I posted this question this morning but I'm not sure of the response Damon provided.
If you were in a rotor style ride and the riders accelerate until the speed of ride is reached- if the radius of the cylinder is 5.0m and the coefficient of friction between clothes and wall is 0.5, how do you find minimum speed you would need to stick to the wall of the ride? Is he saying that I take 2*5(r) *9.8(g)? If not, please explain.Thank you

you need a centripetal acceleration equal to 2g if the mu is 1/2
Force down = m g
friction force up = mu m v^2/r
g = mu v^2/r
v^2/r = g/mu = 2 g
v^2 = 2 r g

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1. Is this correct? I published something else earlier-that has to be wrong but I think this is correct.

Force friction balances weight
Force friction comes from force normal to create centripetal force

mv^2/r = Force normal

Force friction = mu x Force normal = mg

mu x mv^2/r = mg

9.8/05 = 19.6 m/s^2
19.6/5=3.92

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posted by Joey

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