In a classic carnival ride, patrons stand against the wall in a cylindrically sh
ID: 1955429 • Letter: I
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 of the 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 the cylindrical 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?
Explanation / Answer
given all these, Find Volume, V = w*r = 6.8*22.4*2 pi /60 = about 15.95 m/s Then, mv^2 / R = 52(15.95^2) / 6.8 = 1 945.43 N