Minimum Angular Speed for Safe Ride on ROTOR

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To determine the minimum angular speed for the ROTOR ride, the forces acting on a person must be analyzed, particularly the normal force and friction. The coefficient of friction (0.42) and the radius of the cylinder (2.5m) are crucial for calculations. The equation for centripetal acceleration, ac = v²/r, can be used, and the weight of the person (mg) will ultimately cancel out in the equations. By setting up the appropriate equations based on the forces involved, the necessary angular speed can be calculated. Understanding these relationships is key to ensuring riders remain safely against the cylinder wall.
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The ROTOR is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out. If the coefficient of friction is 0.42 and the cylinder has a radius of 2.5m, what is the minimum angular speed of the cylinder so that the people don't fall out?

How would I go about solving this? I don't even know where to start!
 
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let m = mass of a person
let n = normal force exerted on the person by the (spinning) cylinder wall
let μ = the coefficient of friction
let g = acceleration of gravity

Start by thinking about the forces acting on the person along the vertical axis (up and down) in terms of the above. What has to be true if the person doesn't fall? Set up the appropriate equation to express this.

Then, think about what produces the normal force and you should be able to find your answer.
 
I think I know how to find the max cetripetal acceleration with ac = v2/r or
v = sq. root of r(ac). But without knowing the weight of the people inside the ride, I just don't understand how you would find the minimum angular speed.
 
It's the speed that gives you the n that you need.

You do know the weight: mg. The m will eventually cancel out.
 

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