In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." The radius of the room is 5.5 m, The coefficient of static friction is μ=0.1.

Question

What is the minimum rotation frequency such that the people will not slide down the wall when the floor drops out?

Answer 1

If it goes halfway around in one second, then the period of the ride must be 2.0 seconds.

What is going on is that the ride is pushing on the people to make them go in a circle. For every action there is an equal but opposite reaction, and so the people push back on the ride. The centripetal force the ride exerts on people becomes the normal force that causes the friction that keeps them from sliding down the walls, and that friction as such must be larger than their weight, or down the wall they go. There is nothing pushing them to the outside, rather they are accelerating toward the middle of the circle. Remember that you feel "thrown" the opposite direction of the acceleration. (i.e. if your car accelerated forward, you are thrown backward)

So, the normal force is just the centripetal force acting on the people: (I pick the centripetal force with the period in it)
FN = m4ð2r/T2
Since the friction force must be larger than or equal to the weight of the riders, the minimum coefficient of static friction will be:
mg = Ffr = µFN = µ(m4ð2r/T2)
And if you plug the centripetal force in for the normal force, you get:
mg = µ(m4ð2r/T2) - so let's solve for µ:
g = µ(4ð2r/T2) (canceling the mass)
gT2/(4ð2r) = µ (divide both sides by 4ð2r/T2)
µ = gT2/(4ð2r) = (9.80 m/s/s)(2.0 s)2/[4ð2(5.0 m)] = .1986 = .20