Question

A raindrop falls vertically through stationary mist, collecting mass as it falls. The raindrop remains spherical and the rate of mass accretion is proportional toits speed and the square of its radius. Show that, if the drop starts from rest with a negligible radius, then it has constant acceleration g/7. [Tricky ODE.]

Answer 1

In the case where the mass isn't constant you need to use F=(d/dt)(mv). That only simplifies to F=ma when the mass is constant

m: the mass of the raindrop
r: the radius of the raindrop
v: the velocity of the raindrop

2. Relevant equations
Because the raindrop is a sphere:
m=?4/3pr3
for some constant
?

m'(t)=kvr2
for some constant
k


3. The attempt at a solution
Plugging in our equation for m, we have the differential equation
m'=kv(3m4p?)2/3

With initial conditions m = 0 (or close enough) and v = 0.
We need another differential equation, since both m and v are nonconstant.

We need another differential equation, since both m and v are nonconstant.

Suppose the drop falls a dist of dx in time dt, and its mass changes from m to m+dm, and velo from v to v+dv. Then,

mv^2/2 + mg*dx = (m+dm)(v+dv)^2/2 =>
mg*dx = mv*dv + v^2*dm/2 =>
mgv = mv dv/dt + (v^2/2) dm/dt.

The v cancels out, giving you your 2nd diff eqn.