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Centripetal force lab: meaning of slope of radius vs Fc graph

  1. Jul 15, 2011 #1
    1. The problem statement, all variables and given/known data
    I have a graph with the radius on the x axis and Fc on the y axis. I had to then calculate the slope of this linear relationship. I did that, and I got 0.17.

    The problem is that I don't know what this represents. mass? acceleration??

    2. Relevant equations
    Fc = m4π^2rf^2

    btw π = pi = 3.14

    3. The attempt at a solution
    I tried to isolate for the slope of the line, which would be making the left side equal to Fc/r.
    This is what I got:

    Fc/r = m4πf^2

    I know that [STRIKE]Fc[/STRIKE] mass is kept constant.. then I have no idea where to go from there :S
     
    Last edited: Jul 15, 2011
  2. jcsd
  3. Jul 15, 2011 #2
    i dont think it represents something in general,
    its just like change in Fc with unit change in radius ...
     
  4. Jul 15, 2011 #3
    How can [itex]F_c[/itex] be kept constant if you got a slope of 0.17 for an [itex]F_c vs. r[/itex] graph?
     
  5. Jul 15, 2011 #4
    Ah whoops! I made a mistake; Fc isn't constant. I was looking at a graph I extrapolated from. Apologies!

    I mean that mass is constant.

    So does it really not represent anything significant?
     
  6. Jul 15, 2011 #5
    Oh!! I didnt notice OP wrote Fc is constant.
    I Need lenses now ... :biggrin:
     
  7. Jul 15, 2011 #6
    Well, force divided by radius gives units of [itex] \frac{kg}{s^2} [/itex], which, technically, is the unit for surface tension. However, I don't think you wanted to find that surface tension is equal to [itex] 0.17 \frac{kg}{s^2}[/itex].

    You could just note that, as the radius increased from r = a to r = b meters, the centripetal force increase by 0.17 Newtons per meter.
     
  8. Jul 16, 2011 #7
    you mean [itex] \frac{kg}{m^2} [/itex].
     
  9. Jul 16, 2011 #8
    I'm talking about surface tension in fluid dynamics. I was just trying to give an extreme example of how little "kilograms per second squared" means in a Circular Motion context, as [itex] \frac{kg}{s^2} [/itex] is also the unit of [itex]k[/itex], the spring constant, which is much more likely to be the case in the Introductory Physics sub-forum.
     
  10. Jul 16, 2011 #9
    What is "f" in your equation, is it angular velocity? Is it 2*pi or 2*pi^2, you show it both ways? Does your graph have a "y' intercept? What is it? You say the slope is .17 what is the rest of the equation in the form Y=mx+b? Y is Fc, x is the radius in meters, the slope has units of Newtons/meters or kg/s^2. I am guessing the slope is mass* angular velocity^2
     
  11. Jul 16, 2011 #10
    Oh yes ... i got a bit confused :shy:
     
  12. Jul 17, 2011 #11
    The y-intercept is 0.

    f is the frequency.

    The only thing is that we didn't learn about angular velocity. Is that the only thing that it can represent?
     
    Last edited: Jul 17, 2011
  13. Jul 17, 2011 #12

    Redbelly98

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    Welcome back to PF.

    Okay, so the slope equals [itex]4 \pi^2 m f^2[/itex]. If you know m and the value of the slope, you can calculate what the constant f is. Or, if you know f and the slope, you can calculate what m is. But -- was frequency f really a constant throughout your experiment?

    By the way, does your lab apparatus resemble something like the following?

    ucm_app.gif
     
  14. Jul 17, 2011 #13
    frequency f was not constant. The only constant values were Fc (centripetal force), which was 0.5N, and mass, which was 0.03kg.

    are you saying that I can't use the equation if f isn't constant?

    Yes, it is similar.
     
  15. Jul 17, 2011 #14

    Redbelly98

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    Thanks, that helps explain better what is going on.

    Something is wrong here. If Fc is constant, and you plot it on the y-axis, you should get a horizontal line, so slope=0.

    It looks like the variables here are f and r, so the graph should involve those quantities instead. But, just graphing f vs. r will not give a straight line. Did you get any instruction in what you are supposed to actually graph?
     
  16. Jul 17, 2011 #15
    This was similar to our procedure starting on page 4: http://schools.hwdsb.on.ca/highland/files/2011/01/centripetal_force.pdf

    So basically the question I'm trying to answer now is #23.

    The graph of radius and Fc was made after doing #20 and #21 in the PDF. The values of Fc were extrapolated from a graph of f^2 on the x axis and Fc on the y axis, seen in #20.
     
    Last edited by a moderator: Apr 26, 2017
  17. Jul 17, 2011 #16

    Redbelly98

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    Got it.

    Looks like you have already seen that the graph is linear or a direct proportion, and that the slope should equal [itex]4 \pi^2 m f^2[/itex]. You can measure the slope from your graph and, since you know m and f, compare the measured-from-the-graph slope value with what you get by calculating [itex]4 \pi^2 m f^2[/itex].
     
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