Easy but hard problem (loop-the-loop) normal force?

AI Thread Summary
In the discussion about the Gravitron roller coaster's loop-the-loop, the focus is on calculating the normal force experienced by riders at the bottom of the loop. When riders feel weightless at the top, the normal force is zero, leading to the conclusion that centripetal acceleration equals gravitational acceleration (g). However, at the bottom of the loop, the car's speed increases, resulting in a higher centripetal acceleration. The correct relationship shows that the normal force at the bottom is actually twice the weight of the riders, expressed as 2mg, correcting the initial miscalculation. The discussion emphasizes the importance of considering varying speeds and forces at different points in the loop.
daivinhtran
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Homework Statement



The Gravitron single-car roller coaster consists of a single loop-the-loop. The car is initially pushed, giving it just the right mechanical energy so the riders on the coaster will feel "weightless" when they pass through the top of the circular arc. How heavy will they feel when they pass through the bottom of the arc (that is, what is the normal force pressing up on them when they are at the bottom of the loop)? Express the answer as a multiple of mg (their actual weight). Assume any effects of friction or of air resistance are negligible.

Homework Equations



Fc= M(ac)
N + Mg = M(ac) ( at the top)
N - Mg = M(ac) (at the bottom)

The Attempt at a Solution



N + Mg = M(ac)
because the riders feel weightless, the normal force is zero
==>>> g = ac

So at the botoom
N - Mg = M(ac)
N -Mg = Mg
N = 2Mg <==== it's a wrong answer :(

(however the problem is in Work-energy chapters, so I think it has to deal with that)
I solved it without any energy involved
 
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daivinhtran said:

Homework Statement



The Gravitron single-car roller coaster consists of a single loop-the-loop. The car is initially pushed, giving it just the right mechanical energy so the riders on the coaster will feel "weightless" when they pass through the top of the circular arc. How heavy will they feel when they pass through the bottom of the arc (that is, what is the normal force pressing up on them when they are at the bottom of the loop)? Express the answer as a multiple of mg (their actual weight). Assume any effects of friction or of air resistance are negligible.

Homework Equations



Fc= M(ac)
N + Mg = M(ac) ( at the top)
N - Mg = M(ac) (at the bottom)

The Attempt at a Solution



N + Mg = M(ac)
because the riders feel weightless, the normal force is zero
==>>> g = ac

So at the bottom
N - Mg = M(ac)
N -Mg = Mg
N = 2Mg <==== it's a wrong answer :(

(however the problem is in Work-energy chapters, so I think it has to deal with that)
I solved it without any energy involved
It's correct that the centripetal acceleration, ac, is equal to g (and is downward) at the top of the loop. However, ac does not have the same value at the bottom of the loop, because the roller coaster car is moving much faster at the bottom.
 
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