How do Normal Force and Centripetal Acceleration Affect a Ferris Wheel Rider?

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Homework Help Overview

The discussion revolves around the forces acting on a student riding a Ferris wheel, specifically focusing on the normal force and centripetal acceleration at different points of the ride. The subject area includes concepts from mechanics, particularly forces and circular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between normal force and gravitational force at the highest and lowest points of the Ferris wheel. There are attempts to analyze the forces using Newton's second law and centripetal acceleration equations. Questions arise regarding the effects of changing the wheel's speed on the normal force.

Discussion Status

Participants are actively engaging with the problem, attempting to clarify their understanding of the forces involved. Some have provided insights into how the centripetal force changes with speed, while others are revisiting their calculations to ensure accuracy. There is a mix of interpretations regarding the effects of speed on the forces experienced by the rider.

Contextual Notes

There are indications of confusion regarding the application of equations and the interpretation of forces at different points on the Ferris wheel. Participants are also correcting their earlier statements about centripetal acceleration and its implications for the problem.

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Homework Statement



A student of weight 667 N rides a steadily rotating Ferris wheel. At the highest point, the magnitude of the normal force on the student from the seat is 556 N.

a) Does the student feel "light" or "heavy" there?

b) What is the magnitude of Fn at the lowest point?

c) If the wheel's speed is doubled, what is the Fn at the highest point,

d) and at the lowest point?

Homework Equations



Fc = m(v²/R)

Ac = (v²/R)

Fnet = m a



The Attempt at a Solution



Okay this is where I am a bit backwards:

At the top:

Fnet = may
Fn-Fc-mg = may = 0

^^ I think this part is wrong,

I also attempted it like this:

Fn -mg = Fc (Since rotating in a circle, some acceleration due to rotation)

Fn = mg + Fc

570 N = 667 N + Fc

Fc at top = -111 Newtons

So since the seat is being pulled downward, and the normal force is less than the full gravitational force, the student should feel lighter than normal.

at the bottom:

Okay here is where I really get messed up:

Fn -mg = Fc

Fc is now +111 N, still pointing towards the center of the circle.

Fn = mg + Fc = 667 N + 111 N = 778 N.

Now, when the wheel's speed is doubled, how do I calculate the new centripetal force?
 
Last edited:
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***ay = acceleration in the y direction, sorry. ^^
 
***ohhhh wait, hold on, also have equation for Ac wrong, centripetal acceleration should be Ac = v²/R
 
F_{CP}=m\cdot\frac{v^2}{R}
What happens with F when v doubles?
 
Mmm... If (2v)^2 is placed in there, the resulting centripetal force will be four times as large, since the radius is the same for both.
 
So, now I go and repeat the correct calculations, using the equations for Newton's second law where Fnet = Fc.

Thank you!
^.^
 
hagobarcos said:
Mmm... If (2v)^2 is placed in there, the resulting centripetal force will be four times as large, since the radius is the same for both.

Exactly.
 

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