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Static friction coefficient from circular motion prob

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data

    A coin is placed 11.0cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a speed of 36rpm is reached and the coin slides off.

    What is the coefficient of static friction between the coin and the turntable?

    2. Relevant equations

    Normal Force=mg
    g=9.8
    V=(2*pi*r) / T
    a=(v^2)/r
    Max static friction=mu*Normal force
    11.0cm=0.11m



    3. The attempt at a solution

    Normal force=mg
    ma=mg
    a=g
    9.8=(V^2)/r
    9.8r=(V^2)
    (9.8)*(0.11)=V^2
    V=1.04m/s

    1.04m/s=2(3.14)(.11) / T
    1.04T=.6908
    T=0.6642

    I'm stuck and don't know what to do from this point. The max frictional force depends on the normal force, which is equal to mg, so I don't understand how to use the velocity or the T to find Mu, the coefficient? Thank you!
     
  2. jcsd
  3. Oct 20, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good.
    Not good. Replace "mg" with the maximum static friction force.

    The force accelerating the coin (in a circle) is the friction force, not the normal force. Note that you are given the angular speed, which you can convert to tangential speed. (You can also express the centripetal acceleration directly in terms of angular speed.)
     
  4. Oct 20, 2008 #3
    Thanks for the reply. I *think* I'm understanding this a little better, but my next attempt didn't provide a good answer, either. It came out too high a value for Mu.

    Converting rmp to m/s:

    rpm=36
    revs per second=36/60=0.6rps
    T=1/06=1.66

    v=2(pi)(r) / T
    v=2(3.14)(0.11) / 1.66
    v=0.416m/s

    If I'm understanding this correctly, the net force in the "y" direction equals the normal force + gravity. There is no acceleration in the "y" direction, so N=mg. The net force is frictional, so, F(r)=mg. I want to find the maximum frictional force so:

    F(r)=mg
    uN=mg
    u(ma)=mg
    u(m)(v^2/r)=mg
    u(v^2/r)=g
    u=gr/v^2

    Plugging in the values for r, g and v:

    u=(9.8)(0.11)/(0.173)
    u=6.231

    That's too high for u. Can you help me understand where I'm still going wrong? Thank you!
     
  5. Oct 20, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    This part is good.

    This is true. In the y-direction, there is no net force. Thus, N = mg.

    Friction acts horizontally and is the only force in the x-direction. The net force is friction, but it does not equal mg.

    Think this way:

    The net force on the coin equals the friction force. (The vertical forces cancel out.) The maximum value of the friction force is μN, which you know equals μmg.

    Now apply Newton's 2nd law:
    Fnet = ma

    You know Fnet is the friction force. You know how centripetal acceleration works. Set up this equation and solve for μ.
     
  6. Oct 21, 2008 #5
    OK, I think I understand. The net force is the frictional force because the only acceleration is in the horizontal direction. So, I can set it up like this:

    Fr=uN
    Fr=ma
    uN=mg

    ma=umg
    m(v^2/r)=umg
    u=(v^2/r)/g
    u=0.320

    That looks much better!! Thank you!
     
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