# Stoke's Law - Method and how to do it - graph?

Stoke's Law - Method and how to do it - graph???

## Homework Statement

Hi, i have to write a method to do stoke's law on different ball bearings and draw a graph of stoke's law of different ball bearings in glycerol and from the graph i have to find the viscosity of the liquid. How do i do this???

## Homework Equations

I have the top 2 formula's from:
http://en.wikipedia.org/wiki/Stokes'_law

## The Attempt at a Solution

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rl.bhat
Homework Helper

Find radii of different balls. Find their terminal velocities. Draw a graph of R^2 vs terminal . It will be a straight line. From the Slop of this line can find the coefficient of viscosity.

thanks for your reply. so i measure the diameter using a micrometer, then halve to get radius. but how do i find terminal velocity and is the viscosity the gradient of the line? will it be the same for all ball bearings? also what are the formulas for?

rl.bhat
Homework Helper

The terminal velocity will be different for different balls.
Take a glass tube of 5 cm diameter. Fill it with glycerol. From liquid level make 5 cm marks on the tube.
Drop a ball bearing. Start the stop watch when the ball enters the liquid. Note down the time whenever the ball crosses the 5 cm mark. When the time interval (t) becomes constant, the terminal velocity will be 5cm/t. Find the densities of the ball and liquid by any method. Using second formula find coefficient o viscosity. of glycerol.

Find radii of different balls. Find their terminal velocities. Draw a graph of R^2 vs terminal . It will be a straight line. From the Slop of this line can find the coefficient of viscosity.
I think you draw terminal vs r^2 i.e terminal on the y axis and radii on x axis????

rl.bhat
Homework Helper

Yes. Slope of the graph m = 2/9*[rho(b) - rho(gl)]*g/mu, where mu is coefficient of viscosity. From this you can find mu.

Yes. Slope of the graph m = 2/9*[rho(b) - rho(gl)]*g/mu, where mu is coefficient of viscosity. From this you can find mu.
The problem i have with this is that once one has found the gradient of the graph and thus deciphered the value of the constant k, which value would one use for the density of sphere as they all vary

rl.bhat
Homework Helper

The problem i have with this is that once one has found the gradient of the graph and thus deciphered the value of the constant k, which value would one use for the density of sphere as they all vary
If the material of the ball bearings is same, then the density of them will also be the same.

I get that k is the gradient, however this doesn't make sense to me:
where mu is coefficient of viscosity. From this you can find mu.

If the material of the ball bearings is same, then the density of them will also be the same.
ahh i realised after i posted the question how stupid i was, ofc the density doesnt change its the same material !!

I get that k is the gradient, however this doesn't make sense to me:

when you found k, use this equation http://upload.wikimedia.org/math/3/d/b/3dbef61d20c1951c09a727ff8d4a809c.png, where k is equal to Vs to find the value of u which will be the viscosity of the material

when you found k, use this equation http://upload.wikimedia.org/math/3/d...ff8d4a809c.png [Broken], where k is equal to Vs to find the value of u which will be the viscosity of the material
But aren't we looking for the coefficient of viscosity, not viscosity itself? Because the experiment is investigating stokes law, which is F = 6 x Pi x r x coefficient of viscosity x v.

If k = viscosity, we aren't finding the coefficient. Or am I just getting really confused and over complicating it? (I have a feeling I am).

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dfgdfg

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sorry let me start again, the value of the gradiet is equal to the value of Vs in the equation, http://upload.wikimedia.org/math/3/d/b/3dbef61d20c1951c09a727ff8d4a809c.png, the letter U represents coefficient of viscosity, so rearrangement and you have your answer.
Btw are you doing this for some sort of cw?
Ok, that makes more sense; thanks. And yes I am. I'll use this thread as a source for calculating and describing the graph, shouldn't be a problem.