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A square sheet of paper of side 2a has its two opposite corners folded inward to meet at the center O (taken as origin). After folding, the two triangular flaps overlap at the center. Find the center of mass of the resulting shape (the folded paper), assuming uniform mass per unit area.

1. (-a/8, 0)
2. (-a/6, 0)
3. (a/12, 0)
4. (-a/12, 0)

Correct answer: (-a/12, 0)

Solution

Let the square have vertices at (+a, 0), (0, +a), (-a, 0), (0, -a) rotated 45 deg, or more simply at corners (+a, +a), (-a, +a), (-a, -a), (+a, -a) (side 2a, centered at O). Fold the right corner (+a, +a) and... Actually, two corners are folded to the center. Take the square with corners at (+-a, +-a). Fold left corner (-a, 0) in the diamond orientation, or take square with side 2a corners at (+-a, +-a). Fold the two corners: say the right corner (a, 0) in diamond orientation, and left corner (-a, 0) to center O. After folding, a triangular flap that was at positions near the corner is now doubled up near the center. The net CM shift: the unfolded paper has CM at O. After folding corner to center: each triangular piece (mass mₜ) moves its CM from (2a/3, 0) (for right triangle) to a reflected position. For the configuration where two opposite corners of a square sheet are folded to the center, the standard result gives x_cm = -a/12.

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