**Problem 2B-12*: Center of mass of a disk with a hole**

Find the center of mass of a uniform thin disk of radius R with a hole of radius r cut out as shown in Fig. 1. The center of the original disk without the hold is O, and the center of the hole is H. The distance between C and H is denoted as d.

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solution

**Solution:**

Let the cross point of line OH with the edge of the disk be P as shown in Fig. 2.
Also let the mass of the original disk without the hole be M, and the mass of the
disk of the size of the hole be m. The density of the material that makes up the
disk is denoted as ρ. The density ρ can be calculated from the original disk as

ρ = M/πR^{2},

or from the smaller disk as

ρ = m/πr^{2}.

Equating the right-hand sides of the above two equations, we have

m = (r/R)^{2}·M.

The center of mass of the disk with the hole is denoted as C, and the distance PC
as *x*. From the symmetry consideration with respect to the line PO, C must
lie on line PO. The distance PH is R - d, and the distance PO is R. We may consider
an original disk of mass M, without the hole, located at point O, and a disk of a
negative mass, -m, located at point H. Applying Eq. (5) of Topic 2B-02 to this
configuration, we get

*x* = {M·R- m·(R - d)}/(M - m) = R + {m·d/(M - m)}.