In a popular amusement park ride, a rotating cylinder of radius R = 3.40 m is se

Question

In a popular amusement park ride, a rotating cylinder of radius R = 3.40 m is set in rotation at an angular speed of 6.00 rad/s, as in the figure shown below. 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 sn, where n is the normal force—in this case, the force causing the centripetal acceleration.
s =

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 = 3.40 m. Since we don't know the mass, let's write Fc/m = w2r = (6.00 rad/s)2(3.40 m) = 122.4 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)/(122.4 m/s²) = 0.08. Since µ must be greater than or equal to 0.08, the minimum value is 0.08

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