Explanation Draw the schematic diagram of the bent rod as shown in Figure (1). Here, length of the rod is b and angular velocity is ω . Write the eexpression to determine the angular momentum ( H ) H = { ( I x x ω x − I x y ω y − I x z ω z ) i + ( − I y x ω x + I y y ω y − I y z ω z ) j + ( − I z x ω x − I z y ω y + I z z ω z ) k } (I) Here, second moment of inertias along x − x , y − y , and z − z are I x x , I x y , I x z , I y x , I y y , I y z , I z x , I z y , and I z z , and angular velocities are ω x , ω y , and ω z . Substitute 0 for ω y and 0 for ω z in Equation (I). H = { ( − I x z ω z ) i + ( − I y z ω z ) j + ( I z z ω z ) k } (II) Write the expression to determine the moment of inertias of the sections using Figure (I). Part I x z I y z I z z 1 0 0 1 3 ( ρ b ) ( b 2 ) = 1 3 ρ b 3 2 0 ( ρ b ) ( b ) b 2 = 1 2 ρ b 3 ( ρ b ) ( b 2 ) = ρ b 3 3 ( ρ b ) ( b ) b 2 = 1 2 ρ b 3 ( ρ b ) ( b ) ( b ) = ρ b 3 1 12 ( ρ b ) ( b 2 ) + ( ρ b ) [ ( b ) 2 + ( b 2 ) 2 ] ∑ 1 2 ρ b 3 3 2 ρ b 3 8 2 ρ b 3 Here, mass per unit length is ρ . Substitute 1 2 ρ b 3 for I x z , 3 2 ρ b 3 for I y z , 8 2 ρ b 3 for I z z , ω for ω x , ω y , and ω z in Equation (II). H = { ( − 1 2 ρ b 3 ω ) i + ( − 3 2 ρ b 3 ω ) j + ( 8 3 ρ b 3 ω ) k } = ( ρ b 3 ω ) [ − 1 2 i − 3 2 j + 8 3 k ] Thus, the angular momentum about the origin is ( ρ b 3 ω ) [ − 1 2 i − 3 2 j + 8 3 k ] . Write the expression to determine the position of the center of mass of the system of rod. r ¯ = ∑ m i r i ∑ m i = m A r A + m B r B + m C r C m A + m B + m C = r A i + r B j + r C k (III) Here, mass of the rod is m i , position of the rod is r i , masses of the rod in positions are m A , m B , and m C , distance between the rod are r A , r B , and r C

8th Edition

ISBN: 9781118885840

Author: James L. Meriam, L. G. Kraige, J. N. Bolton

Publisher: WILEY

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Chapter 7.8, Problem 57P

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The angular momentum along G(HG).

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The angular momentum along G ( H G ) .

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The angular momentum along G(HG).

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