tailieunhanh - Using the Lagrangian to obtain Equations of Motion

In Section of the textbook, Zak introduces the Lagrangian L = K − U , which is the difference between the kinetic and potential energy of the system. He then proceeds to obtain the Lagrange equations of motion in Cartesian coordinates for a point mass subject to conservative forces, namely, d dt ∂L ∂ xi ˙ − ∂L = 0 i = 1, 2, 3. ∂xi (1) | ECE 680 Selected Notes from Lecture 3 January 14 2008 1 Using the Lagrangian to obtain Equations of Motion In Section of the textbook Zak introduces the Lagrangian L K U which is the difference between the kinetic and potential energy of the system. He then proceeds to obtain the Lagrange equations of motion in Cartesian coordinates for a point mass subject to conservative forces namely ff LL L-0 23 dt ydx dxi 1 Any nonconservative forces acting on the point mass would show up on the right hand side. Here s how the text gets from the definition to the result. We know that for a point mass force is equal to mass times acceleration . . dX F ma mx m dt and work is equal to the integral over distance of the applied force. We can substitute for the force to obtain W iB F dx A 2 B mxx dx A 3 B mX dx A 4 where we have played fast and loose with the derivatives to conclude that xdx dx dt dx dx dx dt x dx. Assuming conservative forces we can then integrate to obtain W m X B 5 so the work is the difference between the kinetic energy at point B and that at point A. Conservation of energy requires that an increase in kinetic energy must be balanced by decrease in potential energy so we can write IB F dx AU J A 6 and thus F w ECE 680 Selected Notes from Lecture 3 January 14 2008 2 where we use the notation VU dU dU dU -1 T We have used the fact that if we measure change in potential energy with respect to a constant reference the derivative of the constant reference is zero so we have V AU VU. We re almost ready to rewrite Newton s equation in its Lagrangian form. We know that dK . mx i ox i so Newton s law becomes F - VU i mXi i 1 2 3. dt dxiJ 7 8 Now with L K U we see that K does not depend on position and U does not depend on velocity so dL _ dK dx i dx i 9 dL _ dU dxị dxị 10 so Newton s equation can be rewritten as 0 i 1 2 3 dt ydxự ơxi 11 asserted earlier. Next in Section Zak extends the above analysis to generalized coordinates by expressing each of the xi in .