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 that

radius R = 3.15 m

angular speed w = 0.650 rev/s

w = 0.650 * 2 * pi rad / s

w = 4.08 rad / s

Now , as Normal force provides the

FN = m * w^2 * R

FN = (W/g) * 4.08^2 * 3.15

FN = 5.34 * W

the normal force is 5.34 * W

Now

for the person to stay at place

coefficient of friction is us

balancing forces in y direction

us * FN = W

us * 5.34 * W = W

us = 0.187

the minimum coefficient of static friction is 0.187

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