Phys 325 Spring 2011 Lecture 18 - Physics 325 Lecture 18...

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Physics 325 Lecture 18 Lagrangian Mechanics For nearly two full semesters we have carefully chosen problems that can be solved via the application of Newton’s laws. In order to make this so, we have to know all the forces that act on the system in question. However, there are many cases in which this is very difficult. For example, if a particle is constrained to move on a non-planar surface, writing down all the forces that constrain the particle to remain on the surface can be very hard. Solving the equations of motion in such a case can be even more difficult. We already know of one alternate method of solving for equations of motion: the energy method. For a conservative system we can write: 0 dE dT dU dt dt dt where T is the kinetic and U is the potential energy. We take a one-dimensional system in which case     U U x T T x (18.1) then 0 dT dx dU dx dT dU x x dx dt dx dt dx dx   (18.2) Taking a specific simple case of a mass in a gravitational field, we have 2 1 2 T mx U mgx which gives mxx mgx mx mg     which is just Newton’s 2 nd law. In general, we could write the equivalent of Equation (18.2) for any “degree of freedom” we wish. That is, I can invent a coordinate q (for instance the angle between the gravitational direction and the moon at the location of the mass) and as long as I’m able to express U and T in terms of and q q I can write 0 dT dq dU dq dT dU q q dq dt dq dt dq dq   This is the basic idea behind generalized coordinates , however our energy technique only works for problems with a single degree of freedom (i.e. one-dimensional) because we
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only get the single equation above. Our goal will be to extend this technique to systems with multiple degrees of freedom. We will sta rt from Newton’s 2 nd and see if we can find a function , , f x x t that satisfies Euler’s equation in the form that we had at the end of Lecture 23: 0 f d f x dt x (18.3) We assume a one-dimensional system (but we will soon expand) with a conservative force. As above (using partial derivatives here for convenience later) we have T mx x so that d T mx dt x we also have U F x   so that Newton’s 2 nd reads U d T x dt x (18.4) Referring to Equation (18.1), we have that 0 and 0 T U x x and we can write and T T U U T U x x x x (18.5) We now define the Lagrangian
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