Normal Force at the bottom of a Ferris Wheel

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Homework Help Overview

The problem involves analyzing the forces acting on a student riding a Ferris wheel, specifically focusing on the normal force experienced at the highest and lowest points of the ride. The subject area includes concepts from dynamics and circular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between normal force, weight, and centripetal acceleration at different points on the Ferris wheel. There is an exploration of the equations governing these forces and the implications of the student's weight and the normal force at the top of the wheel.

Discussion Status

Some participants have provided equations relating the forces at the top and bottom of the Ferris wheel, while others express uncertainty about the necessity of certain variables and how to approach solving for them. There is an ongoing examination of the differences between the normal forces at the top and bottom of the ride.

Contextual Notes

Participants note the challenge of having one equation with two unknowns, which raises questions about the assumptions made in the problem setup. There is also a hint suggesting a relationship between the normal forces at the two points, but no resolution has been reached.

AnkhUNC
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Homework Statement



A student of weight 678 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force N on the student from the seat is 565 N. (a) What is the magnitude of N at the lowest point? If the wheel's speed is doubled, what is the magnitude FN at the (b) highest and (c) lowest point?

Homework Equations





The Attempt at a Solution



So M = 678N, NTop = 565N. Fc = mg - Ntop = 6079.4
So Nbottom = Nbottom - mg = 6079.4 which leads Nbottom to = 12723.8 but this is incorrect. Where am I going wrong?
 
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At the top, normal force, weight, and acceleration all point down:
N + mg = mv^2/r; so N = mv^2/r - mg

At the bottom, normal force and acceleration point up, but weight points down:
N - mg = mv^2/r; so N = mv^2/r + mg
 
I really don't need all that though do I? If I do how am I going to solved for v^2 or r? I only have one equation and two unknowns. At best I'd have Ntop+Nbottom = mv^2/r.
 
AnkhUNC said:
I really don't need all that though do I?
Yep. It's the easy way!
If I do how am I going to solved for v^2 or r?
No need to solve for those.

Examining those expressions for N, how does Nbottom compare to Ntop? (Hint: What's Nbottom - Ntop?)
 

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