At amusement parks, there is a popular ride where the floor of a rotating cylindrical room...

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 4.24 m and the speed of the wall is 16.2 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 53.0 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?

Solutions

Expert Solution

Gravitational acceleration = g = 9.81 m/s²

Mass of the rider = M = 53 kg

Radius of the room = R = 4.24 m

Speed of the wall when the floor falls away = V = 16.2 m/s

Normal force on the rider by the wall = Centripetal force on the rider = F_c

F_c = 3280.5 N

Coefficient of static friction between the rider's back and the wall =

Friction force on the rider = f

The rider remains in place when the floor drops away therefore the weight of the rider is supported by the friction force.

= 0.158

a) Centripetal force acting on a 53 kg rider = 3280.5 N

b) Minimum coefficient of static friction between the rider's back and the wall = 0.158

genius_generous answered 1 year ago

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