1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Falling object in a gravitational field with v^2 drag force

  1. Sep 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider a particle in a gravitational field that is also subject to a resisting force proportional to the velocity squared ( Fdrag = + or - cv2).
    a) Find the terminal velocity, vT, for the object as it falls.

    b) Show that for an object dropped from rest that the velocity is given by v = vTtanh-1(-t/β)
    where β = sqrt( m/gc )

    2. Relevant equations


    3. The attempt at a solution
    I did part a and got vT = sqrt( mg/c )
    but for part b I am not getting the same result that is shown.
    My work is shown in the attachment (the tau in the attachment is the β. I wrote it as β here because its hard to distinguish the t, T, and the tau on here)
     

    Attached Files:

    Last edited: Sep 24, 2016
  2. jcsd
  3. Sep 25, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Your work looks correct to me. Note that your solution has the correct asymptotic behavior for t → ∞; but the answer stated in the problem does not.
     
  4. Sep 25, 2016 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Your integral with respect to v is not correct,
     
    Last edited: Sep 25, 2016
  5. Sep 25, 2016 #4
    I cannot spot the mistake. Do you mind pointing out where I went wrong?
     
  6. Sep 25, 2016 #5
    I used an integral table for this which says that
    int [ dx/(a2-x2)] = (1/a) * tanh-1 (x/a)

    where a = sqrt (mg/c) and this is for x2 < a2 which is the case here
     
  7. Sep 25, 2016 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    To obtain this result, you can change integration variable from x to z, where tanh(z) = x/a.

    If you do the integral by partial fractions using 1/(a2 - x2) = 1/(2a) * [ 1/(a + x) + 1/(a - x) ] , then you get the integral expressed in terms of a logarithm instead of tanh-1.
     
  8. Sep 25, 2016 #7
    Right, I was just trying to express it as was requested in the question details. I'm still wondering if I made an error though as ehild suggests?
     
  9. Sep 25, 2016 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I believe your solution is correct. I thought maybe ehild was thinking of the logarithmic form of the integral.
     
  10. Sep 25, 2016 #9
    Thanks for the help. The result shown in the problem description is definitely wrong for the incorrect asymptotic behavior that you pointed out as t -> infinity. I emailed the professor about this so he can let the class know just in case anybody else was scratching their heads.
     
  11. Sep 25, 2016 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Sorry, I could not read your handwriting properly. Next time type in your derivation, please.
    Your solution is correct. Yes, I thought of the logarithmic form of the integral, but the two forms are equivalent.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Falling object in a gravitational field with v^2 drag force
  1. Free falling object #2 (Replies: 11)

Loading...