Calculating Minimum Coefficient of Friction for Vertical Amusement Park Ride

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SUMMARY

The minimum coefficient of friction required to prevent a rider from slipping in a vertical amusement park ride, where a cylinder of radius 3.00 m rotates at an angular speed of 5.00 rad/s, is calculated to be 0.1308. This calculation is based on the forces acting on the rider, including gravity and the normal force, which is equivalent to the centripetal force. The relevant equations used are Fs = μ * Fn and Fc = mv²/r. The solution confirms that the static friction must counteract the gravitational force acting on the rider.

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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.
 
Last edited:

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