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 0.650 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

given radius of cylinder, R = 3.15 m
angular velocity of cylinder, w = 0.65 rps = 0.65*2*pi = 4.082 rad/s
let weight of the rider be W, coefficient of static friction between the person and the wall be k
then Normal force = m*w^2*R
but m = W/g
N = W*w^2*R/g = 5.3504 W Newtons

for the static equilibrium condition
k*N = W
k = W/N = 1/5.3504 = 0.1869 is the mionimu coefficient of staitc friction required to keep the person in place at this angular velociuty and radius of the cylinder

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