In a popular amusement park ride, a rotating cylinder of radius 3.00 m is set in

Question

In a popular amusement park ride, a rotating cylinder of radius 3.00 m is set in rotation at an angular speed of 6.00 rad/s, as in Figure P7.51. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping? (Hint: Recall that the magnitude of the maximum force of static friction is equal to µn, where n is the normal force - in this case, the force causing the centripetal acceleration.)

Explanation / Answer

fs,max = µn, and in this case the normal force is the centripetal force exerted on the rider to keep her moving in a circle. That centripetal force is provided by the wall of the cylinder, and is Fc = mv2/r = mw2r. We are told that w = 6.00 rad/s, and r = 3.00 m. Since we don't know the mass, let's write Fc/m = w2r = (6.00 rad/s)2(3.00 m) = 108.0 m/s². The force of static friction must balance the gravitational force on each rider, fs = µFc > mg. Therefore, µ > mg/Fc = (9.8 m/s²)/(108.0 m/s²) = 0.091. Since µ must be greater than or equal to 0.091, the minimum value is 0.091.

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