Force Vectors and Circular Motion

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SUMMARY

The discussion focuses on the dynamics of forces acting on a person at the top of a ferris wheel, specifically addressing the concepts of gravitational force (Fg), centripetal force (Fc), and normal force (Fn). It is established that at the top of the ferris wheel, the acceleration is not zero; rather, the centripetal force required for circular motion is provided by gravity, leading to a scenario where the normal force balances part of the weight. The correct relationship is expressed as the unbalanced force being equal to the difference between gravitational force and normal force (mg - N), clarifying the interaction of these forces in perfect circular motion.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of centripetal force and its calculation (Fc = mv²/r)
  • Familiarity with vector components of forces
  • Basic principles of circular motion
NEXT STEPS
  • Study the derivation of centripetal force in circular motion
  • Explore the concept of normal force in varying gravitational contexts
  • Learn about the effects of acceleration on forces in circular motion
  • Investigate real-world applications of circular motion dynamics, such as roller coasters
USEFUL FOR

Physics students, educators, and anyone interested in understanding the mechanics of forces in circular motion, particularly in scenarios involving gravitational effects and normal force interactions.

Ronnin
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This is a case where I know the results but am fighting the math. At the top of a ferris wheel your acceleration in Y is 0. Ol' Newton says all all forces then must cancel, got that. I know my weight at this point will feel less at this point. Here is where I am troubled, vector for gravity points down so does the vector for the circle's acceleration but my Normal points up and must have the same magnitude as Fg and Fc. The answer given subtracts Fc from Fg to get Fn (which feels right, but contradicts my vectors). Any conceptual help is appreciated.
 
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Your acceleration at the top of a ferris wheel is NOT zero. Assuming you are undergoing perfect circular motion, you will have a centripetal force at the top. Gravity will supply the centripetal force, but all of the weight will be too much centripetal force, so therefore the normal force will balance some of the weight.

In perfect circular motion, the only unbalancd force will be centripetal and equal (mv^2)/r . At the top of the ferris wheel the unbalanced force is equal to mg-N.
 

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