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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

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

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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).

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  • $\begingroup$ 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

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