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Circular Motion Derivation

  1. Oct 12, 2008 #1
    Hi for a lab I need to derivate the following formula.
    T=√((4π^2 lm)/Mg)

    I know Fnet= (mg/cos)sin

    So so far I know Fnet= m x (4pi^2r)/T^2
    and I know that r=lcos

    but combining the above I got no where since I get an extra sin in there
     
    Last edited: Oct 12, 2008
  2. jcsd
  3. Oct 12, 2008 #2
    Any help?
     
  4. Oct 12, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi Epsillon! Welcome to PF! :smile:
    No, r = … ? :smile:
     
  5. Oct 12, 2008 #4
    actually r=lsin(pheta)
     
  6. Oct 12, 2008 #5
    Ok so that was actually the mistake preventing me from solving the formula.

    So I finally got it and I typed this up in words and posted it here incase someone needs it. If it doesent make sense I can perhaps attach the file.
    Can someone verify that what I did is right?

    Deriving the formula
    F_net=ma_c=m((4π^2 r)/T^2 )
    F_net= F_c=〖Ft〗_x=Ft×sinθ
    Ft×sinθ= (mg/cosθ)sinθ =mgtanθ
    Now one can combine the two calculated formulas for F_net
    m((4π^2 r)/T^2 ) =(mg/cosθ)sinθ

    r=lsinθ
    ∴ m((4π^2 lsinθ)/T^2 ) =(mg/cosθ)sinθ

    F_t=mg/cosθ F_t=Mg
    Mg=mg/cosθ
    M=m/cosθ cosθ=m/M
    ∴ m((4π^2 lsinθ)/T^2 ) =(mg/(m⁄M ))sinθ
    (m4π^2 lsinθ)/T^2 =Mgsinθ
    (m4π^2 lsinθ)/Mgsinθ = T^2
    (4π^2 lm)/Mg = T^2
    T=√((4π^2 lm)/Mg)
     
    Last edited: Oct 12, 2008
  7. Oct 13, 2008 #6

    tiny-tim

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    A bit long, but basically ok. :smile:
     
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