Consider a simple pendulum, a mass m suspended on a massless string of fixed length. For simplicity assume that the string does not bend and that the motion takes place in a vertical plane. Obtain and solve the equation(s) of motion. The problem can be solved in many different ways and most of these ways offer some lessons. (d) Solve the problem again using the Lagrangian formalism, the Cartesian coordinates x and y, and the constraint x2+y2-2=0. (d) Lagrangian solution - single constraint The simple pendulum has a single dynamical degree of freedom, the angle the pendulum makes with the vertical 0 But we can choose to solve the problem in Cartesian coordinates, in which the Lagrangian is (x points to the right, y points down) Lox, y, x, j)=gmy+m(x²+ y²) The Cartesian coordinates are not independent, they are connected by the constraint that the mass m moves on a circular path -P²+x²+ y²=0 The Lagrangian in Cartesian coordinates with the above constraint added is Lox, y, A. x, y) = g my + (-² + x² + y²) + ½±m (x² +3²) Here A plays the role of another coordinate. With this in mind the Euler-Lagrange equations 21x-mx=0 gm-my+2y=0 -P²+x²+ y²=0 The first two equations are the horizontal and vertical Newton's second laws. Since only gravity and the rope acts on the mass, the à terms must be the horizontal and vertical components of the tension in the rope. (This is even more apparent remembering that x=/ sin(0) and y = cos(0).) The last equa- tion is just the constraint itself. The above three equations can be solved for the three unknown coordinates x, y and A. The simplest way to get to the solution is just expressing the constraint as x=/sin(0) y= cos(0) With this the first two equations of motions become Set 4.nb sin(0) (2x+m)=mö cos(0) m(g+10 sin(0) + cos(0) +21/cos(0)=0 Eliminating A yields the familiar equation of motion for 0 g sin(0)+10=0 After solving this we can calculate A using 3 of 5 x=sin(0) 21x-mx=0
Consider a simple pendulum, a mass m suspended on a massless string of fixed length. For simplicity assume that the string does not bend and that the motion takes place in a vertical plane. Obtain and solve the equation(s) of motion. The problem can be solved in many different ways and most of these ways offer some lessons. (d) Solve the problem again using the Lagrangian formalism, the Cartesian coordinates x and y, and the constraint x2+y2-2=0. (d) Lagrangian solution - single constraint The simple pendulum has a single dynamical degree of freedom, the angle the pendulum makes with the vertical 0 But we can choose to solve the problem in Cartesian coordinates, in which the Lagrangian is (x points to the right, y points down) Lox, y, x, j)=gmy+m(x²+ y²) The Cartesian coordinates are not independent, they are connected by the constraint that the mass m moves on a circular path -P²+x²+ y²=0 The Lagrangian in Cartesian coordinates with the above constraint added is Lox, y, A. x, y) = g my + (-² + x² + y²) + ½±m (x² +3²) Here A plays the role of another coordinate. With this in mind the Euler-Lagrange equations 21x-mx=0 gm-my+2y=0 -P²+x²+ y²=0 The first two equations are the horizontal and vertical Newton's second laws. Since only gravity and the rope acts on the mass, the à terms must be the horizontal and vertical components of the tension in the rope. (This is even more apparent remembering that x=/ sin(0) and y = cos(0).) The last equa- tion is just the constraint itself. The above three equations can be solved for the three unknown coordinates x, y and A. The simplest way to get to the solution is just expressing the constraint as x=/sin(0) y= cos(0) With this the first two equations of motions become Set 4.nb sin(0) (2x+m)=mö cos(0) m(g+10 sin(0) + cos(0) +21/cos(0)=0 Eliminating A yields the familiar equation of motion for 0 g sin(0)+10=0 After solving this we can calculate A using 3 of 5 x=sin(0) 21x-mx=0
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