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Terminal velocity of a raindrop

  1. May 30, 2017 #1
    1. The problem statement, all variables and given/known data

    A raindrop of initial Mass ##M_0## starts to fall from rest under the influence of gravity. Assume that the drop gains mass from the cloud at a rate proportional to the product of its instantaneous mass and its instantaneous velocity ##\dfrac{dM}{dt} = kMV##, where ##k## is constant. show that speed eventually become constant.

    2. Relevant equations


    3. The attempt at a solution

    ##P_i = M_0v##

    ##P_f = \left( M_0 + \left(\Delta t \dfrac{dM}{dt}\right) \right)(v + \Delta v)##

    therefore ##\Delta P = M_0 \Delta v + v \Delta t\dfrac{dM}{dt}##

    Since ##F = \lim_{\Delta t \to 0} \dfrac{\Delta P}{\Delta t}##

    Therefore ##F = M_0 \dfrac{dv}{dt} + v \dfrac{dM}{dt}##

    Substituting ##\dfrac{dM}{dt} = kMV##,

    ##F = M_0 \dfrac{dv}{dt} + k M v^2##

    The velocity will be constant if the forces are balanced,

    therefore, ##Mg = M_0 \dfrac{dv}{dt} + k M v^2##

    Now I don't know what to do ? should I solve for ##v## in this differential equation and then find the time at which it would be constant (I tired but it was very messy with lots of constants, I got ##v = \displaystyle u\dfrac{1- \exp(2ku\alpha t)}{\exp(2ku\alpha t) - 2} ## where ##u = \sqrt{h/k}## and ##\alpha = M/M_0##) ?
     
  2. jcsd
  3. May 30, 2017 #2

    jack action

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    Actually:
    $$F = \frac{d(Mv)}{dt} = M \dfrac{dv}{dt} + v \dfrac{dM}{dt}$$
    Then you just need to find the condition that will set ##\dfrac{dv}{dt} = 0##.
     
  4. May 30, 2017 #3
    But I got ##M_0## instead of ##M##.
     
  5. May 30, 2017 #4

    jack action

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    I know, but it should be the instantaneous mass, not the initial mass:

    ##P_i = Mv##
    ##P_f = \left( M + \left(\Delta t \dfrac{dM}{dt}\right) \right)(v + \Delta v)##

    With the initial mass ##M_0##, it is only valid for the initial instant as with ##M## it is valid at any moment.
     
  6. May 30, 2017 #5
    Then we get,

    ##Mg = M \dfrac{dv}{dt} + k M v^2##

    ##g - kv^2 = \dfrac{dv}{dt}##. Therefore at ##g = kv^2##, velocity is constant. Is this fine ?
     
  7. May 30, 2017 #6

    jack action

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    That would be my answer. When ##v= \sqrt{\frac{g}{k}}##, you will have no more acceleration, so it will stay at this speed constantly.
     
  8. May 31, 2017 #7
    Last edited: May 31, 2017
  9. May 31, 2017 #8

    Nidum

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    and shouldn't the buoyancy and drag forces on the drop be considered ?
     
  10. May 31, 2017 #9
    -
     
    Last edited: May 31, 2017
  11. May 31, 2017 #10
    But in the question no value of relative velocity is given :).

    :nb):nb)
     
  12. May 31, 2017 #11

    haruspex

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    I agree that the equation F=dp/dt=d(mv)/dt=mdv/dt+vdm/dt is not kosher. It treats mass as something that can change in a closed system.
    However, it does give the right answers when the mass entering or leaving the system does so with no momentum in the frame of reference. In this case, we can assume that the moisture being collected had been stationary relative to the ground.
     
  13. Jun 1, 2017 #12
  14. Jun 1, 2017 #13
    Oh! that is my mistake. In this problem u=0
     
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