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Another roller coaster question

  1. Mar 28, 2009 #1
    http://img13.imageshack.us/img13/9297/masteringphysicsq1.jpg [Broken]

    At the bottom it's 510N, Top 666N,

    at bottom Fnormal=netforce= mv^2/r + mg

    http://img13.imageshack.us/img13/8244/masteringphysicsq1g.jpg [Broken]
    -my attempt

    I'm not sure how to figure out the acceleration
    or determine the time for one loop on the graph.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 28, 2009 #2
    great 510 n = 5 seconds....
    666n = 15 seconds
     
  4. Mar 28, 2009 #3

    LowlyPion

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    I don't think you need to determine the period or the velocity.

    Where will the weight be the minimum? And the maximum?

    Presuming that the ferris wheel is not a super spinning Whirl-a-Gig, then you know

    mg - mv2/r = Min
    mg + mv2/r = Max

    Then just solve for m*g.
     
  5. Mar 28, 2009 #4
    I got two different awnsers, for the coaster at the bottom 12.171kg and top 14.52kg
    i think i've done it wrong :\
     
  6. Mar 28, 2009 #5

    LowlyPion

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    Try constructing the equations.

    Then subtract 1 from the other.

    You will determine then what mv2/r is and then you can figure the weight from either of the 2 equations.

    I only get 1 answer.
     
  7. Mar 28, 2009 #6
    What like?
    666-510 = (mg + mv^2/r) - (mg - mv^2/r)
    i can't figure out the velocity for the mv^2/r
     
  8. Mar 28, 2009 #7
    666 = mg + mv^2/r
    510 = mg - mv^2/r

    666+510 = 2mg + - mv^2/4
    = 1176 = 2mg
    1176/9.8 = 120
    120/2 = 60

    60kg?
     
  9. Mar 28, 2009 #8

    LowlyPion

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    That's right. Adding them works too. In fact better as it yields the m*g directly.

    m = 1176/(2*9.8) = 60
     
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