2
$\begingroup$

I'm trying to solve $\displaystyle\int \frac{m \; dv }{mg-kv^2}$. Plugging this into Wolfram Alpha produces $$ \sqrt{\frac{m}{gk}} \tanh^{-1} \left( \sqrt{\frac{k}{gm}}v \right).$$ My question is, what happens when $\displaystyle v> \sqrt{\frac{gm}{k}}$? Then it seems like you're taking the inverse hyperbolic tangent of something greater than $1$, which isn't real.

So is there no real solution to this integral for $\displaystyle v> \sqrt{\frac{gm}{k}}$? Or if there is, what is it?

  • 1
    If $v>\sqrt{\frac{gm}{k}}$, the function $\frac{m}{mg-kv^2}$ has a non-integrable singularity at $v=\sqrt{\frac{gm}{k}}$. What does $\frac{m}{mg-kv^2}$ represent in the physical problem? The endpoints of the integration range matter, really much. – 2017-02-01
  • 0
    It looks like an accelerated motion with friction: you should be aware that in such a case there is a limit speed. – 2017-02-01
  • 0
    Yes...I'm integrating by parts. I have $m \;dv=mg-kv^2$. There are no limits, I'm just trying to find $v(t)$ (I guess you can assume $t \geq 0$). But assuming an initial value $v_0$ greater than the terminal speed, the solution should look like a decaying exponential. How do I get that solution? – 2017-02-01
  • 1
    @Miatrix When your integrand has non-integrable singularities, indefinite integration has a tendency to produce problematic results. (For example, $\int \frac{1}{x} dx = \ln(|x|)+C$ is actually rather misleading.) It is really better to work with limits of integration (perhaps variable ones, but where the variables are constrained such that the integral converges). – 2017-02-01
  • 0
    @Miatrix: according to $v_0v_L$ we have $$v(T) \approx v_L\left(1\pm C e^{-Dt}\right)$$ with $C,D>0$ and the sign $\pm$ depending on the initial situation. We never cross the non-integrable singularity. – 2017-02-01

3 Answers 3

0

Indefinite integration is a useful tool, but it can lead us astray at times, including in cases when the integrand has non-integrable singularities. The proper resolution of this problem is definite integration with symbolic limits. This technique has the bonus of not having to find the integration constant at the end of the calculation.

In this context, we are looking at:

$$\int_{v_0}^v \frac{m}{mg-ku^2} du.$$

Here the integrand, because it has non-integrable singularities at $u_1:=-m^{1/2} g^{1/2} k^{-1/2},u_2:=m^{1/2} g^{1/2} k^{-1/2}$, tells us that $v_0$ and $v$ must be in the same one of the regions $(-\infty,u_1),(u_1,u_2),(u_2,\infty)$, otherwise the integral does not converge. Let us write the partial fraction expansion:

$$\frac{m}{mg-ku^2} = -\frac{m}{k} \frac{1}{u^2-\frac{mg}{k}} = -\frac{m}{k} \frac{1}{u_2-u_1} \left ( \frac{1}{u-u_2} - \frac{1}{u-u_1} \right ).$$

For $v,v_0 \in (u_1,u_2)$, the end result after cancelling minus signs is that we have two positive terms, so that integration gives us

$$\left. \frac{m}{k} \frac{1}{u_2-u_1} \left [ \ln(u_2-u) + \ln(u-u_1) \right ] \right |_{u=v_0}^{u=v}.$$

For $v,v_0 \in (u_2,\infty)$ the integration goes differently, resulting in

$$\left. \frac{m}{k} \frac{1}{u_2-u_1} \left [ -\ln(u-u_2) + \ln(u-u_1) \right ] \right |_{u=v_0}^{u=v}.$$

This seems like a minor algebra quirk at a glance, but actually this has a huge impact on the qualitative behavior of the final solution to the differential equation. Physically, when $v_0 \in (u_1,u_2)$, the object will accelerate to its terminal velocity; when $v_0 \not \in [u_1,u_2]$, the object will decelerate to its terminal velocity. (What happens if $v_0=u_1$ or $v_0=u_2$? Why can't this derivation tell us that?)

(Please do check my minus signs, it's quite possible I made a mistake somewhere.)

0

This formula $$ \sqrt{\frac{m}{gk}} \tanh^{-1} \left( \sqrt{\frac{k}{gm}}v \right).$$ shows the time $t$ elapse for a mass $m$, falls with $v_0=0$ from height $h$ and has velocity $v$.

For natural phenomenons, we can't decide based on pure mathematical formulas and then need to know all physical parameters which interpret their behaviors.

This value of time is determined by velocity $v$ which specifies by nature and we can't describe it by arbitrary values.

It's clear when $\displaystyle v> \sqrt{\frac{gm}{k}}$, from mathematically rules we may deduce some properties of this experiment, but these properties aren't valuable in physics systems.

  • 0
    @Ian Perfectly. So for these reasons we say these aren't valuable. – 2017-02-01
  • 0
    @Ian But what if you fire a projectile with initial velocity greater than the terminal velocity? That is physically achievable, right? So shouldn't there be some mathematical formulation of $v(t)$ assuming a $v^2$ friction force? – 2017-02-01
  • 1
    What I wrote before (that velocities above the terminal velocity aren't admissible) was overstated. What's really true is that the velocity cannot *cross* the terminal velocity, but in fact the object could be thrown faster than its terminal velocity and then it will *decelerate* as it falls. This is why I removed that comment and am writing an answer along these lines. – 2017-02-01
  • 0
    @Ian Right. But how would one find the equations of motion of that deceleration, assuming a $v^2$ friction force? Is it possible? – 2017-02-01
0

Let we re-construct the problem from scratch. We have a free-fall motion ($a=g$) and a dissipative force (friction $-kv^2$). Our body starts at $t=0$ with initial speed $v_0=v(0)$ and we want to understand its speed at $t=T>0$. We have:

$$ a(t) = \frac{dv}{dt} = \frac{mg-kv^2}{m} = g-\frac{k}{m}v^2$$ hence $$ \frac{dt}{dv} = \frac{1}{g-\frac{k}{m}v^2} $$ and by integrating both sides over the interval $(v(0),v(T))$ we get: $$ T = \int_{v(0)}^{v(T)}\frac{dv}{g-\frac{k}{m}v^2} $$ We may notice that $v_L=\sqrt{\frac{mg}{k}}$ is a limit speed: if $v_0v_L$, $v(t)$ is a decreasing function bounded below by $v_L$.
In the first case

$$ T = \sqrt{\frac{m}{gk}}\left[\text{arctanh}\left(\frac{v(T)}{v_L}\right)-\text{arctanh}\left(\frac{v(0)}{v_L}\right)\right]$$ from which: $$ \exp\left(2T\sqrt{\frac{gk}{m}}\right)=\frac{v_L+v(T)}{v_L-v(T)}\cdot\frac{v_L-v_0}{v_L+v_0} $$ that holds also in the second case$^{(*)}$. Solving that with respect to $v(T)$, $$ \frac{1+v(T)/v_L}{1-v(T)/v_L}=\frac{1+v_0/v_L}{1-v_0/v_L}\,\exp\left(2T\sqrt{\frac{gk}{m}}\right)$$ leads to: $$\boxed{\, v(T) = v_L\cdot\frac{\frac{1+v_0/v_L}{1-v_0/v_L}\,\exp\left(2T\sqrt{\frac{gk}{m}}\right)-1}{\frac{1+v_0/v_L}{1-v_0/v_L}\,\exp\left(2T\sqrt{\frac{gk}{m}}\right)+1}.}$$ $(*)$ The crucial fact is that $\frac{v_L-v_0}{v_L-v(T)}$ is positive anyway for $T>0$, and $$\exp\left(2\,\text{arctanh}(z)\right) = \frac{1+z}{1-z} $$ holds for every $z\neq 1$.
Of course we have not studied the case $v_0=v_L$: in such a case the speed is simply constant.

So, long story short: the solution of the paradox is that the non-integrable singularity of $\frac{m}{mg-kv^2}$ is actually never enclosed in the real integration range, because $v(0)$ and $v(T)$ always lie on the same side of $v_L$.

  • 0
    Forgive me if I'm being stupid, but your final expression is of the form $v_L \cdot \frac{\chi - 1}{\chi + 1} < v_L$. That is, $v(T)$ is never greater than $v_L$, so how can $v(0)>v_L$? – 2017-02-01
  • 0
    @Miatrix: look carefully: if $v_0>v_L$, then $1-v_0/v_L$ is negative, hence the speed is surely $v_L\cdot\frac{\chi-1}{\chi+1}$, but $\chi$ is negative. – 2017-02-01