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Speeds on a roller coaster

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data

    The radius of curvature of a loop-the-loop for a roller coaster is 14.0 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.35mg. Find the speed of the roller coaster at the top of the loop. (Enter your answers from smallest to largest.)

    (This question has two answers, one of which, I found = the speed at the top of the loop. I'm guessing the 2nd answer is the speed at the bottom of the loop?)


    2. Relevant equations

    ∑Fr = mar
    Fsp + mg = m(v2/R)

    a = v2/R


    3. The attempt at a solution

    Fsp = m[(v2/R) - g]
    .35mg = m[(v2/R) - g]
    .35g = (v2/R) - g
    1.25g = v2/R
    v_top = √(gR1.35) ≈ 13.62 m/s

    Now to find the 2nd answer, I think I need to calculate a (since I believe this is constant):

    a = v2/R ≈ 13.25 m/s2

    But this can't be correct I'll just get the same answer for velocity.

    Does anyone know how I am supposed to find the 2nd answer?
     
  2. jcsd
  3. Oct 7, 2012 #2

    PhanthomJay

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    1. Acceleration is not constant.
    2. At the top , Fsp and mg are in the same direction. Not so at the bottom.
    3. Try conservation of energy approach.
     
  4. Oct 7, 2012 #3
    We haven't covered conservation of energy yet (we've just finished Newton's laws, friction, uniform circular motion, drag forces, and an intro to work and dot-products).

    I sum the forces at the bottom of the roller coaster:

    ƩF = Fsp - mg = mv2/R

    ∴ Fsp = m[(v2/R) + g]

    I don't know the mass or the velocity though and I don't have enough information to get another equation to solve for either, do I?
     
  5. Oct 7, 2012 #4

    PhanthomJay

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    It seems that you assumed what the question was for the 2nd answer. Have you written down the problem statement correctly?
     
  6. Oct 7, 2012 #5
    Yes. I copied & pasted the problem:

    The "largest" value is the one I found above (≈13.62 m/s).
     
  7. Oct 7, 2012 #6

    PhanthomJay

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    How do you know it is the largest?
     
  8. Oct 7, 2012 #7
    When I entered the answer (we use WebAssign for our homework), it showed as correct in the space that said "largest value" :).
     
  9. Oct 7, 2012 #8
    After getting more information from my teacher, here is the solution for the "smallest" value, i.e. the speed at the bottom of the roller coaster:

    Assume Fsp = .35mg at the bottom of the roller coaster as well the top.

    Fsp - mg = mv2/R
    .35g - g = v2/R
    √|R(.35g - g)| = v

    The question is worded very poorly. Thank you much for all of your assistance, PhanthomJay.
     
  10. Oct 7, 2012 #9

    PhanthomJay

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    Well, that gives the sq rt of a negative number, so I don't think the question asks anything about speed at the bottom. Now last time I rode the loop the loop coaster, I was upside down at the top, the car below the rails. On a 'regular' roller coaster, you are right side up at the top, the car above the rails. In the first case, the normal force and weight both act down. In the second case, the weight acts down but the normal force acts up. You might want to try (g - .35g)R = v^2, and see if webassign gives you a gold star. But something is not right.
     
  11. Oct 7, 2012 #10
    What I wrote above included | & | to indicate taking the absolute value and gave the correct result (at least the answer that the website was looking for).

    I'm not seeing how I can manipulate the sum of the forces so it would be R(g - .35g).
     
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