Derivation of Drag Force in air(D=bv)

  • #1

Main Question or Discussion Point

I have stuck at a step during derivation.
Here it is.
D=bv
mg-bv=ma
a=g-[tex]\frac{bv}{m}[/tex]
[tex]\frac{dv}{dt}[/tex]=g-[tex]\frac{bv}{m}[/tex]
[tex]\frac{dv}{g-[tex]\frac{bv}{m}[/tex]}[/tex] = dt
or,
[tex]\frac{dv}{\frac{mg-bv}{m}}[/tex] = dt
[tex]\frac{dv}{mg-bv}[/tex] = [tex]\frac{dt}{m}[/tex]
let, u=mg-bv
[tex]\frac{du}{dv}[/tex]=0-b
[tex]\frac{du}{dv}[/tex]=-b
multiplying both sides by, "-b"
[tex]\frac{-bdv}{mg-bv}[/tex]=[tex]\frac{-bdt}{m}[/tex]
[tex]\frac{du}{u}[/tex]=[tex]\frac{-bdt}{m}[/tex]
integrating L.H.S. from 0 to u and Integrating R.H.S. from 0 to t
[tex]\int^{u}_{0}\frac{du}{u}[/tex]=[tex]\frac{-b}{m}[/tex][tex]\int^{t}_{0}dt[/tex]
ln(u)|[tex]^{u}_{0}[/tex]=[tex]\frac{-bt}{m}[/tex]
ln(u)=[tex]\frac{-bt}{m}[/tex]
ln(mg-bv)=[tex]\frac{-bt}{m}[/tex]
mg-bv=e[tex]^{\frac{-bt}{m}}[/tex]

Now, I am stuck
L.H.S. must be [tex]\frac{mg-bv}{mg}[/tex]
but I am unable to get mg in denominator.

Where I am wrong?
 

Answers and Replies

  • #2
649
2
I haven't checked everything, but one thing is that you cannot integrate u from 0, since that integral doesn't converge. So you cannot choose to have u=0 when t=0 as you have done by your choice of integration limits.

How about selecting v=0 at t=0 as your initial condition, i.e. you'll do the u-integral with the lower limit 'mg' instead of 0.

What do you think?
 
  • #3
AlephZero
Science Advisor
Homework Helper
6,993
291
The place you went wrong was

integrating L.H.S. from 0 to u and Integrating R.H.S. from 0 to t
If v = 0 when t = 0, what does u equal when t = 0? (Hint, the answer is not 0).
 

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