# Center of Mass of Disk with Square Hole

This Demonstration computes the center of mass of a disk of radius 1 with a square hole of circumradius located at a distance from the center of .
The surface area function integrates the difference between the circular edge of the disk and that of the square hole up to a position on the axis.
The abscissa of the center of mass is computed as the value for which the integral is equal to zero.

### DETAILS

Since the disk and the hole are symmetrical with respect to the axis, the coordinate of the center of mass of the upper half is the coordinate of the center of mass of the entire disk and hole. Therefore, only the upper half is considered here.
The function
defines the difference between the disk's outer circle and the edge of the square hole. This is the net surface area at .
The function is the total mass moment relative to . Solving the equation for gives the abscissa of the center of mass.

### PERMANENT CITATION

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