Calculating Velocity in the Millikan Oil Drop Experiment

[happyparticle]

Homework Statement

Find Millikan oil drop velocity

Relevant Equations

$m\frac{dv}{dt} = \frac{4\pi a^3(\rho - \rho_1)g}{3} - 6\pi n a v$

Hi,
I try to find the velocity for a oil drop.

I found the forces.
$F=ma => m\frac{dv}{dt} = \frac{4\pi a^3(\rho - \rho_1)g}{3} - 6\pi n a v$

with v on the right side, I don't see how to get the solution.

I found the solution on few websites, but without the path to find the solution from the equation above.

This is basically, the Millikan oil drop experiment without the electric field.

 

[kuruman]

Did you mean the terminal velocity? The solution is a function of time.

 

[happyparticle]

I need to find v(t). Probably the velocity at any moment.

 

[Gordianus]

What do you know about differential equations?

 

[happyparticle]

What do you know about differential equations?

I know I have to integrate both side.

$dv = \frac{4\pi a^3(\rho - \rho_1)g}{3m} - \frac{6\pi n a v}{m} dt$

$v = \int (\frac{4\pi a^3(\rho - \rho_1)g}{3m} - \frac{6\pi n a v}{m}) dt$

$v = (\frac{4\pi a^3(\rho - \rho_1)g}{3m}t + C_1 - \frac{6\pi n a v}{m}t + C_2)$

However, this is not the solution.

 

[Gordianus]

It isn't the right Solution. The second term of the RHS has the variable $v$. However, you considered it as a constant.

 

[haruspex]

I know I have to integrate both side.

Not immediately, because you have dv/dt on the left and a v on the right. What will you integrate with respect to?
You first need to get it into the form (function of v only, or constant)dv=(function of t only, or constant)dt.

 

[Delta2]

Search the web for "ordinary differential equations of separable variables". Basically to solve it you do what it is being said at the last line of post #7 and then you integrate both sides and then some minor algebraic manipulation to find the solution.

 

[happyparticle]

I tried different algebraic manipulations, but I can't find the right solution which is $v(t) = (\frac{2a^2(p-p')g}{9n})(1-e^(\frac{-t}{2a^2p/9n}))$

My steps are

$\int \frac{dv}{4a^2/18n\cdot (p-p')g - v} = \int \frac{9n}{2a^2(p-p')} dt$

$-ln|\frac{2a^2g(p-p')g -9v}{9}| + C_1 = \frac{9nt}{2a^2(p-p')} + C_2$

I can't find a way to have the same solution as above.

 

[Delta2]

You are very close to the solution. Use one integration constant $C=C_1-C_2$ and perhaps the solution you typed in post #9 has a typo? Shouldnt the exponential be $e^{-\frac{t}{2a^2(p-p')/9n}}$?

 

[haruspex]

When you know the solution you are supposed to get for a differential equation, the cheat way is to work backwards, differentiating the solution. It is usually easier to get to exactly the DE. Then reverse the steps.

 

[happyparticle]

I don't think I can use reverse the steps since I found the solution on wikipedia.

$V(t) = \frac{2a^2g(p-p')g}{9n} - e^{\frac{-t}{2a^2p/9n}} + C$
It's still not exactly the same solution. Even if C = 1 for a=v=t =0

 

[Gordianus]

Let me simplify a bit:

Let $A=\frac{4\pi a^3(\rho-\rho_1)g}{3m}$ and $B=6\pi\eta a$

Thus $\frac{dv}{dt}=A-Bv$. Now, make the substitution $u=A-Bv$

 

[kuruman]

The substitution suggested by @Gordianus is a good one. I would also suggest that when the time comes to integrate, use definite integrals with upper and lower limits. In posting #5 I would have written $$\int_{v_0}^{v(t)} \frac{dv}{4a^2/18\eta\cdot (p-p')g - v} = \int_0^t \frac{9\eta}{2a^2(p-p')} dt$$ This says that at $t=0$ the velocity is $v_0$ and at $t=t$ the velocity is $v(t)$. Evaluating the definite integrals automatically takes care of the integration constants. Once you integrate, all you have to do is solve for $v(t)$.

Definite integrals also eliminate the annoying (to me) absolute value in the argument of the natural log. If the algebra is correct, the logarithm cannot have a negative argument because it describes a physical situation, not an abstract mathematical possibility. Don't forget that the difference of two logarithms is the logarithm of the ratio of arguments.

 

[happyparticle]

Let me simplify a bit:

Let $A=\frac{4\pi a^3(\rho-\rho_1)g}{3m}$ and $B=6\pi\eta a$

Thus $\frac{dv}{dt}=A-Bv$. Now, make the substitution $u=A-Bv$

Almost got it, but I think my problem is algebraic.$\frac{4\pi a^3(p-p_1)g}{3m} + \frac{6 \pi n av}{m} + c = e^\frac{t\pi n a}{m}$

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

$v(t) = \frac{m}{6 \pi n a}(g - e^\frac{t\pi n a}{m})$

 

[happyparticle]

Don't forget that the difference of two logarithms is the logarithm of the ratio of arguments.

Even if $v_0 \neq v(t)$ ?

 

[kuruman]

Yes your problem is algebraic. In your final expression you are adding the exponential, which is dimensionless, to g which has dimensions of acceleration.

 

[kuruman]

Even if $v_0 \neq v(t)$ ?

Even then.$$\int_{v_0}^{v(t)}\frac{dv}{v}=\ln\left[\frac{v(t)}{v_0}\right]$$There is nothing wrong with that.

 

[Gordianus]

We're so close...

 

[kuruman]

We're so close...

It sometimes is not easy to see the forest for the trees. Two semesters of calculus didn't do it for me but a formal course in differential equations did.

 

[hutchphd]

Millikan only used the terminal velocity in his experiments. Are we not sending the OP on a goose chase here?? To the OP: The terminal velocity is easy to obtain by looking for a solution where $\frac {dv} {dt} =0$ . Of course it is good to be able to solve the full diff eq but the asymptote was all that Millikan used because that's all he could really see by eye.

 

[kuruman]

Millikan only used the terminal velocity in his experiments. Are we not sending the OP on a goose chase here?? To the OP: The terminal velocity is easy to obtain by looking for a solution where $\frac {dv} {dt} =0$ . Of course it is good to be able to solve the full diff eq but the asymptote was all that Millikan used because that's all he could really see by eye.

In posting #2 I did ask OP exactly for that reason whether he/she meant terminal velocity and the answer was no. Besides, shouldn't one have the external force of the electric field also in the equation if that were the case?

 

[happyparticle]

I think this is the way I should find the solution. I just can't get it right.

I really don't see where I made a mistake.

$\frac{4\pi a^3(p-p_1)g}{3m} -\frac{6 \pi n av}{m} + c = e^\frac{t\pi n a}{m}$

$g - \frac{6 \pi n av}{m} + c = e^\frac{t\pi n a}{m}$

Is this right?

 

[hutchphd]

In posting #2 I did ask OP exactly for that reason

I know you did.
But Millikan turned off the field and measured the terminal velocity to find the mass of each of the droplets. Otherwise he could only report the charge to mass ratio.
With the field on I think he tried for perfect balance and no velocity by changing E.
I loved doing this experiment as an undergrad. Transcendental experience.

Note: the links in the above are all screwed up...apparently completely... apologies.

 

[kuruman]

Is this right?

Several reasons why it is not right.
You have a constant $c$. What is its value?
Your expression is dimensionally incorrect.
It is not in the form where $v$ is on side of the equation and everything else on the other.

If you showed your work in some detail, we could perhaps point out where you went wrong.

 

[kuruman]

I know you did.
But Millikan turned off the field and measured the terminal velocity to find the mass of each of the droplets. Otherwise he could only report the charge to mass ratio.
With the field on I think he tried for perfect balance and no velocity by changing E.
I loved doing this experiment as an undergrad. Transcendental experience.

Note: the links in the above are all screwed up...apparently completely... apologies.

I also remember doing the experiment as an undergraduate, but the details are lost in the recesses of my memory. That was in the days when one could not buy a turnkey experiment from PASCO. I do remember, however, two things: trying to stop the drop with the field and having a devil of a time trying to sort out the data. I also remember that we didn't see any quarks $\dots$

 

[happyparticle]

Several reasons why it is not right.
You have a constant $c$. What is its value?
Your expression is dimensionally incorrect.
It is not in the form where $v$ is on side of the equation and everything else on the other.

If you showed your work in some detail, we could perhaps point out where you went wrong.

I mean at this point is it correct? Because I don't really see where is my error. Is it before that?

whatever I do I still get the same answer.

 

[Charles Link]

If I'm not mistaken, post 13 should have $B=6 \pi \eta a/m$.

Meanwhile $\frac{dv}{dt}+Bv=A$ has a straightforward solution to the differential equation that the OP should recognize:
$v_p=\frac{A}{B}$ and $v_h=Ce^{-Bt}$.
It is simple matter for $v(0)=0$ to solve for $C$.
($v_p$ is the particular solution, and $v_h$ is the homogeneous solution. The complete solution is the sum of these, with the constant $C$ determined by the initial conditions).

Note: I don't get the exponents mentioned in posts 9 and 10. I believe those are in error.
Edit: See post 41. Post 9 is correct.

 

[happyparticle]

I'm wondering what I did wrong. I mean I'm confuse and for the moment I want to stick with one way to find the solution. I already tried few way, but I can't find the solution. I'm close, but I don't find my error. I just did the whole thing few times and I can't see what I did wrong.

 

[Charles Link]

I'm wondering what I did wrong. I mean I'm confuse and for the moment I want to stick with one way to find the solution. I already tried few way, but I can't find the solution. I'm close, but I don't find my error. I just did the whole thing few times and I can't see what I did wrong.

What you have in post 12 from Wikipedia I believe is incorrect. That's why it didn't work. Wikipedia has an error. See post 28 for the solution. (Edit: See post 41. Wikipedia has correct solution).
In post 15, you introduce a definition for $m$, where $m$ is already in the problem. You need to use some other letter.

 

[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.