Posted by Sleepless on Saturday, April 12, 2008 at 4:13am.
The moment of inertia of the disc before the hole is drilled is
I0 = (1/2) M R0^2.
From that, SUBTRACT the moment of inertia of what the hole would be, if it were solid and displaced h from the center.
Its mass would be m = M(R1/R2)^2
Using the parallel axis theorem, its moment of inertia would be
I' = (1/2)m R1^2 + m h^2.
Subtract I' from I0 for the answer. Don't forget to substitute M(R1/R0)^2 for m.
much thanks!
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