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Angular momentum of a disk

  1. Apr 12, 2007 #1
    A woman with a mass of 45.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.490 rev/s about an axis through its center. The disk has a mass of 120 kg and a radius of 4.50 m. Calculate the magnitude of the total angular momentum of the woman-plus-disk system. (Assume that you can treat the woman as a point.)

    Ok, well I first converted rev/s into rad/s.

    0.490 rev/s * 2(pi) = 3.079 rad/s

    Next I determined that the moment of inertia for a disk is

    (1/2)mr^2

    Next I used the equation for angular momentum:

    L = I(omega)
    = [(1/2)mr^2]omega
    = [(1/2)(120 kg + 45 kg)(4.50m)^2]*3.079 rad/s
    = 5143.8 kg m^2/s

    This one seemed really straightforward to me, but the program I use for physics says the answer isn't right. Anyone know what I did wrong?
     
  2. jcsd
  3. Apr 12, 2007 #2
    I believe you omitted the radius squared term for the disk in your next to last eqn.
     
  4. Apr 12, 2007 #3

    Dick

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    Aside from the fact that one should never treat women as points, I think the problem is that she has a factor of (1/2) on her. Her moment of inertia is m*r^2.
     
  5. Apr 12, 2007 #4
    right, they hate to be treated as point objects. :mad:

    You're right I didn't see that the R^2 was factored out, and the 1/2 also factored. Its funny you get used to seeing things presented a certain way the errors in perception one can make at a glance. I even looked twice.
     
  6. Apr 12, 2007 #5

    Dick

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    I've looked more than twice and still been wrong. See my most recent fiasco in thermodynamics...
     
  7. Apr 13, 2007 #6
    How is the woman mr^2. Is she considered a hollow cylinder? If so, then, I have to add the momentum of the disk itself and then the momentum of her separately? How do account for the fact that she is on the disk, such that the disk is heavier with her on it?
     
  8. Apr 13, 2007 #7
    Point mass I=mR^2 which you are advised to treat her as, this eqn is one of the few moment of inertias that follows immediately from its definition
     
  9. Apr 13, 2007 #8

    Dick

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    The disk isn't 'heavier'. It just has a normal force acting on it - which doesn't affect its moment of inertia. The moment of inertia of the hollow cylinder and a point mass are both mr^2 for the same reason - all of their mass is concentrated at the rim.
     
  10. Nov 11, 2007 #9
    ive got this same question, and i am clueless on where to go
     
  11. Nov 12, 2007 #10

    Dick

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    Just add the moments of inertia of the point-woman and the disk and multiply by the angular velocity. If that's not enough of a clue, then I'd suggest you repost the question so others can help. Resurrecting old threads is a bad idea. They have too much confusing baggage attached.
     
  12. Nov 12, 2007 #11
    Hello Ph123

    What is the program you use for physics?

    C
     
  13. Nov 12, 2007 #12
    they are both mr^2 right?
     
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