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Max static friction witout mass or normal force?

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data
    The coefficient of static friction between a small coin and a turntable is 0.30. The turntable rotates at 33.3 RPM. What is the max distance from the center of the turntable at which the coin will not slide?

    2. Relevant equations
    Fs max =μs (FN)
    V = ω (r)

    a=(V^2)
    -----
    r

    Fc=m(vt)^2
    ------
    r

    3. The attempt at a solution
    Didn't get very far. How can this be done without the mass of the coin, or the normal force supporting it? The Force of static friction is the only horizontal force acting on the coin, correct? When the inertia of the coin overcomes this Force of static friction, the coin will slide, correct?
     
  2. jcsd
  3. Apr 19, 2010 #2

    Matterwave

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    Science Advisor
    Gold Member

    You can think of the frictional force as the centripetal force which keeps the coin in a circular orbit. Therefore, you will see that F=ma=μN=μmg
    Notice that the m cancels.

    All you need now is just what is the a required?
     
  4. Apr 19, 2010 #3
    Just did some more research and found the following:

    F_cent = m(v^2)/R = F_s_max = (mu_s) · mg ,

    giving us

    mu_s = (v^2) / gR .

    So we figure out what the total horizontal force could be...
    F_cent = m(v^2)/R = F_s_max = (mu_s) · mg
    Come up with an equality that features quantities we are interested in...
    mu_s = (v^2) / gR
    and solve for R?

    mu_s is dimensionless
    g is in m/s^2
    and the answers are all in m.

    How can I go from RPMs to something that will give me an answer in meters?
     
  5. Apr 19, 2010 #4
    The SI unit of rotational velocity is radians per second or s^(-1). RPM is revolutions per minute. How many radians in one revolution?
     
  6. Apr 19, 2010 #5

    Matterwave

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    You have the equation in your first post! V=omega*r (omega in radians)

    You need to convert your RPM into radians/second which will give you omega.
     
  7. Apr 19, 2010 #6
    There are 2pi radians in one revolution and 60 seconds in one minute, so...
    33.3 RPM (2pi rad/1 REV) (1 min/60 s)= 3.49 rad/s!!

    So I have r = (v^2)/(mu_s)(g).....and I'm no longer thinking straight.
     
  8. Apr 19, 2010 #7
    According to the answer key, the correct answer is 0.242 m. I'm still confused as to how we go from radians to meters. What obvious thing am I missing??
     
  9. Apr 19, 2010 #8
    w = 3.49 rad/s is correct. Substitute for V in your equation. Namely V = rw. By the way where does this problem come from? The GRE? I know I have seen it before with these numbers.
     
  10. Apr 20, 2010 #9
    Yep. It's from the GRE.
    So: r =[(3.49rad/s)(r)]^2/[mu_s(g)] ?
     
  11. Apr 20, 2010 #10
    Yep just solve for r. You could have all along used rotational velocity instead of just velocity. So mv^2/r = mrw^2. Also if your studying for the GRE I recommend http://grephysics.net. It is a great site that works out all the problems of the 4 tests which have been released.
     
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