Coefficient of static friction problem

[rykirk]

Hey guys,

I was wondering if you could help me out with the following problem...

In a "Rotor-ride" at a carnival, people pay money to be rotated in a vertical cylindrically walled "room". If the room radius is 5.0 m and the rotation frequency is .60 revoloutions per second when the floor drops out, what is the minimum coefficient of static friction so that people will not slip down.

I keep getting stuck in the same place...I have calculated the velocity using the formula for circular motion that states v=(2*pi*r)/T, getting (10*pi)/2 and could easily get the acceleration of the ride by squaring the velocity and dividing by the radius..but I'm not 100% sure what to do after that. I know I need an equation in which the mass will cancel out on both sides, but I'm not sure how to tie the friction coefficient into the problem.

Any help would be greatly appreciated

Thanks a lot,

Ryan

 

[Pyrrhus]

Remember in the cyclone the cetripetal force will be the normal force, (forces on x-axis) and on the y-axis we got the friction force pointin up and the weight pointing down.

Also a hint $$F_{f} = \mu N$$

 

[eternityras]

perhaps the equation you are looking for is

u (coeffi of static friction)= [(4pi^2)(f^2)(r)]/g