Loop de loop normal force roller coaster

AI Thread Summary
The discussion focuses on the forces acting on a roller coaster during a loop-de-loop, specifically normal force, gravitational force, and centripetal force. The equations provided for the bottom, side, and top of the loop are validated, confirming that the normal force is greatest at the bottom and least at the top. It is clarified that the motion of the cart, rather than an upward force, prevents it from falling off the track at the top of the loop. Additionally, while the normal force provides centripetal acceleration, it is not the only force acting on the coaster. The relationship between centripetal acceleration and linear acceleration is noted as not fundamentally related.
johnj7
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hello, I'm trying to study for the mcat, and I have a conceptual question about normal force, mg, and centripetal force during a loop-de-loop on a roller coaster.

Could you validate these force equations?

1. At the very bottom of the loop:
N - mg = ma = mv^2 / r
N = mg + ma

2. At the side of the loop:
N = ma = mv^2 / r
the normal force is providing all of the centripetal acceleration

3. At the very top of the loop:
N + mg = ma = mv^2 / r
N = ma - mg

Thus the normal force would be the greatest at the bottom of the loop, and least at the top of the loop.

Is all of this correct?

also, for the very top of the loop, since normal force and weight are directed downward, what force prevents the cart from just dropping off the tracts?
in relation to the previous question, what is happening on the side of the loop?

thank you very much !
 
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johnj7 said:
hello, I'm trying to study for the mcat, and I have a conceptual question about normal force, mg, and centripetal force during a loop-de-loop on a roller coaster.

Could you validate these force equations?

1. At the very bottom of the loop:
N - mg = ma = mv^2 / r
N = mg + ma
Good.

2. At the side of the loop:
N = ma = mv^2 / r
the normal force is providing all of the centripetal acceleration
Yes, the normal force provides the centripetal acceleration. But it's not the only force acting on the coaster. So N = ma_c ≠ ma

3. At the very top of the loop:
N + mg = ma = mv^2 / r
N = ma - mg
Good.

Thus the normal force would be the greatest at the bottom of the loop, and least at the top of the loop.
Good.

Is all of this correct?

also, for the very top of the loop, since normal force and weight are directed downward, what force prevents the cart from just dropping off the tracts?
The fact that it's moving is what prevents the cart from falling off, not any upward force. (Just like when you toss a ball in the air. What force makes the ball rise? :wink:)
in relation to the previous question, what is happening on the side of the loop?
Same basic idea, but see my note above regarding #2.
 
Hmm, I don't think its highly important, but when the roller coaster is on the side, and the Normal force is not the only force acting.. what would the force equations look like?

ie:

Fy : N = ma_c
Fx : mg = ma

??

also, are centripetal acceleration and linear acceleration related?

if so, how? what variables would I need?

thank you very much!
 
johnj7 said:
Hmm, I don't think its highly important, but when the roller coaster is on the side, and the Normal force is not the only force acting.. what would the force equations look like?

ie:

Fy : N = ma_c
Fx : mg = ma
Good. (Where x is vertical and y is horizontal.)
also, are centripetal acceleration and linear acceleration related?
Not fundamentally.
 
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