- #36

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Line 8 is also incorrect. You are redefining ## m ##.

The first 6 lines are good. Try working more carefully from there.

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- Thread starter happyparticle
- Start date

In summary, the person was trying to find the velocity of an oil drop and found the forces and found the solution. However, they don't think they can use the inverse of the steps suggested by Gordianus to get the solution. They tried different algebraic manipulations, but they could not find the right solution.

- #36

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Line 8 is also incorrect. You are redefining ## m ##.

The first 6 lines are good. Try working more carefully from there.

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- #37

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Charles Link said:Line 8 is also incorrect. You are redefining ## m ##.

I don't understand. m is the mass of the oil drop. I have to replace m for his mass.

The final solution should be

## v(t) = \frac{2 a^2g(p-p_1)}{9n} (1 - e^{-t\frac{9 n}{2 a^2(p)}}) ##

- #38

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The expression with ## V_s(\rho-\rho_1)g ## has the difference in densities, which makes for the weight minus the archimedes type buoyant force. ##V_s =(4/3) \pi a^3 ## is the volume of the sphere.EpselonZero said:I don't understand. m is the mass of the oil drop. I have to replace m for his mass.

## \rho_1 ## is the density of the air which is the displaced fluid in archimedes terms.

## (4/3) \pi a^3 (\rho-\rho_1) ## has the units of mass, but it is a reduced mass and is not ## m ##.

- #39

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Ah make sense, so ##m = \frac{4 \pi a^3}{3}p##

V(0) = 0 when C = g, right?

V(0) = 0 when C = g, right?

- #40

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You are ok above through line 6. Try completing it with the new info.

- #41

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Going from line 6 to line 7 you made an algebraic error. Once you correct that, you should be able to get the correct answer. (You take ## e^x ## of both sides). ##e^{a+b}=(e^a )(e^b) ##.

and yes, your solution in post 37 (from Wikipedia) is correct, with ## \rho=3m/(4 \pi a^3) ##, and is in agreement with mine.

Note: I can see now why they wrote the solution how they did: It contains simply the radius of the drop, ##a ##, and they wanted to have ## a ## as the one parameter for the size of the drop=thereby, they eliminated the ## m ## in the exponent, and replaced it with ## (4/3) \pi a^3 \rho ##.

and yes, your solution in post 37 (from Wikipedia) is correct, with ## \rho=3m/(4 \pi a^3) ##, and is in agreement with mine.

Note: I can see now why they wrote the solution how they did: It contains simply the radius of the drop, ##a ##, and they wanted to have ## a ## as the one parameter for the size of the drop=thereby, they eliminated the ## m ## in the exponent, and replaced it with ## (4/3) \pi a^3 \rho ##.

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- #42

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I'm still stuck.

I get ##A - Bv = C'e^{-Bt}##

Then I replace A,B and m.

##\frac{(p-p_1)g}{p} - \frac{9nv}{2a^2p} = C' e^{\frac{-t9n}{2a^2p}}##

## v = \frac{2a^2p}{9n} (\frac{(p-p_1)g}{p} - C' e^{\frac{-t9n}{2a^2p}})##

- #43

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## C'=A ## because ## v=0 ## at ## t=0 ##.

I think you almost have it...

Yes, I believe you got it. The p's will cancel in numerator and denominator.

Very good. :)

I think you almost have it...

Yes, I believe you got it. The p's will cancel in numerator and denominator.

Very good. :)

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- #44

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## v(t) = \frac{2a^2pg}{9n} (\frac{(p-p_1)}{p} - \frac{(p-p_1)}{p}e^{\frac{-t9n}{2a^2p}})##

- #45

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Factor out ## A ##, and you have exactly what you need, with ## v(t)=(\frac{2a^2 pg}{9n})(\frac{p-p_1}{p}) (1-e^{etc}) = \frac{2 a^2 g (p-p_1 )}{9n}(1-e^{etc}) ##.EpselonZero said:

## v(t) = \frac{2a^2pg}{9n} (\frac{(p-p_1)}{p} - \frac{(p-p_1)}{p}e^{\frac{-t9n}{2a^2p}})##

The p's cancel. :)

- #46

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You are right, after all this I didn't remember what I was looking for.Charles Link said:Factor out ## A ##, and you have exactly what you need, with ## v(t)=(\frac{2a^2 pg}{9n})(\frac{p-p_1}{p}) (1-e^{etc}) = \frac{2 a^2 g (p-p_1 )}{9n}(1-e^{etc}) ##.

The p's cancel. :)

Thanks Charles and everyone else who help me. Finally! I'm going to sleep.

- #47

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Good, but there is an unresolved question.EpselonZero said:You are right, after all this I didn't remember what I was looking for.

Thanks Charles and everyone else who help me. Finally! I'm going to sleep.

As @hutchphd pointed out, Millikan had no need for such an equation because he only dealt with two scenarios: terminal velocity and stasis. So is the reference to Millikan in the thread title misleading?

- #48

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- #49

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The French page finds the general solution, as you have in this thread, but then takes t to infinity to find the terminal velocity. The English page avoids solving the ODE by just setting the acceleration to zero. I confess to sympathy for the English sloth.EpselonZero said:

Again, you don't need to solve the ODE for that.EpselonZero said:My question is titled the Millikan experiment. Ultimately, I have to find for which value of q, v=0.

So unless you were specifically instructed to find the general solution of the ODE I would say it was unnecessary - but a valuable exercise.

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- #51

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- #52

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Alright, I'll try it.

I have done a course in differential equations, but I forgot a lot of things.

I have done a course in differential equations, but I forgot a lot of things.

- #53

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Don't worry. Practice, practice and practice.

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