Roller Coaster Dynamics: Forces at the Top and Bottom of the Track

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Homework Help Overview

The discussion focuses on the dynamics of forces experienced by a roller coaster or train at the top and bottom of the tracks, particularly in relation to the concepts of acceleration and normal force as described by F=MA.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the differences in forces at the top versus the bottom of the roller coaster track, questioning how these relate to the normal force and the sensation of weightlessness. There are attempts to set up equations involving mass, gravity, and acceleration to analyze the forces acting on a rider.

Discussion Status

The discussion is active, with participants sharing insights about the direction of acceleration and the normal force at different points on the track. Some guidance has been provided regarding the need to consider the direction of acceleration when applying the equations.

Contextual Notes

Participants are working within the constraints of a specific problem involving an 80 kg person and are trying to understand the implications of their calculations for forces at the top and bottom of the roller coaster track.

Jason03
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Im trying to figure when you have a rollercoaster or train, how are the forces different at the top of tracks compared to the bottom of the tracks.

How does this relate to setting up F=MA...and the Normal Force.

I know at the top you would have a feeling of weightlessness and at the bottom you are pushed into your seat greatly.

Thanks
 
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It doesn't relate to any real forces, it relates to the acceleration of the rider. At the top his acceleration is down, at the bottom it's up. The normal force has to balance both gravity and the acceleration of the rider.
 
ok, I am setting up a problem that asks for the force exerted on the seat of an 80 kg person. At the top, and at the bottom


So I am using this equ. ( i have r,v, m and g)

F=MA

N-mg = m (v^2/r)


so what would differ for the top and bottom?
 
The direction of the v^2/r. It's directed towards the center of rotation. At the top, it's down. At the bottom, it's up. You have to keep track of the direction of v^2/r by hand.
 
ok thanks...i will give that a try
 

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