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Terminal velocity and drag coefficient?

  1. Feb 19, 2007 #1
    Ok, I think I'm getting a hold on this problem, but I want to make sure I'm in the direction...

    1. The problem statement, all variables and given/known data
    Let [tex]\inline{v_\infty}[/tex] be the terminal velocity of a falling body in air. If the body is dropped from rest, find the expression for the speed as a function of time (1) if the air resistance is proportional to the speed, and (2) if it is proportional to the square of the speed. In each case, relate the terminal velocity to the drag coefficient (proportionality constant) [tex]\inline\rho[/tex], mass m, and g, the acceleration due to gravity.


    2. Relevant equations
    1.[tex]f=ma=m\frac{dv}{dt}[/tex]

    2.[tex]f=f_a-f_g=\rho v^2-mg[/tex]

    3.[tex]a=\frac{dv}{dt}=\frac{d^2x}{dt^2}[/tex]



    3. The attempt at a solution
    I believe this is the solution for case (2), but I don't really know what it is for case (1). I know that case (1) involves slow moving bodies, and vice versa for case 2.

    If you take eqn. 1 and divide by m, that gives you the acceleration.

    [tex]\frac{f}{m}=\frac{dv}{dt}=\frac{\rho}{m}v^2-g\\
    =g\left(\frac{\rho}{mg}v^2-1\right)\\
    =g\left(\frac{v^2}{v_\infty}-1\right)[/tex]

    where [tex]\inline v_\infty[/tex] is the terminal velocity [tex]\inline{\sqrt{\frac{mg}{\rho}}}[/tex]

    Using separation of variables, I got to this:

    [tex]\frac{v_\infty^2dv}{v^2-v_\infty^2}=gdt[/tex]

    Then after taking the integral of each side, and then doing a bunch of algebra which I won't show here [unless you need me to put it to better understand the process], I eventually got this:

    [tex]v(t)=(v_\infty)\frac{e^{\frac{-2gt}{v_\infty}}-1}{e^{\frac{-2gt}{v_\infty}}+1}[/tex]

    Is this correct for case (2)? If so, what would change for case (1)? Thanks for your help! :smile:
     
    Last edited: Feb 19, 2007
  2. jcsd
  3. Feb 19, 2007 #2
    The physics here is writing the differential equations. The rest is just math.

    For the first, the differential equation is as follows:

    [tex]
    m\frac{dv}{dt} = \rho v - mg
    [/tex]

    For the second:

    [tex]
    m\frac{dv}{dt} = \rho v^2 - mg
    [/tex]

    The terminal velocity is just the limit as [tex]t \rightarrow \infty[/tex].
     
  4. Feb 20, 2007 #3
    So that final answer is correct? I would just have to re-do the math with a v instead of a v²?
     
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