# Problem on a liquid drop free fall

• Abhisek Roy
In summary: So if we could answer that question, we would have found the velocity of the drop.In summary, the water drop continually evaporates. The velocity of the drop as a function of time is determined by gravity and the rate of evaporation.

## Homework Statement

A freely falling water drop continually evaporates. Assuming that the net momentum carried away by the vapor vanishes, calculate the velocity of the drop as a function of time. (Assume that the rate of evaporation varies as the surface area)[/B]

## Homework Equations

dA/dr = rate of evaporation, where A is surface area of water droplet...and radius of spherical water droplet.

## The Attempt at a Solution

dA/dr= d(πr^2)/dr=2πr

Cant understand what "net momentum carried away by the vapor vanishes" means...cud it be that there r no drag forces the the vapor drop...

. Then how do I find the vel. Of the drop?

Possibly you are overthinking this.

My understanding matches yours -- that there are no drag forces on the vapor drop. The momentum carried away by the vapor simply vanishes and can be ignored. With that understanding, the only thing affecting the velocity of the drop is gravity. There is a simple formula for the velocity of a body acted upon only by gravity.

The more difficult part, if you want to tackle it, would be determining the time required (if finite) before the droplet evaporates completely.

• Abhisek Roy
jbriggs444 said:
Possibly you are overthinking this.

My understanding matches yours -- that there are no drag forces on the vapor drop. The momentum carried away by the vapor simply vanishes and can be ignored. With that understanding, the only thing affecting the velocity of the drop is gravity. There is a simple formula for the velocity of a body acted upon only by gravity.

The more difficult part, if you want to tackle it, would be determining the time required (if finite) before the droplet evaporates completely.

Yes the expression for velocity is fairly simple... If no drag forces are involved... If that's what's to be deduced from that statement... Thanks... N yea I'll try to find an expression fr total e CV aporation fr vapor too... But don't know whr to start.

I think the rate of evaporation = variation of area of drop is the main eqn. Then
Abhisek Roy said:
Yes the expression for velocity is fairly simple... If no drag forces are involved... If that's what's to be deduced from that statement... Thanks... N yea I'll try to find an expression fr total e CV aporation fr vapor too... But don't know whr to start.

Your expression for the rate of evaporation is incorrect. It should be that the rate of change of mass of the drop with respect to time is proportional to the surface area.

Chet

jbriggs444 said:
The more difficult part, if you want to tackle it, would be determining the time required (if finite) before the droplet evaporates completely.
Arguably, that is all part of answering the question about velocity as a function of time. The domain of that function will be the time for which the drop exists.