Explanation Given information: The aluminum wire with a weight per unit length is 0.033 lb / ft . The following figure represents the aluminum wire. Figure-(1) Concept used: Write the expression for the mass of aluminum wire section 1. m 1 = 2 π r g × ( W L ) A L ....... (I) Here, the acceleration due to gravity is g , the mass of the aluminum wire section 1 is m 1 , the weight per unit length of aluminum wire section 1 ( W L ) A L and the radius of aluminum wire is r . The mass of the all section is same. Write the expression for the mass of the aluminum section 2. m 2 = m 3 = m 4 Here, the mass of the aluminum section 2 is m 2 , the mass of the aluminum section 3 is m 3 and the mass of the aluminum section 4 is m 4 . Write the expression for the mass of the aluminum section 5. m 5 = m 4 Here, the mass of the aluminum section 5 is m 5 . Write the expression for the mass moment of the inertia of section 1. ( I x ) 1 = m 1 a 2 2 ....... (II) Here, the mass moment of inertia of section 1 is ( I x ) 1 and the radius of the section 1 is a . Write the expression for the mass moment of the inertia of section 2. ( I x ) 2 = m 2 L 2 3 ....... (III) Here, the mass moment of inertia of section 2 is ( I x ) 2 and the length of the section 2 is L . Write the expression for the mass moment of the inertia of section 3. ( I x ) 3 = I ¯ x + m 3 d 2 ....... (IV) Here, the mass moment of inertia of component 3 with respect to centroidal axis is I ¯ x , the mass moment of inertia of section 3 is ( I x ) 3 and the distance of the section 2 is d . Write the expression for the mass moment of the inertia of section 4. ( I x ) 4 = I ¯ x 4 + m 4 d 2 ....... (V) Here, the mass moment of inertia of component 4 with respect to centroidal axis is I ¯ x 4 , the mass moment of inertia of section 4 is ( I x ) 4 and the distance of the section 4 is d . Write the expression for the mass moment of inertia of section 4 with respect to centroidal axis. I ¯ x 4 = m 4 L 2 12 ....... (VI) Write the expression for the square of the distance of location centroidal axis for section 4. d 2 = b 2 + ( L 2 ) ....... (VII) Here, the square of the distance of location of section 1 is b . Write the expression for the mass moment of the inertia of section 5. ( I x ) 5 = I ¯ x 5 + m 5 d 2 ....... (VIII) Here, the mass moment of inertia of component 5 with respect to centroidal axis is I ¯ x 5 , the mass moment of inertia of section 5 is ( I x ) 5 and the distance of the section 5 is d . Write the expression for the mass moment of inertia of section 5 with respect to centroidal axis. I ¯ x 5 = m 5 L 2 12 ....... (IX) Write the expression for the square of the distance of location centroidal axis for section 5. d 2 = L 2 + ( L + b 2 ) ....... (X) Here, the square of the distance of location of section 1 is b . Write the expression for the mass moment of inertia of section 1 with respect to y axis. ( I y ) 1 = m 1 a 2 ....... (XI) Here, the mass moment of inertia of section 1 with respect to y axis is ( I y ) 1 . Write the expression for the mass moment of inertia of section 2 with respect to y axis. ( I y ) 2 = ( I ¯ y ) + m 2 d 2 ....... (XII) Here, the mass moment of inertia of section 2 with respect to y axis is ( I y ) 2 . Write the expression for the mass moment of inertia of section 3 with respect to y axis. ( I y ) 3 = ( I ¯ y ) + m 3 d 2 ....... (XIII) Here, the mass moment of inertia of section 3 with respect to y axis is ( I y ) 3 . Write the expression for the mass moment of inertia of section 3 with respect to centroidal axis. I ¯ y 3 = m 5 L 2 12 ....... (XIV) Write the expression for the moment of inertia with respect to y axis. I y = ( I y ) 1 + 2 ( I y ) 2 + 2 ( I y ) 3 ....... (XV) Write the expression for the moment of inertia with respect to x axis. I x = ( I x ) 1 + ( I x ) 2 + ( I x ) 3 + ( I x ) 4 + ( I x ) 5 ....... (XVI) Write the expression for the mass of aluminum wire section 2. m 2 = L g × ( W L ) A L ....... (XVII) Here, the acceleration due to gravity is g , the mass of the aluminum wire section 2 is m 2 , the weight per unit length of aluminum wire ( W L ) A L and the length of aluminum wire is L . Mass products of inertia I x y is expressed as follows: I x y = ∑ ( I x ′ y ′ + m x y ) ...... (XXVIII) Here. Mass products of inertia is I x y . Mass products of inertia I y z is expressed as follows: I y z = ∑ ( I y z ) ...... (XXVIX) Here. Mass products of inertia is I y z . Mass products of inertia I z x is expressed as follows: I z x = ∑ ( I z x ) ...... (XXX) Here. Mass products of inertia is I z x . Calculation: Substitute 0.033 lb / ft for ( W L ) A L , 32.2 ft / s 2 for g and 16 in for r in Equation (I). m 1 = 2 π ( 16 in ) 32.2 ft / s 2 × 0.033 lb / ft = 2 π ( 16 in ( 1 ft 12 in ) ) 32.2 ft / s 2 × 0.033 lb / ft = 0.27646 lb 32.2 ft / s 2 = 8.5857 × 10 − 3 lb ⋅ s 2 / ft Substitute 0.033 lb / ft for ( W L ) A L , 32.2 ft / s 2 for g and 8 in for L in Equation (XVII). m 2 = 8 in 32.2 ft / s 2 × 0.033 lb / ft = 8 in ( 1 ft 12 in ) 32.2 ft / s 2 × 0.033 lb / ft = 0.022 lb 32.2 ft / s 2 = 6.832298 × 10 − 4 lb ⋅ s 2 / ft Substitute 8.5857 × 10 − 3 lb ⋅ s 2 / ft for m 1 and 16 in for a in Equation (II). ( I x ) 1 = ( 8.5857 × 10 − 3 lb ⋅ s 2 / ft ) ( 16 in ) 2 2 = ( 8.5857 × 10 − 3 lb ⋅ s 2 / ft ) ( 16 in ( 1 ft 12 in ) ) 2 2 = 7.63173 × 10 − 3 lb ⋅ ft ⋅ s 2 Substitute 8.5857 × 10 − 3 lb ⋅ s 2 / ft for m 1 and 16 in for a in Equation (XI). ( I y ) 1 = ( 8.5857 × 10 − 3 lb ⋅ s 2 / ft ) ( 16 in ) 2 = ( 8.5857 × 10 − 3 lb ⋅ s 2 / ft ) ( 16 in ( 1 ft 12 in ) ) 2 = 0.015263 lb ⋅ ft ⋅ s 2 Substitute 6.832298 × 10 − 4 lb ⋅ s 2 / ft for m 2 and 8 in for L in Equation (III). ( I x ) 2 = ( 6.832298 × 10 − 4 lb ⋅ s 2 / ft ) ( 8 in ) 2 3 = ( 6.832298 × 10 − 4 lb ⋅ s 2 / ft ) ( 8 in ( 1 ft 12 in ) ) 2 3 = ( 6.832298 × 10 − 4 lb ⋅ ft ⋅ s 2 ) 3 = 1.01219 × 10 − 4 lb ⋅ ft ⋅ s 2 Substitute 0 for I ¯ y , 6.832298 × 10 − 4 lb ⋅ s 2 / ft for m 2 and 16 in for d in Equation (XII). ( I y ) 2 = 0 + 6.832298 × 10 − 4 lb ⋅ s 2 / ft ( 16 in ) 2 = 6.832298 × 10 − 4 lb ⋅ s 2 / ft ( 16 in ( 1 ft 12 in ) ) 2 = 1.21463 × 10 − 3 lb ⋅ ft ⋅ s 2 Substitute 0 for I ¯ x , 6.832298 × 10 − 4 lb ⋅ s 2 / ft for m 3 and 8 in for d in Equation (IV). ( I x ) 3 = 0 + ( 6.832298 × 10 − 4 lb ⋅ s 2 / ft ) ( 8 in ) 2 = ( 6.832298 × 10 − 4 lb ⋅ s 2 / ft ) ( 8 in ( 1 ft 12 in ) ) 2 = 3.03654 × 10 − 4 lb ⋅ ft ⋅ s 2 Substitute 6.832298 × 10 − 4 lb ⋅ s 2 / ft for m 4 and 8 in for L in Equation (VI). I ¯ x 4 = ( 6.832298 × 10 − 4 lb ⋅ s 2 / ft ) ( 8 in ) 2 12 = ( 6.832298 × 10 − 4 lb ⋅ s 2 / ft ) ( 8 in ( 1 ft 12 in ) ) 2 12 = 2.530 × 10 − 5 lb ⋅ ft ⋅ s 2 Substitute 16 in for b and 8 in for L in Equation (VII). d 2 = ( 16 in ) 2 + ( 8 in 2 ) 2 = 256 in 2 + 16 in 2 = 272 in 2 ( 1 ft 2 144 in 2 ) = 1

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Author: BEER

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Chapter B, Problem B.62P

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The mass moment of inertia of the aluminum wire section with respect to the axis through the origin characterized by the unit vector (−3i−6j+2k)/7.

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Chapter B, Problem B.61P

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The mass moment of inertia of the aluminum wire section with respect to the axis through the origin characterized by the unit vector ( − 3 i − 6 j + 2 k ) / 7 .

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The mass moment of inertia of the aluminum wire section with respect to the axis through the origin characterized by the unit vector (−3i−6j+2k)/7.

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