Cite as: Sanjay Sarma, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 1 Lecture Outline 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Sanjay Sarma 4/2/2007 Lecture 13 Lagrangian Dynamics: Generalized Coordinates and Forces Lecture Outline Solve one problem by Newtonian and Lagrangian Methods. “Lagrangian approach is simple but devoid of insight.” Both methods can be used to derive equations of motion. Figure 1: Wheel on an incline. Figure by MIT OCW. 1. Solve a well-known problem by Newton’s method: Wheel down incline 2. Critique Solution 3. Present Lagrange Equations 4. Solve well-known problem by Lagrange’s MethodCite as: Sanjay Sarma, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � � 2 Example: Wheel Rolling Down Incline Example: Wheel Rolling Down Incline Figure 2: Free body and kinematic diagrams of wheel rolling down incline. The wheel is subject to a normal force, N , a frictional force, F , and a gravitational force, mg. Figure by MIT OCW. What is the acceleration? F = friction 1 degree of freedom, θ. Newton’s Method 3 unknowns: N, F , and θ Equations F = ma F = ma x x y y τ = Iα x-direction: ¨ −F + mg sin φ = mr θ y-direction:Cite as: Sanjay Sarma, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 3 Example: Wheel Rolling Down Incline N − mg cos φ = 0 Torques about center of wheel: F r =1 mr 2θ ¨ 2 Eliminate F : 3 mgr sin φ = mr 2θ ¨ 2 ¨2 gθ = sin φ 3 r Critique of Newton’s Method Critique: Did the reaction forces do any work? Did the reaction force move? N did not move. F applied but point did not move at point of contact. No work done by those forces. 1. Had to explicitly deal with non-contributing forces. Newton’s Laws are de-rived in cartesian coordinates, but problem’s answer was not in cartesian. 2. Newton’s laws are written in Cartesian coordinates, but we really care about configurational parameters. 3. Had to write several kinematic constraints to reduce unknowns x and y. Schematic of Newton versus Lagrangian Approach Figure 3: Diagram depicting the differences between Newtonian and Lagrangian approach. Figure by MIT OCW.Cite as: Sanjay Sarma, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � 4 Example: Wheel Rolling Down Incline Generalized Coordinates Figure 4: Left. Generalized coordinate for bead on wire. For a bead moving along a wire, one generalized coordinate, the distance along the wire, can be used to describe the position instead of the two Cartesian coordinates x and y. Right. Generalized coordinate for a two beam system. In the system of connected bars OP and PQ driven by a motor at Q, there is only one degree of freedom. Thus, an angular coordinate such as theta, angle of OP with the x-axis, or phi, angle between OP and PQ, completely describes the coordinates of the rigid body. If one was not using generalized coordinates, one would need x,y-coordinates at points O, P, and Q. Figure by MIT OCW. Only 1 degree of freedom in both cases Newton: x, y for each of points O, P , and Q. φ would also work. θ is a generalized coordinate; it naturally, completely, and independently describe co-ordinates of the rigid body. The generalized coordinates of a system (of particles or rigid body or rigid bodies) is the natural, minimal, complete set of parameters by which you can completely specify the configuration of that system. Lagrange Method We have already seen a generalized force. Where? Torque. Newton: F = ma # coordinates = # degrees of freedom Lagrange: d ∂L ∂L Qi = − dt ∂q˙i ∂qiCite as: Sanjay Sarma, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � 5 Example: Wheel Rolling Down Incline Qi: generalized force L: Lagrangian qi: generalized coordinates T + V = Constant L = T − V 1 equation per coordinate No reaction forces appear Application of Lagrange Method Figure 5: Wheel rolls down incline. The wheel rotates by an angle, θ and traverses a distance of rθ. Figure by MIT OCW. Generalized Coordinate: θ = q1 Generalized Force: Q1 = 0 11 1 1 1 T = Iω2 + mv 2 = mr 2 q˙12 + mr 2 q˙12 2 2 2 2 2 V = −mgrθ sin φ = −mgrq1 sin φ Gravity accounted for by V Ignore non-contributing forces 1. Internal rigid body forces 2. Reaction forces across slipping surfaces 3. Rolling forcesCite as: Sanjay Sarma, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � 6 Example: Wheel Rolling Down Incline L = T − V =3 mr 2 q˙12 + mgrq1 sin φ 4 ∂L 3 = mr 2 q˙1∂q˙1 2 ∂L = mgr sin φ ∂q1 ¨2 q¨1 = θ = g sin φ 3d ∂L 3 2 ¨= mr q1dt ∂q˙1 2 Non-Contributing Force Comments Lagrange leads to math for 2 and 3 degrees of freedom problems. Figure 6: Block sliding along incline. Two forces are present, a reaction force, R, and a gravitational force, mg. Figure by MIT OCW.Cite as: Sanjay Sarma, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 7 Example: Wheel Rolling Down Incline R can be ignored if frictionless Figure 7: Free body diagram of block sliding along incline. F cannot be ignored if friction did work. Figure by MIT OCW. F cannot be ingored (friction did work) Figure 8: Free body diagram of block sliding along incline with attached spring. Spring contributes because spring lengthens. Figure by MIT OCW. Spring contributes because spring lengthens. Generalized Force Definition Generalized Force: mF ORC ES � ∂rjQi = F j ∂qij=1 Take every force and dot