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 4.57 m and the speed of the wall is 14.8 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 58.7 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

Part A)

The formula for centripetal force is Fc = mv2/r

Fc = (58.7)(14.8)2/(4.57)

Fc = 2813 N (or 2810 N using three significant figures)

Part B)

The formula for friction is Ff = Fn

We know that Fn = 2813 N from Part A. Also, since the rider is not sliding down, the frictional force must balance the rider's weight, so...

mg = Fn

(58.7)(9.8) = ()(2813)

= .204

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