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Roller Coaster Physics

  1. Jul 10, 2013 #1
    Hello,

    When doing simple physics problems involving roller coasters, I'm having trouble finding the centripetal force/acceleration.

    When the object is at the bottom of the loop, I understand the centripetal acceleration relies on the normal force.

    However, when the object is at the top of the loop, why does it also rely on the normal force? Doesn't the gravity vector and normal vector point in the same direction?

    Please help, I am completely confused, and can't find any good explanations on this for intuition on the internet.

    Thanks.
     
  2. jcsd
  3. Jul 10, 2013 #2

    Simon Bridge

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    That is correct.
    Try it out though - tie something to a string and twirl fast it in a vertical circle ... notice how the string tugs on your hand even at the top of the circle?
    In this experiment, the tension in the string plays the role of the normal force.
    There should be a speed where the normal force is zero - and speed below that won't make a full circle.

    Note: gravity contributes to the centripetal force all the time too.
    The contribution due to gravity is the radial component - there is also a tangential component which opposes the upwards part of the circle and favors the downwards part.
     
  4. Jul 11, 2013 #3
    Thanks for the reply.

    Would this mean that when solving for the centripetal force at the top of the ferris wheel, we use:

    [itex] F_g+N = m*a_c[/itex] ?

    Also, how would you easily solve for [itex]a_c[/itex]?
     
  5. Jul 11, 2013 #4

    Simon Bridge

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    Ferris wheels are a bit trickier in that they generally suspend the passengers so they can go more slowly around the circle... they also typically move at a constant angular velocity. You can calculate the centripetal acceleration from that.

    The typical question will ask about the slowest speed that will get a toy car around a loop of track.
    Since N depends on speed, this will happen for the speed where N=0.
     
  6. Jul 12, 2013 #5

    CWatters

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    Roller coasters and ferris wheels are rigid (the object is forced to follow a circular path radius r with velocity v) so the centripetal acceleration must be v2/r at all times.

    The sum of all forces acting on the object must provide that acceleration. What changes between top and bottom is the balance of the forces that provides the acceleration:

    At the top gravity acts in the right direction to accelerate the object towards the centre so the normal force must reduce so the sum is constant.

    At the bottom gravity acts in the wrong direction to accelerate the object so the normal force must increase so the sum is constant.

    The exact balance between the forces can to some extent be chosen by the roller coaster/ferris wheel designer. On most ferris wheels the cabin/chairs are suspended and the wheel turns very slowly. At the top gravity provides more than v2/r on it's own. In which case the normal force must be negative (eg it must act outwards). Since the car is suspended the people inside experience a normal force that it in the same direction as that which they experience standing on the ground.
     
  7. Jul 12, 2013 #6
    Thanks for the amazing reply! Just one quick question: if instead of a roller coaster, this was a ferris wheel, what would be the force at the top? The normal force acts outward, so would it just be gravity? Thanks!
     
  8. Jul 12, 2013 #7

    jbriggs444

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    The previous response by CWatters covers this.

    You know that the acceleration of a person sitting in a car on the Ferris wheel is v2/r. This acceleration must be the result of all the forces acting on the person. The forces acting on the person are the downward force of gravity and the force from the Ferris wheel.

    If the normal force from the Ferris wheel on the person were equal to gravity then the net force would be zero. That would mean that the person would not be accelerating. So the normal force cannot just be gravity. Instead, the normal force must reflect the difference between the acceleration that gravity alone would produce and the actual acceleration that the person experiences.

    Adopting a sign convention where downward is positive...

    F = ma
    Fgravity - Fnormal = ma
    mg - Fnormal = mv2/r
    - Fnormal = mv2/r - mg
    Fnormal = mg - mv2/r
    Fnormal = m ( g - v2/r )

    Where m is the mass of the person, g is the acceleration of gravity, v is the tangential velocity of the Ferris wheel, r is the radius of the Ferris wheel and Fnormal is the normal force of the Ferris wheel supporting the person at the top of the rotation.
     
  9. Jul 12, 2013 #8
    Thanks! So this means that the downward force of gravity must be GREATER than the normal force, to ensure that there is a centripetal force.
     
  10. Jul 12, 2013 #9

    jbriggs444

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    If the [upward] normal force were greater than gravity then there would still be a net centripetal force. It's just that the trajectory of the object would then be curving upward rather than downward. The momentary center of curvature would be above rather than below.
     
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