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Indiana Jones rope swing physics

  1. Jan 25, 2010 #1
    Hi,
    I am preparing for a test and have the following question.
    there is the image attached too.

    Indiana Jones is swinging from a rope. The distance between the pivot point and his center of mass is 31.0 m. He begins swinging from rest at an angle θ = 19.0o as shown in the figure. Assuming that Indiana and the rope can be treated as a simple pendulum, what is the value of θ after 1.27 s (in degrees)?

    so i this is a harmonic motion question. as far as i understand i can use the
    formula of location (x=...) but use angles to represent the location and thus get the answer. but somehow i get the wrong answer all the time.
    could some one show me how it is done so i can see where i am wrong?

    thanks
     

    Attached Files:

  2. jcsd
  3. Jan 25, 2010 #2

    tiny-tim

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    Hi zhenyazh! :wink:
    Show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
     
  4. Jan 25, 2010 #3
    ok.
    so indiana starts from the farthest point and i want to work in degrees and not meters t to find position.
    so i say x=Acos(wt+fi)
    where A=19 deg, t=.127 and w=sqrt(g/l) where l is 31 if i understand the question correctly.
    the answer i get is wrong.
     
  5. Jan 25, 2010 #4

    tiny-tim

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    Hi zhenyazh! :smile:

    (have an omega: ω and a phi: φ and a square-root: √ :wink:)
    Isn't t = 1.27 ? And what did you use for φ ?
     
  6. Jan 25, 2010 #5
    about the time yes you are right, it is a type.
    about fi, i used zero since i use cos for my calculation and start from the right amplitude
     
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