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I am stuck on the Exercise 3.5 of Newtonian Dynamics by R. Fitzpatrick:

A block of mass $m$ slides along a horizontal surface which is lubricated with heavy oil such that the block suffers a viscous retarding force of the form

$$F = - c\,v^n,$$

where $c>0$ is a constant, and $v$ is the block's instantaneous velocity. If the initial speed is $v_0 $ at time $t=0$, find $v$ and the displacement $x$ as functions of time $t$. Also find $v$ as a function of $x$. Show that for $n=1/2$ the block does not travel further than $2\,m\,v_0^{3/2}/(3\,c)$.

The last part of the question asks to show that for $n=1/2$ the block does not travel further than $2mv_0^{3/2}/(3c)$.

We start from Newton's second law $$ m \frac{d^2x}{dt^2} = m \frac{dv}{dt} = m v \frac{dv}{dx}= -cv^n. $$

Separating variables gives $$ \int_{v_0}^{v} \frac{dv'}{(v')^{n-1}} = -\frac{c}{m} \int_0^x dx', $$

$$ v^{-n+2} = v_0^{-n+2} - \frac{(-n+2)cx}{m}. $$

Plugging $n=1/2$, $$ v^{3/2} = v_0^{3/2} - \frac{3cx}{2m}. $$

Setting the velocity to zero (this must be the case if the block stops moving), $$ x =\frac{2m v_0^{3/2}}{3c}, $$ which is the desired result.

The problem arises when I try to solve for $x$ in terms of $t$. Now, $$ m \frac{dv}{dt} = -cv^n, $$ $$ \int_{v_0}^{v} \frac{dv'}{(v')^n} = -\int_0^t \frac{c}{m} dt', $$ $$ \frac{1}{v^{n-1}} = \frac{1}{v_0^{n-1}} - \frac{(-n+1)c}{m} t. $$ Rising everything to $1/(1-n)$ power (of course, assuming that $n \ne 1$), $$ v = \left( \frac{1}{v_0^{n-1}} - \frac{(-n+1)c}{m} t \right)^\frac{1}{1-n}.$$

Plugging $n=1/2$ gives: $$ \frac{dx}{dt} = \left( v_0^{1/2} -\frac{c}{2m} t \right)^2. $$

Let's separate the variables and try to integrate, $$ \int_0^x dx = \int_0^t \left( v_0^{1/2} - \frac{c}{2m} t' \right)^2 dt', $$ $$ x_{\mathrm{f}} = \int_0^{\infty} \left( v_0^{1/2} - \frac{c}{2m} t' \right)^2 dt'. $$ I've plugged $t = \infty$ because it seems to me that the block must stop to this time if it's going to stop at all. The problem is that the integral on the right hand side won't converge! So $x$ has no finishing point, which contradicts the first part of the solution. What's going on here?

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    $\begingroup$ I'm voting to close this question as off-topic because check-my-work questions are off-topic. $\endgroup$
    – ACuriousMind
    Mar 15 '15 at 17:18
  • $\begingroup$ @ACuriousMind You see, it is not just a homework question. There is nothing special about the power n=1/2. I've already got the result. The question arose from the contradiction that I found. $\endgroup$
    – cdlscpmv
    Mar 15 '15 at 17:29
  • $\begingroup$ I agree, the core of this is not really a check-my-work question, though it was hard to tell for a couple of reasons: (1) you had the sentence at the end asking whether you had made a mistake, which is kind of the definition of a check-my-work question (if you're still not sure whether you've made a mistake, you shouldn't be coming to us yet), and (2) there's a lot of math here. You can easily drop most of the details of the derivations because they're not relevant to the question you actually want to ask. I edited to fix point #1 for you but I'd suggest you edit to fix #2. $\endgroup$
    – David Z
    Mar 15 '15 at 17:42
  • $\begingroup$ I've deleted some parts of the derivation. Before posting this question I thought it would be easier for people to follow the derivation if I write it. I will try not to do it again ;). Anyway, thank you for the help! $\endgroup$
    – cdlscpmv
    Mar 15 '15 at 18:31
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From $$\dfrac{dx}{dt}=\left(v_0^{1/2}-\dfrac{c}{2m}{t}\right)^2$$ and $$v(t_f)=\left.\dfrac{dx}{dt}\right|_{t=t_f}=0$$ you should be able to get a finite bound on your last integral.


EDIT: one possible reason for which your final integral doesn't properly converge comes from an earlier step. Indeed, you moved from: $$m\dfrac{dv}{dt}=-cv^n$$ to: $$\dfrac{dv}{v^n}=-\dfrac{c}{m}dt$$ The big caveat here is of course that this is only valid for $v\neq 0$. And in fact, the physical solution tells us that $v=0$ forever when $t_f$ is reached!

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  • $\begingroup$ Thanks, man. It did work. But, anyway, I wonder why I cannot plug in infinity into the integral. The block will not move after reaching that final point, so even after infinite amount of time the block must be there. $\endgroup$
    – cdlscpmv
    Mar 15 '15 at 17:52
  • $\begingroup$ @cdlscpmv Can I get my "accepted solution" back now? :D $\endgroup$
    – Demosthene
    Mar 15 '15 at 18:00
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As you said,

$$ \int_0^x dx = \int_0^t \left( v_0^{1/2} - \frac{c}{2m} t' \right)^2 dt', $$

which means, after the integration, that:

$$ x(t) = x_1(t)=\frac{c^2 t^3}{12m^2} - \frac{ct^2v_0^{1/2}}{2m}+tv_0 $$

The point in time for which $x_1(t_f) = x_f = \frac{2mv_0^{3/2}}{3c}$ is $t_f=\frac{2mv_0^{1/2}}{c}$ which is not infinite. At this point, the velocity is zero, so the force is zero, so the acceleration is zero, so the velocity remains zero and the position remains constant. More specifically:

$$ x(t)= \begin{cases} x_1(t) & t\leq t_f \\ x_f & t \gt t_f \\ \end{cases} $$

This function is not analytic, though it's continuous and has a first derivative. This is induced by the (unrealistic) $v^{1/2}$ force. Non-analytic solutions don't typically appear in realistic scenarios, but they can pop up in toy scenarios like this one or Norton's dome.

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  • $\begingroup$ Thank you for the answer! I didn't know that a non-analytic function could cause such problems. From now on I will be more careful. $\endgroup$
    – cdlscpmv
    Mar 15 '15 at 18:00

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