- #1
jeremy5561
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I know that the moment of inertia of a disk about the axis perpendicular to the plane of the disk is
I=1/2 MR^2
and by the perpendicular axis theorem the moment of inertia about the other axis is
I=1/4 MR^2
I want to get the 2nd result with calculus. I can't get the right answer. What am I doing wrong?
I know the moment of inertia of a rod is 1/12 MR^2
I take the mass of a section of the disc to me pyr dx where y is the depth of the disc
but when I integrate it gets really messy.
What's the correct way to do this?
http://individual.utoronto.ca/jeremyli/moi.jpg
I=1/2 MR^2
and by the perpendicular axis theorem the moment of inertia about the other axis is
I=1/4 MR^2
I want to get the 2nd result with calculus. I can't get the right answer. What am I doing wrong?
I know the moment of inertia of a rod is 1/12 MR^2
I take the mass of a section of the disc to me pyr dx where y is the depth of the disc
but when I integrate it gets really messy.
What's the correct way to do this?
http://individual.utoronto.ca/jeremyli/moi.jpg
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