problem 54) (Chapter 5) At amusement parks, there is a popular ride where the fl

Question

problem 54) (Chapter 5) 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 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 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

as you know where ever we have any kind of force that means we have acceleration which is either change in magnitude or direction of speed. for centripetal force as you see there isn't change in magnitude but direction of motion, so centripetal force which has the direction of center ward exists to change the direction to be circular. to solve any problem in physics you need FBD(free body diagram) that you just consider the forces exerted to body for this case f{f}=µ{s}N friction force upward mg of rider downward F centripetal force because of presence of wall outward (if we had the string inward) N normal force inward, now F = m(v²/R)=1666.6(N)=N f{f} = µ{s}N=mg µ{s} = mg/N=(55×9.8/1666.6)=0.32(unitless) {-} means sub

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