Obtain the equation of motion for a particle falling freely under the action of gravity under a resistance which is directly proportional to the square root of velocity. Describe the motion and find the maximum velocity.

It appears to be a simple problem. We write $$m\frac{dv}{dt}=mg-b\sqrt v$$ $$\Longrightarrow \int_0^v\frac{dv}{mg-b\sqrt v}=\int_0^t\frac{dt}{m}$$ I am assuming that initial velocity is zero. Now, the right integral is straightforward. Let the left one be $I$ Apply substitution $$mg-b\sqrt v=u$$ Then $$\Longrightarrow\Big(\frac{mg-u}{b}\Big)^2=v$$ $$\Longrightarrow 2\Big(\frac{mg-u}{b}\Big)\Big(\frac{-1}{b}\Big)du=dv$$ Using it in the expression for $I$, we get, $$I=\int_{mg}^{mg-b\sqrt v}\frac{2(u-mg)}{b^2(u)}du$$ $$=\frac{2}{b^2}(-b\sqrt v)-\frac{2mg}{b^2}\ln\Big(\frac{mg-b\sqrt v}{mg}\Big)=\frac{t}{m}$$ $$\Longrightarrow\frac{2mg}{b^2}\ln\Big(\frac{mg}{mg-b\sqrt v}\Big)=\frac{t}{m}+\frac{2\sqrt v}{b}$$ $$\Longrightarrow \frac{mg}{mg-b\sqrt v}=e^{\frac{tb^2}{2m^2g}+\frac{b\sqrt v}{mg}}$$ $$\Longrightarrow \frac{mg-b\sqrt v}{mg}=e^{-\Big(\frac{tb^2}{2m^2g}+\frac{b\sqrt v}{mg}\Big)}$$ $$\Longrightarrow \sqrt v=\frac{mg}{b}\Bigg(1-e^{-\Big(\frac{tb^2}{2m^2g}+\frac{b\sqrt v}{mg}\Big)}\Bigg)$$ Now, this is an implicit function of $\sqrt v$, and I can't proceed to find the maximum value of $\sqrt v$ from here. What I think is, this appears to be an increasing function of time. Can I assume that the maximum value will occur at $t\longrightarrow\infty$? In that case, I think the $\sqrt v$ in the power will simply be ignored and maximum value will be $\frac{mg}{b}$.

Also, I thought of differentiating this expression with respect to time, and I got that $\frac{d\sqrt v}{dt}$ is never $0$.

Please help


closed as off-topic by John Rennie, stafusa, Kyle Kanos, Jon Custer, ZeroTheHero Sep 6 at 21:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, stafusa, Kyle Kanos, Jon Custer, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Note that the question is using somewhat incorrect terminology. If the object is freely falling, then technically there cannot be a drag force. $\endgroup$ – Aaron Stevens Sep 5 at 13:37
  • $\begingroup$ Unless my eyes are playing tricks, you missed a $u$ in the denominator of your integral when you make the substitution. You only have the $du$ term, but left out the 1/$u$, n'est ce pas? $\endgroup$ – Bill N Sep 5 at 17:11
  • 1
    $\begingroup$ @BillN, yes I missed the term, corrected now. But that was while copying the partial solution from copy to here, so all the subsequent steps are correct. $\endgroup$ – Martund Sep 5 at 17:33
  • $\begingroup$ From your first equation, the maximum velocity occurs when dv/dt = 0. v(max) =(mg/b)^2 $\endgroup$ – R.W. Bird Sep 5 at 18:58
  • $\begingroup$ May I know the reason for downvote, dear downvoter? $\endgroup$ – Martund Sep 6 at 13:30

I will note that you essentially already have a solution to your problem, but I will supply you with some additional information and techniques that should help in problems like these.

Note that the question does not actually ask you to solve for $v(t)$. We can learn all that we need to learn from the differential equation. First, a change of variables makes this equation easier to deal with by introducing dimensionless variables $$x=\left(\frac{b}{mg}\right)^2\cdot v$$ $$\tau=\frac{b^2}{m^2g}\cdot t$$

Which changes the differential equation to be (check for yourself) $$\frac{\text dx}{\text d\tau}=1-\sqrt x$$

If we want to get an idea of the evolution of $x$, we can just plot $\text dx/\text d\tau$ as a function of $x$ enter image description here

Now notice that when $x<1$ we have $\dot x>1$. This means that $x$ will be "pushed" to larger values when $x<1$. Similarly, when $x>1$ we have $\dot x<1$. This means that $x$ will be "pushed" to smaller values when $x>1$.

Putting this all together, we see that $x=1$ is a stable equilibrium for this system. The system always tends to $x=1$. Furthermore, $x$ cannot move past $x=1$. So if we start at rest ($x=0$), then the maximum value $x$ will obtain is $1$ (although this doesn't tell us how long it will take to get there).

Furthermore, you can see how the system will approach $x=1$ based on the shape of the graph. For example, if $\dot x$ is larger, then $x$ is changing faster. If $\dot x$ is closer to $0$, then $x$ is not changing as quickly.

I will leave the specifics of how qualitatively $x$ changes with time, and how that relates back to $v(t)$ using the change of variables. What I do want you to realize is that you can learn a lot about a system without actually explicitly solving the differential equation. This is useful when the solution is messy (as you have seen).

  • $\begingroup$ I don't have solution to my problem. $\endgroup$ – Martund Sep 5 at 14:37
  • $\begingroup$ @Martund I meant you obtained a relation that solves the differential equation. Although it is very messy. $\endgroup$ – Aaron Stevens Sep 5 at 14:39
  • $\begingroup$ Nice solution, but now when I have solved the differential equation, differentiating it back would give me the original differential equation. But when we differentiate and put $\frac{d\sqrt v}{dt}=0$, we don't get any solution, while for differential equation we directly get the maximum value of $v$. So, have I done something wrong or the maximum value will never be achieved? $\endgroup$ – Martund Sep 5 at 18:17
  • $\begingroup$ @Martund I would assume the maximum value is not reached in finite time. This even happens for the usual drag forces like $-bv$. $\endgroup$ – Aaron Stevens Sep 5 at 18:24

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