Hello! I have this problem I must submit by tomorrow 11 PM and I am getting the wrong answer and cannot figure out what I'm doing wrong. Any help would be great. Here is the problem: "In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." (See Fig. 5-35.) The room radius is 4.7 m, and the rotation frequency is 0.5 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?" My work so far: T = 1/f, so T (period) = 1/.5 =2. To find velocity, 2(pi)(r)/T, so 2(pi)(4.7)/2 = 14.765 m/s 3 Forces on the people in the spinner. A static friction force that goes in toward the circle made by the ride, and a weight force that goes down (mg) and a Normal Force that goes up. No movement on the y axis so mg=Normal Force. Static Friction = coef. of stat fric X Normal Force. or coef X mg. F = ma, a = v^2/r so F = m X v^2/r Set m X v^2/r = coeff. X mg....so mass cancels out leaving v^2/r = coeff X g. Then I plug in numbers (14.765)^2/4.7 = coeff X 9.8 I get for the coefficient 4.73 which is the WRONG ANSWER! :surprised Anyone have any clue what I am doing wrong? I would appreciate help so much, thank you!!