Raindrop Velocity and Mass Dynamics: Finding Space Traversed Over Time

[GregoryGr]

Homework Statement

Water vapor condense in a raindrop with rate μ units of mass per time.
The raindrop starts with 0 velocity with initial mass $$ M_0 $$ and falls horizontally. Find the space it traverses as a function of time (g is given).

Homework Equations

∂m/ ∂t= μ

The Attempt at a Solution

Assuming that the mass dm of vapor to be condesed has a total velocity v0, and using Newton's law, I have $g= \frac{dv}{dt} + \frac{dm*(v-v0)}{dt*m}$ . From there the only thing i could think is that I could write dv/dt as dv/dm*dm/dt but it doesn't seem to help...

 

[etudiant]

Nice problem.
Presumably all the incremental mass gained as the original drop falls under gravity is at 0 velocity as well, so there will be a constant retarding force on the drop. However, because the drop is increasing in mass over time, the slowing effect of the additional mass declines.

 

[tiny-tim]

Hi GregoryGr!

… and falls horizontally.

(That's a good trick! )

Consider the momentum (and force) at times t and t+dt.

(and use m(t) = M_o + µt from the start! )

 

[haruspex]

Assuming that the mass dm of vapor to be condesed has a total velocity v0, and using Newton's law, I have $g= \frac{dv}{dt} + \frac{dm*(v-v0)}{dt*m}$ .

Good start. Can you write m as a function of t?

 

[GregoryGr]

Well, I got the answer without writing the momentum @ t and t+dt, with this:

$$\vec{F}= \frac{d\vec{p}}{dt} \Rightarrow \int_{0}^{t}\mu gdt=\int_{0}^{\mu v}d(\mu v)$$
Anybody want to show me alternative ways to solve it? It seems that the integral you have to calculate is too hard for a physics I exercise.

EDIT: where I write /mu I meant the m(t)

 

[haruspex]

Well, I got the answer without writing the momentum @ t and t+dt, with this:

$$\vec{F}= \frac{d\vec{p}}{dt} \Rightarrow \int_{0}^{t}\mu gdt=\int_{0}^{\mu v}d(\mu v)$$

Not sure how you get that, and it doesn't look to me like it's going to give the right answer. What answer did you get from it?
What was wrong with your initial approach? What ODE did you get after writing m = μt? It didn't look hard to solve to me. Because the terms (gt, v, v't) are dimensionally the same, a natural try is v = At^α.

 

[Rocco1234]

Begin:
dm/dt = u --> m(t) = u * t + M0, where m(0) = M0
F = d/dt(m * v) = m * dv/dt + v * dm/dt = -m * g, assuming no air resistance.
m * dv/dm * u = -v * u - m * g --> dv/dm + (1/m) * v = -g/u --> d/dm(m * v) = -m * g/u --> m * v = -m^2 * g/(2 * u) + C -->
v(m) = -m * g/(2 * u) + C/m, v(M0) = 0 --> C = M0^2 * g/ (2 * u) -->
v(m) = -g/(2 * u) * (m - M0^2/m) -->
v(t) = -g/(2 * u) * (u * t + M0 - M0^2/(u * t + m0))
Integrating we obtain:
x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2 * ln(u * t + m0)) + K

If x(0) = h --> K = h + g/(2 * u^2) * (M0^2/2 - M0^2 * ln(M0)) -->
x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2/2 + M0^2 * ln(M0/(u * t + M0)) + h.
End.

For x(m) we have:
x(m) = -g/(2 * u^2) * ((m^2 - M0^2)/2 + M0^2 * ln(M0/m)) + h.

 

[theodoros.mihos]

I think that $\mu_0,\,M_0$ must satisfy:
$$ \frac{\partial{m}}{\partial{t}} = Cm(t) \,\text{where}\, \left[\frac{\partial{m}}{\partial{t}}\right]_{t=0} = \mu_0 $$

 

[haruspex]

I think that $\mu_0,\,M_0$ must satisfy:
$$ \frac{\partial{m}}{\partial{t}} = Cm(t) \,\text{where}\, \left[\frac{\partial{m}}{\partial{t}}\right]_{t=0} = \mu_0 $$

The problem statement is clear that the rate of mass gain is constant.

 

[theodoros.mihos]

So the force is constant, ok.

 

[ehild]

Homework Statement

Water vapor condense in a raindrop with rate μ units of mass per time.
The raindrop starts with 0 velocity with initial mass $$ M_0 $$ and falls horizontally.

Falls horizontally ? How?
Did you mean it falls vertically? Or the raindrop starts to move horizontally?

 

[theodoros.mihos]

May be "move horizontally", perhaps without friction.

 

[Rocco1234]

Assuming the raindrop falls vertically.
Begin:
dm/dt = u --> m(t) = u * t + M0, where m(0) = M0
F = d/dt(m * v) = m * dv/dt + v * dm/dt = -m * g, assuming no air resistance.
m * dv/dm * u = -v * u - m * g --> dv/dm + (1/m) * v = -g/u --> d/dm(m * v) = -m * g/u --> m * v = -m^2 * g/(2 * u) + C -->
v(m) = -m * g/(2 * u) + C/m, v(M0) = 0 --> C = M0^2 * g/ (2 * u) -->
v(m) = -g/(2 * u) * (m - M0^2/m) -->
v(t) = -g/(2 * u) * (u * t + M0 - M0^2/(u * t + M0))
Integrating we obtain:
x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2 * ln(u * t + M0)) + K

If x(0) = h --> K = h + g/(2 * u^2) * (M0^2/2 - M0^2 * ln(M0)) -->
x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2/2 + M0^2 * ln(M0/(u * t + M0)) + h.
End.

For x(m) we have:
x(m) = -g/(2 * u^2) * ((m^2 - M0^2)/2 + M0^2 * ln(M0/m)) + h.

 

[Rocco1234]

So the force is constant, ok.

The problem seems fairly simple.
Assume that the raindrop falls vertically.
Begin:
dm/dt = u --> m(t) = u * t + M0, where m(0) = M0
F = d/dt(m * v) = m * dv/dt + v * dm/dt = -m * g, assuming no air resistance.
m * dv/dm * u = -v * u - m * g --> dv/dm + (1/m) * v = -g/u --> d/dm(m * v) = -m * g/u --> m * v = -m^2 * g/(2 * u) + C -->
v(m) = -m * g/(2 * u) + C/m, v(M0) = 0 --> C = M0^2 * g/ (2 * u) -->
v(m) = -g/(2 * u) * (m - M0^2/m) -->
v(t) = -g/(2 * u) * (u * t + M0 - M0^2/(u * t + M0))
Integrating we obtain:
x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2 * ln(u * t + M0)) + K

If x(0) = h --> K = h + g/(2 * u^2) * (M0^2/2 - M0^2 * ln(M0)) -->
x(t) = -g/(2 * u^2) * ((u * t + M0)^2/2 - M0^2/2 + M0^2 * ln(M0/(u * t + M0)) + h.
End.

 

[haruspex]

So the force is constant, ok.

No, the 'drag' that results from mass accumulation will be proportional to the speed.
Rocco's analysis looks right, but I've not checked it in detail. Too tedious without LaTeX.