Roller Coaster + Centripetal Force

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For a roller coaster car at the top of a vertical loop, the minimum centripetal acceleration required to maintain contact with the track is equal to 1g, where g is the acceleration due to gravity. At this point, the gravitational force provides the necessary centripetal force, meaning no normal force acts on the car. If the car's speed falls below this minimum, it will lose contact with the track. The discussion clarifies that while the normal force typically counteracts gravity, it is absent when the car is just maintaining contact at the loop's peak. Understanding these forces is crucial for analyzing the dynamics of roller coasters.
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A roller coaster car is on a track that forms a circular loop in the vertical plane. If the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its centripetal acceleration at this point?

g downward
g upward
0.5 g downward
2 g upward

Homework Equations


I assumed Normal force points down and so does gravity, therefore the minimum acceleration is 2g... why is it really 1g.
 
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When the car is going at less than the minimum speed, then there will be too much force, and the car will fall from the track. At the minimum speed the centripetal force at the very top will be exactly equal to the force of gravity, mg, because no normal force will be required.
 
I thought Nf was a contact force and didn't depend on speed or anything like that
 
As I understand, normal force is a reaction to the force that an object exerts on the surface it is in contact with. So if an object near the surface of the Earth (i.e. ordinary life) rests on a surface, gravity is pulling it down with a force equal to its weight, mg, and since the object is on a surface and can't fall through the surface, it conveys that force onto the surface it rests on. The surface pushes back with an equal and opposite force, which is what we call the normal force. In the case of the roller coaster, if the object has just enough acceleration to (barely) maintain contact with the rail, then it is not pressing against the rail, and therefore there is no normal force at that position. The only force acting on the object would be the force of gravity. And if the object is to continue moving in a circle, that gravitational force must provide the centripetal force to maintain circular motion.
 
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I thought Nf was a contact force and didn't depend on speed or anything like that

Does the car exert a force on the track when it is at the top? Does the track exert a force on the car?

In the case of uniform circular motion, the net acceleration of the car is equal to the centripetal acceleration. Since at the top there exists only the gravitational force acting on the car, what does this imply the centripetal acceleration must be?

(Note that a normal force could be introduced into the problem. Suppose, for instance, that the wheels of the car were inside a frictionless track such that a contact force was exerted on the car so that it remained along the track at all times.)
 
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