Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_30

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the -coordinate of the centroid of the volume obtained by revolving the shaded triangular area about the -axis through . Problem 5/30 (a) Calculation of the -coordinate of the centroid of the volume 1. Formula: The -coordinate of the centroid of a volume of revolution about the - axis is given by: where the differential volume element using horizontal disks is . 2. Substitution: From the given diagram, the shaded triangular area has vertices at , , and . The hypotenuse is defined by the equation , which gives the inner radius for a horizontal disk at height as . The outer boundary is the vertical line at , so the outer radius is . The height ranges from to . Substituting these into the volume integral: 3. Calculation: Total Volume : z z360 ∘ z zz =zˉ V zdV∫ dV=π(r − out 2 r )dz in 2 (r,z)=(0,0)(a,0)(a,a/2) z= r = a a/2 2 r zr = in 2z r=ar = out a z0a/2 V= π ( a − ∫ 0 a/2 2 (2z))dz 2 z d V = ∫ z ⋅ ∫ 0 a/2 π(a− 2 4z)dz 2 V V=π ( a − ∫ 0 a/2 2 4z)dz= 2 πaz− z [ 2 3 4 3 ] 0 a/2 Centroid Integral : Final Coordinate: 4. Result: ● Final Conclusion: The -coordinate of the centroid of the generated volume is . This represents a point located at of the total height above the -axis. ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 5/30 V=πa − = [ 2 ( 2 a ) 3 4 ( 2 a ) 3 ]π − ⋅ = [ 2 a 3 3 4 8 a 3 ]π − = ( 2 a 3 6 a 3 ) zdV∫ z d V = ∫π ( az − ∫ 0 a/2 2 4z)dz= 3 π az−z [ 2 1 224 ] 0 a/2 z d V = ∫π a − = [ 2 1 2 ( 2 a ) 2