Explanation Given information: The uniform density and weight is W . Calculation: Sketch the thin bent plate as shown in Figure 1. Express the mass of section 1 and 2 as shown below: m 1 = m 2 = 1 2 W g Find the magnitude of OA as shown below: λ O A = 1 3 ( i + j + k ) Refer to Figure 9.28, “Mass moment of inertia” in the textbook. Find the moment of inertia about x axis as shown below: I x = ( I x ) 1 + ( I x ) 2 = ( 1 12 m 1 c 1 2 + m 1 d 1 2 ) + ( 1 12 m 2 c 2 2 + m 2 d 2 2 ) (1) Here, ( I x ) 1 is moment of inertia about x axis section 1 and ( I x ) 2 is moment of inertia about x axis section 2. Refer to Figure 1. Substitute 1 2 W g for m 1 , a for c 1 , ( a 2 ) for d 1 , ( a 2 + a 2 ) for c 2 2 , and ( a 2 ) 2 + ( a 2 ) 2 for d 2 2 in Equation (1). I x = [ ( 1 12 ) ( 1 2 W g ) a 2 + ( 1 2 W g ) ( a 2 ) 2 + ( 1 12 ) ( 1 2 W g ) ( a 2 + a 2 ) + ( 1 2 W g ) ( a 2 ) 2 + ( a 2 ) 2 ] = 1 2 W g [ ( 1 12 + 1 4 ) a 2 + ( 1 6 + 1 2 ) a 2 ] = 1 2 W g a 2 Find the moment of inertia about y axis as shown below: I y = ( I y ) 1 + ( I y ) 2 = ( 1 12 m 1 c 1 2 + m 1 d 1 2 ) + ( 1 12 m 2 c 2 2 + m 2 d 2 2 ) (2) Here, ( I y ) 1 is moment of inertia about y axis section 1 and ( I y ) 2 is moment of inertia about x axis section 2. Refer to Figure 1. Substitute 1 2 W g for m 1 , ( a 2 + a 2 ) for c 1 , ( a 2 ) 2 + ( a 2 ) 2 for d 1 , ( a 2 ) for c 2 2 , and a 2 + ( a 2 ) 2 for d 2 2 in Equation (2). I y = [ 1 12 ( 1 2 W g ) ( a 2 + a 2 ) + 1 2 W g [ ( a 2 ) 2 + ( a 2 ) 2 ] + 1 12 ( 1 2 W g ) ( a 2 ) + 1 2 W g [ ( a ) 2 + ( a 2 ) 2 ] ] = 1 2 W g [ ( 1 6 + 1 2 ) a 2 + ( 1 12 + 5 4 ) a 2 ] = 1 2 W g [ ( 2 3 ) a 2 + ( 64 48 ) a 2 ] = 1 2 W g ( 2 a 2 ) I y = W g a 2 Find the moment of inertia about z axis as shown below: I z = ( I z ) 1 + ( I z ) 2 = ( 1 12 m 1 c 1 2 + m 1 d 1 2 ) + ( 1 12 m 2 c 2 2 + m 2 d 2 2 ) (3) Here, ( I z ) 1 is moment of inertia about z axis section 1 and ( I z ) 2 is moment of inertia about z axis section 2. Refer to Figure 1. Substitute 1 2 W g for m 1 , a for c 1 , ( a 2 ) for d 1 , ( a 2 ) for c 2 2 , and ( a ) 2 + ( a 2 ) 2 for d 2 2 in Equation (3). I z = [ 1 12 ( 1 2 W g ) a 2 + 1 2 W g ( a 2 ) 2 + ( 1 12 ) ( 1 2 W g ) a 2 + 1 2 W g ( a 2 ( a 2 ) 2 ) ] = 1 2 W g ( 1 12 + 1 4 ) a 2 + ( 1 12 + 5 4 ) a 2 = 1 2 W g ( 16 48 + 72 48 ) a 2 = 1 2 W g ( 88 48 ) a 2 I z = 5 6 W g a 2 As the centroidal products of inertia I x ′ y ′ , I z ′ y ′ , and I x ′ z ′ both components of zero. Because of symmetry. The centroid of section y ¯ 1 is 0. Find the product of inertia about x axis as shown in below: I x y = ∑ ( I ¯ x ′ y + m x ¯ y ¯ ) = ∑ I ¯ x ′ y + ( m x ¯ y ¯ ) 1 + ( m x ¯ y ¯ ) 2 (4) Substitute 0 for ∑ I ¯ x ′ y , 1 2 W g for m 1 , 1 2 W g for m 2 ,0 for y ¯ 1 , a for x ¯ 2 , and ( a 2 ) for y ¯ 2 in Equation (4)

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ISBN: 9781259633133

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 9.167P

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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 9.166P

Chapter 9.6, Problem 9.168P

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

Solution Summary: The author calculates the mass moment of inertia with respect to the line joining the origin O and point A.

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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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