## Homework Statement

In a popular amusement park ride, a cylinder of radius 3.00 m is set in rotation at an angular speed of 5.00 rad/s. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall of the cylinder is needed to keep the rider from slipping?

Fs=mu*Fn
Fc=mv^2/r

## The Attempt at a Solution

I didnt know how to start this problem so any guidance would be helpful. Thank you.

Doc Al
Mentor
What forces act on the rider? Is the rider accelerating? What's the direction of the acceleration? What force provides the centripetal force? What force holds the rider up?

Apply Newton's 2nd law to both vertical and horizontal forces.

The forces acting on the rider is gravity, normal force which is equal to the centripetal force and the static friction is keeping the persons from falling. So if I use mus=fs/fn and fn=mv^2/r.
fs=g g=9.81m/s/s
mu=9.81/(5^2/3)
mu=.1308
and that's what I got. I think this is correct but am unsure.

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