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

2024
Views:

289

Pages:

3
Author:

diamundgalx45p

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 144. m = ρsteel V Have m1 = 7850 kg/m 3 ( 0.200 × 0.120 × 0.600 ) m 3 Then = 113.040 kg m2 = 7850 kg/m 3 × ( 0.200 × 0.080 × 0.360 ) m 3 = 45.216 kg 2 π  m3 = 7850 kg/m 3 ×  ( 0.100 ) ( 0.120 )  m 3 2  = 14.7969 kg 2 m4 = 7850 kg/m 3 × π ( 0.050 ) ( 0.120 )  m 3   = 7.3985 kg Using Figure 9.28 for components and and the equations derived above for components and , have ( )1 + ( I y )2 + ( I y )3 Iy = Iy Now where ( )1 Iy  1  0.600 2  0.200 2   2 2 2 = (113.040 kg )  ( 0.600 ) + ( 0.200 )  +  +   m   2  2     12   = 15.0720 kg ⋅ m 2 continued COSMOS: Complete Online Solutions Manual Organization System ( )2 Iy  1  0.360 2  0.200 2   2 2 2 = ( 45.216 kg )  ( 0.360 ) + ( 0.200 )  +  +   m   2  2     12   = 2.5562 kg ⋅ m 2 ( )3 Iy 2   1  16  4 × 0.100    2 2 2  0.100 0.100 0.600 = (14.7969 kg )   − + + +  ( ) ( )    m  2 3π         2 9π  = 6.3024 kg ⋅ m 2 ( I y )4 = ( 7.3985 kg )  12 ( 0.050 )2  + ( 0.100)2 + ( 0.600 )2   m 2   = 2.7467 kg ⋅ m 2 Then I y = (15.0720 + 2.5562 + 6.3024 − 2.7467 ) kg ⋅ m 2 = 21.1839 kg ⋅ m 2 or I y = 21.2 kg ⋅ m 2 continued COSMOS: Complete Online Solutions Manual Organization System To the Instructor: The following formulas for the mass moment of inertia of a semicylinder are derived at this time for use in the solutions of Problems 9.144–9.147. From Figure 9.28 ( I x ) cyl Cylinder: = 1 mcyla 2 2 ( I y )cyl = ( I z )cyl = 121 mcyl ( 3a2 + L2 ) ( Symmetry and the definition of the mass moment of inertia I = ∫ r 2dm imply ( I )semicylinder ∴ and ( I x )sc = = 1 (I ) 2 cylinder 11 2  mcyla  2 2  ( I y )sc = ( I z )sc = 12 121 mcyl ( 3a2 + L2 ) However, msc = 1 mcyl 2 Thus, ( I x )sc = and 1 msca 2 2 ( I y )sc = ( I z )sc = 121 msc ( 3a 2 + L2 ) Also, using the parallel axis theorem find 16  2 1 I x′ = msc  − a 2 π2  9   1 16  2 1 2 I z′ = msc  − a + L 2 12   4 9π  where x′ and z′ are centroidal axes. )