Explanation Given information: The weight per unit length of 0.033 lb / ft . The unit vector λ is ( − 3 i + 6j + 2k ) 7 . Calculation: Sketch the homogeneous wire as shown in figure 1. Find the mass of each component 1 as shown below: m 1 = W g = 1 g ( W / L ) A L L (1) Here, ( W / L ) is weight per unit length and L is the length of the wire. Substitute 0.033 lb / ft for ( W / L ) , 32.2 ft / s 2 for g , and ( 2 π × 16 in . ) for L in Equation (1). m 1 = 1 32.2 × 0.033 × ( 2 π × 16 in . ) × 1 ft 12 in . = 8.5857 × 10 − 3 lb ⋅ s 2 / ft Find the mass of each component as shown below: The mass of components 2,3,4,5 are equal. m 2 = m 3 = m 4 = m 5 = W g = 1 g ( W / L ) A L L (2) Here, ( W / L ) is weight per unit length and L is the length of the wire. Substitute 0.033 lb / ft for ( W / L ) , 32.2 ft / s 2 for g , and ( 8 in . ) for L in Equation (2). m 2 = m 3 = m 4 = m 5 = 1 32.2 × ( 0.033 ) ( 8 ) × 1 12 = 0.6832 × 10 − 3 lb ⋅ ft / s 2 The centroid products of inertia of each component is zero due to symmetry. I ¯ x ′ y ′ = 0 I ¯ y ′ z ′ = 0 I ¯ z ′ x ′ = 0 The centroidal distances of sections 1, 4 and 5 about the x axis is equal to zero. The centroidal distances of section 1 about the y axis is equal to zero. The centroidal distances of sections 1, 2 and 3 about the z axis is equal to zero. x ¯ 1 = x ¯ 4 = x ¯ 5 = 0 y ¯ 1 = 0 z ¯ 1 = z ¯ 2 = z ¯ 3 = 0 Find the moment of inertia ( I x y ) about x y as shown below: ( I x y ) = ∑ ( I ¯ x ′ y ′ + m x ¯ y ¯ ) = ∑ I ¯ x ′ y ′ + m 2 x ¯ 2 y ¯ 2 + m 3 x ¯ 3 y ¯ 3 (3) Here, m 2 is mass of section 2, m 3 is mass of section 3, x ¯ 2 y ¯ 2 is centroid section about x , y axis. Substitute 0 for ∑ I ¯ x ′ y ′ , 0.6832 × 10 − 3 lb ⋅ ft / s 2 for m 2 , 0.6832 × 10 − 3 lb ⋅ ft / s 2 for m 3 , 16 in . for x ¯ 2 , 4 in . for y 2 , 12 in . x ¯ 3 , 8 in . y ¯ 3 in Equation (3). ( I x y ) = ( 0.6832 × 10 − 3 ( 16 in . ( 1 ft 12 in . ) 4 in . ( 1 ft 12 in . ) ) + 0.6832 × 10 − 3 ( 12 in . ( 1 ft 12 in

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ISBN: 9781260804638

Author: BEER

Publisher: MCG

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Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

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Chapter 9.6, Problem 9.172P

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The mass moment of inertia of the component with respect to the axis through origin.

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Chapter 9.6, Problem 9.171P

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! Required information Problem 09.141 - Mass moment of inertia of machine elements NOTE: This is a multi-part question. The machine element shown is fabricated from steel, where h = 64 mm. The density of steel is 7850 kg/m³. 20 mm 40 mm 80 mm 40 mm -20 mm 40 mm Problem 09.141.b - Mass moment of inertia of machine elements Determine the mass moment of inertia of the assembly with respect to the y axis. The mass moment of inertia of the assembly with respect to the y axis is 22.9 ✪ × 10-³ kg.m². X

Problem 09.086 - Orientation of the principal axes and the corresponding moments of inertia For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia when b= 76 mm and h = 56 mm. b The value of mis The value of 0m2 is The value of Imax is The value of I min is 180 270 1.429 7292 × h x 106 mm4. 106 mm4

For the component described in the problem indicated, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.Prob. 9.141Reference to Problem 9.141:

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The mass moment of inertia of the component with respect to the axis through origin.

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