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Quarter-Circle Centroid & Moment of Inertia

Medium DifficultyFE Statics

Find the centroid and area moment of inertia of a quarter circle with a square cutout.

Concept

For a composite area with holes, treat cutouts as negative areas. The centroid uses the weighted-average formula, and the moment of inertia uses the parallel-axis theorem to transfer each part's I to a common axis before subtracting. \overset{x}{ˉ} = \frac{\sum A _{i} x ˉ _{i}}{\sum A _{i}}, \overset{y}{ˉ} = \frac{\sum A _{i} y ˉ _{i}}{\sum A _{i}} I = I_{c} + A d^{2} Key property: For a quarter circle of radius R in the first quadrant with center at the origin, \overset{x}{ˉ} = \overset{y}{ˉ} = \frac{4 R}{3 π} and I_{x}_{base} = I_{y}_{base} = \frac{π R ^{4}}{16} .

A quarter circle of radius R = 10 in lies in the first quadrant with its center at the origin. A 5 in \times 5 in square is cut out; the square's bottom-left corner is at the origin. Find: The centroid (\overset{x}{ˉ}, \overset{y}{ˉ}) of the remaining area. The centroidal moments of inertia I_{x} and I_{y} .

Quarter circle: R = 10 in, first quadrant, center at origin Square cutout: 5 in \times 5 in, bottom-left corner at origin

A_{1} = \frac{π R ^{2}}{4} = \frac{π ( 10 ) ^{2}}{4} = 25 π \approx 78.54 in^{2} A_{2} = 5 \times 5 = 25 in^{2} A_{net} = A_{1} - A_{2} = 25 π - 25 = 25 (π - 1) \approx 53.54 in^{2}

The centroid of a quarter circle in the first quadrant (center at origin) is at \frac{4 R}{3 π} from each axis. \overset{x}{ˉ}_{1} = \overset{y}{ˉ}_{1} = \frac{4 R}{3 π} = \frac{4 ( 10 )}{3 π} = \frac{40}{3 π} \approx 4.244 in \overset{x}{ˉ}_{2} = \overset{y}{ˉ}_{2} = \frac{5}{2} = 2.500 in

Subtract the cutout's first moment from the solid quarter circle's: \sum A_{i} \overset{x}{ˉ}_{i} = 25 π \cdot \frac{40}{3 π} - 25 (2.5) = \frac{1000}{3} - 62.50 = 270.83 \overset{x}{ˉ} = \frac{270.83}{53.54} \approx 5.06 in \overset{y}{ˉ} = \overset{x}{ˉ} \approx 5.06 in (by symmetry about y = x)

Use \frac{π R ^{4}}{16} for the quarter circle and \frac{b h ^{3}}{3} for the rectangle about its base: I_{x, 1}_{x -axis} = \frac{π R ^{4}}{16} = \frac{π ( 10 ) ^{4}}{16} = 625 π \approx 1, 963.50 in^{4} I_{x, 2}_{x -axis} = \frac{b h ^{3}}{3} = \frac{5 ( 5 ) ^{3}}{3} = \frac{625}{3} \approx 208.33 in^{4} I_{x, net}_{x -axis} = 625 π - \frac{625}{3} \approx 1, 963.50 - 208.33 = 1, 755.17 in^{4}

Apply the parallel-axis theorem in reverse to transfer from the base axis to the composite centroid: I_{x} = I_{x, net} - A_{net} \overset{y}{ˉ}^{2} = 1, 755.17 - 53.54 (5.06)^{2} \approx 1, 755.17 - 1, 370 = 385 in^{4} I_{y} = I_{x} \approx 385 in^{4} (by symmetry)

Centroid: \overset{x}{ˉ} = \overset{y}{ˉ} \approx 5.06 in Moments of inertia: I_{x} = I_{y} \approx 385 in^{4}

\overset{x}{ˉ} = \frac{\sum A _{i} x ˉ _{i}}{\sum A _{i}}, I = I_{c} + A d^{2}, I_{quarter-circle} = \frac{π R ^{4}}{16}

Use PolyCalc calculators to try this problem yourself.