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How heavy does a roller coaster rider feel?

  • Thread starter Walley1
  • Start date
  • #1
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Homework Statement


A roller coaster rider has a mass of 80 kg and is riding in the coaster shown in the figure. If the vehicle has a speed of 10.0 m/s at the top of the first hill (assume same height as the top of the circle above Point A):
a) how "heavy" does the rider feel at point A?
b) how heavy does the rider feel at point B?
http://wildedgedesigns.com/RollerCoaster.PNG [Broken]

Homework Equations


EK=(mv^2)/2
EP=mgh
F=ma
Ac=V^2/r


The Attempt at a Solution


Potential Energy at the top of the first hill: 80 * 9.8 * 20 = 15,680J
Kinetic energy at the top of the first hill: 0.5 * 80 * 10^2 = 4,000J
Total energy: 19,680J
At the bottom of the first hill, all that energy is kinetic, so:
0.5 * 80 * v^2 = 19,680J
v = 22.18m/s
Ac = 22.18^2/10 = 49.20m/s^2
F = ma = 80 * 49.20 = 3,936N - That's the answer to Part A
But then there's an issue. The potential energy at point B is mgh = (80)(9.8)(30) = 23,520J. That's more energy than the system has. I don't know how the rider is going to make it to point B without falling back down, so I certainly don't know how heavy he feels. Is my approach to solving this totally incorrect?
 
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Answers and Replies

  • #2
cepheid
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Umm, so implicitly in your calculations, you decided that potential energy should be measured from point A (i.e. h = 0 at point A). That's totally fine. However, assuming that the centres of the two circles are at the same height, then point B is only 25 m above point A. So, maybe you want to re-evaluate your computation of the potential energy at point B.

EDIT: *Regardless* of the choice of reference point (h = 0), it's always true that point B is 25 m above point A, which means your computation of the change in potential energy from A to B is wrong. I just wanted to clarify that the result doesn't depend on your arbitrary choice of a zero point.
 
  • #3
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Thanks! I'm pretty thick sometimes.
 
  • #4
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One more question: I should still factor in his weight due to gravity, correct? That is to say, since he weights 784 N, I should add that to how heavy he feels. No?
 
  • #5
cepheid
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Science Advisor
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One more question: I should still factor in his weight due to gravity, correct? That is to say, since he weights 784 N, I should add that to how heavy he feels. No?
Yeah, you have to add these two, but that's not true at every point in the circle, only at this lowest point. To see why, draw a free body diagram. Only two forces act on the coaster: 1. gravity and 2. the normal force from the track. The "centripetal force" is a kinematic requirement: i.e. if the object is in circular motion, then combination of the physical forces that are present in this situation must be providing it. In this case, gravity is downward, whereas the centripetal force would be directed upward (towards the centre of the circle). So gravity is no help in providing the centripetal force, it falls on the normal force to do so. The NET force must be the required centripetal force, which tells you that normal force (upward) - weight (downward) = Fc. Or, Normal force = weight + centripetal force.

How heavy you *feel* depends on the normal force, in other words, it depends on how hard the floor pushes up on you. In this case, if the car were just sitting at the bottom of the hill (point A), all the normal force would have to do is support it against gravity. So it would just be pushing up with a force equal to the weight of the car, and things would not feel heavier than usual. However, in this situation, the normal force not only has to support the weight of the car, it also has to provide the additional upward force required to keep the thing moving in its circular path. So the floor of the car pushes up on the rider with a force greater than his weight (greater than what he usually feels) and therefore he feels heavier.
 

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