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Physics 13 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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OpenStudy (jamesj):

Use the parallel axis theorem

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Ryt

OpenStudy (jamesj):

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.

OpenStudy (jamesj):

Yes

OpenStudy (anonymous):

so mR^2 /4 + m R^2

OpenStudy (anonymous):

@JamesJ ??

OpenStudy (jamesj):

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

OpenStudy (anonymous):

So..wat is ur answer

OpenStudy (anonymous):

17mR^2 / 16

OpenStudy (anonymous):

@JamesJ

OpenStudy (jamesj):

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

OpenStudy (anonymous):

R

OpenStudy (jamesj):

No, the *perpendicular* distance

OpenStudy (jamesj):

|dw:1360168373543:dw|

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