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The moment of inertia of a thin sheet of mass m and radius R about AB ?
 one year ago
 one year ago
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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dw:1360165380404:dw
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
Use the parallel axis theorem
 one year ago

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

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
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

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
so mR^2 /4 + m R^2
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
No, the moment of inertia rotating a disc about its center axis is mR^2/16
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
So..wat is ur answer
 one year ago

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17mR^2 / 16
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
Whats the perpendicular distance between the CM and the axis in this problem?
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
No, the *perpendicular* distance
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
dw:1360168373543:dw
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
dw:1360168379873:dw
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
We want the distance which is perpendicular to the axis of rotation, as I shown on my diagram
 one year ago

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Ok..i got it nw.....Thxxxx a lot
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Yup..i got it
 one year ago
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