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Conservation of Energy

  1. Mar 18, 2015 #1
    1. The problem statement, all variables and given/known data

    vlQD6eH.png
    2. Relevant equations

    Conservation of Energy (Potential + Kinetic = Potential + Kinetic)

    3. The attempt at a solution

    At the start of the ramp, potential energy is mgh (gravitational potential) and kinetic is 0, since it's not moving.
    At the bottom of the loop, potential energy is 0 (I consider the bottom to be the zero of gravitational potential to make things easy), and kinetic is 0.5 mv^2.

    The ms cancel to leave me with 2gh = v^2, but that's as far as I got...
     
  2. jcsd
  3. Mar 18, 2015 #2

    RaulTheUCSCSlug

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    Well keep in mind that it will only have to have enough kinetic energy to get to the top of the roller coaster and also since we need to neglect friction, then at the top of the loop, we can have KE equal to zero, since it will only need to have very minimal Kinetic energy to get over the top of the loop (since the top of the loop is actually a very very small straight line, we know this if we examine the tangent line at the top of the circle but it isn't essential for the problem to be honest).

    So take your equation for PE, and set it equal to KE, and make sure you have your h correct. (think about what R means in this case)
     
    Last edited: Mar 18, 2015
  4. Mar 18, 2015 #3
    Err, KE and PE are BOTH zero at the top of the loop? How does that make sense when considering the conservation of energy? (mgh + 0 = 0 + 0)

    Also, I'm not sure I get what you're saying...set PE equal to KE, isn't that what I did above?
     
    Last edited: Mar 18, 2015
  5. Mar 18, 2015 #4

    RaulTheUCSCSlug

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    Woops sorry I meant that we only need just enough kinetic energy to get you through the top of the roller coaster so kinetic energy basically is zero at the top, therefore it is maximum potential energy. So PE is KE. Got ahead of myself ha
     
  6. Mar 18, 2015 #5
    I'm still confused :(
    How does finding the energy at the top help me to solve the problem? If KE is basically zero at the top of the loop, that would imply that it's all potential energy...are you saying that PE at the top of the loop is the same as the PE at the very start? Because that would imply that the answer is 2R...
    Or am I missing another type of PE?
     
  7. Mar 18, 2015 #6

    RaulTheUCSCSlug

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    You have the right answer! The answer is infact 2R! Do you get why though? The PE has to be at the very least the same PE from the beginning so that it may go through the loop.
     
  8. Mar 18, 2015 #7
    But the answer key says it's "D", or 2.5R. I don't understand why :(
     
  9. Mar 18, 2015 #8

    RaulTheUCSCSlug

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    Oh sorry then I don't know how to do it either : /
     
  10. Mar 18, 2015 #9
    Thank you for the help anyway :) I figured out the correct answer.
     
  11. Mar 18, 2015 #10

    RaulTheUCSCSlug

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    Oh I forgot about the centripetal acceleration! The cars can not have 0 KE at the top or else it would just fall straight down! So you want KE energy to be equal to the centripetal acceleration, which is V^2/R, so you want to solve for V, and that will tell you how much KE you need. See where you can go from there.
     
  12. Mar 19, 2015 #11

    haruspex

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    An energy cannot equal an acceleration. Consider the force balance at the top of the loop (##\Sigma F_{y} = ma_{y}##).
     
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