Moment of inertia of disc with a hole

Suppose we have a disc with a hole, when computing moment of inertia of this about the disc's centre. Why do we subtract the moment of inertia of the removed part from the moment of inertia of complete disc?

-
Moment of inertia of a part of body is independent of remaining body and it is a scalar. – evil999man May 23 '14 at 6:12
Rubbish. Moment of inertia is a tensor quantity – user34304 May 23 '14 at 6:55

Because the moment of inertia for a point mass is:

$$I = mr^2$$

When calculating the moment of inertia for continuous bodies we use calculus to build them up from infinitesimal mass elements, so effectively to calculate the moment of inertia of the disk (without hole) we're doing:

$$I_{disk} = \sum_i^{disk} m_ir^2$$

for the collection of infinitesimal masses $m_i$ that make up the disk. When you create a hole you simply subtract off all the point masses in the hole to remove them from the sum:

$$I_{net} = \sum_i^{disk} m_ir^2 - \sum_j^{hole} m_j r^2$$

where the second sum is over all the infinitesimal masses in the bit you're removing to make the hole. So you end up with:

$$I_{net} = I_{disk} - I_{hole}$$

PS as Stefan points out in his comment, the $I_{hole}$ above is the moment of inertia about the axis of the disk not about the axis through the centre of the hole.

-
+1 correct for rotational symmetric case. In general the hole isn't at the center of the disk. In that nonsymmetric case Parallel axis theorem has to be applied to compute moment of inertia of the hole. – Stefan Bischof May 23 '14 at 6:53

Moment of inertia of a composite body is found out by adding together the moments of inertia of constituent parts(of course, taken with respect to the same axis). Formally, we say that moment of inertia is an additive property. Which simply put here, means that:

$$I_{disk(without\ hole)}=I_{cut\ out\ part}+I_{disk\ with\ hole}$$, where, all moments of inertia are taken with respect to the same axis.

-