Explanation Given information: The mass per unit length of wire is m ′ . Calculation: Sketch the aluminum as shown in Figure 1. Express the mass of each component as shown below: m = ( m L ) L (1) Here, ( m L ) is mass per unit length of wire. Substitute m ′ for ( m L ) in Equation (1). m = m ′ L Find the mass of the component 1 and 5 as shown below: m 1 = m 5 = m ′ L (2) Substitute ( π 2 R 1 ) for L in Equation (2). m 1 = m 5 = m ′ ( π 2 R 1 ) = m ′ ( π 2 R 1 ) Find the mass of the component 2 and 4 as shown below: m 2 = m 4 = m ′ L (3) Substitute ( R 2 − R 1 ) for L in Equation (3). m 2 = m 4 = ( ( R 2 − R 1 ) ) L Find the mass of the component 3 as shown below: m 3 = m ′ L (4) Substitute ( π 2 R 1 ) for L in Equation (4). m 3 = m ′ ( ( π 2 R 1 ) ) = m ′ π 2 R 1 As the symmetry of the centroid products of inertia, ( I ¯ x ′ y ′ ) 1 = ( I ¯ x ′ z ′ ) 1 = 0 ( I ¯ y ′ z ′ ) 5 = ( I ¯ x ′ z ′ ) 5 = 0 ( I ¯ x ′ y ′ ) 3 = ( I ¯ y ′ z ′ ) 3 = 0 And component 2 and 4 also zero. The centroid of section 1, 2,3,4,5 sections is zero. x ¯ 1 = x ¯ 2 = 0 y ¯ 2 = y ¯ 3 = y ¯ 4 = 0 z ¯ 4 = z ¯ 5 = 0 Sketch the arc as shown in Figure 2. Refer to Equation 9.47 in section 9.6A Mass product of inertia in the textbook. Applying parallel axis theorem, it follows I x y = I y z = I z x for component 2 and 4. Find the moment of inertia ( I u v ) for one quarter of a circular arc as shown below: Refer to Figure 2. Consider u = a cos θ and v = a sin θ Find the mass element as shown below: Refer to Figure 2. d m = ρ d V = ρ [ A ( a d θ ) ] Here, A is cross section area of the wire. m = m ′ ( π 2 a ) = ρ A ( π 2 a ) d m = m ′ a d θ Find the element ( d I u v ) as shown below: ( d I u v ) = ( a cos θ ) ( a sin θ ) ( m ′ a d θ ) = m ′ a 3 sin θ cos θ d θ (5) Integrate the Equation (5) with respect to θ and apply the limits as shown below: I u v = ∫ d I u v = ∫ 0 π 2 m ′ a 3 sin θ cos θ d θ = m ′ a 3 [ 1 2 sin 2 θ ] 0 π 2 = 1 2 m ′ a 3 Find the moment of inertia section I x y of wire as shown below: I x y = ∑ ( I x y ) = [ ( I ¯ x ′ y ′ ) 1 + m 1 x ¯ 1 y ¯ 1 ] + [ ( I x ′ y ′ ) 3 + m 3 x 3 y 3 ] + ( I x y ) 5 (6) Here, ( I ¯ x ′ y ′ ) 1 is product of inertia section 1 about x y axis, ( I x ′ y ′ ) 3 is product of inertia section 3 about x y axis, ( I x y ) 5 , and product of inertia section 5

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ISBN: 9781259977121

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 9.158P

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Find the mass products of inertia Ixy,Iyz, and Izx of the wire.

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Chapter 9.6, Problem 9.157P

Chapter 9.6, Problem 9.159P

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Find the mass products of inertia I x y , I y z , and I z x of the wire.

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Find the mass products of inertia Ixy,Iyz, and Izx of the wire.

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