Solving for Maximum Speed on a Theme Park Ride: Phys Unit 4 - Swings Question 6

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Homework Help Overview

The discussion revolves around a physics problem related to the dynamics of a theme park ride, specifically focusing on the maximum speed and g-forces experienced by riders at the bottom of a loop. The problem involves concepts from mechanics, including gravitational potential energy, kinetic energy, and forces acting on the rider.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between gravitational potential energy and kinetic energy, and how to calculate the maximum speed at the bottom of the loop. Questions arise regarding the forces acting on the rider, particularly the normal force from the seat and its role in determining g-forces.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations of the forces involved and discussing the implications of changing the ride's dimensions on g-forces. Some guidance has been provided regarding the role of radius in the calculations, leading to further exploration of the relationship between speed and g-forces.

Contextual Notes

There is a specific focus on the implications of changing the radius of the ride and how it affects the g-forces experienced by riders. Participants are referencing a follow-up question that requires them to analyze the statements made by two students regarding the effects of ride size on g-forces.

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Homework Statement


Question 6:
http://www.xtremepapers.com/Edexcel/Advanced%20Level/Physics/2008%20Jan/6754_01_que_20080121.pdf

"On one type of theme park ride, a..."


Homework Equations





The Attempt at a Solution


So at the bottom of the loop the ride is at its fastest
GPE = KE
mgh = 0.5*m*v2
max speed = 18.8

The net force is mv2/r = Tension - mg

How do I continue?
 
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jsmith613 said:
The net force is mv2/r = Tension - mg
Consider the forces on the rider. (Not tension, but what?)
 
Doc Al said:
Consider the forces on the rider. (Not tension, but what?)

reaction force from seat?
 
jsmith613 said:
reaction force from seat?
Right. The reaction or normal force from the seat.
 
ok I think I am working on it...
 
so
R = mv^2/r + mg
R/mg = g-force
g-force = 3
right?
 
but the next question 6b) is

Two students in the queue are having a discussion.
A says: “If they made a new ride twice as big the g-force at the bottom would be
amazing!”
B says: “I think the g-force wouldn’t be any different.”
With reference to your calculation, explain which student is correct.

If we look at our equation r (radius) is clearly a factor
but the answer says B is correct>
how?
 
or would this cancel out due to a higher max speed?
 
jsmith613 said:
so
R = mv^2/r + mg
R/mg = g-force
g-force = 3
right?
Right.
 
  • #10
jsmith613 said:
If we look at our equation r (radius) is clearly a factor
but the answer says B is correct>
how?
mv2 also depends on the radius. Express that term in terms of the radius and see what happens.
 
  • #11
Doc Al said:
mv2 also depends on the radius. Express that term in terms of the radius and see what happens.

yes they then cancel out
so (2g+g)/g = g-force
3 = g-force
 
  • #12
jsmith613 said:
yes they then cancel out
so (2g+g)/g = g-force
3 = g-force
Good.
 

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