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Decreasing diameter of a spehere and drag

  1. Oct 31, 2011 #1
    Hey all,
    I have been mulling this around in my head for awhile and cannot make sense logically.
    In thinking about a sphere falling through some random liquid, how does decreasing the diameter of the sphere affect the rate at which it falls?

    I am thinking of two formulas and the seem to contradict each other in terms of their results.
    First would be Stoke's law dealing with drag force:
    F(d)= 3(pi)μVd
    Here if I reduce the radius, the force of drag decreases which makes me think the sphere would fall faster because it has less resisting its fall.

    The next would be a derived equation using the force of buoyancy, drag force, and weight. This assuming these are the only three forces on a falling sphere. It is solved for velocity.
    V= [2r^2(ρ(sphere)-ρ(fluid))g]/(9μ)
    Here it appears that if the radius is reduced, the velocity is also reduced and it therefore falls slower.
    My "logic" tells me a smaller sphere would fall faster, but it has failed me many times in physics.
    Thanks for the help.
     
  2. jcsd
  3. Oct 31, 2011 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Dust (even spherical dust) takes almost forever to settle out, whether in air or water.

    Do you think that a glass cricket ball would sink through honey faster or slower than a marble?

    If you are near an engineering workshop, you might be able to pick up a huge ball bearing and compare its descent through syrup with that of a smaller one.
     
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