Posted by Joey on .
I posted this question this morning but I'm not sure of the response Damon provided. I did it once but then I updated it later in this at the bottom please check booth of my answers.
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
PhysicsPlease help with formula  Damon, Tuesday, December 7, 2010 at 11:11am
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
PhysicsI think I have itPlease check  Joey, Tuesday, December 7, 2010 at 5:29pm
Is this correct? I published something else earlierthat 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
sqrt 3.92 = 1.98 rads/s

PhysicsPlease check if I did this correctly 
Joey,
Please scroll down for the entire answerI did it once in the first paragraph and than after Damon's response, I did it againI think the 2nd one is correct, maybe
Thanks you