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

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|dw:1360165380404:dw|

JamesJ

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Use the parallel axis theorem

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I (AB) = I (CM) + m * perpendicular Distance

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Ryt

JamesJ

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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.

JamesJ

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Yes

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so mR^2 /4 + m R^2

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@JamesJ ??

JamesJ

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No, the moment of inertia rotating a disc about its center axis is mR^2/16

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So..wat is ur answer

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17mR^2 / 16

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@JamesJ

JamesJ

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Whats the perpendicular distance between the CM and the axis in this problem?

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R

JamesJ

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No, the *perpendicular* distance

JamesJ

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|dw:1360168373543:dw|

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|dw:1360168379873:dw|

JamesJ

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We want the distance which is perpendicular to the axis of rotation, as I shown on my diagram

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Ok..i got it nw.....Thxxxx a lot

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Yup..i got it