At Six Flags, there is a popular ride where the floor of a rotating cylindrical

Question

At Six Flags, 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. (It used to be called Tom's Twister). Suppose the radius of the room is 3.20 m and the speed of the wall is 10.2 m/s when the floor falls away (a) What is the source of the centripetal force acting on the riders? (b) How much centripetal force acts on a 65.0-kg rider? (c) What is the minimum coefficient of static that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away?

Explanation / Answer

Given that

radius of the room r=3.2 m

speed of the walls v=10.2 m/s

basing on the concept of rational dynamices

the source of the centripetal force acting on the rider is because the floor is rotation and it is sphere type of rotation

now we find the centripetal force acting on the rider

the centripetal force Fc=mv^2/r=65*10.2^2/3.2=2113.3 N

now we find the coefficient of static friction on the floor

the coefficient of static friction on the floor =v^2/gr=10.2^2/*9.8*3.2

=1.66

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