Explanation Given: The specific weight of the material, γ = 90 lb / ft 3 . Show the assembly of the material as in Figure 1. From Figure 1, Compute the moment of inertia of the assembly about an axis passing through the center of mass O . I O = I G + m d 2 = I G + m ( a + b ) 2 (I) Here, the mass of the material is m , the diameter is d , and the moment of inertia about G is I G . Compute the moment of inertia of the assembly about an axis passing through the center of mass G . I G = { 1 2 [ γ g π ( a + b ) 2 e ] ( a + b ) 2 − 1 2 [ γ g π b 2 e ] b 2 + 1 2 [ γ g π ( b ) 2 d ] b 2 − 1 2 [ γ g π ( e ) 2 ( d ) ] e 2 } (II) Here, the acceleration due to gravity is g . Compute the mass of the material. m = γ g π ( b 2 − e 2 ) d + γ g π [ ( a + b ) 2 − b 2 ] e (III) Conclusion: Take the value of g as 32 .2 ft / s 2 . Substitute 32 .2 ft / s 2 for g , 0 .5 ft for a , 0 .25 ft for d , 2 ft for b , 1 ft for e , 1 ft for c , and 90 lb / ft 3 for γ in Equation (II). I G = { 1 2 [ 90 32.2 π ( 0.5 + 2 ) 2 ( 1 ) ] ( 0.5 + 2 ) 2 − 1 2 [ 90 32.2 π ( 2 ) 2 ( 1 ) ] ( 2 ) 2 + 1 2 [ 90 32.2 π ( 2 ) 2 ( 0.25 ) ] ( 2 ) 2 − 1 2 [ 90 32.2 π ( 1 ) 2 ( 0.25 ) ] ( 1 ) 2 } = { ( 0

INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

14th Edition

ISBN: 9780133918922

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 10.8, Problem 107P

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The moment of inertia of the assembly about an axis that is perpendicular to the page passing through the center of mass O .

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Chapter 10.8, Problem 106P

Chapter 10.8, Problem 108P

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INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

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The moment of inertia of the assembly about an axis that is perpendicular to the page passing through the center of mass O .

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The moment of inertia of the assembly about an axis that is perpendicular to the page passing through the center of mass O .

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