Calculating Minimum Coefficient of Friction for Safe Rotor-Ride at Carnival

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SUMMARY

The discussion focuses on calculating the minimum coefficient of static friction required for a safe rotor-ride at a carnival, specifically with a room radius of 4.6 m and a rotation frequency of 0.50 revolutions per second. The initial approach incorrectly equated the force of friction and centripetal force to the force of gravity, leading to an erroneous negative coefficient. After clarification and proper application of the forces involved, the correct coefficient of static friction was determined to be 0.22, aligning with the solution provided in the reference material.

PREREQUISITES
  • Understanding of centripetal force and its calculation
  • Knowledge of static friction and its coefficient
  • Ability to draw and interpret free body diagrams
  • Familiarity with basic physics equations involving mass, gravity, and acceleration
NEXT STEPS
  • Study the derivation of centripetal force equations in circular motion
  • Learn about the factors affecting the coefficient of static friction
  • Explore free body diagram techniques for analyzing forces in physics problems
  • Investigate real-world applications of static friction in amusement park rides
USEFUL FOR

This discussion is beneficial for physics students, engineers, and amusement ride designers who are interested in understanding the dynamics of forces in rotating systems and ensuring safety in ride design.

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In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room". The room radius is 4.6 m, and the rotation frequency is 0.50 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?

Attempt at solution

Force of friction + Centripetal force = Force of gravity
μs Fn + mar = mg
μs g + v2/r = g
μs = (g- v2/r)/g
μs = (g-4π2r f2)/g
μs = (9.8 - 4π2(4.6)(0.5)2)/ 9.8
μs = -3.63

This obviously makes no sense because a coefficient of static friction must be less than 1 and positive to begin with and the answer in the back of the book is 0.22

I've rechecked my math several times over so i know my error has to be in my initial step of setting the force of friction and the centripetal force equal to the force of gravity, but this is my best guess as to how the answer could come about

Thank you for any help!
Anthony
 
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This is not true: "Force of friction + Centripetal force = Force of gravity"

For a person to not slip downwards, what forces must be balanced? Draw a free body diagram
 
zhermes said:
This is not true: "Force of friction + Centripetal force = Force of gravity"

For a person to not slip downwards, what forces must be balanced? Draw a free body diagram

In a free body diagram there would be a person with centripetal force going inward, force of gravity vertically down and the normal force to gravity straight up, and to not fall would mean the gravity and normal forces being equal to each other...?
 
Yes, if by "normal force to gravity" you mean "force of friction."
 
so i should try solving Ffr= Fg ?
 
Exactly. What equation do you get force the force of friction?
 
Is it Fn μs = Ffr ?

Okay I think I got it

Ffr=Fg
Fn μs = mg
[STRIKE]m[/STRIKE](v2/r) μs = [STRIKE]m[/STRIKE]g
4 \pi2r f2 μs = g

Then inputting all the given values and solving for μs yields the correct answer of 0.22

Thank you so much!
 
Last edited:
Good work! Happy to help.
 

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