(a) Explanation Given Information: The total mass of the shaft is m . Calculation: Show the shape of the given shaft as in Figure (1). Calculate the length of the rod ( L ): L = r + r + 2 π r = 8.2832 r Write the expression for the angular velocity ( ω ) : ω = ω i For the portion AC and DB , Write the expression for the centroidal moment of inertia about x ( I x ) and xz ( I x z ) : I ¯ x = I ¯ x z = 0 Write the expression for displacement about ( x ) the axis x : x = − r ( r − cos θ ) = − r ( 1 − cos θ ) Write the expression for density per unit length ( ρ ) : ρ = m L Write the expression for displacement about ( z ) the axis z : z = − r sin θ Write the expression mass: m = ρ r θ Differentiate the above equation with respect to θ . d m = ρ r d θ For portion CG , use polar coordinate ( θ ) . Write the expression for the moment of inertia about axis x ( I x ) C G : ( I x ) C G = ∫ z 2 d m Substitute − r sin θ for z and ρ r d θ for dm . ( I x ) C G = ∫ ( − r sin θ ) 2 ρ r d θ Take the limit from 0 to π . ( I x ) C G = ∫ 0 π r 2 sin 2 θ ρ r d θ ( ∵ sin 2 θ = 1 2 − 1 2 cos 2 θ ) = ∫ 0 π r 2 ( 1 2 − 1 2 cos 2 θ ) ρ r d θ = ∫ 0 π ( 1 2 ρ r 3 − 1 2 cos 2 θ ρ r 3 ) d θ = ρ r 3 ∫ 0 π 1 2 d θ − ∫ 0 π 1 2 cos 2 θ d θ Simplify the Equation, ( I x ) C G = ρ r 3 [ 1 2 θ − 1 4 sin ( 2 θ ) ] 0 π = ρ r 3 [ 1 2 ( 0 ) − 1 4 sin ( 2 ( 0 ) ) ] − [ 1 2 ( π ) − 1 4 sin ( 2 ( π ) ) ] = π ρ r 3 2 Write the expression for the product moment of inertia about axis xz ( I x z ) C G : ( I x z ) C G = ∫ x z d m Substitute − r sin θ for z , − r ( 1 − cos θ ) for x and ρ r d θ for dm . ( I x z ) C G = ∫ ( − r sin θ ) − r ( 1 − cos θ ) ρ r d θ = ∫ r 2 ( 1 − cos θ ) sin θ ρ r d θ Take the limit from 0 to π . ( I x z ) C G = ∫ 0 π r 2 ( 1 − cos θ ) sin θ ρ r d θ = ∫ 0 π r 2 ( sin θ − cos θ sin θ ) ρ r d θ = r 3 ρ ∫ 0 π sin θ d θ − ∫ 0 π ( cos θ sin θ ) d θ Simplify the Equation, ( I x z ) C G = r 3 ρ ( [ − cos θ ] 0 π − [ 1 2 sin 2 θ ] 0 π ) = r 3 ρ [ − cos π − ( − cos 0 ) ] − [ 1 2 sin 2 ( π ) − 1 2 sin 2 ( 0 ) ] = r 3 ρ − ( − r 3 ρ ) = 2 r 3 ρ For portion GD : Moment of inertia for portion GD is as same as portion CG (b) To determine The angle ( θ ) formed by H G and the axis AB . (c) To determine The angular momentum H A of the shaft about point A .

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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Chapter 18, Problem 18.149RP

(a)

To determine

The angular momentum HG of the shaft about its mass center G.

(b)

To determine

The angle (θ) formed by HG and the axis AB.

(c)

To determine

The angular momentum HA of the shaft about point A.

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Chapter 18, Problem 18.148RP

Chapter 18, Problem 18.150RP

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Chapter 18 Solutions

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

Ch. 18.1 - A thin, homogeneous disk of mass m and radius r...Ch. 18.1 - Prob. 18.2P Ch. 18.1 - 18.3 Two uniform rods AB and CE, each of weight 3...Ch. 18.1 - A homogeneous disk of weight W = 6 lb rotates at...Ch. 18.1 - Prob. 18.5P Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P Ch. 18.1 - Prob. 18.11P

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The angular momentum H G of the shaft about its mass center G .

Solution Summary: The angular momentum of the shaft about its mass center G is underset_mr2omega

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The angular momentum HG of the shaft about its mass center G.

The angle (θ) formed by HG and the axis AB.

The angular momentum HA of the shaft about point A.

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Chapter 18 Solutions