# Falling Rain Drop (Variable Mass)

1. Apr 20, 2016

### Potatochip911

1. The problem statement, all variables and given/known data
Suppose a rain drop with mass $m_0\neq 0$ is falling due to gravity with initial velocity $v_0$, assume $\frac{dm}{dt}=k=$constant. Solve the differential equation and determine the velocity as $t\to\infty$

2. Relevant equations
$F=\frac{dp}{dt}=\dot{m}v+m\dot{v}$
3. The attempt at a solution

Since $F=mg$ the D.E. is $\dot{m}v+m\dot{v}=mg$, substituting in $\frac{dm}{dt}=k$ we find $kv+m\dot{v}=mg$, since the mass increases with time we have $m=m_0+kt$. One method I've attempted is essentially using the integrating factor but since the LHS is already a product rule derivative I don't need to multiply it by anything so:

$$\frac{d}{dt}(mv)=mg \\ \int \frac{d}{dt}(mv)dt=\int mg dt=\int mg \frac{dt}{dm}dm=\frac{g}{k}\int_{m_0}^{m} m\, dm\\ v(m_0+kt)+C=\frac{g}{2k}((m_0+kt)^2-m_0^2)=\frac{g}{2k}(2m_0kt+k^2t^2)$$

at $t=0$, $v=v_0\Rightarrow$, $v_0(m_0+0)+C=0\Longrightarrow C=-v_0m_0$

therefore...

$$v(m_0+kt)-v_0m_0=gm_0t+\frac{gkt^2}{2}\\ v(m_0+kt)=m_0(v_0+gt)+\frac{gkt^2}{2} \\ v=\frac{m_0(v_0+gt)}{m_0+kt}+\frac{gkt^2}{2(m_0+kt)}$$

Now I find this solution a bit odd since my professor said that it would become asymptotic but I think he's wrong about this? The first term will approach $\frac{m_0g}{k}$ but the second term clearly diverges as $t\to\infty$ and it doesn't make sense that an object would reach a terminal velocity without drag. I suppose you could say it asymptotically goes to infinity but it's increasing linearly...

2. Apr 20, 2016

### DrSteve

3. Apr 20, 2016

### haruspex

By itself that means nothing. What is supposed to tend asymptotically to a constant, or to a certain function of what?
The link DrSteve posted shows the acceleration tends to a constant.

4. Apr 20, 2016

### Potatochip911

Okay I suppose I'm just too used to thinking of functions converging to a constant when that term is mentioned.

5. Apr 20, 2016

### ehild

The velocity tends to a simple function as t tends to infinity. What is that simple function?

6. Apr 20, 2016

### Potatochip911

I believe it's just $v=\frac{m(t)g}{k}$ from setting $\dot{v}=0$ in the D.E.

Simplifying my expression for $v(t)$ would give $v(t)=\frac{m_0g}{k}+\frac{gt}{2}$ though

Sorry for so many edits but just noticed that the first one becomes $v=\frac{m_0g}{k}+gt$ which is almost the same as the result from simplification.

7. Apr 20, 2016

### ehild

You can not set $\dot v = 0$. The acceleration does not tend to zero. At great t, v increases linearly with time with rate g/2 instead of g as it were for a constant-mass body.

8. Apr 20, 2016

### Potatochip911

Whoops, I realize now it doesn't make much sense to set acceleration equal to 0 :)

Interestingly at large t, it appears v always increases by $g/2$ regardless of $\alpha$ for $\frac{dm}{dt}=k^{\alpha}$

9. Apr 20, 2016

### ehild

$k^{\alpha}$ is just an other constant. Why do you mix alpha in? It was said that dm/dt=k.

10. Apr 20, 2016

### ehild

It is interesting the acceleration at long time tends to g/2, regardless of k.