Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_7

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The homogeneous slender rod has a uniform cross section and is bent into a circular arc of radius . Determine the - and -coordinates of the mass center of the rod by direct integration. Problem 5/7 (a) Determination of the x-coordinate of the mass center 1. Formula: For a homogeneous slender rod, the -coordinate of the mass center (centroid) is calculated by: In polar coordinates for a circular arc: and . 2. Substitution: Based on the diagram, the rod spans from to . In radians, these are and . The total length is . 3. Calculation: Using trigonometric values: 4. Result: axy x =xˉ L xdL∫ x=acosθdL=adθ θ = 1 45 ∘ θ = 2 180− ∘ 15= ∘ 165 ∘ 4 π 12 11π L= a(θ − 2 θ )= 1 a(165− ∘ 45) = ∘ 180 ∘ π 3 2πa =xˉ = 3 2πa (acos θ )(ad θ ) ∫ 45 ∘ 165 ∘ cos θ d θ 2π 3a ∫ 45 ∘ 165 ∘ =xˉ [sin θ ] = 2π 3a 45 ∘ 165 ∘ (sin165− 2π 3a ∘ sin45) ∘ sin165= ∘ sin(180− ∘ 15)= ∘ sin15≈ ∘ 0.2588 sin45= ∘ ≈ 2 2 0.7071 =xˉ (0.2588− 2π 3a 0.7071)= (−0.4483)≈ 2π 3a −0.2140a =xˉ−0.214a ● Final Conclusion: The -coordinate of the mass center of the rod is . (b) Determination of the y-coordinate of the mass center 1. Formula: Similarly, the -coordinate is calculated by: where and . 2. Substitution: Using the same limits and total length : 3. Calculation: Using trigonometric values: 4. Result: ● Final Conclusion: The -coordinate of the mass center of the rod is . ✨ Final Answer Summary (a) (b) Mecademy AI Solution · ENGProblem 5/7 x=xˉ−0.214a y =yˉ L ydL∫ y=asinθdL=adθ L