In one line
Treat the full outer shape as a positive area and each hole or cut-out as a negative area, then use the same weighted-average formula.
Explained simply
If you take a full chocolate bar and bite out a piece, the bar becomes lighter on that side and its balancing point shifts away from the bite. Subtracting the bitten area does exactly that in the calculation.
Technically, a hole removes first moment \( A_h x_h \) from the parent shape, so \( \bar{x} = (A_1 x_1 - A_h x_h)/(A_1 - A_h) \).
How it works
- Draw the complete outer shape (no holes) and take it as part 1.
- List each removed region (circle, semicircle, triangle, rectangle) as a separate part with a negative area.
- Find each part's centroid from the common axes.
- Compute \( \Sigma A_i x_i \) and \( \Sigma A_i y_i \) with negative entries for the holes.
- Divide by the net area \( \Sigma A_i \).
- Check: the centroid should shift away from the holes.
Key formula
\( \bar{x} = \dfrac{A_1x_1 - A_2x_2}{A_1 - A_2} \), \( \bar{y} = \dfrac{A_1y_1 - A_2y_2}{A_1 - A_2} \), where part 2 is the removed area (mm\(^2\)) with centroid \( (x_2, y_2) \) in mm.
Quick example
A 100 × 100 mm square plate with a circular hole of radius 20 mm centred at (70, 50) mm. Square: A = 10000, centroid (50, 50). Hole: A = 1256.64, centroid (70, 50). By symmetry about y = 50, \( \bar{y} = 50 \) mm. \( \bar{x} = (10000 \times 50 - 1256.64 \times 70)/(10000 - 1256.64) = 412035.4/8743.36 = 47.13 \) mm, shifted left, away from the hole as expected.
Common mistakes
- Subtracting the hole's moment but dividing by the gross area instead of the net area.
- Using the hole's diameter as the radius.
- Getting the semicircular cut-out's centroid on the wrong side of its diameter.
How MU asks it
Very common 6 to 8-mark numerical: 'Find the centroid of the plate with a circular hole / semicircular cut-out'. Show positive and negative areas clearly in the table (write the minus signs), give the net area, and comment that G shifts away from the cut-out.