An amusement park ride consists of a large vertical cylinderthat spins around it

Question

An amusement park ride consists of a large vertical cylinderthat spins around its axis fast enough such that any person insideis held up against the wall when the floor drops away. Thecoefficient of static friction between person and wall isµs, and the radius of the cylinder isR. (a) Show that the maximum period of revolution necessary tokeep the person from falling is T =(42Rµs/g)1/2. (Do this on paper. Your instructor mayask you to turn in this work.)

(b) Obtain a numerical value for T assuming thatR = 3.50 m andµs = 0.200.
1 s
How many revolutions per minute does the cylinder make?
2 rev/min

(c) If the rate of revolution of the cylinder is made to besomewhat larger, what happens to the magnitude of each one of theforces acting on the person? What happens in the motion of theperson?
3

(d) If instead the cylinder's rate of revolution is made to besomewhat smaller, what happens to the magnitude of each one of theforces acting on the person? What happens in the motion of theperson?
4 An amusement park ride consists of a large vertical cylinderthat spins around its axis fast enough such that any person insideis held up against the wall when the floor drops away. Thecoefficient of static friction between person and wall isµs, and the radius of the cylinder isR. (a) Show that the maximum period of revolution necessary tokeep the person from falling is T =(42Rµs/g)1/2. (Do this on paper. Your instructor mayask you to turn in this work.)

(b) Obtain a numerical value for T assuming thatR = 3.50 m andµs = 0.200.
1 s
How many revolutions per minute does the cylinder make?
2 rev/min

(c) If the rate of revolution of the cylinder is made to besomewhat larger, what happens to the magnitude of each one of theforces acting on the person? What happens in the motion of theperson?
3

(d) If instead the cylinder's rate of revolution is made to besomewhat smaller, what happens to the magnitude of each one of theforces acting on the person? What happens in the motion of theperson?
4

Explanation / Answer

(a) to derive the expression, use: .             normalforce from wall = m v2 / R .                         n = m v2 / R . and           v = 2R /T           (i.e. distance over time) . and               s n =mg            (vertical forces are balanced: friction up, mgdown) . Combine these threeexpressions:       s   [ m ( 2R / T)2 / R ] = mg . Solve for T:           T = ( 4 2 R s / g)1/2 (b)          T = ( 42 3.50 * 0.200 / 9.80)1/2 =     1.68 seconds . If it takes 1.68 s for one rev, then itmakes .                  60 / 1.68 =    35.7 rev permin . .           

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