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Centripetal Force Lab Question

  1. Jul 2, 2007 #1
    I need help with this lab I'm doing involving swinging a mass in a circle (I'm sure you guys have heard of it). I have to plot a graph of Radius vs. FT^2 and then compare the value of the slope to its theoretical value.

    A couple questions:
    Which goes on the y axis? Radius or FT^2?
    How do I ascertain the theoretical value of the slope to compare to?

  2. jcsd
  3. Jul 2, 2007 #2


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    What is FT^2??
  4. Jul 2, 2007 #3
    I have no clue. I'm just multiplying F with T^2.

    I've done some research to help myself, though I still don't know what's going on. Here's the link I found, scroll down to the bottom for the tidbit on FT^2.

  5. Jul 2, 2007 #4


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    Ok then. What are F and T^2. Sorry but I can't access links from where I am.
  6. Jul 3, 2007 #5
    I see...the link already gives you what you need. You should already know that Net Force = ma and Net Centripetal Force = mv^2/r. v = 2(pi)r/T, in which T is time. Therefore, all that was done was that using these formulas and plugging in values, the Net Force = ma = mv^2/r = 4(pi)^2Rm/T^2. I dont think its FT^2, I think you're confusing it for just F.
    ---To answer the question about the axes, F goes on the y-axis and Radius goes on the x-axis; consider it as for you increase or decrease the radius, you'll retrieve another value for F.
    ---To explain about the situation with FT^2 being mistaken for F, is that
    FT^2 = 4(pi)^2Rm, which I cant really identify (its not acceleration or the velocity). Using just F is the centripetal force, and you would graph whether F increases as the radius increases (the slope is I suppose deltaF/deltaT). Then again, I'm not exactly completely sure that this is the answer, its just I can not identify FT^2.
    ---If its any help...if it really is FT^2, then the slope should identify FT^2/R, in which it is equal to 4(pi)^2m; if you could find what that means, then you would be able to find out what the slope refers to.
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