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Conservation of Mechanical Energy and Centripetal Acceleration Problem

  1. Sep 28, 2008 #1
    Show that a roller coaster with a circular vertical loop. The difference in your apparent weight at the top of the circular loop and the bottom of the circular loop is 6 g's-that is, six times your weight. Ignore friction. Show also that as long as your speed is above the minimum needed, this answer doesn't depend on the size of the loop or how fast your go through it.

    My working for the first half of the problem, the 6g's part is as follows

    Radius if loop is R
    Hieght from which it is released is h

    The speed at bottom of the loop is determined by the conservation of mechanical energy
    1/2 mvbottom2=2mgh

    Apparent weight at the bottom of the loop is obtained by the below:

    mvBot2= FNorm. Bot.-mg
    Therefore apparent weight at bottom is
    FNorm. Bot.=mvbot2/R+mg
    FNorm. Bot.=2mgh/R+mg (using result obtained via conservation of energy)

    To find speed at top of the loop we have from Conservation of Energy
    1/2 mvtop2+mg(2R)=mgh
    mvtop2=2mg(h-2R)
    Therefore using the above the apparent weight at the top of the loop is

    mvTop2/R = FNorm. Top.+mg
    Therefore Apparent weight is :
    FNorm. Top. = (2mg(h-2R))/R - mg

    Hence
    FNorm. Bot. - FNorm. Top. =
    2mgh/R + mg - [((2mg(h-2R))/R - mg)]=
    2mgh/R + mg - 2mgh/R + 4mg + mg=
    6mg

    That's how i proved the first section. Can anyone please tell me how to complete the problem, namely proving that as long as your speed is above the minimum needed, the answer doesn't depend on the size of the loop or how fast your go through it. This part of the problem is relaly bugging me and I've tried heaps of ways but can't come up with a definitive, good answer
     
    Last edited: Sep 28, 2008
  2. jcsd
  3. Sep 28, 2008 #2
    the minimum speed needed is solved by mg = mv2 / R
    Therefore the minimum speed in this instance is the square root of (gr)

    But i can't find a way to tie this into the problem and prove the last part
     
  4. Sep 28, 2008 #3
    please clarify here :D:D:D
     
  5. Sep 28, 2008 #4
    help guyz please. desperate
     
  6. Sep 28, 2008 #5
    It looks like you already solved it. You arrived at your final equation without any restrictions on R or V (other than the speed must be sufficient to make it around the loop.)

    Incidentally there are a few errors which I assume are typos in what you wrote, starting with your first equation. You should go back over your work.
     
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