3
$\begingroup$

If a particle moves in a place with air resistance (but no other forces), will it ever reach a zero velocity in finite time? The air resistance is proportional to some power of velocity - $v^\alpha$, and I have to try it with different $\alpha$. I've solved for the function of position for several alphas and all functions I've gotten decay to $v=0$ as $t \to\infty$, but none ever reach exactly $v=0$ for a finite value of $t$. This should be the case, right?

$\endgroup$
1
  • $\begingroup$ "Air resistance", but "no other forces"?? Sooo... no brownian motion? Zero velocity.. please define what you mean by "zero velocity"? Is one planck distance per second "zero"?. there are a lot of assumptions needed before this can be answered satisfactorily. $\endgroup$
    – PcMan
    Dec 20, 2020 at 12:01

2 Answers 2

1
$\begingroup$

Yes. Without any force it indeed would reach zero speed only in $t=\infty$.

There is no contradiction with the real world. Here we have not only the resistance force, but more and more forces, and a dependance on size and shape of body.

$\endgroup$
1
  • 1
    $\begingroup$ Instead of zero and infinity, if you set "any" minimum velocity that you approximate as zero for practical considerations, then it would reach that velocity in a finite time. The problem with zero is that beyond a certain point Brownian motion will be significant. $\endgroup$
    – Prathyush
    Oct 14, 2012 at 9:12
0
$\begingroup$

Have you considered the full range of values of $\alpha$?

For $\alpha\ge1$, your conclusion is correct: the velocity approaches $v=0$ asymptotically at large times. If you consider $\alpha<1$, you can find solutions which reach $v=0$ in finite time. I'll leave the explicit solutions to you, but I find the time at which the particle stops to be

$$ T=\frac{v(0)^{1-\alpha}}{1-\alpha} \quad {\rm for}\quad \alpha<1. $$

Edit: I should just point out that my constant of proportionality was fixed to $1$. That is, I solved $v^\prime = -v^\alpha$. To get the answer in physical units, you'll have to reinstate it.

$\endgroup$
2
  • $\begingroup$ Ah, ok, I didn't even think of $\alpha$ < 0, because the 3 alphas that I had to try were 1 , 3/2 , 2. Thanks, though! $\endgroup$ Oct 14, 2012 at 19:00
  • $\begingroup$ No problem. It's not just $\alpha<0$ though! For example, try $\alpha=\frac{1}{2}$. $\endgroup$ Oct 15, 2012 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.