Roller coaster circular motion normal forces

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SUMMARY

The discussion focuses on calculating the normal force acting on a roller-coaster car with a mass of 1200 kg as it passes over the top of a circular hill with a radius of 17 m. For a speed of 8.2 m/s, the normal force (FN) is calculated to be -4746.35 N, while at a speed of 17 m/s, FN is -20,400 N. The calculations utilize the formula FN = m(-v²/R), emphasizing the relationship between speed, radius, and the forces acting on the car. The participants express uncertainty about the correctness of these results, prompting further analysis of the forces involved.

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Homework Statement


A roller-coaster car has a mass of 1200 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 17 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 8.2 m/s? (b) What is FN if v = 17 m/s?

Homework Equations


-Fn= m (-v^2 / R)

The Attempt at a Solution



A) 1200(- 8.2^2/ 17)
-Fn= -4746.352941

B) 1200(- 17^2/ 17)
-Fn= -20,400 N
do not believe these answers are correct, but cannt determine any other way to solve, thanks for the help
 
Last edited:
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Consider the forces acting on the car at the top of the loop. The normal reaction,R, upwards and the weight,W. The resultant of theses 2 provide the centripetal force needed to keep the car on the track. So that resultant force=Weight-Normal Reaction.
 

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