Question

In a classic carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast enough, the floor drops from the bottom ofthe room! Friction between the walls of the room and the people on the ride make them the “stick” to the wall so they do not slide down. In one ride, the radius of thecylindrical room is R = 6.8 m and the room spins with a frequency of 22.4 revolutions per minute.

The first question asked: What is the speed of a person “stuck” to the wall? I found this answer to be 15.8 m/s.

This is where I have gotten stuck.
2)What is the normal force of the wall on a rider of m = 52 kg?

Answer 1

m*v^2 / R, where v is the solution to the first question.

52*(15.8^2) / 6.8

Solve; answer will be in Newtons (N)