1. Not finding help here? Sign up for a free 30min 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!

Coin on Turntable

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data
    A coin is placed on a turntable that is rotating at 45.0 rpm. If the coefficient of static friction between the coin and the turntable is 0.186, how far from the center of the turntable can the coin be placed without having it slip off?


    2. Relevant equations

    mv^2/r <= usmg
    us= coefficient of static friction

    3. The attempt at a solution
    I tried using the equation and I converted 45 rpms to rad/sec and I got r=12.17 but this is wrong so I don't know what to do. Maybe im using the wrong equation?
     
  2. jcsd
  3. Feb 27, 2009 #2

    hage567

    User Avatar
    Homework Helper

    Can you show more details of your calculations? We can't see what you've done wrong if you don't show your work.
     
  4. Feb 27, 2009 #3
    mv^2/r =usmg
    v^2/r = usg
    (4.71^2)/r <= .186g
    22.184 = 1.8228r
    r= 12.17

    This seems logical to me but it says it is wrong.
     
  5. Feb 27, 2009 #4

    hage567

    User Avatar
    Homework Helper

    Note: the 4.71 rad/s, is angular velocity. You need to convert that to m/s, which is linear velocity. Do you know the relationship between these two? It should be in your textbook.
     
  6. Feb 27, 2009 #5
    I think v = omega*r

    Where v is the velocity, omega is angular speed (rad/s), and r is the radius.
    so v= 4.71r
    then (4.71r)^2/r= 22.18r = 1.8228
    so r= .082

    which is right! thank you!!
     
  7. Feb 27, 2009 #6

    hage567

    User Avatar
    Homework Helper

    You're welcome. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook