理论力学双语教学第十七章拉格朗日方程.ppt

1 　 积分，得： Integration gives 代入初始条件，t =0 时， 得 Using as the initial conditions t =0, we obtain . 故： Therefore, 　 [Example 2] A slider A is connected to a spring of stiffness k and mass m1. They can slip on the smooth horizontal plane. Further, a simple pendulum is connected to the slider A, the length is l, the mass of the pendulum bob is m2. Determine the differential equation of motion of this system. [例2] 与刚度为k 的弹簧相连的滑块A，质量为m1,可在光滑水平面上滑动。滑块A上又连一单摆，摆长l ， 摆锤质量为m2 ，试列出该系统的运动微分方程。 　 解：将弹簧力计入主动力，则系统成为具有完整、理想约束的二自由度系统。保守系统。取x , ?为广义坐标，x 轴 原点位于弹簧自然长度位置， ? 逆时针转向为正。 Solution This system has two degree of freedom. All constraints are ideal ones. The system is a potential system. We can choose x and ? as generalized coordinates, The zero point of the axis x is at the place of the common length of the spring. For an anticlockwise displacement of B ? is chosen to be positive. 　 　 系统动能： The kinetic energy of the system is 　 The potential energy of the system is (choosing the common length of the spring as the zero point of the potential energy) 系统势能：（以弹簧原长为弹性势能零点，滑块A所在平面为重力势能零点） 　 拉格朗日函数： The Lagranges function is 　 代入： 并适当化简得： results after simplification in Substitution into 　 These are the differential equations of motion of the system. 系统的运动微分方程。 If the system is moving with a small amplitude around the equilibrium position, we can use ? 1o, cos? 1and sin? ?. Neglecting all terms of higher then first order, we get 若系统在平衡位置附近作微幅运动，此时? 1o, cos? 1, sin? ?，略去二阶以上无穷小量，则 　 上式为系统在平衡位置(x =0, ? =0)附近微幅运动的微分方程。 These are the differential equations of motion for small amplitude around the equilibrium position (x =0, ? =0). §17-3 Integrals of the Lagranges equations of the second kind §17-3 拉格朗日第二类方程的积分 For a potential system a first integration of the Lagranges equa