1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Carnival ride, Friction and Normal Forces

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    F=m*a
    Fs= (meu)s*N (Static friction)
    Fk = (meu)k*N (Kinetic Friction)
    Circumfrence = 2*Pi*r
    Weight = m*g

    3. 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.
     
  2. jcsd
  3. Oct 6, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    N= mv^2/r is correct. Why do you think it's wrong? (Some online system?)
     
  4. Oct 6, 2011 #3
    Hmm, I just redid the calculations and for some reason I got a different answer this time. Yes it's an online system. And after redoing the calculations I got the right answer. I must have missed a decimal, or number, or parenthesis. Thanks for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook