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

  • #1

Homework Statement


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 )

Homework Equations




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)
 

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Answers and Replies

  • #2
TSny
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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.
 
  • #3
ehild
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Your integral with respect to v is not correct,
 
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  • #4
Your integral with respect to v is not correct,
I cannot spot the mistake. Do you mind pointing out where I went wrong?
 
  • #5
How did you integrate ##\frac{dv}{\frac{mg}{c}-v^2}##?
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
 
  • #6
TSny
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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
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.
 
  • #7
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.
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?
 
  • #8
TSny
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I believe your solution is correct. I thought maybe ehild was thinking of the logarithmic form of the integral.
 
  • #9
I believe your solution is correct. I thought maybe ehild was thinking of the logarithmic form of the integral.
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.
 
  • #10
ehild
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I cannot spot the mistake. Do you mind pointing out where I went wrong?
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.
 

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