Explanation Given information: The density of the steel is 7850 kg / m 3 and the thickness of the steel is 2 mm . Draw the diagram for the component of the body. Figure-(1) Write the expression for the mass of the component 1. m 1 = ρ t d 1 L 1 ...... (I) Here, the density of the steel is ρ , the length of the component 1 is L 1 , the depth of the component 1 is d 1 and the thickness is t . Write the expression for the mass of the component 2. m 1 = ρ t d 2 L 2 ...... (II) Here, the length of the component 2 is L 2 , the depth of the component 2 is d 2 and the thickness is t . The product moment of inertia of component 1 is zero in all the planes due to symmetry. ( I ¯ x ′ y ′ ) 1 = 0 ( I ¯ y ′ z ′ ) 1 = 0 ( I ¯ z ′ x ′ ) 1 = 0 Write the expression for the x- coordinate of the component 1. x ¯ 1 = ( L 2 ) ...... (III) Write the expression for the y- coordinate of the component 1. y ¯ 1 = − d 2 2 ...... (IV) Write the expression for the x- coordinate of the component 2. x ¯ 2 = ( L 1 ) ...... (V) Write the expression for the y- coordinate of the component 2. y ¯ 2 = − d 1 2 ...... (VI) Write the expression for the z- coordinate of the component 1. z ¯ 1 = L 2 ...... (VII) Write the expression for the z- coordinate of the component 2. z ¯ 2 = L 2 2 ...... (VIII) Write the expression for the mass moment of inertia of the plane x y . I x y = I ¯ x y + m x ¯ y ¯ ...... (IX) Write the expression for the mass moment of inertia of the plane y z . I y z = I ¯ y z + m y ¯ z ¯ ...... (X) Write the expression for the mass moment of inertia of the plane z x . I z x = I ¯ z x + m z ¯ x ¯ ...... (XI) Calculation: Substitute 7850 kg / m 3 for ρ , 200 mm for L 1 , 120 mm for d 1 and 2 mm for t in Equation (I). m 1 = ( 7850 kg / m 3 ) ( 2 mm ) ( 200 mm ) ( 120 mm ) = ( 7850 kg / m 3 ) ( 2 mm ( 1 m 10 3 mm ) ) ( 200 mm ( 1 m 10 3 mm ) ) ( 120 mm ( 1 m 10 3 mm ) ) = ( 7850 kg / m 3 ) ( 4.8 × 10 − 5 m 3 ) = 0.3768 kg Substitute 7850 kg / m 3 for ρ , 100 mm for L 2 , 120 mm for d 2 and 2 mm for t in Equation (II). m 2 = ( 7850 kg / m 3 ) ( 2 mm ) ( 100 mm ) ( 120 mm ) = ( 7850 kg / m 3 ) ( 2 mm ( 1 m 10 3 mm ) ) ( 100 mm ( 1 m 10 3 mm ) ) ( 120 mm ( 1 m 10 3 mm ) ) = ( 7850 kg / m 3 ) ( 2.4 × 10 − 5 m 3 ) = 0.1884 kg Substitute 100 mm for L 2 in Equation (III). x ¯ 1 = ( 100 mm ) = ( 100 mm ( 1 m 10 3 mm ) ) = 0.1 m Substitute 120 mm for d 2 in Equation (IV). y ¯ 1 = − ( 120 mm 2 ) = − ( 120 mm 2 ( 1 m 10 3 mm ) ) = − 0

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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 B, Problem B.44P

To determine

The mass product of the inertia Ixy.

The mass product of the inertia Iyz.

The mass product of the inertia Izx.

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Chapter B, Problem B.43P

Chapter B, Problem B.45P

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