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

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

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Use the parallel axis theorem
I (AB) = I (CM) + m * perpendicular Distance

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Other answers:

Ryt
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.
Yes
so mR^2 /4 + m R^2
No, the moment of inertia rotating a disc about its center axis is mR^2/16
So..wat is ur answer
17mR^2 / 16
Whats the perpendicular distance between the CM and the axis in this problem?
R
No, the *perpendicular* distance
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We want the distance which is perpendicular to the axis of rotation, as I shown on my diagram
Ok..i got it nw.....Thxxxx a lot
Yup..i got it

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