In an old-fashioned amusement park ride, passengers stand inside a 4.6-m-diamete

Question

In an old-fashioned amusement park ride, passengers stand inside a 4.6-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 0.65 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed." What is the minimum angular speed, in rpm, for which the ride is safe?

Explanation / Answer

The centripetal force created by the spinning is the normal force of the wall on the rider. This normal force times the minimum coefficient must equal the weight of the rider to hold him up so we have F = µ*N = m*g and N= m*r*?^2 where ? is the angular velocity. So the minimum ? occurs when the minimum coefficient is employed. The static friction should be sufficient but for safety use the minimum kinetic coefficient So µ*N = µ*m*r*?^2 = m*g So ? = sqrt(g/(µ*r)) = sqrt(9.8m/s^2/(0.40*2.3m)) = 3.26 rad/s

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