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Terminal velocity of falling drop with increasing mass

  1. Sep 3, 2010 #1
    I m solving the problem in which the drop is falling from some height and its size is increasing with the rate dm/dt=kMv where v is the velocity of the drop at that instant , so we have to find the terminal velocity of the drop gained due to air drag,
    The problem is that if the drop gains the terminal velocity then the mass will increase with its rate as defined , then the forces balanced are disturbed and there is no any terminal velocity more , the same thing is going to happen all the time so will it gain the terminal velocity any time ? If possible then how?
     
  2. jcsd
  3. Sep 3, 2010 #2
    Why do you think that "forces balanced are disturbed"? I suggest that you sit down and do some math, you may find the answer :smile: Don't forget using the general form of Newton's 2nd law: F = dp/dt.
     
  4. Sep 3, 2010 #3
    What is the air drag formula?
     
  5. Sep 3, 2010 #4
    air drag I mean ,,,,viscous force,,
     
  6. Sep 3, 2010 #5
    ya tried newtons 2 law
    F=dp/dt
    F=vdm/dt+mdv/dt
    here in above equation dm/dt and dv/dt are variable but at the terminal velocity net force is zero but this is not possible as both the terms are variable,,,,,
     
  7. Sep 3, 2010 #6
    Well, I'm not sure how to model this problem if we take into account the viscous force (notice that normally viscous force refers to the drag force on a body whose mass remains unchanged. In this case, the drop gets bigger by sweeping through the air and absorbing water vapors; its mass changes with time). I'll leave it aside.

    So from Newton's law: F=dp/dt, where dp = mdv + vdm, as you deduced.
    Notice that F=mg and dm/dt = kmv, we arrive at this equation: g = dv/dt + kv2. When the drop reaches its terminal velocity, its speed is unchanged, i.e. dv/dt = 0 (net force = 0 is a wrong condition!). From here, you obtain vterminal.

    The subtle thing I mentioned earlier lies in the equation: F=vdm/dt+mdv/dt. At terminal speed, you can see that the impulse that the external force gives to the drop does NOT accelerate the drop, but instead, gives the additional mass dm the momentum vdm.

    So external force has 2 roles: one is to accelerate the drop, the other is to provide momentum to additional mass. F is not necessarily proportional to dv/dt. This is why F cannot be 0 when terminal speed is reached.
     
    Last edited: Sep 3, 2010
  8. Sep 4, 2010 #7
    thanx,,,,,, i did the same thing but i was not sure of that
    i got the answer was v=g/k^1/2 it is correct as u said ,,
     
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