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Horizontal Motion with Quadratic Drag

  • #1

Homework Statement



Consider an object that is coasting horizontally (positive x direction) subject to a drag force f = -bv - cv^2 . Write down Newton's second law for this object and solve for v by separating variables. Sketc the behaviour of v as a function of t. Explain the time dependence for t large. (Which force term is dominant when t is large?)

Homework Equations





The Attempt at a Solution



I start with Newton's second law:

m(dv/dt) = -bv -cv^2

m(dv/dt) = -v(cv + b)

(-m/v)(dv/dt) -cv = b

dv/v + (cv/m)dt = (-b/m)dt

Then... do I integrate with respect to v for the dv terms and t for the dt terms? I can't figure out what this will give me with the second term on the right...
 

Answers and Replies

  • #2
674
2
You need all the 'v' terms together with 'dv'. But you subtracted 'cv' in line 3 which is a mistake. Instead divide the whole equation in line 2 by 'v(cv+b)'. Then solve for it.
 
  • #3
I get to the point where all my v terms are on one side so that I have:

(m dv)/[v(cv+b)] = -dt

From here when I try and integrate with respect to v on the left and t on the right, I'm doing a definite integral setting the limits between v and v0 and t and t0 so I get:

m[ln(cv^2 + bv) - ln(cv0^2 - bv0^2)] = -t

ln(cv^2 + bv) = -t/m + ln(cv0^2 - bv0^2)

Then if I take the exponential of both sides I still end up in a situation where I can't solve for v because there will be v on both sides... i.e

cv^2 + bv = e^(-t/m) + cv0^2 - bv0^2

How can I solve this for v?
 
  • #4
674
2
You need to check your integration, it is incorrect.
 
  • #5
How exactly do I integrate m(dv)/[v(cv+b)] ? The way I do it I consistently get natural log expressions which I am guessing is wrong because the function should tend towards 0 when I plot it...
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
6
How exactly do I integrate m(dv)/[v(cv+b)] ? The way I do it I consistently get natural log expressions which I am guessing is wrong because the function should tend towards 0 when I plot it...
[tex]\frac{d}{dv}\ln|cv^2+bv|=\frac{2cv+b}{cv^2+bv}\neq\frac{1}{cv^2+bv}[/tex]

To integrate your expression properly, try completing the square on the denominator and then making an appropriate substitution.
 
  • #7
674
2
Another trick is to break the fraction up into to fractions that look like this:

[tex]\frac{1}{v(cv+b)} = \frac{A}{v} + \frac{B}{cv+b}[/tex]

Then solve for A and B. Integrating the two fractions separately will be much easier.
 
  • #8
Ah that clears things up. I see where I went wrong there, thanks for your help!
 

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