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How does kinetic friction affect a spinning body?

  1. Aug 23, 2004 #1
    For example, there is a rotating cylinder with a moment of intertia of 1.5, a speed of 10 rad/sec and a force perpendicular to the rim of 5 newtons and they have a kinetic friction coeffecient of .5.

    How long does it take to stop?
     
  2. jcsd
  3. Aug 23, 2004 #2

    chroot

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    Gmaximus,

    Please post homework help problems here, in the Homework Help forum.

    Coefficients of friction always relate normal force (force perpendicular to a surface) to lateral force (force across the surface of an object due to the friction).

    In this case you have a 5N normal force and a 0.5 coefficient of friction, so the force due to friction is:

    [tex]F_f = \mu_k F_N[/tex]

    Where [itex]F_f[/itex] is the force due to the friction and [itex]F_N[/itex] is the normal force.

    Just do that multiplication to find the force (due to friction) that's slowing down the cylinder. Once you have that force, you can use the angular form of Newton's second law to find the acceleration:

    [itex]\tau = I\alpha[/tex]

    where [itex]\tau[/itex] is the torque due to the friction force, I is the moment of inertia, and [itex]\alpha[/itex] is the angular acceleration.

    Then you can just use standard kinematics to find the time it takes the acceleration [itex]\alpha[/itex] to dissipate the initial velocity of 10 rad/sec.

    Let me know if you need any more help.

    - Warren
     
  4. Aug 23, 2004 #3

    Doc Al

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    Welcome to PF!

    Our policy here is for you to present your work so we can see where you are getting stuck. (We don't do the work for you! :smile: )

    Here's a hint to get you started: Find the torque due to the frictional force, then apply Newton's 2nd law (for rotational motion).
     
  5. Aug 23, 2004 #4
    Homework? I laugh at the thought. School hasn't started yet :tongue:

    I was just never taught how friction works with stuff thats rotating, and it was bugging me.

    So, the only force slowing down the rotation is the force of friction?

    If the answer is yes, then that is fantastic.


    On a related note does L=P (Angular vs Linear momentum?) or is it L=PR (R=radius)?
     
    Last edited: Aug 23, 2004
  6. Aug 23, 2004 #5
    It just doesn't make sense that the force of friction is affected by the moment arm...


    Lastly: Do all measurements of length have to be in m?
     
    Last edited: Aug 23, 2004
  7. Aug 23, 2004 #6

    chroot

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    Definition of angular momentum:

    [tex]\mathbf{L} = \mathbf{r} \times \mathbf{p}[/tex]

    The force of friction does not depend on the moment arm -- it depends only on the normal force and the coefficient of friction.

    The torque resulting from that friction force depends on the moment arm, however, because

    [tex]\mathbf{\tau} = \mathbf{r} \times \mathbf{F}[/tex]

    - Warren
     
  8. Aug 23, 2004 #7
    Ok.... thanks alot. You are awesome.

    Will the kinematics give me a wrong answer if i use cm instead of m?
     
  9. Aug 23, 2004 #8

    chroot

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    And no, you can use whatever units of length you want, so long as you are consistent. In other words, if you want to use feet, you have to change your unit of force (because the Newton is defined in terms of meters) and so on.

    All in all, I'd recommend never using anything other than SI units in physics problems unless there's some specific overriding reason.

    - Warren
     
  10. Aug 23, 2004 #9

    chroot

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    Do you mean this equation?

    [tex]\omega = \omega_0 + \alpha t[/tex]

    There are no units of length in this equation -- the angular velocity is in rad/sec, and the acceleration is in rad/sec2.

    - Warren
     
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