Explanation Given information: The density of steel is 7 , 850 kg / m 3 . Calculation: Sketch the circular ring as shown in Figure 1. Find the mass of component 1 as shown below: m 1 = ρ V 1 (1) Here, ρ is density of steel and V 1 is volume of component 1. Modify Equation (1). m 1 = ρ ( π r 2 h ) (2) Substitute 7 , 850 kg / m 3 for ρ , 80 mm for r , and 40 mm for h in Equation (2). m 1 = 7 , 850 × π ( 80 mm × 1 m 10 3 mm ) 2 ( 40 mm × 1 m 10 3 mm ) = 7 , 850 × π × 0.08 2 × 0.04 = 6.31334 kg Find the mass of component 2 as shown below: m 2 = ρ V 2 (3) Here, ρ is density of steel and V 2 is volume of component 2. Modify Equation (3). m 2 = ρ ( π r 2 h ) (4) Substitute 7 , 850 kg / m 3 for ρ , 20 mm for r , and 60 mm for h in Equation (4). m 2 = 7 , 850 × π ( 20 mm × 1 m 10 3 mm ) 2 ( 60 mm × 1 m 10 3 mm ) = 7 , 850 × π × 0.02 2 × 0.06 = 0.59188 kg Find the mass of component 3 as shown below: m 3 = ρ V 3 (5) Here, ρ is density of steel and V 3 is volume of component 3. Modify Equation (5). m 3 = ρ ( π r 2 h ) (6) Substitute 7 , 850 kg / m 3 for ρ , 20 mm for r , and 40 mm for h in Equation (6). m 3 = 7 , 850 × π ( 20 mm × 1 m 10 3 mm ) 2 4 ( 60 mm × 1 m 10 3 mm ) = 7 , 850 × π × 0.02 2 × 0.04 = 0.39458 kg As the product of inertia about y z and z x as symmetry. It is implies, I y z = 0 I z x = 0 ( I x y ) 1 = 0 As the moment of inertia of section 2 and 3 as symmetry. It is implies, ( I ¯ x ′ y ′ ) 2 = 0 ( I ¯ x ′ y ′ ) 3 = 0 Find the product of the inertia x y as shown in below: I x y = ∑ ( I ¯ x ′ y ′ + m x ¯ y ¯ ) = m 2 x ¯ 2 y ¯ 2 − m 3 x ¯ 3 y ¯ 3 (7) Substitute 0.59188 kg for m 2 , 40 mm for x ¯ 2 , and 30 mm for y ¯ 2 , 0.39458 kg for m 3 , 40 mm for x ¯ 3 , and 20 mm for y ¯ 3 in Equation (7)

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ISBN: 9781259633133

Author: BEER

Publisher: MCG

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Moment of Inertia

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Chapter 9.6, Problem 9.165P

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Find the mass moment of inertia with respect to the line joining the origin O and point A.

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Chapter 9.6, Problem 9.164P

Chapter 9.6, Problem 9.166P

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Find the mass moment of inertia with respect to the line joining the origin O and point A .

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