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Homework Help: Roller Coaster Question

  1. Feb 19, 2007 #1
    1. The problem statement, all variables and given/known data

    the roller coaster starts with the first hill being 50 m high. When the coaster gets to the lowest point, starting from the first hill, it has already entered the circular loop. The passengers should be subject to a maximum of 2.5 g's.

    Figure out the maximum radius the circular loop can have.

    2. Relevant equations

    v = squ.root of 2gh
    v = squ.root of Rg

    3. The attempt at a solution

    What is the velocity at the bottom of the first hill?
    KE + PE = KE +PE
    (1/2)mv2 + mgh = (1/2)mv2 + mgh
    (1/2)v2 + gh = (1/2)mv2 + mgh (The masses cancel out because it is the same
    coaster at the top and bottom.)
    (1/2)v2 + gh = (1/2)v2 + gh (Substitute the numbers at each location)

    ( 1\2 ) (0) + (9.8) ( 50 ) h = ( 1\ 2 ) v^2 + 9.8 ( 0 ) (The height at the bottom is zero because it is the lowest point when comparing to the starting height.)

    = 31 m/s
  2. jcsd
  3. Feb 20, 2007 #2
    Now that you have the velocity of the roller coaster, then you can figure out the centripital acceleration. According to the question, the net acceleration is 2.5g.

    Therefore, [tex]\frac{mv^2}{r} +g=2.5g[/tex]
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