Center of mass of a uniform density square centered at the origin. Offset by width/4?

[InvisibleMan1]

Homework Statement

To get used to finding the center of mass of an object, I have decided to start with a uniform density square. The square is centered at the origin. The center of mass should be at the center of the square, and thus at the origin.

When I tried to solve this however, my answer was offset by width/4. I've gone through the integration a couple times and I cannot find any errors in my work...

Homework Equations

Definitions:
M = Total mass of the object
A = Area of the object
M/A = The uniform density of the object.
r = Vector from the origin to a point on the object.

Original equation for the center of mass of a 2D object with uniform density.
M/A*int(int(rdy)dx)*1/M


w = The width/height of the square.
A = w^2
r = xi+yj

Inner integral:
Interval: [-w/2 w/2]
int(xi+yj)dy = xwi+(w^2)/4j

Outer integral:
Interval: [-w/2 w/2]
int(xwi+(w^2)/4j)dx = (w^3)/4i+(w^3)/4j

Substituting into the original equation:
M/(w^2)*((w^3)/4i+(w^3)/4j)*1/M = 1/(w^2)*((w^3)/4i+(w^3)/4j) = w/4i+w/4j

That answer should be 0i+0j, but it isn't.

The Attempt at a Solution

Using r=1i+ij instead of r=xi+yj looks like it would solve the problem, but that breaks the definition, since every point in the square is not located at <1, 1>.

 

[Mark44]

Homework Statement

To get used to finding the center of mass of an object, I have decided to start with a uniform density square. The square is centered at the origin. The center of mass should be at the center of the square, and thus at the origin.

When I tried to solve this however, my answer was offset by width/4. I've gone through the integration a couple times and I cannot find any errors in my work...

Homework Equations

Definitions:
M = Total mass of the object
A = Area of the object
M/A = The uniform density of the object.
r = Vector from the origin to a point on the object.

Original equation for the center of mass of a 2D object with uniform density.
M/A*int(int(rdy)dx)*1/M


w = The width/height of the square.
A = w^2
r = xi+yj

Inner integral:
Interval: [-w/2 w/2]
int(xi+yj)dy = xwi+(w^2)/4j

Outer integral:
Interval: [-w/2 w/2]
int(xwi+(w^2)/4j)dx = (w^3)/4i+(w^3)/4j

Substituting into the original equation:
M/(w^2)*((w^3)/4i+(w^3)/4j)*1/M = 1/(w^2)*((w^3)/4i+(w^3)/4j) = w/4i+w/4j

That answer should be 0i+0j, but it isn't.

The Attempt at a Solution

Using r=1i+ij instead of r=xi+yj looks like it would solve the problem, but that breaks the definition, since every point in the square is not located at <1, 1>.

Why are you using r? Since your square is centered at (0, 0) and is of uniform density, the moments about the x and y axes are both going to be zero, so the center of mass will be at (0, 0). You can set up integrals for M_x and M_y, but you should be able to convince yourself that each integral will be zero, due to the symmetry of the figure and the constant density.

 

[InvisibleMan1]

I know it will be (0,0). I said as much in the post. I chose this because it is a simple example of finding the center of mass. I don't need to find this center of mass. I need experience finding the center of mass. Starting with known results is a very good place to start.

At any rate, the problem has been solved. (-w/2)^2 != -w^2/4