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

  1. Apr 2, 2009 #1
    Stoke's Law - Method and how to do it - graph???

    1. The problem statement, all variables and given/known data
    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???


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


    3. The attempt at a solution
    Don't know where to start, so please help. thanks in advance.
     
  2. jcsd
  3. Apr 2, 2009 #2

    rl.bhat

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

    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.
     
  4. Apr 2, 2009 #3
    Re: Stoke's Law - Method and how to do it - graph???

    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?
     
  5. Apr 2, 2009 #4

    rl.bhat

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

    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.
     
  6. Apr 11, 2009 #5
    Re: Stoke's Law - Method and how to do it - graph???

    I think you draw terminal vs r^2 i.e terminal on the y axis and radii on x axis????
     
  7. Apr 12, 2009 #6

    rl.bhat

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

    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.
     
  8. May 15, 2009 #7
    Re: Stoke's Law - Method and how to do it - graph???

    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
     
  9. May 15, 2009 #8

    rl.bhat

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

    If the material of the ball bearings is same, then the density of them will also be the same.
     
  10. May 16, 2009 #9
    Re: Stoke's Law - Method and how to do it - graph???

    I get that k is the gradient, however this doesn't make sense to me:
    Can you explain further please?
     
  11. May 16, 2009 #10
    Re: Stoke's Law - Method and how to do it - graph???

    ahh i realised after i posted the question how stupid i was, ofc the density doesnt change its the same material !!

    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
     
  12. May 16, 2009 #11
    Re: Stoke's Law - Method and how to do it - graph???

    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).
     
    Last edited by a moderator: May 4, 2017
  13. May 17, 2009 #12
    Re: Stoke's Law - Method and how to do it - graph???

    dfgdfg
     
    Last edited: May 18, 2009
  14. May 17, 2009 #13
    Re: Stoke's Law - Method and how to do it - graph???

    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.
     
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