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Roller coaster problem

  1. Aug 9, 2006 #1
    hi there, i'm planning to take university physics in the fall, but i'm lousy at it and doing some practise problems in advance :( i was doing some loop problems, and i'm extremely confused and disheartened and was wondering if anyone could help me?

    The set up is pretty simple:
    - there is a ramp of height, H
    - A block is released from H
    - At the bottom of the ramp there is a loop with radius of 10m, and R<H
    - The block is 1kg

    The question I got stuck on is:
    1. How fast does the block need to be going at the bottom of the ramp so that the acceleration of the block at the top of the loop is 4g?

    - I thought this was simply A=v^2/r, where A=4g, giving me v=(4gr)^(1/2).
    - I put in R=10, my answer is wrong, and I guess it wasn't as simple I thought. :| How should I actually go about this?

    I also don't get this question:
    2. What is the speed of the block as it exits the loop if the normal force at the top of the loop was 80N.
    -I made Fnet=normal force + mg, then substituted to get Fnet=80+(1kg)(9.8N)=898N.
    - Made Fnet=(mv^2)/r, where r=10. Again, my answer is wrong, so am I just oversimplifying these problems? Is there a better way to do them?

    Thanks so much in advance!

    Lori-Anne
     
  2. jcsd
  3. Aug 9, 2006 #2

    rcgldr

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    Kind of a trick question. The block always gets 1g of downwards acceleration from gravity, so the block only needs 3g's of acceleration from the tracks at the top of the loop.

    Update they didn't take the 1g downwards of accleration into account as part of the 4 g's of acceleraion.
    Unless I'm missing something here, you have extra information. Assuming no friction losses, speed exiting the loop is the same as the speed entering the loop.

    Update My mistake, a different problem, 80N at the top is ~8.16g's, different than the 4g's from the first problem.
     
    Last edited: Aug 10, 2006
  4. Aug 9, 2006 #3
    hmm, even if i do v=(3gr)^(1/2), solving for r=10, it still does not match the answer. (i'm not sure if that is what you meant for me to do, however :| ) anyway, the given answer is 28 m/s.

    for question #2, the answer is 39.5m/s.

    thanks for you help; but more feedback would be awesome!
     
  5. Aug 9, 2006 #4
    for #2, there is aditional information and no friction loss. so you are right, speed entering would equal speed exiting, but how do i calculate that speed? my way of doing it is not yielding the right answer.
     
  6. Aug 9, 2006 #5

    rcgldr

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    Don't forget the speed at the bottom of the loop increases by the change in gravitational energy. In this case the block ends up 20 meters lower than at the top of the loop. However I'm not getting good answers either.
     
  7. Aug 10, 2006 #6

    rcgldr

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    If the diameter is 10 meters (radius of 5 meters), and the track applies 4g's of force to the block (for a total of 5 g's downards on the block at the top), then I get 14 m/sec at the top, 14m/sec increase in velocity for a 10 meter decrease in altitude, so 28 m/sec at the bottom, which matches the given answer.

    Solving for #2 also assuming a 5 meter radius, 80N at the top translates into ~8.16g's and a velocity of 20 m/sec. At the bottom of the loop, the 10 meter decrease in altitude results in a 14 meter increase in speed as before, so the answer should be 34m/sec (not 39.5).

    The questions / answers are just plain wrong or worded very poorly.
     
    Last edited: Aug 10, 2006
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