Explanation Given information: The sides of the solid cube ( c ) is 80 mm. The spin rate ( ψ ˙ ) is 40 rad/s. The angular velocity ( ϕ ˙ ) is 5 rad/s. The angle ( β ) is 30 ° . Calculation: Show the diagram of the system as in Figure (1). The axes is taken in such a way that z axis passes through the diagonal of the body BC and x axis lies in the plane of A, B, C, and D . Here, G is the center of gravity of the cube and the length of GB is 3 2 c . Write the expression for the moment of inertia of the cube in the x direction ( I x ) as follows: I x = 1 6 m c 2 Here, m is the mass a is the side of the cube. The moment of inertia in x , y , and z directions for a cube is the same. The moment of inertia in y and z directions is equal to 1 6 m c 2 . Write the expression for the angular velocity of the system ( ω ) as follows: ω = − ϕ ˙ sin θ i + ( ψ ˙ + ϕ ˙ cos θ ) k (1) Here, ϕ ˙ is the rate of precession, ψ ˙ is the rate of spin and θ is the rate of nutation. Calculate the angular momentum about the mass center G ( H G ) using the formula: H G = I x ω x i + I y ω y j + I z ω z k Here, ω x is the angular velocity in x direction, ω y is the angular velocity in y direction and ω z is the angular velocity in the z direction. From Equation (1), Substitute − ϕ ˙ sin θ for ω x , 0 for ω y and ( ψ ˙ + ϕ ˙ cos θ ) for ω z . H G = I x − ϕ ˙ sin θ i + I y ( 0 ) j + I z ( ψ ˙ + ϕ ˙ cos θ ) k = − I ′ ϕ ˙ sin θ i + I ( ψ ˙ + ϕ ˙ cos θ ) k Write the expression for the angular velocity of the reference frame Gxyz as follows: Ω = − ϕ ˙ sin θ i + ϕ ˙ cos θ k Here, Gxyz is the angular velocity for the frame of reference. Calculate the rate of change of angular momentum ( H ˙ G ) using the formula: H ˙ G = ( H ˙ G ) x y z + Ω × H G Here, ( H ˙ G ) x y z is the angular momentum about the frame of reference Gxyz . Angular momentum about the frame of reference Gxyz is 0. Substitute 0 for ( H ˙ G ) x y z , − ϕ ˙ sin θ i + ϕ ˙ cos θ k for Ω , and − I ′ ϕ ˙ sin θ i + I ( ψ ˙ + ϕ ˙ cos θ ) k for H G . H ˙ G = 0 + ( − ϕ ˙ sin θ i + ϕ ˙ cos θ k ) × ( − I ′ ϕ ˙ sin θ i + I ( ψ ˙ + ϕ ˙ cos θ ) k ) (2) Calculate the vector multiplication: Ω × H G = [ i j k − ϕ ˙ sin θ 0 ϕ ˙ cos θ I ′ ϕ ˙ sin θ 0 I ( ψ ˙ + ϕ ˙ cos θ ) ] = [ i [ 0 − 0 ] − j [ ( − ϕ ˙ sin θ ) I ( ψ ˙ + ϕ ˙ cos θ ) − ( I ′ ϕ ˙ sin θ ) ( ϕ ˙ cos θ ) ] + k [ 0 − 0 ] ] = − j [ − I ψ ˙ ϕ ˙ sin θ + I ϕ ˙ 2 sin θ cos θ − I ′ ϕ ˙ 2 sin θ cos θ ] = j [ I ψ ˙ ϕ ˙ sin θ − ( I ′ − I ) ϕ ˙ 2 sin θ cos θ ] (3) Substitute equation (3) in equation (2)

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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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 18.3, Problem 18.115P

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The angle (θ) that the diagonal BC forms with the vertical.

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Chapter 18.3, Problem 18.114P

Chapter 18.3, Problem 18.116P

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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 angle ( θ ) that the diagonal BC forms with the vertical.

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The angle (θ) that the diagonal BC forms with the vertical.

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