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.
A curcial part of a piece of machinery starts as a flat uniform cylindrical disk of radius R0 and mass M. It then has a circular hole of radius R1 drilled into it. The hole's center is a distance h from the center of the disk. Find the moment of inertia of this disk (with off-center hole) when rotated about its center, C.
Hint: Consider a solid disk and subtract the hole; use parallel-axis theorem.
2 answers
much thanks!
2 answers