certain amusement park ride consists of a large rotating cyl

certain amusement park ride consists of a large rotating cylinder of radius R = 3.05 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 0.630 rotations per second, what is the magnitude of the normal force F_N between a rider and the wall, expressed in terms of the rider\'s weight W? F_N = What is the minimum coefficient of static friction mu_s required between the rider and the wall in order for the rider to be held in place without sliding down? mu_s

Solution

here,

radius , r = 3.05 m

f = 0.63 rot/s

w = 2 * pi * f = 3.96 rad/s

Fn = centripital force

Fn = m * r * w^2

Fn = (W/g) * 3.05 * 3.96^2

Fn = 4.88 W

the normal reaction force is 4.88 W

let the coefficient of friction be us

for the person not to slide down

Fn * us >= W

4.88 * W * us >= W

us >= 0.205

the coefficient of friction must be >= 0.205