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Provincial Exam: Circular Motion

  1. Jan 3, 2010 #1
    1. The problem statement, all variables and given/known data

    A frictionless 3.0 kg roller coaster cart rolls down an incline, and then "loops the loop."

    At what height, h, should the cart be released so that it does not fall off the circular track.

    The loop has a radius of 6.0m and a height of 12.0 m.

    2. Relevant equations

    [tex] a_c = \frac{v^2}{r} [/tex]

    3. The attempt at a solution

    This is an old practice exam, so if this has anything to do with escape velocities, it has been taken out of our curriculum.

    I don't know quite where to start.

    I know that you would require a certain velocity, conceptually, to be able to make it around the loop.

    And [tex] v \propto h [/tex] because of the law of energy conservation. but i dont know what minimum velocity i would need?

    I could kinda see a relation of using a centripetal acceleration to find out the velocity required and therefore the height, and find out the centripetal accelleration using Newtons Second, i just don't know how i can work it out.
     
  2. jcsd
  3. Jan 3, 2010 #2

    Doc Al

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    Staff: Mentor

    Figure out the minimum speed at the top of the loop by examining the forces acting. Apply Newton's 2nd law.
     
  4. Jan 3, 2010 #3
    Hi Doc Al, haven't seen you for a while.

    At the top of the loop, their are two forces, both the normal force and the gravitational force are in the same direction.

    But i don't know the normal force? I would assume that their is one because something has to keep it in its circular path. But i have no way to calculate it.

    Once i have the normal force.. i would assume

    [tex] F_c = F_N + F_g [/tex]

    [tex] m\frac{v^2}{r} = F_N + mg [/tex]

    Again though, i am unsure how to find the normal force, two unknowns and i only have one equation. Is their another equation im neglecting?
     
  5. Jan 3, 2010 #4

    Doc Al

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    Staff: Mentor

    What will the normal force be when the cart just barely loses contact with the track?
     
  6. Jan 3, 2010 #5
    zero, so then gravity will F centrepetal? and calculate from there?
     
  7. Jan 4, 2010 #6

    Doc Al

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    Staff: Mentor

    Exactly.
     
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