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# ch08 - Chapter 8 Kinematics and Kinetics of Rigid Bodies in...

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Chapter 8 Kinematics and Kinetics of Rigid Bodies in Three-dimensional Motion 8.1 Spherical coordinates 8.2 Angular Velocity of Rigid Bodies in Three-Dimensional Mo- tion 8.3 Angular Acceleration of Rigid Bodies in Three-Dimension Mo- tion 8.4 General Motion Of and On Three-Dimensional Bodies 87

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8.4.1 GOAL: Determine the angular velocity and acceleration of one arm of the illustrated mechanism. GIVEN: Constant angular velocity of inner arm, and constant angular velocity of outer arm with respect to inner arm DRAW: The figure shows the mechanism with the original coordinate axes and some newly at- tached unit vectors. Unit vectors ı ,  , k are aligned with the ground-fixed X, Y, Z axes. Unit vectors b 1 , b 2 , b 3 are attached to the inner arm OA . FORMULATE EQUATIONS: We’ll use the expressions for angular velocity and acceleration on a rotating body. SOLVE: The angular velocity of arm AB is equal to the angular velocity of arm OA plus the relative angular velocity of arm AB with respect to arm OA . The angular velocity of OA is ˙ ψk , and the relative angular velocity of AB with respect to OA is - ˙ θb 2 . Thus: ω AB = ˙ ψk - ˙ θb 2 (1) This can be written is the b -frame as: ω AB = ˙ ψb 3 - ˙ θb 2 To determine the angular acceleration of AB we can differentiate (1): α AB = d dt ω AB = d dt ˙ ψk - ˙ θb 2 = ¨ ψk =0 + ˙ ψ d dt k =0 - ¨ θb 2 =0 - ˙ θ d dt b 2 Since the angular speeds are constant, and k is fixed in space, the only term remaining is - ˙ θ d dt b 2 . The tip of unit vector b 2 sweeps in the - b 1 direction with speed ˙ ψ . So we have: α AB = - ˙ θ d dt b 2 = - ˙ θ ˙ ψ ( - b 1 ) α AB = ˙ θ ˙ ψb 1 88
Alternatively, we could have used the expression: d dt ω AB = d dt S ω AB + ω OA × ω AB = 0 + ˙ ψk × ( ˙ ψk - ˙ θb 2 ) = ˙ θ ˙ ψb 1 89

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8.4.2 GOAL: Determine the angular velocity of a rotating disk. GIVEN: Inner shaft’s angular velocity and the angular velocity of the disk with respect to the shaft. DRAW: FORMULATE EQUATIONS: We’ll use the expression for angular velocity on a rotating body. SOLVE: The angular velocity of the disk D is equal to the angular velocity of inner shaft AB plus the relative angular velocity of disk D with respect to shaft AB . The angular velocity of AB is - ω 1 , and the relative angular velocity of disk to shaft is ω 2 ı . Thus: ω D = ω 2 ı - ω 1 90
8.4.3 GOAL: Determine the angular acceleration of a rotating caster. GIVEN: Angular velocity of the caster’s frame is ω 1 b 3 and the angular velocity of the caster with respect to the frame is - ω 2 b 1 . DRAW: FORMULATE EQUATIONS: We’ll use the expression for acceleration on a rotating body: d dt N ω C = d dt F ω C + ω F × ω C (1) SOLVE: The caster C has a constant rotation rate with respect to the frame F and so (1) simplifies to d dt N ω C = ω F × ω C = ω 1 b 3 × ( - ω 2 b 1 ) = - ω 1 ω 2 b 2 α C = - ω 1 ω 2 b 2 91

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8.4.4 GOAL: Determine the angular velocity and acceleration of one wheel of the illustrated mechanism.
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ch08 - Chapter 8 Kinematics and Kinetics of Rigid Bodies in...

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