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

2019
Views:

277

Pages:

2
Author:

salcignofi14

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 158. γ st First note m = ρstV = Then  490 lb/ft 3   1ft  m1 =  0.08 in.) ( 6 × 3.6 ) in 2    2 (  12in.   32.2 ft/s  g tA 3 = 15.2174 × 10−3 lb ⋅ s 2 /ft  490 lb/ft 3  2   1 ft  π m2 =  0.08 in.)  (1.8 in.)    2 ( 2   12 in.   32.2 ft/s  3 = 3.5855 × 10−3 lb ⋅ s 2 /ft  490 lb/ft 3  2   1 ft  π m3 =  0.08 in.)  ( 3.6 in.)    2 ( 4   12 in.   32.2 ft/s  3 = 7.1710 × 10−3 lb ⋅ s 2 /ft Note that symmetry implies (I ) x′y ′ 1,2 ( ) = I y′z′ 1,2 (I ′ ′ ) = (I ′ ′ ) xy Now 3 yz 3 ( ) = I z′x′ 1,2 =0 =0 I uv = I u′v′ + mu v continued COSMOS: Complete Online Solutions Manual Organization System Thus I xy = Σmx y = m1x1 y1 − m2 x2 y2 + m3 x3 y3 0  0.6   = (15.2174 × 10−3 lb ⋅ s 2 /ft )  − ft  ft   12  12   4 × 1.8  1 ft   1.8  − ( 3.5855 × 10−3 lb ⋅ s 2/ft )  2.4 − ft    in. × 3π  12 in.   12   = ( −114.131 − 73.326 ) × 10−6 lb ⋅ ft ⋅ s 2 0 or I xy = −187.5 × 10−6 lb ⋅ ft ⋅ s 2 0 I yz = Σmy z = m1 y1 z1 − m2 y2 z2 + m3 y3 z3 Now 0 or I yz = 0 I zx = ( I zx )1 − ( I zx )2 + ( I zx )3 Also 0 0 = m1 z1x1 − m2 z2x2 + ( I zx )3 Now determine ( I zx )3 0 ( dI zx )3 = ( dI z′x′ )3 + z x dm Have   x  γ = ( z )  −   st t x dz   2  g  =− γ st or g ( I zx )3 Therefore, =− I zx = −  t  a32  g 4  t = 4m3 π a32 ( a ) 1 m3a32 2π 1  3.6  7.1710 × 10−3 lb ⋅ s 2 /ft  ft  2π  12  ( ) 2m3 a 2 2m  1 1  a3 z − z 3 dz = − 23  a32 z 2 − z 4  2 ∫0 4 0 π a3 π a3  2 =− Finally ( γ st  π m3 = Now 1 γ st tz a32 − z 2 dz 2 g ) 2 or I zx = −102.7 × 10−6 lb ⋅ ft ⋅ s 2