Calculating Centrifugal Force in a "Rotor-ride

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SUMMARY

The discussion focuses on calculating the minimum coefficient of static friction required to prevent individuals from slipping down in a "Rotor-ride" with a radius of 4.6 meters and a rotation frequency of 0.4 revolutions per second. The centripetal acceleration is determined to be 45.4 m/s² using the formula a = (4π²r)/T². The centripetal force is expressed as F = mass * centripetal acceleration, highlighting that the mass of the individual is not provided but may cancel out in the calculations. Participants emphasize the importance of understanding the relationship between centripetal acceleration, force, and friction in this context.

PREREQUISITES
  • Understanding of centripetal acceleration and its calculation
  • Familiarity with Newton's Second Law of Motion
  • Knowledge of static friction and its role in circular motion
  • Ability to manipulate equations involving mass and forces
NEXT STEPS
  • Research the derivation of centripetal acceleration formulas
  • Study the relationship between centripetal force and friction in circular motion
  • Explore examples of static friction calculations in physics problems
  • Learn about the effects of varying mass on centripetal force equations
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Physics students, educators, and anyone interested in understanding the dynamics of rotational motion and forces in amusement park rides.

JC Polly
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In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled room (think of a ride like the gravitron). The room radius is 4.6m and the rotation frequency is .4 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?

I understand that i need to solve for centripetal acceleration and solve for a normal force and fictional force, but I'm having trouble finding the process to get to that solution. Any help would be greatly appreciated.
 
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Hi JC Polly, welcome to PF! :smile:

What is the centripetal acceleration using the formula you know for it? What is the equation for the centripetal force in that direction that results from that cenrtripetal acceleration, using Newton's 2nd law? What then is the min friction force required to keep the person from falling?
 
I found the centripetal acceleration to be 45.4 m/s^2 a=(4π^2(r))/T^2. The equation I have for centripetal force is F=mass*centripetal acceleration, but my problem is the problem didn't give a mass for a human so that's where I'm stuck at. I know that some problems we did in class the masses of the the object canceled out through manipulating the equations but that was only in examples when the object was moving in the vertical direction.
 
Well leave the centripetal force with the m in it, and maybe you will find it cancels out!
 

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