Speed of falling water drop

[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$$\pi$$r$$^{2}$$


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$$\pi$$r$$^{2}$$

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

 

[paranormal]

Any guidance would be appreciated.





Hello m.g.,

I understand your struggle in applying calculus and differential equations in this mechanics class. Let me guide you through the steps to solve this problem.

Firstly, we need to identify the variables involved in this problem. We have the mass of the water drop, which is increasing at a rate proportional to its area. We also have the speed of the drop, which we are trying to find as a function of time. Lastly, we have the gravitational force acting on the drop, which is causing it to accelerate.

Next, we can write the equation for the increase in mass as:

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

Where k is the proportionality constant and r is the radius of the drop. We can also rewrite this equation in terms of the drop's volume, V, as:

\frac{dm}{dt} = k\frac{dV}{dt}

Since the volume of a sphere is given by V = \frac{4}{3}\pi r^3, we can substitute this into the above equation to get:

\frac{dm}{dt} = \frac{k4}{3}\pi r^2\frac{dr}{dt}

Now, we can use the chain rule to write this in terms of the drop's speed:

\frac{dm}{dt} = \frac{k4}{3}\pi r^2\frac{dr}{dt} = \frac{k4}{3}\pi r^2\frac{dr}{dt}\frac{dt}{dt} = \frac{k4}{3}\pi r^2\frac{dr}{dt}\frac{1}{v}

Substituting this into our original equation for the forces, we get:

\frac{dv}{dt} = g - (\frac{v}{m})\frac{dm}{dt} = g - (\frac{v}{m})\frac{k4}{3}\pi r^2\frac{dr}{dt}\frac{1}{v}

Now, we can rearrange this equation to isolate the variables and solve for v:

\frac{dv}{dt} = g - (\frac{k4}{3}\pi r^2)\frac{dr}{dt}

\frac{dv}{g - (\frac{k4}{3}\pi r^2)\frac{dr}{dt}} = \frac{1}{m