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Roller Coaster Loop

  1. Jan 17, 2004 #1
    Hi, part my project involves designing a vertical loop on a roller coaster that is not to exceed 5 G. The weight of the coaster is 4.5 x 10^4 N. But, I don't know how high I need to start the coaster so that there is enough mechanical energy to make it through the loop. I read online that the coaster must start 1/2R taller than the loop, but that always generates 6Gs for some reason, which is too much. My question is, how do I calculate the initial velocity required to get through a loop of R radius? The part that confuses me is that I don't know how to handle gravity in a vertical loop.
     
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  3. Jan 18, 2004 #2

    HallsofIvy

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    Gravity in a vertical loop is pretty much the same as gravity anywhere!
    I don't know where you got that "(1/2)R taller". Perhaps that's allowing for friction. If you ignore friction, then "conservation of energy" gives the answer. It should be pretty easy to see that the car will go back up to its original height. You can calculate the speed the car must have at the bottom: since the top of the loop is 2R high, it has potential energy 2RMg. At the bottom the potential energy would be 0, of course, so the kinetic energy must be (1/2)M v2= 2RMg. As usual in such problems M cancels out and we have v= 2&sqrt;(Rg). That would be the speed necessary for the car to just make it to the top of the loop. At any speed greater than that, the car will go around the loop.

    Of course, in a real situation, you will have to use a higher speed to allow for energy lost through friction. That "(1/2)R taller" looks pretty good to me. How did you arrive at the "6G"s figure? By experiment? Seems to me the best way to arrive at the correct speed and height is to experiment with different heights between 2R (just enough to get to the top- not enough to make it around) and (5/2)R (too much apparently) and see what happens.
     
  4. Jan 18, 2004 #3
    Thanks, that helps a lot. I got the 1/2R information at http://physics.about.com/cs/gravit1/a/010703.htm, so I assume the information to be valid. His calculations assume a frictionless track as well, so maybe you can shed some light on why he arrived at 1/2R.
     
  5. Jan 19, 2004 #4

    Doc Al

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    The "1/2R taller" represents the additional (potential) energy you need to keep the car pressed against the track at the top of the loop. Without sufficient speed the car will lose contact with the track. (Assuming the car is not attached to the track.)

    To calculate this, apply F=ma at the top of the loop. Set the normal force equal to zero to find the minimum speed to maintain contact:
    mg = mv2/R
    This implies that the KE at the top of the loop must equal mgR/2; thus the car must start at a point R/2 higher than the top of the loop.
     
  6. Jan 20, 2004 #5
    Ok, I understand now. So in reality, starting at a height equal to the height of the loop isn't enough to completely make it through the loop. That will only get you to the top of the loop. In order not to fall once you get to the top of the loop, you need to start R/2 higher than the loop. The 6g figure was also a miscalculation, so ignore that... Thanks for the clarification.
     
    Last edited: Jan 20, 2004
  7. Feb 10, 2004 #6
    hay, on a similar subject, im doin maths A2 at the min and were jus startin coursework on roller coasters - the relation between the hight you start at and the radius of the loop (ie how high you need to be to get round a loop radius R) The equations are fine, but as part of it we need some real figures.. dont suppose any1 knows of any good sites t get hight and loop radius' from..?
     
  8. Feb 10, 2004 #7

    HallsofIvy

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    Ah! In order to be pressed up against the track with a force of one gravity, the starting point has to be 1/2 R taller. Exactly the same height will get you through the top of the loop with "zero gravity" and anything larger will press you to the rail with some force.
     
  9. Feb 10, 2004 #8

    Doc Al

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    No, that's not right. There is a minimum speed required to maintain contact (non-zero normal force) with the track. That minimum speed is attained when the starting point is 1/2 R above the top of the loop. Anything less and you won't make it to the top of the loop.
     
  10. Feb 11, 2004 #9

    HallsofIvy

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    Odd. About 99% of the time I agree with your posts and then you come up with something I consider outrageous (which means I come up with something you consider outrageous, of course).

    Neglecting friction, if the car starts from exactly the same height as the loop, it will reach the top of the loop with 0 speed (conservation of energy). Starting any slight distance above the height of the loop, it will have some non-zero kinetic energy at the top of the loop and will continue through.

    Why is that not correct?
     
  11. Feb 11, 2004 #10

    Doc Al

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    Well, I guess that means you are correct 99% of the time. Not bad!
    Conservation of energy holds, of course, but it does not allow you to predict the height that the car will reach. Remember that the track affects the car's direction.

    An analogy: I roll a ball down a frictionless hill of height "h". At the bottom of the hill is a ski jump which shoots the ball into the air. While it's true that if the ball flies to a height "h" it will get there with 0 speed, you can't know if it will get there without looking at the details of the ramp. (In fact for any angle other than straight up, it won't get there.)

    Since the car is traveling underneath the track as it goes through the loop, at any point that the track ceases to provide a normal force the car will leave the track and become a projectile.

    Consider the forces on the car at the top of the loop (underneath the track!): mg (down) and any normal force (N) also down. If the car is still on the track with some speed V, then its acceleration is V2/R downward. Thus to have the normal force just go to 0 will require: mg = mV2/R

    From this you can conclude the the KE of the car must be:
    mV2/2 = mgR/2

    This is the minimum KE that the car must have to just maintain contact with the track at the top of the loop.
     
  12. Mar 8, 2004 #11
    Hay, am i invisible???!?! does anyone know where i can find rollercoaster specs like loop radius etc? i really need this help, caus ive looked all over and i cant find anything.
     
  13. Dec 5, 2009 #12
    Hi to everybody!!

    Hey I have to do a physics proyect about building a roller coaster and I´m trying to figure out how different does the high of my first loops have to be with respect of the first hill of my roller coaster...

    I saw some of your formulas, and I´m new in the physics world so can anybody please tell me
    what does R means?? in 1/2R for example... what do you mean with R??? :S :confused:
    i really hope u could help...
     
  14. Dec 5, 2009 #13
    R is the radius of the loop. The energy supplied is that from gravity which equals
    mass x gravity x height. As the car accelerates downhill, some of the potential energy becomes kinetic which is given by 0.5x mass x velocity squared.

    So if no friction is present we can calculate the velocity at any point by knowing how much lower the car is than its starting point.

    The other consideration is loop de loops. To maintain contact with the tracks requires sufficient speed which is what the conversation above is about. This help?
     
  15. Dec 7, 2009 #14
    Yes!! I think that´s woking fine for now..
    hey if i have doubts can i ask you about them? about formulas and all that stuff?
     
  16. Dec 7, 2009 #15
    sure, just post whatever on this thread, may be a couple of days before I get back to ya tho.
     
  17. Dec 7, 2009 #16
    thnx!!!! :smile:
     
  18. May 26, 2010 #17
    Hey guys I have an assignment and I need to design a roller coaster with banked corners, a loop, and a hill.

    Anyway I'm having trouble finding a radius for my loop the velocity going in to the loop is 39.61817765m/s

    formulae:

    ac{centripetal accelerational} = (v{velocity}2)/r{radius}

    thanks guys:biggrin:
     
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