COSMOS Chapter 9 Solution 132 - Edubirdie

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

2019
Views:

349

Pages:

1
Author:

Madelyn Bradshaw

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 132. For solid cylinder: I = ∴I = 1 2 ma 2 m = ρV = ρπ a 2h 1 ρπ a 4h 2 For ring: I AA′ = 1 1 1 ρπ a24h − ρπ a14h = ρπ h a24 − a14 2 2 2 ( ( m = ρπ h a22 − a12 ) ) (a) By parallel-axis theorem I BB′ = I AA′ + ma12 = 1 ρπ h a24 − a14 + ρπ h a22 − a12 a12 2 ( ) ( ) I BB′ = 1 ρπ h a24 + 2a22a12 − 3a14 2 ( ) (b) For Maximum I BB′ : dI BB′ 1 = ρπ h 4a22a1 − 12a13 = 0 da1 2 ( a1 = 0; ) ( ) 4 a22 − 3a12 = 0 a12 = 1 2 a2 3 a1 = 1 a2 3 (c) Maximum I BB′ : I BB′ = 2 4  1 a  a   ρπ h  a24 + 2a22  2  − 3  2   2   3  3   = 1 2 1   ρπ h  a24 + a24 − a24  2 3 3   = 1 4  ρπ h  a24  2 3  I BB′ = 2 ρπ ha24 3

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