1
$\begingroup$

The standard first-course free-fall with linear drag situation posits a particle falling with a constant acceleration (typically due to approximation of gravity), but with a retarding force that is proportional to the velocity. The equation of motion for the particle is written as

$$m\frac{dv}{dt}=-mg-\lambda v \,\,\,\,\,\,\,\,\textrm{or}\,\,\,\,\,\,\,\,m\frac{d^2y}{dt^2}=-mg-\lambda\frac{dy}{dt}$$

The solution for the velocity as a function of time, $v(t)=\frac{dy}{dt}$, is simply a decreasing exponential plus a constant. On the other hand, the solution for the position as a function of time, $y(t)$, is a decreasing exponential plus a linear term plus a constant, which is a bit more complicated. It's complicated in the sense that the inverse function, $t(y)$, is transcendental (involving the Lambert $W$-function), so solving for the time at which the particle is at a certain height is typically not easy (analytically).

My goal is to somehow change this problem so that, given initial conditions $v(t=0)=0$ and $y(t=0)=h$, I can solve for the velocity at the time/position when/where the particle "hits the floor" (i.e. $v(y=0)=v(t_{\textrm{hit}})$) analytically.

If we multiply the original differential equation by $v^{-1}$, we get

$$m\frac{dv}{v\,dt}=m\frac{dv}{dy}=-\frac{mg}{v}-\lambda$$

This is a nonlinear differential equation with an obvious singularity at $v=0$. I don't know how to solve this, and I don't know how to get a sensible closed-form expression for my desired quantity $v(t_{\textrm{hit}})$. I made a hand-waved plot of what the solutions would look like.

enter image description here

Is it possible to find a nice closed-form solution for $v(t_{\textrm{hit}})$? If so, am I on the right track? Please give suggestions.

$\endgroup$
  • $\begingroup$ The linear term has nothing to do with the problem, at all. It's just a transformation into a completely equivalent inertial system, which is physically irrelevant. $\endgroup$ – CuriousOne Sep 15 '15 at 0:16
  • $\begingroup$ I can't see how that would simplify the problem. Maybe you could expand on it? The linear term has coefficient $-mg/\lambda$, so from what I understood, you are suggesting to make the transformation $y'=y+(mg/\lambda)t$, $t'=t$. In that inertial system, there would be no linear term, but now we would be solving for $y'(t_{\textrm{hit}})$, and that isn't necessarily equal to 0 as it is in the original inertial frame. $\endgroup$ – Arturo don Juan Sep 15 '15 at 0:33
  • 2
    $\begingroup$ @ArturodonJuan: you seem to want to needlessly want to complicate things. $m\frac{d^2y}{dt^2}=-g-\lambda\frac{dy}{dt}$ is easily solved. Form there any information you want should be deducible. $\endgroup$ – Gert Sep 15 '15 at 0:55
  • $\begingroup$ @Gert I know that that equation is easily solvable. I even stated what the solutions will behave like. What I'm saying is that to solve for the velocity of the particle when it has "hit the ground" (i.e. $y=0$) doesn't seem to be an easy task. I can't see how to get a closed-form solution for that particular quantity. $\endgroup$ – Arturo don Juan Sep 15 '15 at 1:04
  • 1
    $\begingroup$ @ArturodonJuan - If by "closed form solution" you mean "expressible as a finite combination of elementary functions", the answer is you can't. You already showed that the inverse function is non-elementary. End of story. $\endgroup$ – David Hammen Sep 15 '15 at 1:10
1
$\begingroup$

Your equation $$ m\frac{dv}{dy} = -\frac{mg}{v}- \lambda $$ is solvable, but it doesn't lead to a closed expression v=v(y). It gives instead y=y(v), which will leave you with a simple looking, but still transcendental equation for v(y=0).

To solve, separate your variables before formal integration: $$ \frac{v}{v + \frac{mg}{\lambda}}dv = -\frac{\lambda}{m}dy $$ Integration obtains then $$ v - \frac{mg}{\lambda}\ln\left( v + \frac{mg}{\lambda} \right) = -\frac{\lambda}{m}y + C $$ From the initial condition $v=0$ for $y = h$ the integration constant is $$ C = \frac{\lambda h}{m} - \frac{mg}{\lambda}\ln\frac{mg}{\lambda} $$ so the final expression becomes $$ y(v) = h + \frac{m^2 g}{\lambda^2}\left( \ln\left( \frac{\lambda v}{mg} + 1\right) - \frac{\lambda v}{mg} \right) $$ Note that since $v<0$ and $\frac{\lambda |v|}{mg} < 1$, the term in $v$ on the rhs is negative as it should. For y=0 we are left with $$ \frac{\lambda^2 h}{m^2 g} + \frac{\lambda v_{hit}}{mg} = \ln\left( \frac{\lambda v_{hit}}{mg} + 1\right) $$

As I said, it's still a transcendental equation, but at least it's only logarithmic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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