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Physics 430: Lecture 17 Examples of Lagrange’s Equations Dale E. Gary NJIT Physics Department.

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Presentation on theme: "Physics 430: Lecture 17 Examples of Lagrange’s Equations Dale E. Gary NJIT Physics Department."— Presentation transcript:

1 Physics 430: Lecture 17 Examples of Lagrange’s Equations Dale E. Gary NJIT Physics Department

2 October 28, 2010  This section is a long one, where several examples are worked out in detail. In addition, there are many problems at the end of the chapter for you to work on. You should do as many as possible.  I will not be doing every example, but let’s do a couple of them: Example 7.4: A Particle Confined to Move on a Cylinder  Statement of the problem: Consider a particle of mass m constrained to move on a frictionless cylinder of radius R, given by the equation  = R in cylindrical polar coordinates ( , , z). Besides the force of constraint (the normal force on the cylinder), the only force on the mass is a force F =  kr directed toward the origin. Using z and  as generalized coordinates, find the Lagrangian L. Write down and solve Lagrange’s equations and describe the motion.  Solution: We are already told what the generalized coordinates are, so we can now proceed to find the Lagrangian L = T – U. To get the kinetic energy, we need the velocity in terms of the generalized coordinates. The z velocity is easy, just The  velocity (azimuthal) is just Thus, the kinetic energy is just 7.5 Examples

3 October 28, 2010  Solution, cont’d: Now we need the potential energy (due to this central spring force), which is: Notice that we could drop the R 2 term, since it is constant, but it will not survive the Lagrange equation anyway. Putting these together, the Lagrangian is Now we form the Lagrange equations (one for z and one for  ) and solve. The z equation is The  equation is This just says that the angular momentum mR 2  = constant. Thus, the particle executes simple harmonic motion in z, but rotates in  at a constant rate around the cylinder. Example 7.4, cont’d r  = R z

4 October 28, 2010  Let’s now do a non-trivial two-coordinate problem—the block sliding down a wedge that we showed in Phun last time.  We will take the generalized coordinates as shown in the figure. Notice that the first coordinate q 1 is along the plane, so that it has no component in the constraint force direction. Also, q 1 is measured relative to the (moving) wedge.  The velocity of the wedge is just, but the velocity of the block has components from both q 1 and q 2 :  Thus, the total kinetic energy is  The potential energy of the wedge can be taken as zero, and the block is just Example 7.5: A Block Sliding on a Wedge q2q2 q1q1 

5 October 28, 2010  Putting these together, the Lagrangian is  There are two generalized coordinates, and so there are two Lagrange equations. The second is simplest:  Note that this is conservation of momentum in the x direction.  The first one is  This has two unknowns, but we can get another equation for the accelerations by differentiating the q 2 relation: or  Finally, plug this into the above equation and solve Example 7.5, cont’d

6 October 28, 2010  Let’s do another one. How many coordinates? Since it is a wire, there is only one degree of freedom, angle . What is the velocity to insert into the kinetic energy? This is a little tricky. The bead can slide up and down on the wire, which clearly has a velocity But the hoop is also rotating, so there is another motion with velocity.  The kinetic energy is then  The potential energy relative to its position at the bottom of the hoop (when the hoop is not rotating and  = 0 ), is  The Lagrangian is  The sole Lagrange equation is then  Solving for the angular acceleration: Example 7.6: Bead on a Spinning Wire Hoop   R 

7 October 28, 2010  We cannot solve: in terms of elementary functions, but we can look at the behavior to see if we can find an equilibrium and what the frequency of oscillation about any equilibria are.  This requires remembering all the way back to chapter 5! Think physically about what will happen. For no rotation (  = 0), the bead will act like a pendulum—equation  But for a rotating hoop, the bead will move up the wire and we might expect a new equilibrium. Recall that at equilibrium the bead will be at rest, i.e. which occurs at either  = 0 (which we have already examined) or at  Solving for the equilibrium position  We can find the oscillation frequency for small displacements   . Example 7.6, cont’d   R 

8 October 28, 2010  Recall that we said the left-hand side of the Lagrange equation is called the generalized force, while the right-hand side is the time derivative of the generalized momentum.  For normal coordinates like x, y, z, the generalized force is really the force, and the generalized momentum is really the momentum.  However, for angular coordinates we saw that the generalized force was the torque, and the generalized momentum was the angular momentum.  In any case, when the generalized force is zero (i.e. the Lagrangian is independent of the generalized coordinate q i, then the generalized momentum is conserved.  In such a case, we can say that the coordinate q i is ignorable. Another way to say it, which leads to an important idea in Physics, is to say that L is invariant under transformations (translations, rotations, etc.) in coordinate q i. 7.6 Generalized Momenta and Ignorable Coordinates

9 October 28, 2010  We will not study section 7.8, but I do want to introduce an idea of central importance to quantum mechanics—the Hamiltonian.  It comes about by considering the time derivative of the Lagrangian or  Now, if the Lagrangian does not depend on time, the last term on the right vanishes, so we have  Bringing the Lagrangian over to the right, the result is called the Hamiltonian: 7.8 Hamiltonian When the Lagrangian does not depend explicitly on time, the Hamiltonian H is conserved.

10 October 28, 2010  The key point at the moment that I would like you to remember is that when the relationship between generalized coordinates and Cartesian coordinates is time independent (so that the Lagrangian is time independent), then the Hamiltonian is just the total energy H = T + U, and it is conserved.  The rest of section 7.8 goes on to show that when the Lagrangian is independent of time, then  Subtracting the Lagrangian (T – U) from this obviously gives T + U. Hamiltonian-cont’d


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