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)/(A1-A2)=
=(A1•0-A2•x2)/(A1-A2)
= - πr²•2/π(R²-r²) =
= - 1•2/(16-1) =2/15 = - 0.133 m
Center of mass x=- 0.133 m, y = 0
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 x-coordinate 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)
2 answers
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.
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.
0 answers