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Force of friction and Torque problem

  1. Jul 20, 2008 #1
    Hi all, I'm a new member here. I've encountered a lot of help from searching through this website! I've searched through several threads dealing with friction and torque on here, but still can't quite figure out this problem...

    1. The problem statement, all variables and given/known data
    A plot of [tex]\tau[/tex] vs. [tex]\alpha[/tex] has a vertical intercept of 0.00176 [Nm]. If the bearings are 2 [mm] from the center of the wheel, then what is the force of friction?

    2[mm] = 0.002 [m]

    2. Relevant equations
    [tex]\tau[/tex]=I[tex]\alpha[/tex] + [tex]\tau[/tex][tex]_{friction}[/tex]
    F[tex]_{friction}[/tex] = [tex]\mu[/tex]n (Though is this cannot be directly applied since we are dealing with rotational forces, right?)
    [tex]\tau[/tex] = Force x radius
    3. The attempt at a solution
    I know [tex]\tau[/tex][tex]_{friction}[/tex][Nm] = 0.00176 [Nm]
    I am confused about the relation between frictional force and torque...

    [tex]\tau_{friction}[/tex] = F[tex]_{friction}[/tex][N] * radius [m]
    F[tex]_{friction}[/tex][N] = [tex]\tau_{friction}[/tex] / radius [m]
    F[tex]_{friction}[/tex][N] = 0.00176 [Nm] / 0.002 [m] = 0.88 [N]



    (Btw, I get "Database error
    The Physics Help and Math Help - Physics Forums database has encountered a problem." after pushing Preview Post once in a while, this can't be normal right?)
     
  2. jcsd
  3. Jul 20, 2008 #2

    Andrew Mason

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    This is correct. But you have not given us enough information. If you know the geometry of the wheel, you could work out I. You could then work out the torque due to friction. With that you can then determine the retarding force (assume the retarding force is applied to the ball bearings at 2 mm from the centre).

    AM
     
  4. Jul 21, 2008 #3
    Hi Andrew, thanks for the response.

    There is no other information given, just the vertical intercept of [tex]\tau[/tex] vs [tex]\alpha[/tex], and that the ball bearings are 2mm from the center of the wheel. I believe it may be inferred that the wheel is of uniform mass, thus I=1/2 MR[tex]^{2}[/tex] though neither M nor R are given...

    Is it correct to assume that the vertical intercept is equal to the torque from friction?
     
  5. Jul 21, 2008 #4

    Andrew Mason

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    Sorry. I was seeing "intercept" and thinking "slope". You have enough information.

    The intercept gives you the value of [itex]\tau[/itex] when [itex]\alpha[/itex] = ?

    AM
     
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