Direct Dynamics_Newton&acirc;€“Euler Equations of Motion

# Direct Dynamics_Newton&acirc;€“Euler Equations of...

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Chapter 5 Direct Dynamics: Newton–Euler Equations of Motion The Newton–Euler equations of motion for a rigid body in plane motion are m ¨ r C = F and I Czz α = M C , or using the Cartesian components m ¨ x C = F x , m ¨ y C = F y , and I Czz ¨ θ = M C . The forces and moments are known and the differential equations are solved for the motion of the rigid body (direct dynamics). 5.1 Compound Pendulum Exercise Figure 5.1a depicts a compound pendulum of mass m and length L . The pendulum is connnected to the ground by a pin joint and is free to swing in a vertical plane. The link is moving and makes an instant angle θ ( t ) with the horizontal. The local acceleration of gravity is g . Numerical application: L = 3 ft, g = 32 . 2 ft/s 2 , G = mg = 12 lb. Find and solve the Newton–Euler equations of motion. Solution The system of interest is the link during the interval of its motion. The link in rota- tional motion is constrained to move in a vertical plane. First, a reference frame will be introduced. The plane of motion will be designated the ( x , y ) plane. The y -axis is vertical, with the positive sense directed vertically upward. The x -axis is horizontal and is contained in the plane of motion. The z -axis is also horizontal and is perpen- dicular to the plane of motion. These axes de±ne an inertial reference frame. The unit vectors for the inertial reference frame are ı , j , and k . The angle between the x and the link axis is denoted by θ . The link is moving and hence the angle is chang- ing with time at the instant of interest. In the static equilibrium position of the link, 183

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184 5 Direct Dynamics: Newton–Euler Equations of Motion A θ L O x y A θ x y C A θ x y C 0 1 O O j ı F 01 x F 01 y G (a) (b) (c) n t n t I α C m a C Fig. 5.1 Compound pendulum the angle, θ , is equal to π / 2. The system has one degree of freedom. The angle, θ , is an appropriate generalized coordinate describing this degree of freedom. The system has a single moving body. The only motion permitted that body is rotation about a Fxed horizontal axis ( z -axis). The body is connected to the ground with the rotating pin joint (R) at O . The mass center of the link is at the point C . As the link is uniform, its mass center is coincident with its geometric center. Kinematics The mass center, C , is at a distance L / 2 from the pivot point O and the position vector is r OC = r C = x C ı + y C j , (5.1) where x C and y C are the coordinates of C
5.1 Compound Pendulum 185 x C = L 2 cos θ and y C = L 2 sin θ . (5.2) The link is constrained to move in a vertical plane, with its pinned location, O , serving as a pivot point. The motion of the link is planar, consisting of pure rotation about the pivot point. The directions of the angular velocity and angular acceleration vectors will be perpendicular to this plane, in the z -direction.The angular velocity of the link can be expressed as ω = ω k = d θ dt k = ˙ θ k , (5.3) ω is the rate of rotation of the link. The positive sense is clockwise (consistent with the x and y directions deFned above). This problem involves only a single moving rigid body and the angular velocity vector refers to that body. ±or this reason, no

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## This note was uploaded on 10/28/2009 for the course MECH 320 taught by Professor D.a during the Spring '09 term at American University of Beirut.

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Direct Dynamics_Newton&acirc;€“Euler Equations of...

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