be quantified.

    What force holds drops of water together? What force is there at the surface of a fluid? also , I guess the surface tension at the outside of the sphere is enough to balance the relatively small forces involved in the slow rotation of a small object.


    Surface Tension. But I can't find a force to quantify it. I found a formula for Capillary Action, but I'm not sure if that will work here.

    Quantitative is key here, and I couldnt find anything about mechanics of particles in a centrifuge. It seems like such problems would've been solved a long time ago, and would be homework problems in some book or paper by now, but I can't find those.
     
  7. Oct 7, 2013 #6

    Bandersnatch

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    You don't replace ma with anything. When writing equations of motion, you put ma on one side and list all the forces present on the other. ma is the resultant net force that governs the motion of your particle. It's what you want to find.
    Also, don't decide on its sign yourself, let it do it for you.
    We begin with:
    [itex]ma=ƩF[/itex]
    The liquid rotates because some force(be it surface tension or whatever) serves as a centripetal force.
    In a frame of reference rotating with the liquid, it changes to centrifugal. For a particle inside the bubble, it'll serve the same role as the force of gravity, pulling it "down" - towards the outside.
    [itex]ma=mω^2r[/itex]

    Buoyancy is the result of the primary force(just as with normal forces). Whatever causes objects to have weight also causes buoyancy. On Earth it's the force of gravity, and the buoyancy depends on g, here it'll depend on the centrifugal acceleration.
    [itex]ma=mω^2r-ρ_fVω^2r[/itex]
    Negative values will point towards the centre of rotation("up", or r=0).
     
  8. Oct 7, 2013 #7
    Ah, I thought your question was "how [to] tell if a particle in a centrifuge in deep space will move towards the rotation axis or to the rim of the fluid" to which the answer is clearly displayed in the video. Your question is actually "how to calculate the forces on a particle...".

    No, Ma is equal to the sum of all the forces, but MRω2 is certainly equal to one of the forces as Bandersnatch says.

    Hang on, you said "deep space" not "in orbit". It doesn't matter though, providing the rotating sphere is in free fall you can ignore gravity.

    Well OK, and as others pointed out my statement that the bouyant force doesn't exist was a bit over the top, but what I was trying to point out was that buoyancy is not an inherent property of an object, it is the result of differences in pressure caused by some external force.

    You wrote rω2 above, I'm not sure why you don't think this is the expression you want.

    Calculating surface tension is non-trivial, but all that matters here is that it is large enough to hold the sphere together. You don't need to know the magnitude of the surface tension to work out the buoyant force here because it doesn't provide any pressure gradient, any more than you need to know what altitude a lake is above sea level to work out the bouyant force in the lake (ignoring any variation in g).

    This isn't a centrifuge. Centrifuges work by accelerating suspended particles at much greater than g in order to overcome the effects of Brownian motion which keeps the particles in suspension. The rotating sphere held together by surface tension has a much lower acceleration than g so is only capable of separating much larger particles. If you really want to look at centrifuges, google "centrifuge equation".
     
    Last edited: Oct 8, 2013
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