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# ilovepdf.com_split_2 - 88 Chapter 3 Kinetics of Particles...

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88 Chapter 3. Kinetics of Particles Question 3–12 A particle of mass m is attached to a linear spring with spring constant K and un- stretched length r 0 as shown in Fig. P3-12. The spring is attached at its other end at point P to the free end of a rigid massless arm of length l . The arm is hinged at its other end and rotates in a circular path at a constant angular rate ω . Knowing that the angle θ is measured from the downward direction and assuming no friction, determine a system of two differential equations of motion for the particle in terms of r and θ . ωt l m r K O P θ Figure P3-12 Solution to Question 3–12 Kinematics First, let F be a fixed reference frame. Then, choose the following coordinate system fixed in reference frame F : Origin at O E x = Along OP When t = 0 E z = Out of Page E y = E z × E x Next, let A be a reference frame fixed to the arm. Then, choose the following coordinate system fixed in reference frame A : Origin at O e x = Along OP e z = Out of Page ( = E z ) e y = e z × e x Finally, let B be a reference frame fixed to the direction along which the spring lies (i.e., the direction Pm ). Then, choose the following coordinate system fixed in reference

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89 frame B : Origin at O u r = Along Pm u z = Out of Page ( = E z = e z ) u θ = u z × u r The geometry of the bases { E x , E y , E z } , { e x , e y , e z } , and { u r , u θ , u z } is shown in Fig. 3- 10. Using Fig. 3-10, we have the following relationship between the basis { e x , e y , e z } e ωt ωt u r u θ e y u z , e z , E z E x E y x θ θ Figure 3-10 Geometry of Bases { E x , E y , E z } , { e x , e y , e z } , and { u r , u θ , u z } for Question 3–12 . and the basis { u r , u θ , u z } : e x = cos - ωt) u r - sin - ωt) u θ e y = sin - ωt) u r + cos - ωt) u θ (3.319) Next, observing that the basis { e x , e y , e z } rotates with angular rate ω relative to the basis { E x , E y , E z } , the angular velocity of reference frame A in reference frame F is given as F ω A = ω E z = ω e z (3.320) Next, using Eq. (3.319), we observe that the angle formed between the basis vectors
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