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Pendulum - energy

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data
    80kg mass man is swinging on a swing. His amplitude is 1m. in 60 seconds he does 15 swings. Find the potential and kinetic energy after 1/12 period.


    2. Relevant equationsW(p)=mgh or kx^2/2 W(k)=mv^2/2 W(max)=Fr^2mg/2l



    3. The attempt at a solution i Found out that maximum energy is 100J. But when i am calculating energy of petencial or kinetic i get 1/3 or 2/3 of maximum...or something even more horrid. =]
     
  2. jcsd
  3. Apr 29, 2010 #2
    oh c'mon!!! there must be someone out there knowing how to do this...
     
  4. Apr 29, 2010 #3

    tiny-tim

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    Hi Patrickas! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    erm :redface: … you can't post around 2am New York time and expect an answer 4 hours later!! :rolleyes:

    anwyay, show us your full calculations, and then we'll see what went wrong, and we'll know how to help. :smile:
     
  5. Apr 29, 2010 #4

    yea, but there are alot more people out there not from new york, who might know.
     
  6. May 1, 2010 #5
    hmm. Anyone..?
     
  7. May 1, 2010 #6

    tiny-tim

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    Show us your full calculations, and then we'll see what went wrong, and we'll know how to help. :smile:
     
  8. May 1, 2010 #7
    T=60/15=4s
    t=4/12=1/3s

    So i know that W=kx^2/2 Well then i tried getting k, k=mg/l , now i needed the lengh of swings so i figured mv^2/r=mgr/l l=r^2g/v^2, now i need speed =D. v=s/t is s= 4*r? , or s=2*3.14*r? and t=1/3. Anyways i sence that something here is wrong. The numbers don't add up properly.

    Assuming that s=2*3.14*r i can get T=2*3.14*[tex]\sqrt{\frac{l}{g}}[/tex] and get that l(t=4s)=4m Then i get that maximum energy is 100J. Now what...

    Ahh... i used to calculate s=4r and not 2*3.14*r. I think it works out.

    W=[tex]\frac{mgt2*3.14r}{l2T}[/tex]

    When i put t=1/3 i get that potencial or kinetic(don't know which..) is 25J !!! Which means that other is 75J Now which is which...?
     
    Last edited: May 1, 2010
  9. May 1, 2010 #8
    Hey thanks for the help! I have been trying to solve this one for days! hehe
     
  10. May 1, 2010 #9

    tiny-tim

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    Hi Patrickas! :smile:

    You're making this very complicated. :redface:

    You know T = 4, A = 1, and so distance x = 1*cos(2πt/4).

    So you can find v … then KE = 1/2 mv2, and PE = … ? :smile:
     
  11. May 1, 2010 #10

    O yea! Thanks! It should be much faster and easier to solve those problems now.
     
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