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Homework Help: Centripetal Force Proportionality Check

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Write a proportionality statement for F, m, f and r using your three proportionality statements.

    2. Relevant equations
    F = centripetal force
    f = frequency
    m = mass
    r = radius

    F \propto f^2

    r \propto F^2

    m \propto F^2

    3. The attempt at a solution

    m \propto F^2 & r \propto F^2
    therefore: m \propto r
    m/r \propto F^2/F^2 \propto some constant
    F \propto m/r . f^2
    F \propto mf^2/r

    which matches F \propto mv^2/r as f^2 \propto v^2

    Edit: I removed all the tex tags because they messed everything up, hope this is still understandable

    Thank you in advance:smile:
    Last edited: Nov 9, 2008
  2. jcsd
  3. Nov 9, 2008 #2


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    Not quite sure I get this. If the orbital frequency is to remain constant, then r ~ F.

    If the orbital speed is to remain constant, then F ~ 1/r

    not sure I get this either...
  4. Nov 9, 2008 #3
    We did a lab in class and those were the proportionality statements I came up with.

    They might be wrong, we used very poor equipment and the data is nopt very good, nor is there much to work from.

    Here are the two graphs that I based my last two proportionality statements off of:
    attachment.php?attachmentid=16291&d=1226259979.gif attachment.php?attachmentid=16292&d=1226259979.gif

    See what I mean about the bad data? I was pretty sure about the mass graph, though...

    I guess I should have clarified that the rest of the variables were constant for the proportionality statements.

    Thanks again.

    Attached Files:

  5. Nov 9, 2008 #4
    I suppose you know how the centripetal force is expressed in terms of m, v, and r?

    All you need now is to rewrite v in terms of f and r. That will tell you how F depends on f,m, and r.

    Your answer, F\propto m f^2/r, is certainly not correct.
  6. Nov 9, 2008 #5


    2nr~d therefore v~f right?

    which would give me F~mf^2/r

    Where did I go wrong then?
  7. Nov 9, 2008 #6
    You went wrong in your definition of f.
  8. Nov 9, 2008 #7
    Ah, I see. To bad I have to go to work now, anyway, thanks.

    How does radius relate to Force though, Is my graph correct?
  9. Nov 9, 2008 #8
    I cannot see your graphs....they are pending approval.
  10. Nov 9, 2008 #9
    Sorry, didn't look at that.

    Is r~F or r~F^2? (if everything else is constant)
  11. Nov 9, 2008 #10
    if by "everything else" you mean f and m, then r ~ F
  12. Nov 9, 2008 #11
    How's this look:

    [tex] v=\frac{d}{t} [/tex]
    [tex] f=rot/t [/tex]
    [tex] 1rot=2 \pi r^2 [/tex]
    [tex] f2 \pi r^2=v [/tex]
    [tex] F= \frac{4 \pi^2 m r^2 f^2}{r} = 4 \pi^2 m r f^2 [/tex]

    Edit: I changed the graphs and the lines were just as poor, so I don't know why I decided to square them in the first place, thanks for pointing that out.
    Last edited: Nov 9, 2008
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