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Loop the loop kinetic energy

  1. Feb 13, 2009 #1

    TG3

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    1. The problem statement, all variables and given/known data
    A cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 15 m. Friction can be ignored. In order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.5 times the weight of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.) You may treat the cart as a point particle.
    (I have calculated the speed necessary to be 22. 52 m/s)
    The cart is launched horizontally along a surface at the same height as the bottom of the loop by releasing it from rest from a compressed spring with spring constant k = 10000 N/m. What is the minimum amount X that the spring must be compressed in order that the cart "safely" negotiate the loop?


    2. Relevant equations
    Potential Energy = MGH
    K=1/2 MV^2

    3. The attempt at a solution

    9.81 x 500 x 30 = 147150 = the amount of energy lost on the way up.

    V=22.52
    M= 500
    22.52^2 x 500 x 1/2 = 126787.6

    126787.6+147150= 273937.6
    273937.6 =1/2 500 x v^2
    1095.75 = v^2
    33.1 = v
    But this is wrong... where did I mess up?
     
  2. jcsd
  3. Feb 13, 2009 #2

    AEM

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    It appears that when you found the total energy you needed:

    total energy = 126787.6+147150= 273937.6

    That you turned around and substituted back into [tex] \frac{1}{2} m v^2 [/tex]

    when what you wanted to do was to find the distance the spring was compressed for which you need to use

    [tex] \frac{1}{2} k x^2 [/tex]
     
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