At amusement parks, there is a popular ride where the floor of a rotating cylind

Question

At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 4.00 m and the speed of the wall is 10.4 m/s when the floor falls away. The source of the centripetal force on the riders is the normal force provided by the wall. (a) How much centripetal force acts on a 68.7 kg rider? (b) What is the minimum coefficient of static friction that must exist between the rider's back and the wall, if the rider is to remain in place when the floor drops away?

Explanation / Answer

a. The wall of the ride is the source of the centripetal force. b.Centripetal force = (mv^2/r) = ((55.0 kg)*(10.0 m/s)^2)/3.30 m = 1670 N c. Normal force = (mv^2/r) Normal force = 1670 N, which is the same as the centripetal force, Force of friction = m*g Force of friction = (55.0 kg)*(9.81 m/s^2) Force of friction = 540. N Force of Friction = Coefficient of static friction * Normal force 540. N = Coefficient of static friction * 1670 N Coefficient of static friction = 0.323

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