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 = 3.15 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 0.470 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? What is the minimum coefficient of static friction s required between the rider and the wall in order for the rider to be held in place without sliding down?

Explanation / Answer

Normal force FN = m*R*w^2

FN = W*r*w^2/g

W is the weight

and w is the angular speed = 0.51*2*3.142 = 3.20484 rad/s

frictional force fs = s*FN = s*m*R*w^2

m*g = s*m*R*w^2

s = g/(R*w^2) = 9.81/(3.15*3.2048^2) =

Normal force FN = m*R*w^2

FN = W*r*w^2/g

W is the weight

and w is the angular speed = 0.51*2*3.142 = 3.20484 rad/s

frictional force fs = s*FN = s*m*R*w^2

m*g = s*m*R*w^2

s = g/(R*w^2) = 9.81/(3.15*3.2048^2) = 0.303

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