Calculating Velocity in the Millikan Oil Drop Experiment

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.
  • #1
happyparticle
405
20
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.
 
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  • #2
Did you mean the terminal velocity? The solution is a function of time.
 
  • #3
I need to find v(t). Probably the velocity at any moment.
 
  • #4
What do you know about differential equations?
 
  • #5
Gordianus said:
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.
 
  • #6
It isn't the right Solution. The second term of the RHS has the variable ##v##. However, you considered it as a constant.
 
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  • #7
EpselonZero said:
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.
 
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  • #8
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.
 
  • #9
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.
 
  • #10
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}}##?
 
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  • #11
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.
 
  • #12
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
 
  • #13
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##
 
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  • #14
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.
 
  • #15
Gordianus said:
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})##
 
  • #16
kuruman said:
Don't forget that the difference of two logarithms is the logarithm of the ratio of arguments.

Even if ##v_0 \neq v(t)## ?
 
  • #17
Yes your problem is algebraic. In your final expression you are adding the exponential, which is dimensionless, to g which has dimensions of acceleration.
 
  • #18
EpselonZero said:
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.
 
  • #19
We're so close...
 
  • #20
Gordianus said:
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.
 
  • #21
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.
 
  • #22
hutchphd said:
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?
 
  • #23
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?
 
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  • #24
kuruman said:
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.
 
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  • #25
EpselonZero said:
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.
 
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  • #26
hutchphd said:
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##
 
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  • #27
kuruman said:
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.
 
  • #28
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.
 
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  • #29
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.
 
  • #30
EpselonZero said:
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.
 
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  • #31
Charles Link said:
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.
 
  • #32
EpselonZero said:
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.
 
  • #33
I'm using the method in post 13. It should work I guess. But I can't get the solution.
 
  • #34
EpselonZero said:
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.
 
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  • #35
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 ##
 

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