Calculating Minimum Coefficient of Friction for Vertical Amusement Park Ride

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To determine the minimum coefficient of friction required to prevent riders from slipping in a vertical amusement park ride, the forces acting on the rider must be analyzed. The centripetal force needed for the circular motion is provided by the normal force, while static friction counteracts gravity. Using the equations for static friction and centripetal force, the calculation yields a coefficient of friction of approximately 0.1308. This value indicates the necessary friction between the rider's clothing and the wall to maintain their position. The solution approach involves applying Newton's second law to both vertical and horizontal forces.
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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?

Homework Equations


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.
 
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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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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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