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

2024
Views:

368

Pages:

2
Author:

Quintin Ware

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: Complete Online Solutions Manual Organization System For a cylindrical ring: ( I AA′ )ring = 2 1  d2  1 d  m2   − m1  1  2  2  2  2 2 1π γt 2 2 1π γt 2 2 d2  d2 −  d1  d1  8 4 g 8 4 g   π γt 4 = d 2 − d14 32 g = ( = = π γt 32 g ) (d 2 2 )( − d12 d 22 + d12 1 mring d 22 + d12 8 ( ) ) Then: I AA′ = ∑ ( I AA′ )ring  1ft  1 lb ⋅ s 2  2 2 2 =  2.7221 × 10−3  0.7 + 0.5 in ×   8 ft   12 in.  ( ) 2  1 ft  1 lb ⋅ s 2  2 2 2 + 196.072 × 10−3  4.4 + 0.7 in ×   8 ft   12 in.  ( − ) 2  1 ft  1 −3 lb ⋅ s  2 2 2 110.945 × 10   4 + 1.2 in ×   8 ft   12 in.  ( )  1 ft  1 lb ⋅ s 2  2 2 2 +  8.8730 × 10−3  5 + 4.4 in ×   8 ft   12 in.  ( ( ) 2 2 2 ) = 1.74857 × 10−6 + 3.3785 × 10−3 − 1.67958 × 10−3 + 341.67 × 10−6 lb ⋅ ft ⋅ s 2 = 2.0423 × 10−3 lb ⋅ ft ⋅ s 2 I AA′ = 2.04 × 10−3 lb ⋅ ft ⋅ s 2 And k AA′ = = I AA′ m 2.04 × 10−3 lb ⋅ ft ⋅ s 2 96.722 × 10−3 lb ⋅ s 2/ft = 0.145311 ft k AA′ = 1.744 in.