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

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SUMMARY

The Gravitron roller coaster problem involves calculating the normal force experienced by riders at the bottom of a loop-the-loop. When riders feel weightless at the top, the normal force is zero, leading to the equation g = ac. At the bottom, the correct equation is N - Mg = M(ac), where ac is greater than g due to increased speed. The final conclusion is that the normal force at the bottom is N = 3Mg, not 2Mg as initially calculated.

PREREQUISITES
  • Understanding of centripetal acceleration (ac) and gravitational force (g)
  • Familiarity with Newton's second law of motion
  • Knowledge of forces acting on objects in circular motion
  • Basic principles of energy conservation in mechanical systems
NEXT STEPS
  • Study the relationship between centripetal acceleration and velocity in circular motion
  • Learn about the work-energy theorem and its application in roller coaster dynamics
  • Explore the effects of friction and air resistance on motion in loops
  • Investigate the physics of roller coasters, focusing on energy transformations
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of roller coasters and circular motion in mechanics.

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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