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

2020
Views:

334

Pages:

2
Author:

lankstoetp13

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 140. First compute the mass of each component m = ρV = ρ At Have Now ( ) = ( 7530 kg/m ) ( 0.002 m )( 0.045 m )( 0.020 m ) = 0.013554 kg m1 = 7530 kg/m 3 ( 0.002 m )( 0.045 m )( 0.070 m ) = 0.047439 kg m2 3 ( ) m3 = 7530 kg/m 3 ( 0.002 m ) × 1 ( 0.04 m )( 0.095 m ) = 0.028614 kg 2 Using Fig. 9.28 for components and and the equations derived above for components , have I x = ( I x )1 + ( I x )2 + ( I x )3 2 2 2 2  1 1 = ( 0.047439 kg )  ( 0.07 )  m 2 + ( 0.013554 kg )  ( 0.020 ) + ( 0.07 ) + ( 0.01)   m 2   3  12 1 2 2 2 2  1 + ( 0.028614 kg )  ( 0.095 ) + ( 0.04 )  + ( 2 × 0.095) + ( 0.040 )   m 2     9 18 = ( 0.077484 + 0.068222 + 0.136751) × 10 −3  kg ⋅ m 2 = 0.282457 × 10 −3 kg ⋅ m 2 or I x = 0.2825 × 10−3 kg ⋅ m 2 continued COSMOS: Complete Online Solutions Manual Organization System continued ( )1 + ( I y )2 + ( I y )3 Iy = Iy  2 2 2  1  1 = ( 0.047439 kg )  ( 0.045 )  m 2 + ( 0.013554 kg )  ( 0.045 ) + ( 0.02 )     m 2    3  3   1 2 2 2 1 + ( 0.028614 kg )  ( 0.04 ) + ( 0.045 ) + ( 0.04 )  m 2 9  18  = ( 0.03202 + 0.010956 + 0.065574 ) × 10 −3  kg ⋅ m 2 = 0.10855 × 10−3 kg ⋅ m 2 or I y = 0.1086 × 10−3 kg ⋅ m 2 I z = ( I z )1 + ( I z )2 + ( I z )3   2 2  1  21 2 = ( 0.047439 kg )  ( 0.045 ) + ( 0.070 )     m 2 + ( 0.013554 kg ) ( 0.045 )   + ( 0.070 )  m 2   3 3        2 1 2 2   + ( 0.028614 kg )  ( 0.095 ) + 0.0452 +  0.095   m 2 3   18 = ( 0.0109505 + 0.075564 + 0.187064 ) × 10−3  kg ⋅ m 3 = 0.37213 × 10−3 kg ⋅ m 2 or I z = 0.372 × 10−3 kg ⋅ m 2