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Vertical Loop the Loop Problem

  1. Dec 19, 2006 #1
    Loop the Loop problem help.

    The Problem says that there is a little car going at an initial speed of 4 m/s while entering a vertical loop the loop. Then all it wants is the largest possible radius of the loop that the car can still remain on the track at all times.
    I don't understand, the speed is constantly changing and all you have is the initial. You can't find the centripital force at the top of the loop because the speed has changed from the 4m/s to get to that point. Anyone please help.!
     
  2. jcsd
  3. Dec 19, 2006 #2
    Loop the Loop Problem

    1. The problem statement, all variables and given/known data
    If a little car is traveling at an initial speed of 4m/s entering a vertical loop the loop, what is the largest possible radius the loop can have which the car remains in contact with the track at all times?


    2. Relevant equations
    The only thing i can think of is (Centripital Force)=(mv^2)/r

    3. The attempt at a solution
    Well all i have are 4 points of the loop the starting(Cf=(mv^2)/r=NormalForce-mg), 1/4 the loop(Cf=(mv^2)/r=Fn), At the top of the loop(Cf=(mv^2)/r= Normal Force+mg), And 3/4 the loop(Cf=(mv^2)/r=Fn).
     
  4. Dec 19, 2006 #3
    1. The problem statement, all variables and given/known data
    If a little car is traveling at an initial speed of 4m/s entering a vertical loop the loop, what is the largest possible radius the loop can have which the car remains in contact with the track at all times?



    2. Relevant equations
    The only thing i can think of is (Centripital Force)=(mv^2)/r


    3. The attempt at a solution
    Well all i have are 4 points of the loop the starting(Cf=(mv^2)/r=NormalForce-mg), 1/4 the loop(Cf=(mv^2)/r=Fn), At the top of the loop(Cf=(mv^2)/r= Normal Force+mg), And 3/4 the loop(Cf=(mv^2)/r=Fn).
     
  5. Dec 19, 2006 #4

    OlderDan

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    If the car can make it to the top of the loop without leaving the track, it can make it the rest of the way. You can prove that if you must be comparing the parabolic path of a projectile that reaches the top of the loop with the circular path of the car when the projectile and the car both have the same speed at the top. What is the smallest possible normal force you can have at the top if the car does in fact stay in contact with the track?
     
  6. Dec 19, 2006 #5
    Well the only normal force that it could have at the top is the weight. But if it was only the weight at that point, there is no actual normal force so for that slight moment it is not in contact with the track.
     
  7. Dec 20, 2006 #6
    I am extremely confused by this problem...
     
  8. Dec 20, 2006 #7

    OlderDan

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    The weight is not the normal force. At the top of the track, the normal force is any force the track exerts on the car in addition to its weight. The least possible normal force is zero. That does not mean the car loses contact with the track, even for an instant. It just means that at that one moment the only force required to keep the car moving in the circular path is the weight of the car. In the next instant, more force will be required because the car will start speeding up and because gravity is not in the right direction to provide all of the centripetal force.
     
  9. Dec 20, 2006 #8
    Oh ok. But since there is no mass and no given Velocity at that point, how can you find anything?
     
  10. Dec 20, 2006 #9
    Can't do forces? How 'bout energy?
     
  11. Dec 20, 2006 #10
    A common way of solving these problems is:
    first, just assume the object has a mass m!
    second, analyze what happens when the car falls off from the track.

    If a car loses contact with the track, what would the normal force be? (yes, you guess it, the normal force would be less than zero!)

    Third, now you got the condition, so what other variables are related to this condition? what is the normal force in terms of other informations (energy, momentum, height, net force...etc)?
     
  12. Dec 20, 2006 #11

    chroot

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    Admin note: I have merged all the threads into one. Please do not post the same question multiple times.

    - Warren
     
  13. Dec 20, 2006 #12

    ShawnD

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    I think the key to this problem is understanding the conditions at the start point and at the very top of the loop. You start with kinetic energy. At the top of the loop you still have kinetic energy (the car doesn't just drop straight down once it reaches the top), maximum potential energy, and your gravity acceleration cannot exceed your centripetal acceleration (or the car will break contact).

    start = final
    (0.5)(m)(v1)^2 = (m)(g)(2)(r) + (0.5)(m)(v2)^2
    m factors out, v1 is start velocity, g is gravity, 2r is your height (in terms of r because r is the final answer), v2 is velocity at the top (which we can calculate)
    sub a few things in to simplify it
    (0.5)(4)^2 = (9.8)(2)(r) + (0.5)(v2)^2
    finding (v2)^2 is done by comparing centripetal acceleration to gravity
    (v2)^2/(r) = 9.8
    (v2)^2 = 9.8r
    Put that back in the original equation
    (0.5)(4)^2 = (9.8)(2)(r) + (0.5)(9.8r)
    Now just tidy it up
    8 = 19.6r + 4.9r
    8 = 24.5r

    r = 0.3265m


    I don't think you can just say it's when centripetal force equals the gravity force without taking into account the energies involved. Doing that does not take into account the way gravity slows the car on the way up the loop. Just take a quick run through the numbers.
    (v^2)/r = 9.8
    (v^2)/9.8 = r
    (4^2)/9.8 = r
    r = 1.6325m
    This means the height would be 3.26m. Is it physically possible for the car to even go that high, let alone keep contact?
    (0.5)(m)(v^2) = mgh
    (0.5)(4^2) = (9.8)(3.26)
    8 = 31.948 :yuck:
    Nope, not possible
     
  14. Dec 20, 2006 #13
    no, they would be in contact. think of it this way; if at the top of the track, the force of gravity is equal to the force of centrip. acceleration then there is no net force. this means there would be no movement in the y direction, which means the car would have to remain in it's original position; in contact with the track.
     
  15. Dec 20, 2006 #14
    hmmm, i got a different answer. it's close, but different.
     
  16. Dec 20, 2006 #15

    ShawnD

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    Is your answer 0.408m? I got that one when I did a simple mistake to start with :tongue:
     
  17. Dec 20, 2006 #16
    nope, 0.306m.
    i have a feeling i made a mistake...
     
  18. Dec 20, 2006 #17
    yup, it was me....little calculator error.
     
  19. Dec 20, 2006 #18
    Thanks Lots guys! You were very helpful!
     
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