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Raindrop variable mass

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Water vapor condense in a raindrop with rate μ units of mass per time.
    The raindrop starts with 0 velocity with initial mass $$ M_0 $$ and falls horizontally. Find the space it traverses as a function of time (g is given).


    2. Relevant equations

    ∂m/ ∂t= μ

    3. The attempt at a solution

    Assuming that the mass dm of vapor to be condesed has a total velocity v0, and using newton's law, I have [itex]g= \frac{dv}{dt} + \frac{dm*(v-v0)}{dt*m}[/itex] . From there the only thing i could think is that I could write dv/dt as dv/dm*dm/dt but it doesn't seem to help...
     
  2. jcsd
  3. Nov 9, 2013 #2

    etudiant

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    Nice problem.
    Presumably all the incremental mass gained as the original drop falls under gravity is at 0 velocity as well, so there will be a constant retarding force on the drop. However, because the drop is increasing in mass over time, the slowing effect of the additional mass declines.
     
  4. Nov 9, 2013 #3

    tiny-tim

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    Hi GregoryGr! :smile:
    (That's a good trick! :biggrin:)

    Consider the momentum (and force) at times t and t+dt.

    (and use m(t) = Mo + µt from the start! :wink:)
     
  5. Nov 9, 2013 #4

    haruspex

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    Good start. Can you write m as a function of t?
     
  6. Nov 10, 2013 #5
    Well, I got the answer without writing the momentum @ t and t+dt, with this:

    $$\vec{F}= \frac{d\vec{p}}{dt} \Rightarrow \int_{0}^{t}\mu gdt=\int_{0}^{\mu v}d(\mu v)$$
    Anybody want to show me alternative ways to solve it? It seems that the integral you have to calculate is too hard for a physics I exercise.

    EDIT: where I write /mu I meant the m(t)
     
    Last edited: Nov 10, 2013
  7. Nov 10, 2013 #6

    haruspex

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    Not sure how you get that, and it doesn't look to me like it's going to give the right answer. What answer did you get from it?
    What was wrong with your initial approach? What ODE did you get after writing m = μt? It didn't look hard to solve to me. Because the terms (gt, v, v't) are dimensionally the same, a natural try is v = Atα.
     
  8. May 29, 2015 #7
    Begin:
    dm/dt = u --> m(t) = u * t + M0, where m(0) = M0
    F = d/dt(m * v) = m * dv/dt + v * dm/dt = -m * g, assuming no air resistance.
    m * dv/dm * u = -v * u - m * g --> dv/dm + (1/m) * v = -g/u --> d/dm(m * v) = -m * g/u --> m * v = -m^2 * g/(2 * u) + C -->
    v(m) = -m * g/(2 * u) + C/m, v(M0) = 0 --> C = M0^2 * g/ (2 * u) -->
    v(m) = -g/(2 * u) * (m - M0^2/m) -->
    v(t) = -g/(2 * u) * (u * t + M0 - M0^2/(u * t + m0))
    Integrating we obtain:
    x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2 * ln(u * t + m0)) + K

    If x(0) = h --> K = h + g/(2 * u^2) * (M0^2/2 - M0^2 * ln(M0)) -->
    x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2/2 + M0^2 * ln(M0/(u * t + M0)) + h.
    End.

    For x(m) we have:
    x(m) = -g/(2 * u^2) * ((m^2 - M0^2)/2 + M0^2 * ln(M0/m)) + h.
     
    Last edited: May 29, 2015
  9. May 29, 2015 #8
    I think that ##\mu_0,\,M_0## must satisfy:
    $$ \frac{\partial{m}}{\partial{t}} = Cm(t) \,\text{where}\, \left[\frac{\partial{m}}{\partial{t}}\right]_{t=0} = \mu_0 $$
     
  10. May 29, 2015 #9

    haruspex

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    The problem statement is clear that the rate of mass gain is constant.
     
  11. May 29, 2015 #10
    So the force is constant, ok.
     
  12. May 29, 2015 #11

    ehild

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    Falls horizontally ???? How?
    Did you mean it falls vertically? Or the raindrop starts to move horizontally?
     
  13. May 29, 2015 #12
    May be "move horizontally", perhaps without friction.
     
  14. May 29, 2015 #13
    Assuming the raindrop falls vertically.
    Begin:
    dm/dt = u --> m(t) = u * t + M0, where m(0) = M0
    F = d/dt(m * v) = m * dv/dt + v * dm/dt = -m * g, assuming no air resistance.
    m * dv/dm * u = -v * u - m * g --> dv/dm + (1/m) * v = -g/u --> d/dm(m * v) = -m * g/u --> m * v = -m^2 * g/(2 * u) + C -->
    v(m) = -m * g/(2 * u) + C/m, v(M0) = 0 --> C = M0^2 * g/ (2 * u) -->
    v(m) = -g/(2 * u) * (m - M0^2/m) -->
    v(t) = -g/(2 * u) * (u * t + M0 - M0^2/(u * t + M0))
    Integrating we obtain:
    x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2 * ln(u * t + M0)) + K

    If x(0) = h --> K = h + g/(2 * u^2) * (M0^2/2 - M0^2 * ln(M0)) -->
    x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2/2 + M0^2 * ln(M0/(u * t + M0)) + h.
    End.

    For x(m) we have:
    x(m) = -g/(2 * u^2) * ((m^2 - M0^2)/2 + M0^2 * ln(M0/m)) + h.
     
    Last edited: May 29, 2015
  15. May 29, 2015 #14
    The problem seems fairly simple.
    Assume that the raindrop falls vertically.
    Begin:
    dm/dt = u --> m(t) = u * t + M0, where m(0) = M0
    F = d/dt(m * v) = m * dv/dt + v * dm/dt = -m * g, assuming no air resistance.
    m * dv/dm * u = -v * u - m * g --> dv/dm + (1/m) * v = -g/u --> d/dm(m * v) = -m * g/u --> m * v = -m^2 * g/(2 * u) + C -->
    v(m) = -m * g/(2 * u) + C/m, v(M0) = 0 --> C = M0^2 * g/ (2 * u) -->
    v(m) = -g/(2 * u) * (m - M0^2/m) -->
    v(t) = -g/(2 * u) * (u * t + M0 - M0^2/(u * t + M0))
    Integrating we obtain:
    x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2 * ln(u * t + M0)) + K

    If x(0) = h --> K = h + g/(2 * u^2) * (M0^2/2 - M0^2 * ln(M0)) -->
    x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2/2 + M0^2 * ln(M0/(u * t + M0)) + h.
    End.
     
  16. May 29, 2015 #15

    haruspex

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    No, the 'drag' that results from mass accumulation will be proportional to the speed.
    Rocco's analysis looks right, but I've not checked it in detail. Too tedious without LaTeX.
     
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