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Speed of the pendulum bob?

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A simple pendulum consisting of a small heavy bob attached to a light string of length 40cm is released from rest with the string at 60 degrees to the downwards vertical. Find the speed of the pendulum as it passed through its lowest point.
    2. Relevant equations
    Can't find an equation which is related to the question.
    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2008 #2

    Doc Al

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    Staff: Mentor

    Don't look for "equations", look for applicable principles. You can try to analyze the forces acting on the pendulum and apply Newton's laws (the hard way) or you can look for a conservation law that applies (the easy way).
     
  4. Jan 30, 2008 #3
    x = 40/cos60
    x = 80*10^-2 m

    v = √(2gh)
    v = √(2*10*80*10^-2)
    v = 4ms^-1

    but the answer in the text book is 2ms^-1
     
  5. Jan 30, 2008 #4

    Doc Al

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    This caculation for the change in height of the bob is incorrect. Draw yourself a diagram showing the initial and final positions of the bob and recalculate the initial height.
     
  6. Jan 30, 2008 #5
    Thanks Doc Al

    x = cos60*40
    x = 20*10^-2m

    v = √(2gh)
    v = √(2*10*20*10^-2)
    v = 2ms^-1
     
  7. Jan 30, 2008 #6

    Doc Al

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    Very good, but be careful:
    cos60*40 is the vertical distance below the pivot point. To find the change in height, you must subtract this from the length of the string. Luckily, 40 - 20 = 20. :wink:

    (What if the angle was 30 degrees instead of 60?)
     
  8. Jan 30, 2008 #7
    If the angle was 30 degrees:
    x = cos30*40
    x = 35

    h = 40 - 35
    h = 5

    v = √(2gh)
    v = √(2*10*5*10^-2)
    v = 1ms^-1

    Right?
     
  9. Jan 30, 2008 #8

    Doc Al

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    Staff: Mentor

    Excellent!

    My only suggestion: Don't round off your calculations until the very last step. (Cos30*40 = 34.64, not 35.)
     
  10. Jan 30, 2008 #9
    :wink: Ok! Thanks...
     
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