# HW16 - PROBLEM 5.90 The composite body shown is formed by...

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Unformatted text preview: PROBLEM 5.90 The composite body shown is formed by removing a hemisphere of radius r from a Cylinder of radius R and height 2R. Determine (a) the y coordinate of the centroid when r : 3R/4, (b) the ratio n'R for which )7 : 71.2R. SOLUTION Note, for the axes shown )7 V 7R 723R: *21‘ lIIYA s 4 .1 —2.'r{R4 — L] s R4—1r4 =__ 8 Iii—1r} 3 4 1-1[L] =L 1- W 3 R “1(3)“ (0) r=§R Ez— 341R 4 1-1 E' 3 4 or §=—1.113R C AF)“ (in) ;:—1.2R: —1.2R=— 8 R 3R _ 1F] 3 R r 4 3 or [i] 73.2[i +1.6=0 R R Solving numerically %= 0.8844 PROBLEM 5.104 A ZOO-mm-diameter cylindrical duct and a 100 x ZOO-mm rectangular duct are to be joined as indicated. Knowing that the ducts are fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides yith the centroid of the corresponding area. Also note that symmetry implies Z = 0 { z—eemxm 6: \ Removers *4» mm». at s REMM': l A, m2 I Sclm Rm 4 mm3 )7A,m3 | 1 ”(0.2)(03) = 0.1885 0 0.15 0 0.028274 7! __ , _ = — . 2 0.1 2 2 (0 2)(° 1) O 0314 ( ) = 0.06366 0.25 —0.02000 —0.007854 72' It 2 4(0'1) — 0 04244 A 3 2(0'1) : 0.01571 3,, ‘ ‘ ' 0.30 .0.000667 0.004712 4 (0.3)(02) = 0.060 0.15 0.30 0.00900 0.001800 5 (0.3)(0.2) = 0.060 0.15 0.20 0.00900 0.001200 —1r(0.1)z = «0.1571 40“) 0.20 —0.000667 —0.003142 2 37r (0.3)(0.1) = 0.030 0.15 0.25 0.004500 0.007500 (0.3)(0.1) = 0.030 0.15 0.25 0.004500 0.007500 0.337080 0.023667 0.066991 Have X’ZA = 22A: )‘((0.337080 mmz) = 0.023667 mm3 or X = 0.0702 In )7 = 70.2 mm 4 172A 2 WA: 17(0337080 m2) = 00669911111113 or Y = 0.19874m )7 = 198.7 mm 4 PROBLEM 5.125 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R. SOLUTION First note that symmetry implies f 2 0 4 The element of area (1A of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now M = (m)(Rd¢9) A 2r yEL : ‘— where r = Rsint9 so that dA : ”R2 sin 0‘16 YEL = igsinﬁ It Then A = £2:er2 sinﬂdﬂ : KR2 [—0056]? = ”R2 and IFELdA = §(—3551n9](z122sin9d6) n l = #2R3[2 _ sinlﬂ]2 2 4 0 = —£R3 2 _ , _ It Now yA z ijLdA: y(ﬂR2) = -ER3 or - —lR< y 2 Symmetry implies ...
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