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Ball dropped in viscous oil

  1. Jan 13, 2012 #1
    1. The problem statement, all variables and given/known data
    In an experiment to determine the viscosity of some oil, a ball is dropped into some oil. The position of the ball is given by the formula y=Y+At+Be-Ct. At t=0, y=0.070 m, the velocity dy/dt is 0 and the acceleration is 0.0983 m/s2. As t→∞ the velocity approaches 0.480 m/s. What is the value of Y in MKS units?


    2. Relevant equations
    y=Y+At+Be-Ct
    vy(t)=A+-CBe-Ct
    ay(t)=C2Be-Ct

    3. The attempt at a solution
    With four unknowns (Y,A,B,C), I came up with three equations:
    B+Y=0.070m
    A=CB
    BC2=0.0983

    I wasn't sure what to do as t→∞
     
  2. jcsd
  3. Jan 13, 2012 #2
    Looking at your velocity equation, what happens to the exponential term as t approaches infinity? What does the exponential function look like as its exponent approaches infinity? What if the exponential is in the denominator? Try using a graphing program or wolfram alpha to plot it.
     
  4. Jan 13, 2012 #3
    as t→∞ the exponent would turn to negative, therefore it would go to 0. But that would give me
    vy(∞)=A-CBe-C∞
    0.480=A-CB(1)

    if A=CB

    0.480=CB-CB

    and

    0.480≠0
     
  5. Jan 13, 2012 #4
    You said the exponential term would go to zero but you plugged a 1 into your equation. Yes [itex]e^{-t}[/itex] goes to zero as t approaches [itex]\infty[/itex].
     
  6. Jan 13, 2012 #5
    I skipped a step there, sorry. Yea I figured e^-Ct would go to 0 as t goes to infinity, but the math did not work out. Any more suggestions?
     
  7. Jan 13, 2012 #6
    If the exponential term goes to 0 in your velocity equation, then you end up with 0.480=A correct? This gives you a 4th equation to solve for your 4 unknowns. Now you just have to do some algebra and solve the system of equations for Y.
     
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