2.003SC Recitation 8 Notes: Cart and Pendulum (Lagrange) Cart and Pendulum - Problem Statement A cart and pendulum, shown below, consists of a cart of mass, m1 , moving on a horizontal surface, acted upon by a spring with spring constant k. From the cart is suspended a pendulum consisting of a uniform rod of length, l , and mass, m2 , pivoting about point A . Derive the equations of motion for this system by Lagrange. Speciﬁcally, • Find T , the system’s kinetic energy • Find V , the system’s potential energy • Find vG 2 , the square of the magnitude of the pendulum’s center of gravity Cart and Pendulum - Solution Generalized Coordinates q1 = x, q2 = θ Kinematics The linear velocity of the pendulum’s center of mass, vG , is given by lθ̇ lθ̇ lθ̇ vG = ẋIˆ + ĵ = (ẋ + cosθ)Iˆ + ( sinθ)Jˆ 2 2 2 The square of its magnitude is given by 2 vG = (ẋ + Expanding and simplifying, lθ̇ lθ̇ cosθ)2 + ( sinθ)2 2 2 2 vG = ẋ2 + 2 lθ̇ lθ̇ lẋθ̇ cosθ + ( )2 cos2 θ + ( )2 sin2 θ 2 2 2 l2 θ̇2 2 ˙ = ẋ2 + ẋθlcosθ + vG 4 Kinetic Energy 1 1 1 T = m1 ẋ2 + m2 vG 2 + IG θ̇2 2 2 2 1 ml2 2 1 1 l2 θ̇2 2 2 ˙ T = m1 x˙ + m2 (x˙ + x˙ θlcosθ + )+ ( )θ̇ 2 2 4 2 12 Potential Energy 1 l V = kx2 + m2 g (1 − cosθ) 2 2 Lagrangian L=T −V l 1 1 l2 θ̇2 1 ml2 2 1 2 ˙ L = m1 ẋ2 + m2 (ẋ2 + ẋθlcosθ + )+ ( )θ̇ − kx − m2 g (1 − cosθ) 2 2 4 2 12 2 2 Lagrange’s equation for the ﬁrst generalized coordinate, d ∂L ∂L ( )− = Qx dt ∂ẋ ∂x yields the ﬁrst equation of motion. (m1 + m2 )ẍ + m2 l ¨ m2 l 2 θcosθ − θ̇ sinθ + kx = 0 2 2 (1) Lagrange’s equation for the second generalized coordinate, d ∂L ∂L ( )− = Qθ dt ∂θ̇ ∂θ yields the second equation of motion. ( m2 l2 m2 l2 ¨ m2 l l + )θ + xcosθ ¨ + m2 g sinθ = 0 4 12 2 2 (2) Cart and Pendulum - Problem Statement Assume that the cart and pendulum system now contain a damper/dashpot of constant b between the cart and ground, as well as an external force, F (t) , applied to the cart. Derive the equations of motion for this system by Lagrange. Speciﬁcally, show the generalized forces. Cart and Pendulum - Solution Both the damper and the external force are non-conservative (forces). Consequently, they enter the Lagrange formulation as generalized forces. Generalized Forces δW nc = N N finc · δRi = i n N Qj δξj j=1 δW nc = [−bẋ + F (t)]δx + [0]δθ Qx = −bẋ + F (t) Qθ = 0 which changes the ﬁrst equation to become, (m1 + m2 )ẍ + m2 l ¨ m2 l 2 θcosθ − θ̇ sinθ + kx = −bẋ + F (t) 2 2 So our equations of motion become (m1 + m2 )ẍ + m2 l ¨ m2 l 2 θcosθ − θ̇ sinθ + bẋ + kx = F (t) 2 2 (3) and m2 l2 m2 l2 ¨ m2 l l ( + )θ + xcosθ ¨ + m2 g sinθ = 0 4 12 2 2 (4)