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Yahoo! Group Title

The moment of inertia of a thin sheet of mass m and radius R about AB ?

  • one year ago
  • one year ago

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  1. Yahoo! Group Title
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    |dw:1360165380404:dw|

    • one year ago
  2. JamesJ Group Title
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    Use the parallel axis theorem

    • one year ago
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    I (AB) = I (CM) + m * perpendicular Distance

    • one year ago
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    Ryt

    • one year ago
  5. JamesJ Group Title
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    in other words, start with the moment of inertia of the disc rotated through an axis running through the center of the disc, then apply the theorem to shift the rotation axis.

    • one year ago
  6. JamesJ Group Title
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    Yes

    • one year ago
  7. Yahoo! Group Title
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    so mR^2 /4 + m R^2

    • one year ago
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    @JamesJ ??

    • one year ago
  9. JamesJ Group Title
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    No, the moment of inertia rotating a disc about its center axis is mR^2/16

    • one year ago
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    So..wat is ur answer

    • one year ago
  11. Yahoo! Group Title
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    17mR^2 / 16

    • one year ago
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    @JamesJ

    • one year ago
  13. JamesJ Group Title
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    Whats the perpendicular distance between the CM and the axis in this problem?

    • one year ago
  14. Yahoo! Group Title
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    R

    • one year ago
  15. JamesJ Group Title
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    No, the *perpendicular* distance

    • one year ago
  16. JamesJ Group Title
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    |dw:1360168373543:dw|

    • one year ago
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    |dw:1360168379873:dw|

    • one year ago
  18. JamesJ Group Title
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    We want the distance which is perpendicular to the axis of rotation, as I shown on my diagram

    • one year ago
  19. Yahoo! Group Title
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    Ok..i got it nw.....Thxxxx a lot

    • one year ago
  20. Yahoo! Group Title
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    Yup..i got it

    • one year ago
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