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Centrifugal Forces

  • Thread starter JC Polly
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  • #1
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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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  • #2
PhanthomJay
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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?
 
  • #3
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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.
 
  • #4
PhanthomJay
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Well leave the centripetal force with the m in it, and maybe you will find it cancels out!
 

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