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From torque to acceleration

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]
    Find acceleration a given the torque, T, about the axle.

    2. Relevant equations
    [tex]\tau = I\frac{d\omega}{dt} = I\alpha[/tex]

    Moment of Inertia of a disk (about the axle):
    [tex]I_z = \frac{1}{2}MR^2[/tex]

    3. The attempt at a solution

    [tex]\alpha = Ra[/tex] in this context, so
    [tex]\tau = I Ra[/tex]
    [tex]a = \frac{\tau}{IR} = \frac{2\tau}{MR^3}[/tex]
    I think this is correct, however:

    [tex]\tau = R \times F[/tex], since F = ma, [tex]a = \frac{\tau}{mR}[/tex].
    What does this a represent? Does it represent the linear acceleration of a particle of mass m that happens to be on the edge of the disk?

    - Thanks
     
  2. jcsd
  3. Nov 20, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi saltine! Welcome to PF! :smile:

    (i assume the wheel rolls without slipping?)
    No, [itex]R\alpha = a[/itex] … just consider the dimensions! :redface:
    No, I don't understand that :confused: … tau doesn't come from a force at a distance R … it probably comes from two forces very close together, on either side of the axle. :wink:
     
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