A popular amusement park ride, shown on WB p. 8, operates as follows: riders enter the cylindrical structure when it is stationary with the floor at the point marked "a". They then stand against the wall as the cylinder then begins to rotate. When it is up to speed, the floor is lowered to the position marked "b", leaving the riders "suspended" against the wall high above the floor. Important values for this question:

Question

radius of ride = 1.5 m
μs = 0.53
Include units in all numerical answers.

a. Draw a force diagram for the 82 kg rider after the floor moves down (this will not be collected but should help you answer the rest of the question.

b. What type of force (from your force diagram) is causing the passenger not to slide down the wall?
Correct: Your answer is correct.


c. How big does this force need to be for the person not to fall?


d. What type of force is causing a net force to the center (or centripetal force)?

e. How big (numerical value) does the force in "d" need to be so the force in part "c" is big enough (remember your old equations relating these two types of forces)?


f. What is the minimum tangential speed necessary to make sure that the force in "e" is big enough? In other words, how fast must the ride be moving before the floor drops out?

Answer 1

well, the vertical force of friction has to be greater than person weight.

mu*m v^2/r > mg
mu*v^2/r>g

Answer 2

Your FBD should have weight down, friction up, and a normal force towards the center of the ride. The normal force is equal to the centripetal i.e. v^2/r. Mu (Fn) and as Bob showed this must be greater than mg. And v = (sqrt)(rg/mu). Important idea is that the normal force is causing the circular motion.