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Yahoo!

  • one year ago

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

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  1. Yahoo!
    • one year ago
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    |dw:1360165380404:dw|

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

  3. Yahoo!
    • one year ago
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    I (AB) = I (CM) + m * perpendicular Distance

  4. Yahoo!
    • one year ago
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    Ryt

  5. JamesJ
    • one year ago
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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.

  6. JamesJ
    • one year ago
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    Yes

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

  8. Yahoo!
    • one year ago
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    @JamesJ ??

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

  10. Yahoo!
    • one year ago
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    So..wat is ur answer

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

  12. Yahoo!
    • one year ago
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    @JamesJ

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

  14. Yahoo!
    • one year ago
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    R

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

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

  17. Yahoo!
    • one year ago
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    |dw:1360168379873:dw|

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

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

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

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