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Homework Help: Quadratic drag as v^3/2

  1. Sep 30, 2010 #1
    1. The problem statement, all variables and given/known data
    A block of mass m slides on a horizontal surface that's been lubricated with a heavy oil so that the block suffers a viscous resistance that varies as the 3/2 power of the speed.

    If the initial speed of the block is Vo at x=o, show that the block cannot travel further than 2m(Vo^(1/2))/c

    c is the drag constant.

    2. Relevant equations

    The viscous resistance is defined as F(v)=-cv^(3/2)


    3. The attempt at a solution

    So I defined my axis so that F = ma = -F(v) = cv^(3/2)

    It's obvious that I need to calculate the limit of t when t goes to infinity from the position equation. So I integrate my F=ma equation twice and I don't get the answer.

    For my first integration, I have dv/(v^(3/2)) = (cm)dt which gets me 1/(V^(1/2)) - 1/(Vo^(1/2)) = cmt

    Now I isolate V=dx/dt in order to integrate a second time. I find that V=1/((Vo^(1/2)) + (ct/2m))^2

    Let u = the denominator, du = cdt/2m --> dt = 2mdu/c. The integral becomes dx=1/(u)^2du
    and I get x = -1/u. After I replace u for it's value, my limit when t-->infinity isn't what the question demands.

    Could someone tell me where I might have made a mistake?
     
  2. jcsd
  3. Sep 30, 2010 #2

    fzero

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    The viscous force acts opposite to the direction of motion. It was already defined with the correct sign, but you negated it. If you take your solution and replace c -> -c you might get the correct result.
     
  4. Sep 30, 2010 #3
    I just tried it, and my limit is still wrong. My second integral is now x = m/c((1/Vo^(1/2)) - (ct/m))
     
  5. Sep 30, 2010 #4

    fzero

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    You're missing some minus signs and factors of 2. You're also missing the t=0 term in the integration of the velocity. For the first integration I find

    [tex]-\frac{2}{\sqrt{v}} + \frac{2}{\sqrt{v_0}} = - \frac{ct}{m}[/tex]

    The 2nd integration gives

    [tex] x = \frac{2m v_0 }{c\sqrt{v_0}} \left( - \frac{1}{1+\frac{c\sqrt{v_0}}{2m}t} + 1\right).[/tex]

    After simplifying, you get something that yields

    [tex]x(t\rightarrow\infty) \rightarrow \frac{2m\sqrt{v_0}}{c}.[/tex]
     
  6. Sep 30, 2010 #5
    thank you very much, I understand now
     
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