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Unformatted text preview: 123 Question 3–25 A particle of mass m slides without friction along a track in the form of a parabola as shown in Fig. P325. The equation for the parabola is y = r 2 2 a where a is a constant, r is the distance from point O to point Q , point Q is the pro jection of point P onto the horizontal direction, and y is the vertical distance. Further more, the particle is attached to a linear spring with spring constant K and unstretched length x . The spring is always aligned horizontally such that its attachment point is free to slide along a vertical shaft through the center of the parabola. Knowing that the parabola rotates with constant angular velocity Ω (where Ω = bardbl Ω bardbl ) about the vertical direction and that gravity acts vertically downward, determine the differential equation of motion for the particle in terms of the variable r . g m r y O P Q Ω Figure P325 Solution to Question 3–25 Kinematics For this problem it is convenient to define a fixed inertial reference frame F and a noninertial reference frame A . Corresponding to reference frame F , we choose the following coordinate system: Origin at Point O E x = Along OP When t = E y = Along Oy When t = E z = E x × E y 124 Chapter 3. Kinetics of Particles Furthermore, corresponding to reference frame A , we choose the following coordinate system: Origin at Point O e x = Along OP e y = Along Oy e z = e x × e y The position of the particle is then given in terms of the basis { e x , e y , e z } as r = x e x + y e y = x e x + (x 2 /a) e y (3.554) Furthermore, since the parabola spins about the eydirection, the angular velocity of reference frame A in reference frame F is given as F ω A = Ω = Ω e y (3.555) The velocity in reference frame F is then found using the rate of change transport...
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 Spring '07
 Chakravorty
 Dynamics, Force, Trigraph, reference frame, NB, EY

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