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

2024
Views:

29

Pages:

2
Author:

hekeldigzjbz

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 150. First compute the mass of each component. Mass of each component is indentical m= Have = ( m/L ) L g ( 0.041 lb/ft )(1.5 ft ) 32.2 ft/s 2 = 0.00190994 lb ⋅ s 2 /ft Using the equations given above and the parallel axis theorem, have ( I x )1 = ( I x )3 = ( I x )4 = ( I x )6 and ( I x ) 2 = ( I x )5 I x = 4 ( I x )1 + 2 ( I x )2 Then 2 1 I x = 4 0.00190994 lb ⋅ s 2 /ft  (1.5 ft )  3  ( ( ) ) 2 + 2 0.00190994 lb ⋅ s 2 /ft 0 + (1.5 ft )    = 0.0143246 lb ⋅ ft ⋅ s 2 or I x = 14.32 × 10−3 lb ⋅ ft ⋅ s 2 Now ( I y )1 = 0 ( I y )2 = ( I y )6 ( I y ) 4 = ( I y )5 continued COSMOS: Complete Online Solutions Manual Organization System Then ( )2 + ( I y )3 + 2 ( I y )4 Iy = 2 Iy    1  2 2 = 0.0019094 lb ⋅ s 2 /ft   2   (1.5 ft )  + 0 + (1.5 ft )    3      ( ) 2 2 2  1 + 2  (1.5 ft ) + (1.5 ft ) + ( 0.75 ft )   12   = 0.0019094 (1.5 + 2.25 + 6 ) lb ⋅ ft ⋅ s 2 = 0.0186219 lb ⋅ ft ⋅ s 2 I y = 18.62 × 10−3 lb ⋅ ft ⋅ s 2 By symmetry Iz = I y I z = 18.62 × 10−3 lb ⋅ ft ⋅ s 2

COSMOS Chapter 9 Solution 150

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