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Roller coaster - uniform circular motion

  1. Jan 12, 2016 #1
    Hi, I have been studying up on circular motion in the vertical plane and I am happy with all the math/theory that I have read explaining the normal force and weight force and the feeling of being "heavier" and "lighter". My question is about the top of the hill/ride, The forces acting on you at the top of the ride are the weight force pulling you down and the normal reaction force pointing up. Obviously the normal reaction force here is smaller and the normal reason given for this is that it needs to be for the net force to be directed to the centre of the circle providing the centripetal acceleration. My question is, what causes the normal to be smaller than the weight force? My understanding is that the centripetal force is the net force resulting from the weight and normal forces so what is it that causes the normal to be smaller than the weight force? Is it to do with the inertia of the car? Any help would be greatly appreciated.

    Cheers.
     
  2. jcsd
  3. Jan 12, 2016 #2

    PeroK

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    At the top of the circle the normal force on you is downwards. There are no forces pushing you up at that point.

    If you are travelling fast enough the normal force may be larger than your weight at that point.
     
  4. Jan 12, 2016 #3
    Sorry, perhaps I needed to be clearer, I mean the roller coaster going over a hill, or say a car driving over a hill. not going upside down in a loop.
     
  5. Jan 12, 2016 #4

    PeroK

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    If the car is going fast enough, it will leave the surface, and follow a curved path in the air.

    The normal force cannot be greater than gravity on a hill because it's a reaction to gravity.
     
  6. Jan 12, 2016 #5
    Yes i know, my question is, what causes the normal reaction force to be smaller than the weight force?
     
  7. Jan 12, 2016 #6

    A.T.

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    The normal ground reaction force adjusts to whatever is necessary to prevent the car from sinking into the ground. On the hill some of the weight force goes into accelerating the car downwards, so the ground reaction force doesn't have to balance that part.
     
  8. Jan 12, 2016 #7

    PeroK

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    In a word "acceleration". If the car is not accelerating, then the normal force is the weight. But, if the car is accelerating, which happens on a vertically curved road, the normal force will vary according to the direction of the acceleration.
     
  9. Jan 12, 2016 #8
    Ok, can you potentially explain that further? when the car is exactly on top of the hill, if it were stationery then the normal reaction force would be equal to the weight force, yet when moving it is less than the weight force. Doesn't all the weight force go into accelerating the car down?
     
  10. Jan 12, 2016 #9

    PeroK

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    You're thinking: If a car is "sitting" on a flat road, then the normal force must be the weight. The velocity and acceleration of the car are irrelevant.

    To see the problem with that argument consider two cars "sitting" instantaneously on the road. One has, however, just fallen from a height and has hit the road. It is decelerating rapidly and has a much higher normal force at that instant.
     
  11. Jan 12, 2016 #10
    Yep that makes sense to me, the car that has fallen from a height is exerting a greater force on the road and therefore the force the road exerts back is higher. But in the case of the car on the hill are you suggesting that the car's acceleration in the vertical direction is decreasing? and therefore the normal is less?
     
  12. Jan 12, 2016 #11

    PeroK

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    Here's a way to look at it. Imagine the car reaching the top of a hill with different speeds.

    At high speed, the car leaves the surface and the normal force is zero.

    Get the speed just right and the car stays on the surface but with instantaneously 0 normal force.

    Design the track just the right shape - parabola - and the car will stay in contact with the track but exert zero normal force. The track matches the natural parabolic path of the car under gravity at that velocity.

    Slow the car down and it will try to take a different parabolic path. The car would like to go below the road, but the solid surface prevents it. A normal force is created, keeping the car in a forced path.

    The slower the car travels, the greater the difference between the natural curved path it wants to take and the path along the track it is forced to take. Hence the greater the normal force to keep the car on the track.
     
  13. Jan 12, 2016 #12

    A.T.

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    No, because the car doesn't accelerate down at 1g, but less than that, depending on it's speed and curvature of the hill.
     
  14. Jan 12, 2016 #13

    sophiecentaur

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    I can't find a comment in this thread about the (very relevant) sign of the curvature of the track. When the car is going over a 'hill' the curvature is negative so the centripetal force will be of opposite sign . In a valley, thr curvature is positive so the centripetal force will be positive. The net normal force on the car can be zero, on a hump, at which point. the car will no longer follow the track and 'fly off' on a parabolic trajectory. (The design of a real ride would never allow that - with a pretty wide margin, I'd bet)
     
  15. Jan 12, 2016 #14

    A.T.

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    Did you mean "negative too" rather than "opposite sign"?
     
  16. Jan 12, 2016 #15

    sophiecentaur

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    I just meant that the sign of the force is diferent for the valley and for the top of the hill. It can add or subtract from the 'g force' that's experienced.
     
  17. Jan 12, 2016 #16

    Yes, this all makes sense conceptually (really well explained) and I am more than happy that the maths agrees nicely, my question is what is it that causes the normal force to be less than the weight force at the top of the hill? I understand that if the car is travelling slowly more normal force will exist because the car wants to head into the surface, but how would i finish this sentence? When the car is travelling quicker the normal reaction force gets smaller because .....
    Thanks heaps for all the help. I'm having real trouble with normal reaction forces in situations like this.
     
  18. Jan 12, 2016 #17

    A.T.

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    Read up on constraint forces, which adjust to whatever is needed to enforce certain kinematics. On the atomic level the adjustment is varying amount of EM repulsion, depending on the distance between atoms.
     
  19. Jan 13, 2016 #18

    PeroK

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    ... At that point I'd be tempted to trust the mathematics. This situation at the top of the hill exists only instantaneously and even then only in an idealised fashion. It's because of the centripetal acceleration and the curvature of the slope. More of gravity is deployed making the car change direction and less is left to push it into the track.

    An easier situation is your weight on an elevator accelerating down. As long as the elevator accelerates the normal force is reduced by precisely the elevator's acceleration times your mass.

    Centripetal acceleration is also acceleration. If that's isn't easy to grasp, then trust the mathematics.
     
  20. Jan 13, 2016 #19

    A.T.

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    Well, he agrees that it makes sense conceptually and mathematically. So I don't really know what the remaining issue is.
     
  21. Jan 13, 2016 #20
    ... the downwards acceleration is greater so the net downwards force must be greater; the net downwards force is the weight (which is constant) less the normal reaction force so in order for the net downwards force to increase the normal reaction force must decrease.
     
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