Pyrowolf
- 2
- 0
Homework Statement
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 = 6.3 m and the room spins with a frequency of 23.5 revolutions per minute.
1. Speed of the rider. I used circumfrence * revolutions/ 60 to get 15.5. (Already known to be correct)
2. What is the normal Force?
3. What is the minimum coefficient of friction needed between the wall and the person?
4)If a new person with mass 92 kg rides the ride, what minimum coefficient of friction between the wall and the person would be needed?
6) To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.7 times each persons weight - and therefore adjust the frequency of revolution accordingly. What is the minimum coefficient of friction now needed?
Homework Equations
F=m*a
Fs= (meu)s*N (Static friction)
Fk = (meu)k*N (Kinetic Friction)
Circumfrence = 2*Pi*r
Weight = m*g
The Attempt at a Solution
1. I already did
2. Well Weight is down and friction is up. (It's a circle on the horizontal for a ride). The Normal force points inward of the circle. With no force pointing outward (I think?) So I tried N= mv^2/r which came up wrong. So I'm not sure what I'm missing here.
3. Once I know the N, I can calculate the friction as follows: weight = friction (no movement in the Y axis) so mg = (meu) N -> (meu) = mg/N (this is my thought process anyway)
4. Same as problem 3, just a different mass
6. Change N to be 92*1.7 then solve like 3 and 4.
Also I get no feedback on 4 and 6 so I can't know if I'm right or not.