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Homework Help: Torque and Rotational Kinematics

  1. Nov 3, 2007 #1
    1. The problem statement, all variables and given/known data
    A grindstone in the shape of a solid disk with diameter 0.490 m and a mass of m = 50.0 kg is rotating at omega = 890 rev/min. You press an ax against the rim with a normal force of F = 170 N, and the grindstone comes to rest in 7.20 s.

    2. Relevant equations


    3. The attempt at a solution
    [tex]\tau =I\alpha[/tex]

    [tex]{\omega }_z={\omega }_{0z}+{\mu }_k\alpha t[/tex]


    [tex]\tau =\frac{1}{2}MR^2\alpha[/tex]

    [tex]{\omega }_z=0\alpha =\ \frac{{\omega }_{0z}}{{\mu }_kt}[/tex]

    [tex]\tau =-\frac{1}{2}Mr^2\frac{{\omega }_{0z}}{{\mu }_kt} [/tex]

    [tex]\tau =r\times F[/tex]

    [tex]r\times F=\frac{1}{2}Mr^2\frac{{\omega }_{0z}}{{\mu }_kt}[/tex]

    [tex]{\mu }_k=\frac{Mr{\omega }_{0z}}{2tF}[/tex]

    [tex]F=f{\mu }_k[/tex]

    [tex]{\mu }_k=\frac{Mr{\omega }_{0z}}{2tf{\mu }_k}[/tex]

    [tex]{\mu }^2_k=\frac{Mr{\omega }_{0z}}{2tf}[/tex]

    [tex]{\mu }_k=\sqrt{\frac{Mr{\omega }_{0z}}{2tf}} [/tex]
    Last edited: Nov 3, 2007
  2. jcsd
  3. Nov 3, 2007 #2
    Okay, although not provided in the question, I am assuming that you are supposed to find coefficient of kintic friction. (Your attempt at the solution tells me that.)
    You erred in this statement:
    Note that, torque produced by F is not the one which is producing deceleration in the disc (It is in a perpendicular direction to ωz!), rather it is f, torque due to friction which is producing deceleration. Keep this in mind and I hope you would be able to solve it, as you know other things, it seems.
  4. Nov 6, 2007 #3
    [tex]F=f{\mu }_k[/tex] <---isn't that the force due to friction? Or am I confused? Realizing I kind of messed up my notation, [tex]f = 170N[/tex] and [tex]F_{friction}=F[/tex]
    Last edited: Nov 6, 2007
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