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Yahoo!

  • 3 years ago

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

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  1. Yahoo!
    • 3 years ago
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    |dw:1360165380404:dw|

  2. JamesJ
    • 3 years ago
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    Use the parallel axis theorem

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

  4. Yahoo!
    • 3 years ago
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    Ryt

  5. JamesJ
    • 3 years 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
    • 3 years ago
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    Yes

  7. Yahoo!
    • 3 years ago
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    so mR^2 /4 + m R^2

  8. Yahoo!
    • 3 years ago
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    @JamesJ ??

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

  10. Yahoo!
    • 3 years ago
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    So..wat is ur answer

  11. Yahoo!
    • 3 years ago
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    17mR^2 / 16

  12. Yahoo!
    • 3 years ago
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    @JamesJ

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

  14. Yahoo!
    • 3 years ago
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    R

  15. JamesJ
    • 3 years ago
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    No, the *perpendicular* distance

  16. JamesJ
    • 3 years ago
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    |dw:1360168373543:dw|

  17. Yahoo!
    • 3 years ago
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    |dw:1360168379873:dw|

  18. JamesJ
    • 3 years 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!
    • 3 years ago
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    Ok..i got it nw.....Thxxxx a lot

  20. Yahoo!
    • 3 years ago
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    Yup..i got it

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