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Rotation of a spool about rough ground

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A spool of thread comprises a cylinder of radius R1 and is capped with two disks of radius R2, where R2 > R1. Some thread is tightly wound over the cylinder. The whole spool is laid stationary, side down on the ground and the thread is pulled. The spool rolls without slipping. Show that the magnitude of friction force is

    [tex]f = (\frac{I + mR_1R_2}{I + mR_2^2}T)[/tex], where the tension of the thread pulled is T and the moment of inertia of the spool is I.

    3. The attempt at a solution

    This is not a really difficult problem. I don't really have a big problem with the general method, but I do have some problems arriving at the exact final answer.

    Applying Newton's second law,

    [tex]T - f = ma_{cm}[/tex] for the centre of mass of the spool.

    Applying torques about the centre of the spool

    [tex]TR_1 - fR_2 = I\alpha[/tex]

    Also, given pure rolling, [tex]\alpha = \frac{a_c}{R_2}[/tex]. Substituting this, I get roughly the same expression [tex]f = (\frac{I - mR_1R_2}{I - mR_2^2}T)[/tex].

    Assuming I got the general steps right, the main thing I got wrong here is the torque equation. Had I reversed the signs I would have gotten [tex]-TR_1 + fR_2 = I\alpha[/tex], and the correct answer, but I can't understand how this is so. Can someone help me to see the light??
     
  2. jcsd
  3. Feb 27, 2009 #2

    Doc Al

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    Staff: Mentor

    It's a sign convention issue. To apply the condition for rolling without slipping, the rotational and translational accelerations must have the same sign. Since the torque due to friction acts to increase the rotational acceleration, it should have the same sign in your torque equation.
     
  4. Feb 27, 2009 #3
    Thanks Al.
     
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