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

2022
Views:

20

Pages:

2
Author:

gaarmaakbn9h

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 142. Have Then m = ρcuV = ρcu tA ( ) m1 = m2 = 8940 kg/m 3 ( 0.0008 m )(1.2 m )( 0.15 m ) = 1.28736 kg Using Fig. 9.28 for components and , have I x = ( I x )1 + ( I x )2 and ( I x )1 = ( I x )2 Then 2 1 I x = 2  (1.28736 kg )( 0.15 m )  = 0.0193104 kg ⋅ m 2 3  I x = 19.31 × 10−3 kg ⋅ m 2 ( )1 + ( I y )2 Also Iy = Iy Where ( I y )1 = 13 (1.28736 kg ) (1.2 m )2 + ( 0.15 m )2  = 0.62759 kg ⋅ m 2 and ( I y )2 = ∫ ry2 dm ry2 = x 2 + z 2 = x 2 + (ζ cos 30° ) 2 dm = ρ dV = ρ cu tdζ dx Then ( I y )2 = ρcu t ∫ 0L ∫ 0a ( x2 + ζ 2 cos2 30°) dζ dx continued COSMOS: Complete Online Solutions Manual Organization System 1 L  = ρcu t ∫ 0  ax 2 + a3 cos 2 30°  dx 3   Thus = 1 ρcu t aL3 + a3L cos 2 30° 3 = 1 m2 L2 + a 2 cos 2 30° 3 = 1 (1.28736 kg ) (1.2 m )2 + ( 0.15 m )2 cos2 30° = 0.62517 kg ⋅ m 2 3 ( ( ) where V = aLt ) I y = ( 0.62759 + 0.62517 ) kg ⋅ m 2 I y = 1.253 kg ⋅ m 2 Also I z = ( I z )1 + ( I z )2 Where ( I z )1 = (1.28736 kg )(1.2 m )2 and ( I z )2 1 3 = 0.61793 kg ⋅ m 2 = ∫ rz2dm rz2 = x 2 + y 2 = x 2 + (ζ sin 30° ) 2 dm = ρ dV = ρ cu tdζ dx Then ( I z )2 L a ( ) = ρcu t ∫ 0 ∫ 0 x 2 + ζ 2 sin 2 30° dζ dx 1 L  = ρcu t ∫ 0  ax 2 + a3 sin 2 30°  dx 3   = 1 ρcu t aL3 + a3L sin 2 30° 3 = 1 m2 L2 + a 2 sin 2 30° 3 = 1 (1.28736 kg ) (1.2 m )2 + ( 0.15 m )2 sin 2 30° 3 ( ( ) where V = aLt ) = 0.62035 kg ⋅ m 2 Thus I z = ( 0.61793 + 0.62035 ) kg ⋅ m 2 I z = 1.238 kg ⋅ m 2