A certain amusement park ride consists of a large rotating cylinder of radius R

Question

A certain amusement park ride consists of a large rotating cylinder of radius R = 2.85 m. As the cylinder spins, riders inside feel themselves pressed against the wall. If the cylinder rotates fast enough, the frictional force between the riders and the wall can be great enough to hold the riders in place as the floor drops out from under them. If the cylinder makes 0.490 rotations per second, what is the magnitude of the normal force FN between a rider and the wall, expressed in terms of the rider's weight W?

Explanation / Answer

Angular velocity (w) = 0.490 rpm = 2*Pi * 0.490 radian

w = 3.078 rad

Velocity = wR = 3.078 * 2.85 = 8.7723 m/s

Normal force = Mv^2 /R { centrifugal force }

FN = W * 8.7723^2 / 2.85 = 27.0*W N

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