Terminal velocity of a raindrop

[Buffu]

Homework Statement

A raindrop of initial Mass $M_0$ starts to fall from rest under the influence of gravity. Assume that the drop gains mass from the cloud at a rate proportional to the product of its instantaneous mass and its instantaneous velocity $\dfrac{dM}{dt} = kMV$, where $k$ is constant. show that speed eventually become constant.

Homework Equations

The Attempt at a Solution

$P_i = M_0v$

$P_f = \left( M_0 + \left(\Delta t \dfrac{dM}{dt}\right) \right)(v + \Delta v)$

therefore $\Delta P = M_0 \Delta v + v \Delta t\dfrac{dM}{dt}$

Since $F = \lim_{\Delta t \to 0} \dfrac{\Delta P}{\Delta t}$

Therefore $F = M_0 \dfrac{dv}{dt} + v \dfrac{dM}{dt}$

Substituting $\dfrac{dM}{dt} = kMV$,

$F = M_0 \dfrac{dv}{dt} + k M v^2$

The velocity will be constant if the forces are balanced,

therefore, $Mg = M_0 \dfrac{dv}{dt} + k M v^2$

Now I don't know what to do ? should I solve for $v$ in this differential equation and then find the time at which it would be constant (I tired but it was very messy with lots of constants, I got $v = \displaystyle u\dfrac{1- \exp(2ku\alpha t)}{\exp(2ku\alpha t) - 2}$ where $u = \sqrt{h/k}$ and $\alpha = M/M_0$) ?

 

[jack action]

Actually:
$$F = \frac{d(Mv)}{dt} = M \dfrac{dv}{dt} + v \dfrac{dM}{dt}$$
Then you just need to find the condition that will set $\dfrac{dv}{dt} = 0$.

 

[Buffu]

Actually:
$$F = \frac{d(Mv)}{dt} = M \dfrac{dv}{dt} + v \dfrac{dM}{dt}$$
Then you just need to find the condition that will set $\dfrac{dv}{dt} = 0$.

But I got $M_0$ instead of $M$.

 

[jack action]

I know, but it should be the instantaneous mass, not the initial mass:

$P_i = Mv$
$P_f = \left( M + \left(\Delta t \dfrac{dM}{dt}\right) \right)(v + \Delta v)$

With the initial mass $M_0$, it is only valid for the initial instant as with $M$ it is valid at any moment.

 

[Buffu]

I know, but it should be the instantaneous mass, not the initial mass:

$P_i = Mv$
$P_f = \left( M + \left(\Delta t \dfrac{dM}{dt}\right) \right)(v + \Delta v)$

With the initial mass $M_0$, it is only valid for the initial instant as with $M$ it is valid at any moment.

Then we get,

$Mg = M \dfrac{dv}{dt} + k M v^2$

$g - kv^2 = \dfrac{dv}{dt}$. Therefore at $g = kv^2$, velocity is constant. Is this fine ?

 

[jack action]

That would be my answer. When $v= \sqrt{\frac{g}{k}}$, you will have no more acceleration, so it will stay at this speed constantly.

 

[zwierz]

lly:

F=d(Mv)dt=Mdvdt+vdMdt​

It seems to be wrong. This equation does not take in respect that the particles of the fluid in the cloud have nonzero velocity relative to the raindrop and thus those particles bring momentum as they meet the raindrop. For details seehttp://www3.nd.edu/~powers/ame.20231/ajp.1962.pdf
see also http://www.lcurve.org/writings/PT-NewtonsLawsForVariableMass.pdf

 

[Nidum]

and shouldn't the buoyancy and drag forces on the drop be considered ?

 

[zwierz]

-

 

[Buffu]

It seems to be wrong. This equation does not take in respect that the particles of the fluid in the cloud have nonzero velocity relative to the raindrop and thus those particles bring momentum as they meet the raindrop. For details seehttp://www3.nd.edu/~powers/ame.20231/ajp.1962.pdf
see also http://www.lcurve.org/writings/PT-NewtonsLawsForVariableMass.pdf

But in the question no value of relative velocity is given :).

and shouldn't the buoyancy and drag forces on the drop be considered ?

 

[haruspex]

This equation does not take in respect that the particles of the fluid in the cloud have nonzero velocity relative to the raindrop

I agree that the equation F=dp/dt=d(mv)/dt=mdv/dt+vdm/dt is not kosher. It treats mass as something that can change in a closed system.
However, it does give the right answers when the mass entering or leaving the system does so with no momentum in the frame of reference. In this case, we can assume that the moisture being collected had been stationary relative to the ground.

 

[zwierz]

look here: http://www3.nd.edu/~powers/ame.20231/ajp.1962.pdf
I marked the missed term with the red frame

 

[zwierz]

Oh! that is my mistake. In this problem u=0