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Confusion over Stoke's Law

  1. Nov 7, 2008 #1
    I'm not sure if this is the right place to put this; if it isn't, could someone point me in the right direction... thanks.

    Anyhow, today in Physics (I am currently studying A-levels in England) we came across Stoke's Law. We were taught that, essentially it meant that the greater a spheres radius, the faster it falls through a fluid, and that weight does not factor in. However, I do not understand this; we have been taught that a larger object results in greater resistance from whatever it is moving in.
    And I also wondered how it was possible to rule out weight as a factor. Surely, it is impossible to test the correlation between weight and the speed (keeping volume / radius the same) without changing density, which is something not being changed in the original experiment, therefore you can't really rule out weight (or can you...?).
    These were my issues. I quickly wore out my teacher with my questions... perhaps I am missing some obvious point.

    Anyhow, is there anyone here who could explain, in layman's terms (if that's not asking for too much), why a larger radius makes it descend faster (assuming weight does not matter), and how Stoke managed to rule out weight as the independent variable.

    Have I missed something?

    Thanks in advance

    Riga-b
     
  2. jcsd
  3. Nov 7, 2008 #2

    olgranpappy

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    Your teacher, I think, was talking about the *terminal* velocity of a sphere that is falling through a fluid under the influence of gravity. In this case, consideration of
    [tex]
    m\bold{a}=0=\bold{F}_{\rm gravity}+\bold{F}_{\rm buoyancy}+\bold{F}_{\rm drag}\;,
    [/tex]
    where F_drag is given by Stoke's law, will show that the terminal velocity is proportional to the square of the radius of the sphere. So, with all other things fixed, the larger the sphere, the larger the terminal velocity.
     
  4. Nov 9, 2008 #3
    Ah, I see. So mass is a factor in determining the terminal velocity of the sphere. Or have I misread the equation?

    If it is not, how is it possible to keep all other things fixed (mass and density included) to determine that it is indeed the volume / radius of the sphere?

    Thanks for your reply. Sorry if I'm not getting something.
     
  5. Nov 10, 2008 #4

    olgranpappy

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    F_grav is proportional to the mass of the sphere, F_buoyancy is proportional to the volume of the sphere, F_drag is proportional to the radius of the sphere.
     
  6. Nov 10, 2008 #5

    olgranpappy

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    F_drag is also proportional to the velocity of the sphere.

    By definition, that we are at the terminal velocity mean that the sum of the forces is zero.
     
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