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Motion in retarding force

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass M falls under gravity (force mg) through a liquid with decreasing viscosity so that the retarding force is -2mv/(1+t). If it starts from rest, what is the speed, acceleration, and distance fallen at time t=1.
    2. Relevant equations
    F=ma
    3. The attempt at a solution
    F = 2mv/(1+t) - mg
    F = m[- g + 2v/(1+t)]
    ma = m[-g + 2v/(1+t)]
    a = -g + 2v/(1+ t)

    At this point, I see two paths to a solution, and I don't know which is correct.
    I can let a=x'', v=x'
    x'' - (2/(1+ t)) x' = -g
    where the solution according to mathmatica is (I'm not sure how to solve this exactly, my guess is to use frobenius)
    x =1/3 - C1 T^3 + 1/2(2 C1 + g) T^2 + g T + C1 T + C2

    or I can let
    a=v'
    v' = -g + 2v/(1+ t)
    v' - 2v/(1+ t)= -g
    Where the solution would be
    v = (1 + t) (C1 t + C1 + g) - Edit (I see this solution might be wrong, checking now)

    I'm not sure which is the correct result, and I am concerned about the units on each,
     
    Last edited: Oct 31, 2016
  2. jcsd
  3. Oct 31, 2016 #2

    cnh1995

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    I would prefer this.
    But I think the equation should be
    F=mg-2v/(1+t).
    Negative sign of the retarding force indicates that it is directed opposite to the gravity.
     
  4. Oct 31, 2016 #3
    I see, I was thinking in terms of gravity pulling down, retarding force pushing up.
     
  5. Oct 31, 2016 #4
    Okay, flipping the signs around I get the following:
    a = g - 2v/(1+ t)
    let a=v'
    v' = g - 2v/(1+ t)
    v' + 2v/(1+ t) = g
    upload_2016-10-31_13-14-51.png
    Does this look correct? And what should I do with this constant C?
     

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  6. Oct 31, 2016 #5

    cnh1995

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    I think that won't make any difference as long as both the forces are directed opposite to each other.
     
  7. Oct 31, 2016 #6

    cnh1995

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    I believe this could be solved using the integrating factor method.
    If dy/dx+Py=Q, the integrating factor I=e∫Pdx. Look up this method.
     
  8. Oct 31, 2016 #7
    I did that, and I got a solution that matched the more complicated case of us x''=a, x'=v.
    upload_2016-10-31_13-42-14.png

    The problem now Is that I am unsure of what to do with those constants.
     
  9. Oct 31, 2016 #8

    cnh1995

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    First find the solution for v as a function of time. Here, you will have only one constant. Use the initial condition i.e. v=0 at t=0 to get this constant. So you'll have a complete v vs t relationship. Now differentiate it w.r.t time and put t=1 to get the acceleration at t=1 and integrate v(t) w.r.t time from t=0 to t=1 to get the displacement. I don't think you need three constants for this.
     
  10. Oct 31, 2016 #9
    Oh my god, I'm such and Idiot. I don't know why I didn't see that before!

    Thanks!
     
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