Explanation Given information: The mass per unit length of wire is m ′ . Calculation: Sketch the brass wire as shown in Figure 1. Express the mass of each component as shown below: m = W g = 1 g w L Here, w is a weight per unit length. Find the mass of component 1 as follows: m 1 = W g ( L ) (1) Substitute ( 2 π × a ) for L in Equation (1). m 1 = w g ( ( 2 π × a ) ) = 2 π × w g a Find the mass of component 2 as follows: m 2 = W g ( L ) (2) Substitute ( a ) for L in Equation (2). m 2 = w g ( a ) = w g a Find the mass of component 3 as follows: m 3 = W g ( L ) (3) Substitute ( 2 a ) for L in Equation (3). m 3 = w g ( 2 a ) = w g 2 a Find the mass of component 4 as follows: m 4 = W g ( L ) (4) Substitute ( 2 π × 3 2 a ) for L in Equation (4). m 4 = w g ( 2 π × 3 2 a ) = 3 π w g a Find the centroid of section 1 about x axis ( x 1 ) as shown below: ( x 1 ) = 2 a Find the centroid of section 1 about y axis ( y 1 ) as shown below: ( y 1 ) = a Find the centroid of section 1 about z axis ( z 1 ) as shown below: ( z 1 ) = − a Find the total section ( m x ¯ y ¯ ) 1 as shown below: ( m x ¯ y ¯ ) 1 = 2 π w g a ( 2 a ) ( a ) = 4 π w g a 3 Similarly calculate the remaining values as shown in Table 1. Section m x ¯ y ¯ z ¯ m x ¯ y ¯ m y ¯ z ¯ m x ¯ z ¯ 1 2 π w g a 2 a a − a 4 π w g a 3 − 2 π w g a 3 − 4 π w g a 3 2 w g a 2 a 1 2 a 0 w g a 3 0 0 3 2 w g a 2 a 0 a 0 0 4 w g a 3 4 3 π w g a 2 a − 3 2 a 2 a − 9 π w g a 3 − 9 π w g a 3 12 π w g a 3 ∑ w g ( 1 − 5 π ) a 3 − 11 π w g a 3 4 w g ( 1 + 2 π ) a 3 Find the mass products of inertia I x y of the machine component. I x y = ∑ ( I x ′ y ′ + m x ¯ y ¯ ) = ( ( I x ′ y ′ ) 1 + ( I x ′ y ′ ) 2 + ( I x ′ y ′ ) 3 + ( m x ¯ y ¯ ) ) (5) Substitute 0 for ( ( I x ′ y ′ ) 1 + ( I x ′ y ′ ) 2 + ( I x ′ y ′ ) 3 ) and w g ( 1 − 5 π ) a 3 for ∑ ( m x ¯ y ¯ ) in Equation (5)

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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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 9.159P

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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 9.158P

Chapter 9.6, Problem 9.160P

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Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - Prob. 9.3P Ch. 9.1 - Prob. 9.4P Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - Prob. 9.7P Ch. 9.1 - Prob. 9.8P Ch. 9.1 - 9.9 through 9.11 Determine by direct integration...Ch. 9.1 - Prob. 9.10P

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

Solution Summary: The author explains the mass products of inertia and the weight of each component.

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

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Chapter 9 Solutions