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)
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
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.
To find the x-coordinate of the center of gravity of the system, we can calculate the weighted average of the x-coordinates of the disk and the hole.
First, let's consider the disk. Since it is a uniform disk, the center of gravity of the disk is at the center point of the disk, which is (0, 0).
Next, let's consider the hole. The center of the hole is given as (2, 0). Since the hole has a circular shape, we can assume that its center of gravity is also at its center point.
Now, let's find the weighted average of the x-coordinates of the disk and the hole, taking into account their respective areas.
The area of the disk can be calculated using the formula for the area of a circle: A_disk = π * r^2, where r is the radius of the disk. So, the area of the disk is A_disk = π * (4.00 m)^2.
The area of the hole can be calculated in the same way: A_hole = π * r^2, where r is the radius of the hole. So, the area of the hole is A_hole = π * (1.00 m)^2.
To calculate the weighted average of the x-coordinates, we need to multiply the x-coordinate of each component by its corresponding area, and then divide the sum of these products by the total area.
x_avg = (x_disk * A_disk + x_hole * A_hole) / (A_disk + A_hole)
Since the x-coordinate of the disk is 0 and the x-coordinate of the hole is 2, the equation becomes:
x_avg = (0 * A_disk + 2 * A_hole) / (A_disk + A_hole)
Substituting the values into the equation, we have:
x_avg = (0 * π * (4.00 m)^2 + 2 * π * (1.00 m)^2) / (π * (4.00 m)^2 + π * (1.00 m)^2)
Simplifying the expression, we get:
x_avg = (0 + 2 * π) / (π * 16 + π)
x_avg = (2π) / (17π)
x_avg = 2 / 17
Therefore, the x-coordinate of the center of gravity of the system is 2/17.