Centripetal Acceleration Problem: Finding Angular Speed

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The discussion revolves around a centripetal acceleration problem involving an amusement park ride where passengers must stick to the wall as the floor drops away. The key equation used is ω=sqrt(g/rµ), where the angular speed is influenced by the coefficient of friction. The initial calculation yielded an angular speed of 18.2 rpm, which was incorrect according to the homework site, indicating the correct answer is 23 rpm. It was determined that using a lower friction coefficient of 0.62 leads to the correct solution. If passengers wear clothing with a friction coefficient less than 1 at 18.2 rpm, they would not have sufficient centripetal acceleration to remain against the wall.
Nahtee
Hi all, I've been lurking around the forums for a while to get help with homework but I figured I'd finally make an account to get direct feedback.

I'm having problems with this centripetal acceleration problem,

Homework Statement


"In an old-fashioned amusement park ride, passengers stand inside a 5.4-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.62 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed." What is the minimum angular speed, in rpm, for which the ride is safe?"

Homework Equations


F=ma, Ff=µN, ω=v^2/r

The Attempt at a Solution


So, I started out drawing a FBD with Ff up, Fg down, and Fn towards the center of the circle. Applying N2L gave me 0=µN-mg=µω^2r-g, so ω=sqrt(g/rµ). From this equation its clear that the angular velocity would be lowest when µ=1, since its in the denominator. So after I plugged in the values for g and r I got sqrt(9.81/2.7)= 1.906rad/s. Then dividing that by 2pi to go from radians to revolutions gave me 0.3033rps, x60 to convert to minutes = 18.20rpm.
My online homework site marked this as wrong and says the answer is 23rpm, but even after double checking my work and trying to work backwards from the 23rpm I can't figure out where I went wrong. Any thoughts?
 
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What would happen if someone was wearing clothing with the friction coefficient less than 1 and the ride was rotating at 18.2 rpm?
 
PhanthomJay said:
What would happen if someone was wearing clothing with the friction coefficient less than 1 and the ride was rotating at 18.2 rpm?

Ah I see, using the 0.62 coefficient gives me the correct answer. If someone had clothing with the coeff of friction less than 1 and the ride was rotating at 18.2rpm there wouldn't be enough centripetal acceleration to hold them to the wall, so they would fall off. Thanks for you're help!
 
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