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Pendulum motion and pythagorean theorem

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the relation between the horizontal and vertical components of the ball's position is given by the equation: y = L - [(L^2 - x^2)^1/2]

    http://www.flickr.com/photos/94066958@N08/8553595522/in/photostream/

    2. Relevant equations

    y = L - [(L^2 - x^2)^1/2]

    3. The attempt at a solution

    I know the solution must involve Pythagorean theorem and drawing a second triangle.
    The first triangle has hypotenuse of length L, and other sides are L-y and x. That identity is given by L^2 = (L-y)^2 - x^2.
    Drawing a second triangle, the sides are x and y but I don't know what the identity of the hypotenuse is. that equation would be x^2 + y^2 = hypotenuse^2

    I don't know where to go from there...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 12, 2013 #2

    SammyS

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    Hello sunnyday01. Welcome to PF !

    If a & b are legs of a right triangle with hypotenuse, c, then
    c2 = a2 + b2

    In your equation, L^2 = (L-y)^2 - x^2, why do you have the sign between. (L-y)2 and x2 as a minus sign ?
     
  4. Mar 12, 2013 #3
    formula correction

    It should be:
    L2 = (L-y)2 + x2
     
  5. Mar 12, 2013 #4

    SammyS

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    Yes.

    ... and that should be all you need to get the desired result.
     
  6. Mar 12, 2013 #5
    I think I figured it out!

    so, I think:
    L2 = (L-y)2 + x2
    L2-x2 = (L-y)2
    (L2-x2)1/2 = L - y
    -(L2-x2)1/2 = - L + y
    L -(L2-x2)1/2 = y

    I think that's it! Having to crop a picture for this question made me focus only on the variables I needed in the diagram.

    I see the solution now is really simple, I feel silly for not seeing it sooner. Thank you for being so nice about it!
     
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