Explanation Given information: The weight of the rod is 10 lb and the angular velocity is 12 rad / s . Write the expression of mass of the shaft. m = W g ...... (I) Here, the weight of the shaft is W and the acceleration due to gravity is g . Write the expression of angular momentum along x direction. ( H G ) x = I ¯ x ω ....... (II) Here, the moment of inertia along x axis is I ¯ x and the angular speed is ω . Write the expression of angular momentum along y direction. ( H G ) y = − I ¯ x y ω ....... (III) Here, the moment of inertia along y axis is I ¯ x y and the angular speed is ω . Write the expression of angular momentum along w z direction. ( H G ) z = − I ¯ x z ω ....... (IV) Here, the moment of inertia along z axis is I ¯ x z and the angular speed is ω . Write the expression of moment of inertia along x direction. I ¯ x = 4 ( 1 3 m a 2 ) + 2 ( m a 2 ) ...... (V) Here, the length of the rod is a . Write the expression of moment of inertia about z axis. I ¯ x z = 2 ( m a 2 ) ....... (VI) Write the expression of total angular momentum. ( H G ) total = ( H G ) x i + ( H G ) y j + ( H G ) z k ....... (VII) Here, the unit vector in x direction is i , the unit vector in y direction is j and the unit vector in z direction is k . Write the expression of magnitude of angular momentum. ( H G ) mag = ( H G ) x 2 + ( H G ) y 2 + ( H G ) z 2 ........ (VIII) Write the expression of moment of angular momentum of the shaft about its mass center G . H G = ( H g ) total ( ω i ) ....... (IX) Calculation: Substituting 10 lb for W and 32.2 ft / s 2 for g in Equation (I): m = 10 lb 32.2 ft / s 2 = 0.3105 lb ⋅ s 2 / ft Substituting 0.3105 lb ⋅ s 2 / ft for m and 10 in for a in Equation (V: I ¯ x = 4 { 1 3 ( 0.3105 lb ⋅ ft / s 2 ) ( 10 in ) 2 } + 2 { ( 0.3105 lb ⋅ ft / s 2 ) ( 10 in ) 2 } = [ 4 3 { 0.3105 lb ⋅ s 2 / ft ( ( 10 in ) × ( 1 ft 12 in ) ) 2 } + 2 { 0.3105 lb ⋅ s 2 / ft ( ( 10 in ) × ( 1 ft 12 in ) ) 2 } ] = ( 0.2875 lb ⋅ ft ⋅ s 2 ) + ( 0.43125 lb ⋅ ft ⋅ s 2 ) = 0.71875 lb ⋅ ft ⋅ s 2 Substituting 0.3105 lb ⋅ s 2 / ft for m and 10 in for a in Equation (VI): I ¯ x z = 2 { ( 0.3105 lb ⋅ s 2 / ft ) ( 10 in ) 2 } = 2 { ( 0.3105 lb ⋅ s 2 / ft ) ( ( 10 in ) × ( 1 ft 12 in ) ) 2 } = 0 To determine (a) The angle formed by H G and the axis A B .

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

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

Publisher: McGraw-Hill Education

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Chapter 18.1, Problem 18.17P

To determine

(a)

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

To determine

(a)

The angle formed by HG and the axis AB.

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Chapter 18.1, Problem 18.16P

Chapter 18.1, Problem 18.18P

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Chapter 18 Solutions

Vector Mechanics for Engineers: Dynamics

Ch. 18.1 - Prob. 18.1P Ch. 18.1 - A thin rectangular plate of weight 15 Ib rotates...Ch. 18.1 - Prob. 18.3P Ch. 18.1 - A homogeneous disk of weight W=6 lb rotates at the...Ch. 18.1 - Prob. 18.5P Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Solve Prob. 18.6, assuming that the solid...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P

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