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Falling Rain Drop (Variable Mass)

  1. Apr 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose a rain drop with mass ##m_0\neq 0## is falling due to gravity with initial velocity ##v_0##, assume ##\frac{dm}{dt}=k=##constant. Solve the differential equation and determine the velocity as ##t\to\infty##

    2. Relevant equations
    ##F=\frac{dp}{dt}=\dot{m}v+m\dot{v}##
    3. The attempt at a solution

    Since ##F=mg## the D.E. is ##\dot{m}v+m\dot{v}=mg##, substituting in ##\frac{dm}{dt}=k## we find ##kv+m\dot{v}=mg##, since the mass increases with time we have ##m=m_0+kt##. One method I've attempted is essentially using the integrating factor but since the LHS is already a product rule derivative I don't need to multiply it by anything so:

    $$\frac{d}{dt}(mv)=mg \\
    \int \frac{d}{dt}(mv)dt=\int mg dt=\int mg \frac{dt}{dm}dm=\frac{g}{k}\int_{m_0}^{m} m\, dm\\
    v(m_0+kt)+C=\frac{g}{2k}((m_0+kt)^2-m_0^2)=\frac{g}{2k}(2m_0kt+k^2t^2)
    $$

    at ##t=0##, ##v=v_0\Rightarrow##, ##v_0(m_0+0)+C=0\Longrightarrow C=-v_0m_0##

    therefore...

    $$v(m_0+kt)-v_0m_0=gm_0t+\frac{gkt^2}{2}\\
    v(m_0+kt)=m_0(v_0+gt)+\frac{gkt^2}{2} \\
    v=\frac{m_0(v_0+gt)}{m_0+kt}+\frac{gkt^2}{2(m_0+kt)}
    $$

    Now I find this solution a bit odd since my professor said that it would become asymptotic but I think he's wrong about this? The first term will approach ##\frac{m_0g}{k}## but the second term clearly diverges as ##t\to\infty## and it doesn't make sense that an object would reach a terminal velocity without drag. I suppose you could say it asymptotically goes to infinity but it's increasing linearly...
     
  2. jcsd
  3. Apr 20, 2016 #2

    DrSteve

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  4. Apr 20, 2016 #3

    haruspex

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    By itself that means nothing. What is supposed to tend asymptotically to a constant, or to a certain function of what?
    The link DrSteve posted shows the acceleration tends to a constant.
     
  5. Apr 20, 2016 #4
    Okay I suppose I'm just too used to thinking of functions converging to a constant when that term is mentioned.
     
  6. Apr 20, 2016 #5

    ehild

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    The velocity tends to a simple function as t tends to infinity. What is that simple function?
     
  7. Apr 20, 2016 #6
    I believe it's just ##v=\frac{m(t)g}{k}## from setting ##\dot{v}=0## in the D.E.

    Simplifying my expression for ##v(t)## would give ##v(t)=\frac{m_0g}{k}+\frac{gt}{2}## though

    Sorry for so many edits but just noticed that the first one becomes ##v=\frac{m_0g}{k}+gt## which is almost the same as the result from simplification.
     
  8. Apr 20, 2016 #7

    ehild

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    You can not set ##\dot v = 0##. The acceleration does not tend to zero. At great t, v increases linearly with time with rate g/2 instead of g as it were for a constant-mass body.
     
  9. Apr 20, 2016 #8
    Whoops, I realize now it doesn't make much sense to set acceleration equal to 0 :)

    Interestingly at large t, it appears v always increases by ##g/2## regardless of ##\alpha## for ##\frac{dm}{dt}=k^{\alpha}##
     
  10. Apr 20, 2016 #9

    ehild

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    ##k^{\alpha}## is just an other constant. Why do you mix alpha in? It was said that dm/dt=k.
     
  11. Apr 20, 2016 #10

    ehild

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    It is interesting the acceleration at long time tends to g/2, regardless of k.
     
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