Circular Motion: Ferris Wheel Dynamics

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Homework Help Overview

The discussion revolves around the dynamics of a Ferris wheel, specifically focusing on the apparent weight of a passenger at different points during the ride. The problem involves concepts from circular motion and forces acting on the passenger.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between apparent weight at the highest and lowest points of the Ferris wheel, questioning how to express this difference as a fraction of the passenger's weight. They discuss the equations governing the forces at play and express uncertainty about the implications of their calculations.

Discussion Status

Some participants have made progress on parts A and B of the problem, while others are seeking clarification on part C. There is an ongoing exploration of how to handle mass in the equations, with suggestions made to express results in terms of the weight of the passenger. The discussion reflects a mix of interpretations and attempts to clarify the problem's requirements.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on understanding the concept of apparent weight and its dependence on the forces involved in circular motion.

Mehta29
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I'm on the right track, but I'm stuck here...

The radius of a Ferris wheel is 5 m and it makes one rev in 10 sec

a Find the difference b/w the apparent weight of a passenger at the highest and lowest points, expressed as a fraction of his weight, W

b What would the time for one rev be if the apparent weight at the top were zero?
c What would be the apparent weight at the low point??

I have at the top that

mv^2/r = mg - Fn

and at the bottom

mv^2/r = Fn -mg

I really don't undestand what a is asking...(Fnbottom - Fntop)/ W ??
 
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Mehta29 said:
I really don't undestand what a is asking...(Fnbottom - Fntop)/ W ??
Yes you do. It's exactly what you said. Write your answer as Fnb - Fnt = _______W
 
but I am confused as if anything else would be needed...liek that blank befor e the W...would i need to expand any further or would i just keep it
Fnbottom - Fntop = xW

b would just be v^2/r = g? and then t = 2pir/v

and c would be Fnb = mv^2/r + mg...but how would i eliminate m?
 
Last edited:
Mehta29 said:
but I am confused as if anything else would be needed...liek that blank befor e the W...would i need to expand any further or would i just keep it
Fnbottom - Fntop = xW

b would just be v^2/r = g? and then t = 2pir/v

and c would be Fnb = mv^2/r + mg...but how would i eliminate m?
Solve your earlier top and bottom equations for Fn. Take the difference between the two. Your result will be of the form
Fnb - Fnt = mA where A is a number that can be computed from the given information. You can do that. Once you have that form, multiply and divide by g
mA = mgA/g = WA/g
You can calculate A/g to express your answer as
Fnb - Fnt = (A/g)W with A/g replaced by a number.
 
i got A and B...but how would i manage part c?

im not seein anyway to cancel out the mass...
 
Mehta29 said:
i got A and B...but how would i manage part c?

im not seein anyway to cancel out the mass...
What is your understanding of "apparent weight". All the forces in the problem are proportional to mass. It will cancel out.
 

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