University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2020
Views:

252

Pages:

2
Author:

Dixie Summers

COSMOS: Complete Online Solutions Manual Organization System Iζ = ∫ x 2dm Now and a Iζ = ∫ x 2dm′ = 2 ∫ 02 x 2 ( ρ tζ dx ) a h  = 2 ρ t ∫ 02 x 2  ( a − 2 x )  dx a   a h a 1 2 = 2 ρ t  x3 − x 4  a 3 4 0 3 h a  a  1a = 2ρ t    −   a  3  2  4 2 = (b) Have 4   1 1 ρ ta3h = ma 2 48 24 2 I y = ∫ ry2dm = ∫  x 2 + (ζ sin θ )  dm   = ∫ x 2dm + sin 2 θ ∫ ζ 2dm Now I x = ∫ ζ 2dm ⇒ I y = Iζ + I x sin 2 θ = 1 1 ma 2 + mh 2 sin 2 θ 24 6 or I y = (c) Have ( m 2 a + 4h 2 sin 2 θ 24 ( ) ) I z = ∫ rz2dm = ∫ x 2 + y 2 dm 2 = ∫  x 2 + (ζ cosθ )  dm   = ∫ x 2dm + cos 2 θ ∫ ζ 2dm = Iζ + I x cos 2 θ = 1 1 ma 2 + mh 2 cos 2 θ 24 6 or I z = m 2 a + 4h 2 cos 2 θ 24 ( ) COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 129. Mass of cylindrical ring: m = ρV = = γ π π 2 2  d 2 − d1  t g4 4  π γ 4 g (d 2 2 ) − d12 t Now treat the wheel as a series of 4 concentric rings. (Note - the steel is treated as a large ring minus two smaller rings.) mwheel = ∑ mring = π 0.310 lb/in 3 4 32.2 ft/s + π 0.284 lb/in 3 4 32.2 ft/s 2 −2 × + ( ) × 1.5 in. × 0.7 2 − 0.52 in 2 2 ( π 0.284 lb/in 3 4 32.2 ft/s π 0.043 lb/in 3 4 32.2 ft/s 2 ) × 1.5 in. × 4.42 − 0.7 2 in 2 2 × 1.1 in. × 42 − 1.22 in 2 2 ( ( ) ) × 1.5 in. × 52 − 4.42 in 2 ( mwheel = 2.7221 × 10−3 + 196.072 × 10−3 − 110.945 × 10−3 + 8.8730 × 10−3 = 96.722 × 10−3 ) lbft⋅ s 2 lb ⋅ s 2 ft continued

COSMOS Chapter 9 Solution 128

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