1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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?

  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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook