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

2024
Views:

197

Pages:

3
Author:

marthamanayay9qr13

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 148. First compute the mass of each component. m = ρL Have Then m1 = ( 0.049 kg/m )  2π ( 0.32 m )  = 0.09852 kg m2 = m3 = m4 = m5 = ( 0.049 kg/m )( 0.160 m ) = 0.00784 kg Using the equation given above and the parallel axis theorem, have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 + ( I x )5  1   1  2 2 = ( 0.09852 kg )   ( 0.32 m )  + ( 0.00784 kg )   ( 0.160 m )   2    3   2 + ( 0.00784 kg ) 0 + ( 0.160 m )     1  2 2 2 + ( 0.00784 kg )   ( 0.16 m ) + ( 0.08 m ) + ( 0.32 m )   12    2 2 2 1 + ( 0.00784 kg )  ( 0.16 m ) + ( 0.16 m ) + ( 0.32 m − 0.08 m )   12 = ( 5.0442 + 0.06690 + 0.2007 + 0.86972 + 0.66901) × 10−3  kg ⋅ m 2 = 6.8505 × 10−3 kg ⋅ m 2 or I x = 6.85 × 10−3 kg ⋅ m 2 continued COSMOS: Complete Online Solutions Manual Organization System Iy = (Iy ) + (Iy ) + (Iy ) + (Iy ) + (Iy ) Have 1 (I ) = (I ) where Then y 2 y 4 2 and 3 4 5 (I ) = (I ) y 3 y 5 2 2 I y = ( 0.09852 kg ) ( 0.32 m )  + 2 ( 0.00784 kg ) 0 + ( 0.32 m )      2 2 1 + 2 ( 0.00784 kg )  ( 0.16 m ) + ( 0.24 m )  12  = 10.088 + 2 ( 0.80282 ) + 2 ( 0.46831)  × 10−3 kg ⋅ m 2 = 12.6303 × 10−3 kg ⋅ m 2 or I y = 12.63 × 10−3 kg ⋅ m 2 By symmetry continued Iz = Ix or I z = 6.85 × 10−3 kg ⋅ m 2 COSMOS: Complete Online Solutions Manual Organization System To the Instructor: The following formulas for the mass moment of inertia of wires are derived or summarized at this time for use in the solutions of Problems 9.148–9.150. Slender Rod Ix = 0 I y′ = I z′ = 1 mL2 (Fig. 9.28) 12 1 2 mL (Sample Problem 9.9) 3 I y = Iz = Circle Have I y = ∫ r 2dm = ma 2 Now Iy = Ix + Iz Ix = Iz And symmetry implies ∴ Ix = Iz = 1 2 ma 2 Semicircle Following the above arguments for a circle, have I x = Iz = 1 2 ma 2 I y = ma 2 Using the parallel-axis theorem I z = I z′ + mx 2 or x= 4  1 I z′ = m  − 2  a 2 2 π  2a π