Millikan oil drop velocity

  • #36
Charles Link
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Line 7 is incorrect. ## e^{C-Bt}=C'e^{-Bt} ## is what you get on the right side.
Line 8 is also incorrect. You are redefining ## m ##.
The first 6 lines are good. Try working more carefully from there.
 
  • #37
happyparticle
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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
Charles Link
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I don't understand. m is the mass of the oil drop. I have to replace m for his mass.
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.
## \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
happyparticle
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Ah make sense, so ##m = \frac{4 \pi a^3}{3}p##

V(0) = 0 when C = g, right?
 
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  • #40
Charles Link
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I don't think ## C=g ##, but otherwise correct.
You are ok above through line 6. Try completing it with the new info.
 
  • #41
Charles Link
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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 ##.
 
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  • #42
happyparticle
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I hope I don't waste your time.

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}})##
 
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  • #43
Charles Link
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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. :)
 
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  • #44
happyparticle
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Wow, really? Sadly, I don't see how you can cancel the p's.

## v(t) = \frac{2a^2pg}{9n} (\frac{(p-p_1)}{p} - \frac{(p-p_1)}{p}e^{\frac{-t9n}{2a^2p}})##
 
  • #45
Charles Link
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Wow, really? Sadly, I don't see how you can cancel the p's.

## v(t) = \frac{2a^2pg}{9n} (\frac{(p-p_1)}{p} - \frac{(p-p_1)}{p}e^{\frac{-t9n}{2a^2p}})##
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. :)
 
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  • #46
happyparticle
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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}) ##.
The p's cancel. :)
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.
 
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  • #47
haruspex
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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.
Good, but there is an unresolved question.
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
happyparticle
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For whatever reason I don't have this in the wikipedia english page, but in french you can see here what I was looking for.

My question is titled the Millikan experiment. Ultimately, I have to find for which value of q, v=0.
 
  • #49
haruspex
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For whatever reason I don't have this in the wikipedia english page, but in french you can see here what I was looking for.
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.
My question is titled the Millikan experiment. Ultimately, I have to find for which value of q, v=0.
Again, you don't need to solve the ODE for that.
So unless you were specifically instructed to find the general solution of the ODE I would say it was unnecessary - but a valuable exercise.
 
  • #50
happyparticle
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Yeah, I had to find the general solution and as you said it's probably because it's a valuable exercise.
 
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  • #51
Charles Link
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Good job by @EpselonZero for staying with it until it was solved. Suggestion is to also understand the solution given in post 28. In general, this is a very standard differential equation with a solution to the homogeneous equation, (where the right side is set to zero), that comes with an arbitrary constant ## C ##, along with the particular solution. If you haven't yet had a course in differential equations, you would do well to take one as soon as you can.
 
  • #52
happyparticle
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Alright, I'll try it.
I have done a course in differential equations, but I forgot a lot of things.
 
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  • #53
Gordianus
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Don't worry. Practice, practice and practice.
 
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