Explanation Given information: The mass of the homogeneous square is m . The side of the plate is a . The angle of the shaft rotates ( θ ) is 45 ° . The angular velocity ( ω ) is ω j . The angular acceleration ( α ) is α j . Calculation: Show the diagram of the system as in Figure (1). Show the diagram showing the angle of the axes as in Figure (2). Calculate the angular velocity by resolving along y ′ ( ω y ′ ) using the relation: ω y ′ = ω cos θ Here, ω is the angular velocity. Substitute 45 for θ . ω y ′ = ω cos 45 ° = ω 2 Calculate angular velocity by resolving along z ′ ( ω z ′ ) using the relation: ω z ′ = ω cos θ Substitute 45 for θ . ω z ′ = ω cos 45 ° = ω 2 The angular velocity along the x ′ axis is zero. Write the expression for the moment of inertia about x ′ axis ( I x ′ ) as follows: I x ′ = 1 3 m a 2 (1) Write the expression for the moment of inertia about y ′ axis ( I y ′ ) as follows: I y ′ = 1 12 m a 2 (2) Calculate the moment of inertia about z ′ axis ( I z ′ ) using the relation: I z ′ = I x ′ + I y ′ (3) Substitute Equation (1) and (2) in Equation (3). I z ′ = 1 3 m a 2 + 1 12 m a 2 = 4 m a 2 + m a 2 12 = 5 12 m a 2 Calculate the angular momentum about A ( H A ) using the relation: H A = I x ′ ω x ′ i ′ + I y ′ ω y ′ j ′ + I z ′ ω z ′ k ′ Substitute 1 3 m a 2 for I x ′ , 0 for ω x ′ , 1 12 m a 2 for I y ′ , ω 2 for ω y ′ , 5 12 m a 2 for I z ′ , and ω 2 for ω z ′ . H A = ( 1 3 m a 2 ) ( 0 ) i ′ + ( 1 12 m a 2 ) ( ω 2 ) j ′ + ( 5 12 m a 2 ) ( ω 2 ) k ′ = ( 1 12 m a 2 ) ( ω 2 ) j ′ + ( 5 12 m a 2 ) ( ω 2 ) k ′ (4) Substitute ( cos 45 ° j − sin 45 ° k ) for j ′ and ( sin 45 ° j + cos 45 ° k ) for k ′ in equation (4). H A = [ ( 1 12 m a 2 ) ( ω 2 ) ( cos 45 ° j − sin 45 ° k ) + ( 5 12 m a 2 ) ( ω 2 ) ( sin 45 ° j + cos 45 ° k ) ] = [ ( m a 2 ω 12 2 ) ( 1 2 j − 1 2 k ) + ( 5 m a 2 ω 12 2 ) ( 1 2 j + 1 2 k ) ] = m a 2 ω 24 j − m a 2 ω 24 k + 5 m a 2 ω 24 j + 5 m a 2 ω 24 k = 6 m a 2 ω 24 j + 4 m a 2 ω 24 k = 3 m a 2 ω 12 j + 2 m a 2 ω 12 k = m a 2 ω 12 ( 3 j + 2 k ) Calculate the angular acceleration by resolving along y ′ ( ω ˙ y ′ ) using the relation: ω ˙ y ′ = α cos θ Substitute 45 for θ . ω ˙ y ′ = α cos 45 ° = α 2 Calculate angular velocity by resolving along z ′ ( ω ˙ z ′ ) using the relation: ω ˙ z ′ = α cos θ Substitute 45 for θ . ω ˙ z ′ = α cos 45 ° = α 2 The angular velocity along the x ′ axis is zero

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977251

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 18.2, Problem 18.63P

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The rate of change (H˙A) of the angular momentum (HA) of the plate assembly.

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Chapter 18.2, Problem 18.62P

Chapter 18.2, Problem 18.64P

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Chapter 18 Solutions

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 - Prob. 18.3P Ch. 18.1 - A homogeneous disk of weight W = 6 lb rotates at...Ch. 18.1 - A homogeneous disk of mass m = 8 kg rotates at the...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 - Determine the angular momentum HO of the disk of...

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The rate of change ( H ˙ A ) of the angular momentum ( H A ) of the plate assembly.

Solution Summary: The author calculates the angular velocity by resolving along yprime using the relation:

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The rate of change (H˙A) of the angular momentum (HA) of the plate assembly.

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Chapter 18 Solutions