An amusement park ride consists of a large vertical cylinder that spins about it

Question

An amusement park ride consists of a large vertical cylinder that spins about its axis sufficiently fast that any person inside is held up against the wall when the floor drops away. The coefficient of static friction between the person and the wall is µs, and the radius of the cylinder is R.
(a) Show that the maximum period of revolution necessary to keep the person from falling is T = (4?2Rµs/g)1/2.
(b) Obtain a numerical value for T assuming that R = 4.00 m and µs = 0.400. How many revolutions per minute does the cylinder make?

Explanation / Answer

When you see the drawing, you will realize that the weight of the person has to equal the friction the person makes with the cylinder, so Weight = Friction Because of Newton's law, you get that Weight = mg where m is the mass of the person and g is the gravity acceleration. you know that friction = Mu*N N = m*ac, where ac is the centripetal acceleration ac = W^2R where W is the angular velocity W = 2*pi/T Now that you have all the necesary equations you do as follows f=mg Mu*N=mg Mu*m*ac=mg Mu*ac=g Mu*W^2*R=g Mu*4*pi^2*R/T^2=g and solving for T T = (Mu*4*pi^2*R/g)^1/2 There you have the answer for a now for b you just substitute the values you have in the equation you got previously T = (0.4*4*3.1416^2*4m/9.81m/s^2)^1/2 T = 2.537s since the frecuency = 1/T F = 0.394 rev/s*60s/1min = 23.6 Rev/min

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