Calculating Velocity in the Millikan Oil Drop Experiment

[happyparticle]

What you have in post 12 from Wikipedia I believe is incorrect. That's why it didn't work.
In post 15, you introduce a definition for $m$, where $m$ is already in the problem. You need to use some other letter.

In post 15 I didn't change m to lighten the equation. m is only the mass.
The issue is g.
the solution should be

$v(t) = \frac{mg}{6\pi n a }(1 -e)$

I don't know if you see what I mean.

 

[Charles Link]

In post 15 I didn't change m. The issue is g.
the solution should be

$v(t) = \frac{mg}{6\pi n a }(1 -e)$

I don't know if you see what I mean.

It is not clear at all what you are computing in post 15. This differential equation is a rather simple one that can be solved in a routine manner by a couple of different methods. The method I like is given in post 28. The method in post 13 will also work.

 

[happyparticle]

I'm using the method in post 13. It should work I guess. But I can't get the solution.

 

[Charles Link]

I'm using the method in post 13. It should work I guess. But I can't get the solution.

$-(du/dt)/B=u$ is the modified D.E. with the substitution $u=A-Bv$, so we get $du/u =-B \, dt$, so that
$\ln|u|=-Bt$, so that $u=Ce^{-Bt}$.
$u=A-Bv$, so that $v=(A-u)/B$.
Note: $B=6 \pi \eta a /m$.

The solution in Wikipedia is in error, (Edit: See post 41,Wikipedia solution is correct). so you may have been trying to get a solution that is incorrect.
The solution I get, by both methods is $v(t)=\frac{A}{B}(1-e^{-Bt})$. See if you can get your calculations to agree with this.

 

[happyparticle]

I'll retype my steps

$\int \frac{-du}{Bu} = \int dt$
$u = A-Bv, du -B dv$

$\frac{-ln |A -Bv|}{B} + C_1 = t + C_2$

$C_1 - C_2 = C$
$A = \frac{4 \pi a^3(p-p_1)g}{3m}, B = \frac{6 \pi na}{m}$

$-ln|\frac{ 4 \pi a^3(p-p_1)g}{3m} -\frac{6 \pi na}{m}v| + C = t\frac{6 \pi na}{m}$

$\frac{ 4 \pi a^3(p-p_1)g}{3m} -\frac{6 \pi na}{m}v + C = e^{-t\frac{6 \pi na}{m}}$

$m = \frac{4 \pi a^3(p-p_1)}{3}$

$\frac{ 4 \pi a^3(p-p_1)g}{3 \frac{4 \pi a^3(p-p_1)}{3}} -\frac{6 \pi na}{ \frac{4 \pi a^3(p-p_1)}{3}}v + C = e^{-t\frac{6 \pi na}{ \frac{4 \pi a^3(p-p_1)}{3}}}$

$g -\frac{9 n}{2 a^2(p-p_1)}v + C = e^{-t\frac{6 \pi na}{ \frac{4 \pi a^3(p-p_1)}{3}}}$

$-\frac{9 n}{2 a^2(p-p_1)}v + C = -g + e^{-t\frac{6 \pi na}{ \frac{4 \pi a^3(p-p_1)}{3}}}$

$v(t) = \frac{2 a^2(p-p_1)}{9n} (g - e^{-t\frac{9 n}{2 a^2(p-p_1)}}) + C$

 

[Charles Link]

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.

 

[happyparticle]

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

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

 

[Charles Link]

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 $m$.

 

[happyparticle]

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

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

 

[Charles Link]

I don't think $C=g$, but otherwise correct.
You are ok above through line 6. Try completing it with the new info.

 

[Charles Link]

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 $(4/3) \pi a^3 \rho$.

 

[happyparticle]

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.

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

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

 

[Charles Link]

$C'=A$ 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. :)

 

[happyparticle]

Wow, really? Sadly, I don't see how you can cancel the p's.

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

 

[Charles Link]

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. :)

 

[happyparticle]

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.

 

[haruspex]

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?

 

[happyparticle]

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.

 

[haruspex]

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.

 

[happyparticle]

Yeah, I had to find the general solution and as you said it's probably because it's a valuable exercise.

 

[Charles Link]

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.

 

[happyparticle]

Alright, I'll try it.
I have done a course in differential equations, but I forgot a lot of things.

 

[Gordianus]

Don't worry. Practice, practice and practice.