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: 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/πR2,

or from the smaller disk as

ρ = m/πr2.

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)}.