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

2024
Views:

388

Pages:

2
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 167. First compute the mass of each component m = ρstV = Have 0.284 lb/in 3 V = 0.008819 lb ⋅ s 2 /ft ⋅ in 3 V 2 32.2 ft/s ( ) Then 2 m1 = 0.008819 lb ⋅ s 2 /ft ⋅ in 3 π ( 4 in.) ( 2 in.)  = 0.88667 lb ⋅ s 2 /ft   2 m2 = 0.008819 lb ⋅ s 2 /ft ⋅ in 3 π (1 in.) ( 3 in.)  = 0.083125 lb ⋅ s 2 /ft   m3 = 0.008819 lb ⋅ s 2 /ft ⋅ in 3 π (1 in.) ( 2 in.)  = 0.055417 lb ⋅ s 2 /ft   2 I yz = I zx = 0 Symmetry implies ( I xy )1 = 0 ( I x′y′ )2 = ( I x′y′ )3 = 0 and Now ( ) I xy = Σ I x′y′ + mx y = m2 x2 y2 − m3 x3 y3  1 ft 2  = ( 0.083125 lb ⋅ s 2 /ft ) ( 2 in.)(1.5 in.)  ×  2   144 in   1 ft 2  − ( 0.055417 lb ⋅ s 2 /ft ) ( −2 in.)( −1 in.)  ×  2   144 in  = 0.96209 × 10−3 lb ⋅ ft ⋅ s 2 From the solution to Problem 9.143: I x = 34.106 × 10−3 lb ⋅ ft ⋅ s 2 I y = 50.125 × 10−3 lb ⋅ ft ⋅ s 2 I z = 34.876 × 10−3 lb ⋅ ft ⋅ s 2 continued COSMOS: Complete Online Solutions Manual Organization System 1 ( 2i + 3j) 13 λ OA = By observation Then 0 0 0 I OA = I xλx2 + I y λ y2 + I z λz2 − 2I xy λxλ y − 2I yz λ y λz − 2I zxλz λx  2  = 34.106 × 10−3 lb ⋅ ft ⋅ s 2    13  ( ) 2  3  + 50.125 × 10−3 lb ⋅ ft ⋅ s 2    13  ( ) 2  2  3  − 2 0.96209 × 10−3 lb ⋅ ft ⋅ s 2     13  13  ( ) = (10.4942 + 34.7019 − 0.8881) × 10−3 lb ⋅ ft ⋅ s 2 = 44.308 × 10−3 lb ⋅ ft ⋅ s 2 or I OA = 44.3 × 10−3 lb ⋅ ft ⋅ s 2