I hope this is a suited question for this site since it contains a mix of physics and mathematics. In case I should post this on the physics stackexchange site, please let me know.

A spherical raindrop is falling from the sky. Because of the humid atmosphere the raindrop will gain mass during his fall. The increase in mass per time is proportional to the current surface area.

  1. Find an equation for the radius of the drop as a function of time ($r(t)$). ($r(0) =r_0$)
  2. Find and solve the equation of motion for the raindrop. The equation of motion should depend on $r_0$

My work so far:

Mass: $m(t)=\frac{4}{3} \pi r(t)^3 \rho $

Sufrace Area: $A(t)=4\pi r(t)^2$

$$\begin{aligned} & \implies \frac{dm(t)}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r(t)^3 \rho)=4 \pi r(t)^2 r'(t)\rho\\ & \iff 4 \pi \rho r(t)^2 r'(t)=\lambda 4 \pi r(t) ^2 \iff r'(t)=\frac{\lambda}{\rho} \iff \frac{dr}{dt}=\frac{\lambda}{\rho} \\ & \iff \int dr = \frac{\lambda}{\rho} \int dt \iff r(t)=\frac{\lambda}{\rho}t+c \\ & \color{blue}{\implies r(t)=\alpha t+r_0} \space \space \space \space \space \space \space \space \space \space \space \text{where}\space \space\alpha=\frac{\lambda}{\rho} \end{aligned}$$

Equation of motion:

Gravitational Force: $mg$

Air Friction: $kv(t)$

$$\begin{aligned} &F=mg-kv \iff \frac{d}{dt}(mv)=m'v+v'm=mg-kv \\ & \end{aligned}$$

I tried substituting $m'$ and $m$ into my equation and I am left with:

$$\begin{aligned}& mv'+\frac{3m \alpha}{r}=mg-kv\end{aligned}$$

I tried solving this in wolfram alpha and I get the following result: LINK

It looks super messy and I don't think that is the solution I am supposed to get. Am I doing something wrong? Is there some substitution I can make to make it easier to solve?

  • $\begingroup$ I just recently ran across this: physics.harvard.edu/uploads/files/undergrad/probweek/prob5.pdf and solution: physics.harvard.edu/uploads/files/undergrad/probweek/sol5.pdf Not sure if it helps though. $\endgroup$ – turkeyhundt Oct 31 '15 at 14:39
  • $\begingroup$ The first part is correct as $ \frac {dm}{dr} = k A $. Is the resistance proportional to velocity or its square? $\endgroup$ – Narasimham Oct 31 '15 at 14:46
  • $\begingroup$ @Narasimham It isn't specified in the question. I should have problably used "newtonian friction" $=kv^2$ and not "stokes friction" $=kv$. $\endgroup$ – qmd Oct 31 '15 at 14:51
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    $\begingroup$ You can simplify your life a bit by just noting the following about your model and what you have already solved about it: $r$ is linear in time with a positive slope, $m$ is proportional to $r^3$ with a positive slope, $F$ is proportional to $m$ (and hence $r^3$) with a positive slope and proportional to $v$ with a negative slope. So $r=c_1t+c_2$, $m=c_3r^3$, $F=c_4r^3-c_5v=mv'+m'v=c_3r^3v'+3c_3r^2r'v$. (Written this way, all constants are positive.) $\endgroup$ – Ian Oct 31 '15 at 15:00
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    $\begingroup$ Plugging everything in you get a first order linear equation with non-constant coefficients for $v$, namely $c_4r^3-c_5v=c_3r^3v'+3c_3r^2r'v=c_3r^3v'+c_6r^2v$ (this equation defines $c_6$). This can be solved by the method of integrating factors. $\endgroup$ – Ian Oct 31 '15 at 15:02

If you assume additionally that the friction is proportional to surface area and to velocity, then you can run the following logic to get a pretty simple ODE for $v$:

\begin{align} r & =c_1t + c_2 \\ m & =c_3 r^3 \\ F & =c_4 m - c_5 r^2 v = c_6 r^3 - c_5 r^2 v \\ F & =m'v+mv' \\ & = 3c_3 r^2 r' v + c_3r^3 v' \\ & = 3c_3 c_1 r^2 v + c_3 r^3 v' \\ & = c_6 r^3 - c_5 r^2 v \\ & \Rightarrow 3 c_3 c_1 v + c_3 r v' = c_6 r - c_5 v. \end{align}

In normal form for first order linear ODEs:

$$v' + \frac{3 c_3 c_1 + c_5}{c_3 r} v = \frac{c_6}{c_3}.$$

Actually plugging in what all these constants are in your notation:

$$v'+\frac{4 \pi \rho \alpha + k}{\frac{4}{3} \pi \rho (\alpha t + r_0)} v = g.$$

This is quite tractable using the method of integrating factors.

  • $\begingroup$ Thank you! I used variation of parameters but I guess that also works. $\endgroup$ – qmd Oct 31 '15 at 16:26
  • $\begingroup$ @qmd The two methods are essentially equivalent. :) $\endgroup$ – Ian Oct 31 '15 at 16:38

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