In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5

Question

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will “stick” to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70.

Part A

What is the minimum rotational frequency, in rpm, for which the ride is safe?

Express your answer using two significant figures.

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will “stick” to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70.

Part A

What is the minimum rotational frequency, in rpm, for which the ride is safe?

Express your answer using two significant figures.

Explanation / Answer

The frictional force, f, must equal force due to gravitational acceleration.

f = m x ac x us, where ac = centrifugal acceleration = v2 / r, and us is coef. of static friction.

m(v2 / 2.5m) x 0.6 = mg. The masses cancel so

0.6(v2 / r) = 9.81 m/s^2

v2 = 2.5m(9.8 m/s2) / 0.6

v2 = 40.833; v = 6.39 m/s

v = rev/s(pi)d; rev/s = 6.39 m/s / (pi x 5.0 m) = 0.41 rev / s

or 0.41 x 60 = 24.4 rev / min.

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