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 3.30 m and the speed of the wall is 10.0 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 55.0 kg rider?
c) What is the minimum coefficient of static friction 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

a)the centripetal force acting on the riders is

F = (Mv^2/r)

where M is mass of wall,v = 10.0 m/s and r = 3.30 m

b)mass of rider m1 = 55.0 kg

centripetal force acting on rider is

F1 = (m1v^2/r)

c)the minimum coefficient of static friction is

u_s = (v^2/r x g)

where g = 9.8 m/s^2

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