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Centrifugal forces + pressure and CG

  1. Nov 13, 2012 #1
    Take half torus with solid part of density 1, other half torus with liquid of density 1. The torus turn at W rd/s enough for maintain liquid at external circle. Now, for 0 to 180° we accelerate torus, this increase forces (blue in drawing) from pressure of water and from 180° to 360° we deccelerate torus. Rotationnal speed change from W to W+w to W. It's like torus change the CG without external forces. I thought solid give pressure too but I can imagine 2 liquids with different density, it's the same. I have some questions:

    1/ water don't give pressure so no force exist ?
    2/ solid give same pressure ?
    3/ when we rotate the forces from rotation, this cancel pressure forces ?

    Edit: I add another drawing for show different positions

    Attached Files:

    Last edited: Nov 13, 2012
  2. jcsd
  3. Nov 14, 2012 #2
    Maybe my questions are not precise enough ?
    When I drawn solid in black, it's possible to imagine it maintain from center of circle with solid like this no pressure can appear from solid (imagine a lot of smal ropes for maintain all part of solid). So if solid don't give pressure, it is liquid that don't give pressure ?
  4. Nov 14, 2012 #3
    Does this attachment help?

    There is certainly a tangential ring pressure developed in a rotating ring. In a solid it is a stress and in a liquid it is pressure. Of course these are equal in your ring because the densities are equal. For unequal densities you would have to relocate the centre of mass.

    Attached Files:

    Last edited: Nov 14, 2012
  5. Nov 14, 2012 #4


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    I don't understand exactly what you are doing here, but one difference between the solid and liquid parts is that there must be a container that holds the liquid, and the container will apply a force to the liquid. If you ignore that fact, you will get the wrong forces and stresses.
  6. Nov 14, 2012 #5
    I'll try to understand this evening. That I can't understand: if I imagine solid half torus like small parts, if each part is attached by a rope, the pressure could be limited a lot I think ?

    ok, but this force (the force the container apply) will put on support that take system ?
  7. Nov 14, 2012 #6
    This is actually a very complicated problem. Based on lots of experience in fluid and solid mechanics modelling, I can tell you that the best way to approach a problem like this is to start with some much simpler versions of the problem, and solve them first, and gradually add complexity until you get to the actual problem you are trying to solve. Why? First of all, if you can't solve the simpler versions of the problem, you certainly won't be able to solve the more complex problem. Secondly, you will gain understanding along the way from the solutions to the simpler problems.

    In the system you describe, neither the solid nor the liquid will behave as a rigid body. Both will experience transient non-homogeneous deformations. The liquid will exhibit viscous flow that has to be accounted for, and the solid will exhibit elastic deformation that has to be accounted for. I suggest starting out with the following very simple problem:

    Consider an infinitely long straight pipe containing a viscous fluid. Initially the pipe and the fluid are at rest. At time zero, the pipe is set in motion with a steady velocity V along its axis. What is the velocity and pressure distribution within the fluid as a function of time and radial position? If you can't solve this problem, you will never be able to do the torus problem.

    As a second version of the problem, consider the same pipe and fluid, but this time make the imposed pipe velocity a function of time.

    As a third problem, consider the case where there are rigid plugs within the pipe, and a finite length of fluid between the plugs. The plugs move with the pipe velocity.
  8. Nov 14, 2012 #7
    With W = constant, it's a complicated problem too ? or it's because W change in time ?

    With W=constant, are there forces from liquid ? I think forces from solid can't be near 0 if there is a lot of parts (imagine infinite number of parts and ropes). Even if there are forces from liquid is strange because even on a round forces cancel themselves if we place this system on another object in a part of time (60° for example), this change the CG of object I think.

    For change velocity, we can change the rotation of the system or place a lot of stems for help solid and liquid to gain velocity very quickly. In theory, if velocity is apply on each atom of liquid and solid, forces are centripetal forces only.
  9. Nov 14, 2012 #8


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    Compresibiitiy or elasticity doesn't have any effect on the stress field, unless you want to model this as a large displacement nonlinear problem. The body forces that cause the centripetal acceleration are known from the geometry and density, and the body forces are also the gradient of the stress tensor. The stresses are independent of the material properties, for a given set of the boundary conditions.

    The boundary conditions are the key to the actual stress distrubition. For a rotating solid thin ring or cylinder with no radial constraints imposed, the main stress compoent is tension around the circumference of the ring. But a fluid can't sustain any significant tensile stress (or negative hydrostatic pressure if you prefer to call it that), so there must be some radial constraint forces (for example the fluid pressure on the solid structure that is containing it) to make the situation physically realistic.
  10. Nov 14, 2012 #9
    If the angular velocity is constant, then, as you indicate, then liquid and the solid can be treated as rigid, and the system will be statically determinate. However, if the angular velocity varies with time, then you have to consider the deformational mechanics of both the solid and the liquid.
  11. Nov 14, 2012 #10
    If W is constant, then, as AlephZero points out, the analysis is much simpler than if W varies with time. But this is not the problem that you posed.
  12. Nov 14, 2012 #11
    Could any of these forces apply to the earths core?

    Attached Files:

  13. Nov 14, 2012 #12
    1/ I would like to know if there are forces from pressure of liquid when W = constant ? AlephZero said that container would give the same force but that force come from the support of the system, right ? So the system oscillate around a point, the CG move ?

    2/ If W varies with time, it's easy to change the system for accelerate/deccelerate not the support of the system but a lot of parts of solid and liquid for reduce problem from deformation I think.

    Where come from this tangential force ? Could you explain, I don't understand ?

    @Blind light:
    Do you know an effect not explain on Earth and forces can explain ?
    Last edited: Nov 14, 2012
  14. Nov 14, 2012 #13
    I don't think AZ mentioned the system support, however you did in post#5.

    AZ did say

    and I concur exactly.

    So let us take things one step at a time.

    Do you understand the idea that any rotating object will be subject to internal forces that can be resolved in two directions at right angles - radially and tangentially?
  15. Nov 14, 2012 #14
    No, I don't understand that. I see only centripetal forces. Could you explain ?

    you don't understant the system or the goal of the system ? If it's the goal, it's only for study it. I know the CG can't move but I don't see that.
  16. Nov 14, 2012 #15
    Which direction do you think angular momentum acts?
  17. Nov 14, 2012 #16
    (I want to understand first when W = constant) it's tangential but I don't see how forces come from on your drawing at message #3
  18. Nov 14, 2012 #17
  19. Nov 14, 2012 #18
    Maybe for me it's easier to understand with small solid deformable material. I imagine a part of sector like drawing show. The material is solid but deformable. The deformation put pressure at right and at left, where is the C force in your message #3 ? The deformation is more and more high at external circle.

    Attached Files:

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  20. Nov 14, 2012 #19
    Your last sketch is not far off, so we are getting somewhere. I have drawn something similar in fig4 of the new attachment.

    Starting with the static situation of a ring or section of a cylinder under internal pressure, Fig1 shows the internal pressure tending to burst apart the ring. This would break it across some diameter, say AA.

    The bursting pressures and forces act radially.

    Fig2 shows how this bursting is resisted. There must be a forces F-F acting across the section AA, holding the ring together.
    These forces exactly equal the bursting forces and are equivalent to the forces F in my earlier diagram in post#3

    Fig 3 shows how the forces F-F must be the same all the way round the ring to preserve equilibrium.

    The resisting forces are therfore tangential or circumferential.
    They are called hoop stresses or forces, after the iron hoops used to hold wooden barrels together.

    Now in Fig4 we proceed to a rotating situation. Here the internal pressure is replaced by the a stress due to the increasing radius in the inner and outer faces of a small segment. This is like your last sketch.
    The circumferential stresses are considered constant and equal to the hoop stresses described before.

    (Note I have moved from forces to stresses the force is obviously stress time cross sectional area.)

    Now a rotating disk is not in equilibrium.

    However we may use equilibrium analysis if we introduce the fictitious centrifugal force to balance the real centripetal one. This is equal and opposite to the centripetal force and the technique is known as D'Alembert's Principle.

    Alternatively we may analyse the motion by Newton's laws of motion using the centripetal acceleration.

    I have done this in the previous attachment and called the centrifugal force C.
    This equals mass times distance to the centre of mass of the rotating half ring times the square of the angular velocity.

    Using D'Alembert allows me to then use the laws of equilibrium to equate this to the two hoop forces acting across the cut line forming the diameter of the half ring.

    Attached Files:

    Last edited: Nov 15, 2012
  21. Nov 15, 2012 #20
    Ok, like that I understand :)

    If I change liquid and take gas, gas can pressure liquid at 2 surfaces of half torus but gas can't pressure itself because temperature prevent this, how does it works in this case ?
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