Circular Motion and carnival ride

[all_relative]

Homework Statement

An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8m/s^2.
Given g=9.8 m/s^2, the coefficient of static friction between the person and the wall = 0.337. The radius of the cylinder, R=5.8m. For simplicity, neglect the person's depth and assume he or she is just a physical point on the wall. The person's speed is v= (2{pi}R)/T where T is the rotation period of the cylinder (the time to complete a full circle).
Find the maximum rotation period T of the cylinder which would prevent a 47kg person from falling down. Answer in units of seconds.

*For clarity this ride is one that spins and the person is pressed against the wall*
-attached is an image of the ride-

Homework Equations

v= (2{pi}R)/T
Ff=(coefficient of friction)(Fn)
Ff=m*((v^2)/R)

The Attempt at a Solution

M=coefficient of Friction
m=mass

Ff=M(Fn)
Ff=(0.337)(460.6)
Ff=155.22

Ff=m*((v^2)/R)
155.22=47*v^2/5.8
v=4.3766m/s^2

v= (2{pi}R)/T
4.3766=2{pi}(5.8)/T
T=8.326 seconds

-this answer is wrong-

 

[mgb_phys]

Almost - you are trying to find what normal force is needed so that the frictional force is equal to their weight.

 

[all_relative]

Thanks for the reply.

However, I am still confused as to what I need to do.

Does Ff=47?