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Velocity under a drag force

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data
    An object moving in a liquid experiences a linear drag force: D⃗ =(bv, direction opposite the motion), where b is a constant called the drag coefficient. For a sphere of radius R, the drag constant can be computed as b=6πηR, where η is the viscosity of the liquid.

    Find an algebraic expression for vx(t), the x-component of velocity as a function of time, for a spherical particle of radius R and mass m that is shot horizontally with initial speed v0 through a liquid of viscosity η.
    Express your answer in terms of the variables v0, η, R, t, m, and appropriate constants.

    2. Relevant equations



    3. The attempt at a solution
    Thinking of typical dynamics, I divided the drag force, bv by m to get the acceleration. Then I subtracted acceleration times time from the inivial velocity, v0. So it looked like this:

    v0- (6πηRv0t)/m.

    Obviously it wasn't right, as my teacher today told me that I have to integrate the acceleration function to get the velocity function. I have no idea how to integrate the acceleration function in which it looks like every single variable are constants.

    Help would be appreciated! Thanks in advance.
     
  2. jcsd
  3. Mar 18, 2014 #2

    tiny-tim

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    hi playoff! :smile:
    no, a = dv/dt is a function of v, not a constant :wink:
     
  4. Mar 18, 2014 #3
    Draw a free body diagram, and apply Newton's second law to the mass. Don't forget to include the buoyant force.

    Chet
     
  5. Mar 18, 2014 #4
    Ugh, I have a very shallow understanding in calculus. So if I would integrate it with v in the acceleration function, wouldn't it give me the position function in the velocity function? And the only variables I can use are v0, η, R, t, m, and appropriate constants.

    Thanks for pointing it out though :D

    @Chestermiller: I thought the only force acting in the x-axis is the drag force itself. Would the buoyant force also be acting against the velocity?
     
  6. Mar 18, 2014 #5
    Oops. I should have read the problem statement more carefully. Sorry about that.

    Chet
     
  7. Mar 19, 2014 #6

    tiny-tim

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    (just got up :zzz:)
    i don't understand this :confused:

    to integrate dv/dt = f(v),

    write it dv/f(v) = dt, then integrate both sides :smile:
     
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