Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_27

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the -coordinate of the centroid of the shaded area. Problem 5/27 (a) Determination of the -coordinate of the centroid ● Calculation Process 1. Present Final Formula: The shaded area is a circular sector of outer radius with a smaller concentric circular sector of radius removed. The area is symmetric about the -axis, so the centroid lies on the -axis (). The formula for the -coordinate of the centroid of a circular sector with radius and central angle (where is bisected by the -axis) is: The -coordinate of the composite area is found using: where index 1 refers to the outer sector and index 2 refers to the inner sector. 2. Substitute Values: From the diagram, the total central angle is . Thus, radians. For Area 1 (outer sector): , y y R = 2 a R = 1 a/2 yy=xˉ0 yr2α 2αy =yˉ sector 3α 2rsinα y =yˉ = A ∑ i A ∑ i yˉ i A −A 12 A −A 1 yˉ 12 yˉ 2 2α=45+ ∘ 45= ∘ 90 ∘ α=45= ∘ 4 π R = 2 aα= 4 π A = 1 R α= 2 2 a = 2 ( 4 π ) 4 πa 2 =yˉ 1 = 3(π/4) 2asin(π/4) = 3π/4 2a( /2) 2 3π 4 a 2 For Area 2 (inner sector): , 3. Partial Operations: Calculate the numerator () and the denominator ( ): 4. Final Calculation: 5. Result: ● Final Conclusion: The -coordinate of the centroid of the shaded annular sector area is . ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 5/27 R = 1 a/2α= 4 π A = 2 R α= 1 2 (a/2) = 2 ( 4 π ) 16 πa 2 =yˉ 2 = 3(π/4) 2(a/2)sin(π/4) = 3π/4 a( /2) 2 3π 2 a 2 A ∑ i yˉ i A net A = net A − 1 A = 2 − 4 πa 2 = 16 πa