Minimum friction coefficient for rotating cylinder ride

[gtn013]

Homework Statement

At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered against the wall. Suppose the radius of the room is 3.30 m and the speed of the wall is 10.0 m/s when the floor falls away. What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away if the rider has a mass of 55 kg?

Homework Equations

Centripetal Force = mv^2/r = 1667

friction = u*Fn

The Attempt at a Solution

centripetal force = friction = coefficient of friction * Normal Force

Normal force = mg (weight)

coefficient of friction = centripetal force/Normal Force = 51.5?

Is this correct?

 

[Clever-Name]

You're missing a force. Friction will be resisting the motion of the person from falling, but Gravity is the one doing the pulling, you need to consider that in your force calculations

 

[Stucky]

centripetal force = friction = coefficient of friction * Normal Force

That is incorrect. The centripetal force is equal to the normal force, not the frictional force.

Normal force = mg (weight)

This too is incorrect. The weight of the person should be equal to the frictional force.
Now try finding the coefficient of static friction.