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Angular Acceleration of a Disc

  1. Jun 7, 2004 #1
    I think I'm starting to abuse this forum. Anyways, it seems that I haven't grasped the concept of angular acceleration yet since I'm having trouble solving this problem:

    Suppose I have a disc laying on the x-y plane with center at the origin. Suppose the disc is rotating about the z-axis with angular velocity [itex]\vec{w}[/itex] and about the y-axis with angular velocity [itex]\vec{u}[/itex]. What is the angular acceleration of the disc?

    The resultant angular velocity of this disc is [itex]\vec{w} + \vec{u}[/itex] right? I guess since this resultant angular velocity is changing direction as the disc spins, it gives rise to an angular acceleration. I can't really picture the angular acceleration though (i.e. where is it at?).
     
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  3. Jun 7, 2004 #2
    What the heck am I saying!? The resultant angular velocity vector doesn't change direction. Then where is this angular acceleration I'm asked for comming from? This is weird.
     
  4. Jun 8, 2004 #3

    Gza

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    Remember from [tex] \vec{\tau} = I \vec{\alpha} [/tex] that a torque is required to cause an angular acceleration. This body can't somehow cause a force(or torque for that matter) on itself by newton's third law, since all forces(or torques) are interactions involving more than one body. Therefore the author of your problem must be high. :tongue:
     
  5. Jun 8, 2004 #4

    arildno

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    Where is it stated that the resulting angular velocity [tex]u\vec{j}+w\vec{k}[/tex] is a constant in time??
    The angular acceleration is simply [tex]\vec{\alpha}=\dot{u}\vec{j}+\dot{w}\vec{k}[/tex]
     
  6. Jun 8, 2004 #5
    I'm stating it now then: the angular velocity vectors are constant (i.e. do not change with time). If I applied your formula, then I would get an angular acceleration of zero. I'm starting to loose confidence with this problem.
     
  7. Jun 8, 2004 #6

    arildno

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    Well, then the angular acceleration is zero.
    Let [tex]\vec{\omega}[/tex] be the resultant angular velocity vector.
    We may write:
    [tex]\vec{\omega}=\omega\vec{i}_{\omega}[/tex]

    where:
    [tex]\vec{i}_{\omega}=\frac{u}{\sqrt{u^{2}+w^{2}}}\vec{j}+\frac{w}{\sqrt{u^{2}+w^{2}}}\vec{k}[/tex]
    [tex]\omega=\sqrt{u^{2}+w^{2}}[/tex]

    In your case, [tex]\vec{i}_{\omega}[/tex] is a constant vector defining the rotation plane (for which it is the unit normal)
    [tex]\omega[/tex] is the scalar angular velocity.
     
    Last edited: Jun 8, 2004
  8. Jun 8, 2004 #7
    Maybe the correct answer is zero?
     
  9. Jun 8, 2004 #8

    arildno

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    Sorry about the Latex mess; it should be fixed by now.
     
  10. Jun 8, 2004 #9
    Funny, the book has a different answer. I didn't give a verbatim copy of the problem since it's lengthy and has a picture. Maybe my summary of the problem I gave here is not accurate. No matter, since my doubts have been cleared.
     
  11. Jun 8, 2004 #10

    arildno

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    Could you post the answer from the book?
     
  12. Jun 8, 2004 #11
    The answer is numerical, so I don't think that will help unless I quote the problem exactly. Don't worry about it. Thanks again.
     
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