Physics!
posted by Chris on .
The figure shows a uniform disk with a circular hole cut out of it. The disk has a radius of 4.00 m. The hole's center is at 2.00 m, and its radius is 1.00 m. What is the xcoordinate of the center of gravity of the system?
The center of the disk is (0,0)
The center of the hole is at (2,0)

x1=0, y1=0, x2=2m, y2=0.
R=4 m, r=1 m.
y(c.m.) =0
x(c.m.) = (A1•x1 –A2•x2)/(A1A2)=
=(A1•0A2•x2)/(A1A2)
=  πr²•2/π(R²r²) =
=  1•2/(161) =2/15 =  0.133 m
Center of mass x= 0.133 m, y = 0 
Hint: If you take the equation for the cente of the mass, and you multiply that by the total mass, you get:
Integral of rho(r) r d^3r
This satisfies the superposition principle, the result for a density function rho1 + rho2 is the same as that for rho1 and rho2 separately and then added together.
This means that you can add up the what you get for a disk without a hole to what you get for a hypothetical disk of negative mass. You then divide the result by the total mass.