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Torque on a point on a sphere in a fluid/finding pressure?

  1. Jul 7, 2013 #1
    If there is a rotating sphere (falling through a fluid) a) is the torque the same at every point on the sphere's surface, and b) how would I use said torque to work out the pressure exerted by opposite 'sides' of the sphere on the fluid?



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 7, 2013 #2

    haruspex

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    What equations do you have for drag?
     
  4. Jul 8, 2013 #3
    drag = 1/2 ρf v^2 Cd A

    where pf is the fluid density, v is the velocity of the sphere, Cd is the drag coefficient for the sphere (0.1 for smooth, 0.6 for rough) and A is the reference (cross-sectional) area
     
  5. Jul 8, 2013 #4

    haruspex

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    I forgot to ask what the relationship is between the axis of rotation and the vertical (the direction of travel). Are they the same or orthogonal? If orthogonal, think about the v term of the drag for different points on the surface.
    I assume what this is leading to is an explanation of the Magnus effect. Intuitively, I think I see how the difference in shear forces leads to the required pressure difference, but I'm not an expert in this area.
     
  6. Jul 17, 2013 #5
    They're the same.
     
  7. Jul 17, 2013 #6

    haruspex

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    In that case, the torque at a point on the surface will clearly depend on its latitude, and you could write the equation down fairly easily. But I've no idea how this connects with pressure.
     
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