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I am studying for a test and this practice question has me a bit puzzled. Which equation do i need to solve this problem.

THANKS FOR THE HELP


In a 'Rotor-ride" at a carnival, people are rotated in a cylindrically walled room. The room radius is 4.6 meters, and the rotation frequence is .5 revolutions per second after the floor drops out. What is teh minimum coefficient fo static friction so the people will not slip down.

2007-09-26 10:23:48 · 1 answers · asked by Krazyk78 1 in Science & Mathematics Physics

1 answers

The rotational speed of the 'Rotor-ride" is 0.5 revolutions per second or w = pi/s (in radian/s). Therefore the centripetal accelaration is w^2*r = 4.6*(pi)^2 (m/s^2). Let the man/woman's mass to be m and the minimum coefficient fo static friction to be k, we have:
mk*4.6*(pi)^2 >= mg
or: k >= g/{4.6*(pi)^2} = 0.22

2007-09-26 16:40:50 · answer #1 · answered by Hahaha 7 · 0 0

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