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Doc Al
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#12
Dec4-06, 12:06 PM
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Quote Quote by O.J. View Post
BUT, when i divide it, remember we are taking the limit as dm->0 which means there are infinite no. of pie pieces (the more the pie pieces, the straighter they get which makes my assumption for their centre of mass to be R/2 from the centre valid).
Two problems here:
(1) Even in the limit, the pie pieces are still triangles, not straight segments. The center of mass of these triangles is not at R/2. (This should make intuitive sense, since there's more mass between R/2 and R than between 0 and R/2.)
(2) Regardless of the location of the center of mass, what you need to be adding is the rotational inertia of each pie piece. What's the rotational inertia of each pie piece?

Thats the method used by my book in calculating I for a thin ring.
A thin ring has all its mass at the same distance from the axis--very different from your pie slices.

Bottom line: Dividing the disk into pie pieces is not a good plan for caclulating the rotational inertia.