Calculating Displacement with Air Drag: A Differential Equation Approach

[Espen]

This is not a homework problem, just an idea I had.

If we have a sphere which is being accelerated in space by a force [tex]\vec{F}[/tex] and we take the drag into account, what would the displacement be after t seconds? If the drag force is [tex]\vec{D}=-c_1v-c_2v^2[/tex], then the acceleration would be [tex]\vec{a}=\frac{\vec{F}+\vec{D}}{m} \,,\, a=\frac{F-c_1v-c_2v^2}{m}[/tex] such that the displacement [tex]s=\frac{1}{2}\left(\frac{F-c_1v-c_2v^2}{m}\right)t^2[/tex]. Is this correct? Could someone help me solve this differential equation, if it has a solution?

Thanks for any help.

 

[3DM@rk]

No, that's not the correct formula for the displacement. The well-known [tex]s=\frac{1}{2}at^{2}[/tex] holds only if the acceleration is constant. In this case, the acceleration is a quadratic function of velocity, so we have one scalar differential equation:
[tex]a=\frac{dv}{dt}=A-Bv-Cv^{2}[/tex]
where A= F/m, B=C1/m and C=C2/m are constants. The equation can be solved using separation of variables, as
[tex](1) \int dt=\int\frac{dv}{A-Bv-Cv^{2}}[/tex]

I can solve this easily if either B or C is equal to zero.

• [tex]D=-c_{1}v[/tex]
  Then by solving (1), and then another integration with respect to time, we get (for v(0) = 0)
  [tex]s(t)=\frac{F}{c_{1}}\left(t+\frac{exp(-\frac{c_{1}}{m}t)-1}{\frac{c_{1}}{m}}\right)[/tex]
  where s(t) is the displacement after time t.
  
• [tex]D=-c_{2}v^{2}[/tex]
  
  [tex]v_{m}=\sqrt{\frac{F}{c_{2}}}[/tex]
  is the terminal velocity (where D=F). Then using partial fraction decomposition on (1), integration and then another integration with recpect to time, we get (again for zero velocity at the begining):
  [tex]s(t)=\frac{v_{m}m}{\sqrt{Fc_{2}}}ln cosh\left(t\frac{\sqrt{Fc_{2}}}{m}\right)[/tex]
  where s(t) is the displacement at time t.

I'm too lazy at the moment to solve the problem for generally non-zero C1 and non-zero C2, which should be possible, but some difficulties might be encountered when inverting the function. I hope you got the picture how it works. Also, you can use numerical solution of the equation, which is the common approach in physics.

And finally, I apologize for any inconvenience caused by my English or incorrect English math terms.

 

[Espen]

I solved the integral to

[tex]\int \frac{dv}{A-Bv-Cv^2}=-\frac{2\tan^{-1}\left(\frac{B+2Cv}{\sqrt{-B^2-4AC}}\right)}{\sqrt{-B^2-4AC}}[/tex], so then [tex]t=-\frac{2\tan^{-1}\left(\frac{B+2Cv}{\sqrt{-B^2-4AC}}\right)}{\sqrt{-B^2-4AC}}[/tex]

Working with this, I get that

[tex]v=\frac{\sqrt{-B^2-4AC}\tan\left(\frac{\sqrt{-B^2-4AC}}{2}t\right)-B}{2C}[/tex]

If this is correct, I want to integrate it in order to get an expression for the displacement over time.

[tex]s=\int \frac{\sqrt{-B^2-4AC}}{2C}\cdot\tan\left(\frac{\sqrt{-B^2-4AC}}{2}t\right)-\frac{B}{2C}\rm{d}t[/tex]

I will simplify it be setting [tex]\frac{\sqrt{-B^2-4AC}}{2}=k_1[/tex] and get

[tex]s=\frac{k_1}{C}\cdot\int\tan\left(k_1\cdot t\right)\rm{d}t-\int\frac{B}{2C}\rm{d}t=\frac{k_1}{C}\left(-\frac{\log\left(\cos\left(k_1t\right)\right)}{k_1}\right)-\frac{B}{2C}t+s_0[/tex]

How does this look?

 

[3DM@rk]

[tex]t=-\frac{2\tan^{-1}\left(\frac{B+2Cv}{\sqrt{-B^2-4AC}}\right)}{\sqrt{-B^2-4AC}}[/tex]

By [tex]tan^{-1}[/tex] you mean [tex]tanh^{-1}[/tex], right? Also note that the expression [tex]\sqrt{-B^2-4AC}[/tex] would be imaginary, since constants A, B and C are positive.
My solution of the integral is:

[tex]t=\frac{1}{\sqrt{B^2+4AC}}ln\left(\frac{\sqrt{B^2+4AC}+B+2Cv}{\sqrt{B^2+4AC}-B-2Cv}\right)[/tex]

which could be written like yours considering the fact that:
[tex]\frac{1}{2}ln\left(\frac{1+x}{1-x}\right)=tanh^{-1}[/tex]

The rest seems fine, it is even consistent with my solution for B=0.

 

[arildno]

Do remember, though, that the proper quadratic law for the air drag is [tex]-cv|v|[/tex], rather than [tex]-cv^{2}[/tex] for some constant c.

 

[3DM@rk]

Do remember, though, that the proper quadratic law for the air drag is [tex]-cv|v|[/tex], rather than [tex]-cv^{2}[/tex] for some constant c.

But since the drag force acts in opposite direction to the velocity, for an accelerating force F we can use scalar equations with [tex]F_{d}=-cv^{2}[/tex]. So the above calculations are correct.

 

[arildno]

Well, that would depend upon the problem you were to model with this.
If you think of, say, launched projectile problem within a constant gravity field, then the general form is necessary.

The force law on the way up (velocity "positive") wll go as -cv^2, whereas on its way down (velocity "negative") as +cv^2.

 

[3DM@rk]

The better general formula then would be
[tex]\vec{F_{d}}=-cv^{2}\vec{\hat{v}}[/tex],
where [tex]\vec{\hat{v}}[/tex] is a unit vector codirectional with velocity.
But anyway, with some common sense you can always decide the correct direction of the drag force, so it doesn't accelerate the projectile.
Simple [tex]}-cv^{2}[/tex] is a much better function to calculate with, than some expression with an absolute value.
I agree with you and understand your point. Thanks for the comment.