Example Problem
A thin sheet with a uniform mass per unit area measures 8.0 m x 8.0 m. It is placed with its center at the center of an x-y coordinate system so that the sides of the sheet are aligned with the coordinate axes. A 2.0 m x 2.0 m piece is then removed from the top right corner of the sheet. Where is the center-of-mass of the piece that remains?
Here we make use of symmetry, and the fact that we know the center-of-mass of the original sheet is located at the origin.
The basic idea is that the original sheet is the sum of the two pieces it was broken into. Let's call the larger of these two pieces piece 1, and the smaller one will be piece 2. Write out an equation to find the x-coordinate of the center-of-mass of the original sheet:
| Xcom |
= |
| x1m1 + x2m2
|  |
| m1 + m2
|
|
If we use the symbol σ to represent the mass per unit area of the sheet, m1 = σ A1 and m2 = σ A2.
| Xcom |
= |
| x1 σ A1 + x2 σ A2
| |
| σ A1 + σ A2
|
|
= |
| x1 A1 + x2 A2
| |
| A1 + A2
|
|
We know Xcom = 0, so:
| 0 |
= |
| x1 A1 + x2 A2
| |
| A1 + A2
|
|
| This gives: x1 |
= |
| -x2 A2
| |
| A1
|
|
What are these values?
x2 = 3.0 m
A2 = 2.0 x 2.0 = 4.0 m2
A2 = A - A2 = 64.0 - 4.0= 60.0 m2
| So: x1 |
= |
| -3.0 * 4.0
| |
| 60.0
|
|
= |
| -12.0
| |
| 60.0
|
|
= -0.20 m |
Because of the symmetry of the situation y1 will have the same value.