How Does a Water Drop's Speed Change as It Falls and Gains Mass?

[martyg314]

This is for a mechanics class. I'm sort of re-learning how to apply calculus and differential equations in this class, so I get stuck trying to figure out how to apply the math I learned 2 or more semesters ago. Any help would be greatly appreciated.

Thanks,

m.g.

Homework Statement

"A spherical water drop falls through the atmosphere and its mass increases at a
rate proportional to its area. Find the speed of the drop as a function of time, assuming
that the drop started with zero initial speed."

- disregard drag in this problem
- starting mass is a finite mass
- for the area, cross sectional area or surface area can be used, but cross-sectional area will probably be easier

Homework Equations

I modeled this as a momentum problem, similar to a rocket losing mass as it accelerates, but instead as a drop gaining mass as it falls

The Attempt at a Solution

the increase in mass is proportional to the area (I used the cross section):

\frac{dm}{dt} = k4\pir^{2}


The forces are gravity and the changing momentum:

F=mg-\frac{dp}{dt}

which simplifies to:
\frac{dv}{dt} = g - (\frac{v}{m})\frac{dm}{dt}

Substituting in, I get the differential equation which is where I'm stuck:

\frac{dv}{dt} = g - (\frac{v}{m})k4\pir^{2}

I think I need to remove the m from the equation to solve for v but since r changes as a function of time as well, that doesn't seem to help.

 

[gneill]

m = (4/3)πr³ρ ?

 

[sgd37]

could you not deduce from the mass differential equation that dr/dt is a constant

 

[martyg314]

Thanks for the tips. I think I finally got it.

Knowing that \frac{dR}{dt} is constant, R = kt + R0

I also used the mass/volume/density equation suggested above.

I switched variables to get the equation:

\frac{dv}{dR} = \frac{g}{k} - \frac{3v}{R}

(\rho omitted since we are dealing with water)

And I solved from there.

Thanks for the help. I can see how relatively simple it is now.

-mg