- #1
cubejunkies
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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
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