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"" -1 322 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 323 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 324 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 325 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times " 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 3" -1 5 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 }{PSTYLE "Author" -1 19 1 {CSTYLE "" -1 -1 "Times " 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 8 8 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 4" -1 20 1 {CSTYLE "" -1 -1 "Times" 1 10 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 30 "Classical Mechanics with \+ Maple" }}{PARA 256 "" 0 "" {TEXT -1 66 "Section 5.2: Worked Examples o f the Synthetic and Analytic Methods" }}{PARA 19 "" 0 "" {TEXT -1 41 " Dr. Harald Kammerer\nmaple@jademountain.de" }}}{SECT 0 {PARA 3 "" 0 " " {TEXT -1 14 "Initialisation" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "libname:= \"C:/mylib/m6dynlib\",\"C:/mylib/m6dynfig\",libname:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "with(linalg):with(plots):with(plottools): with(dynamics);with(figures_chapter_5);" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 57 "5.2 Worked Examples of the Synthetic and Analytic Methods " }}{SECT 0 {PARA 4 "" 0 "EX 2" {TEXT -1 15 "5.2.1 Example 1" }}{PARA 0 "" 0 "" {TEXT -1 232 "Consider a single mass, hanging by two springs from the ceiling and guided at both sides in the vertical direction w ithout friction (Fig. 3). No outer forces are active. Here the displac ement in the vertical direction is denoted by " }{TEXT 258 1 "x" } {TEXT -1 18 ", the velocity by " }{TEXT 261 2 "xd" }{TEXT -1 25 " and \+ the acceleration by " }{TEXT 259 3 "xdd" }{TEXT -1 80 ". The displacem ent, the velocity and the acceleration are downward positve. For " } {TEXT 260 3 "x=0" }{TEXT -1 40 " the springs should be free of stress. " }{TEXT 323 62 "The exercise is to find the equation of motion for \+ the system." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display([Fig_ 5_3()],scaling=constrained,axes=none,title=\"Figure 3\");" }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 16 "Synthetic Method" }}{PARA 0 "" 0 "" {TEXT -1 182 "Consider the equilibrium of the free body (Fig. 4). Forc es in the horizontal directions at the vertical guide rails have no in fluence on the motion. On the mass the two spring forces" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Fs1(t):=k1*x(t):" "6#>-%$Fs1G6#%\"tG* &%#k1G\"\"\"-%\"xG6#F'F*" }}}{PARA 0 "" 0 "" {TEXT -1 3 "and" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Fs2(t):=k2*x(t):" "6#>-%$Fs2G6 #%\"tG*&%#k2G\"\"\"-%\"xG6#F'F*" }}}{PARA 0 "" 0 "" {TEXT -1 29 "as we ll as the inertial force" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Fi (t):=m*xdd(t):" "6#>-%#FiG6#%\"tG*&%\"mG\"\"\"-%$xddG6#F'F*" }}}{PARA 0 "" 0 "" {TEXT -1 9 "act where" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xdd(t) := diff(x(t),`$`(t,2)):" "6#>-%$xddG6#%\"tG-%%diffG6$-%\" xG6#F'-%\"$G6$F'\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 41 "At last the wei ght force acts on the body" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 " Fg:=m*g:" "6#>%#FgG*&%\"mG\"\"\"%\"gGF'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Fig. 4 shows all acting forces on the free body ." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display([Fig_5_4()],scaling= constrained,axes=none,title=\"Figure 4\");" }}}{PARA 0 "" 0 "" {TEXT -1 33 "The equilibrium of all forces in " }{TEXT 262 1 "x" }{TEXT -1 22 "-direction yields the " }{TEXT 256 18 "equation of motion" }{TEXT -1 21 " (eom) for the system" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eom_s:=Fi(t)+Fs1(t)+Fs2(t)-Fg=0;" }}}{PARA 0 "" 0 "" {TEXT -1 35 " Or after collecting derivatives of " }{TEXT 263 1 "x" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eom_s:=collect(eom_s,x);" }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 15 "Analytic Method" }}{PARA 0 "" 0 "" {TEXT -1 178 "Here as the first step the kinetic and the potential energy ha ve to be written down. Non-potential forces are not active. It is not \+ necessary to cut the supports for this method." }}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 16 "Potential Energy" }}{PARA 0 "" 0 "" {TEXT -1 49 "The potential energy of the first spring is (see " }{HYPERLNK 17 "spring_ energy" 2 "spring_energy" "" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 27 "V1:=spring_energy(k1,x(t));" }}}{PARA 0 "" 0 "" {TEXT -1 37 "and for the second spring accordingly" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 27 "V2:=spring_energy(k2,x(t));" }}}{PARA 0 "" 0 "" {TEXT -1 64 "The potential energy caused by the position of the b ody is (see " }{HYPERLNK 17 "gravitational_energy" 2 "gravitational_en ergy" "" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 34 "(notice the po sitive direction of " }{TEXT 264 1 "x" }{TEXT -1 12 " downwards!)" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Vg:=gravitational_energy(m,g ,-x(t));" }}}{PARA 0 "" 0 "" {TEXT -1 29 "The total potential energy i s" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "V:=V1+V2+Vg;" }}}{PARA 0 "" 0 "" {TEXT -1 43 "Or after collecting all the same powers of " } {TEXT 265 1 "x" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "V:=collect (V,x);" }}}}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 14 "Kinetic Energy" }} {PARA 0 "" 0 "" {TEXT -1 115 "There is only one mass which can only tr anslate in one direction. Rotations are imposible. The velocity is giv en by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xd(t):=diff(x(t),t): " "6#>-%#xdG6#%\"tG-%%diffG6$-%\"xG6#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 38 " So the kinetic energy is simply (see " }{HYPERLNK 17 "kinetic_ energy" 2 "kinetic_energy" "" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "T:=kinetic_energy(m,xd(t));" }}}}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 21 "Necessary derivatives" }}{PARA 0 "" 0 "" {TEXT -1 17 "According to the " }{TEXT 257 19 "Lagrangian equation" }{TEXT -1 106 " several derivatives have to be calculated. To do this with Ma ple it is useful to make some substitutions:" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs1:=\{x(t)=s,diff(x(t),t)=v\}:" "6#>%&subs1G<$/-% \"xG6#%\"tG%\"sG/-%%diffG6$-F(6#F*F*%\"vG" }}}{PARA 0 "" 0 "" {TEXT -1 39 "With this we get for the kinetic energy" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "T:=subs(subs1,T);" }}}{PARA 0 "" 0 "" {TEXT -1 35 "and for the potential of the system" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "V:=subs(subs1,V);" }}}{PARA 0 "" 0 "" {TEXT -1 23 "No w the kinetic energy " }{TEXT 266 1 "T" }{TEXT -1 56 " has to be diffe rentiated with respect to the velocity v" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "L1:=diff(T,v);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "Next \+ the kinetic energy " }{TEXT 267 1 "T" }{TEXT -1 59 " has to be differe ntiated with respect to the displacement " }{TEXT 269 1 "s" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "L2:=diff(T,s);" }}}{PARA 0 "" 0 "" {TEXT -1 87 "At last the potential energy has to be differentiated wit h respect to the displacement " }{TEXT 268 1 "s" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "L3:=simplify(diff(V,s));" }}}{PARA 0 "" 0 "" {TEXT -1 40 "Now the substitutions are made backwards" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs2:=\{v=diff(x(t),t),s=x(t)\}:" "6#>%&su bs2G<$/%\"vG-%%diffG6$-%\"xG6#%\"tGF./%\"sG-F,6#F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "L1:=subs(subs2,L1);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 19 "L2:=subs(subs2,L2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "L3:=subs(subs2,L3);" }}}{PARA 0 "" 0 "" {TEXT -1 38 "The equation of motion is now given by" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 26 "eom_a:=diff(L1,t)-L2+L3=0;" }}}{PARA 0 "" 0 "" {TEXT -1 48 "Or after collecting all the same derivatives of " }{TEXT 270 1 "x" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eom_a:=collect(e om_a,x);" }}}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 23 "Conclusion of Exam ple 2" }}{PARA 0 "" 0 "" {TEXT -1 384 "In this example it seems to be \+ much easier to use the synthetic method, because all active and all pa ssive forces are very clear and simple to describe. To formulate the e quilibrium is not very time-consuming. To solve the problem by use of \+ the analytic method, the solution is more formal but needs more effort . But this is a very simple system, so this result cannot be generaliz ed." }}}}{PARA 0 "" 0 "" {TEXT -1 40 "In the next example we use the f unction " }{HYPERLNK 17 "LAGRANGE" 2 "LAGRANGE" "" }{TEXT -1 10 " from the " }{HYPERLNK 17 "dynamics" 2 "dynamics" "" }{TEXT -1 40 " package to get the Lagrangian equations" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 15 "5.2.2 Example 2" }}{PARA 0 "" 0 "" {TEXT -1 93 "Consider a system \+ of two bodies as given in Fig. 5. The first body is a cylinder with ra dius " }{TEXT 272 1 "r" }{TEXT -1 10 " and mass " }{TEXT 273 1 "M" } {TEXT -1 42 ". The second body is a cube with the mass " }{TEXT 275 1 "m" }{TEXT -1 143 ". The cube hangs by an inextensible rope that is fi xed on the ground and lies over the cylinder. The cylinder hangs by a \+ spring with stiffness " }{TEXT 274 1 "k" }{TEXT -1 18 " from the ceili ng." }}{PARA 0 "" 0 "" {TEXT -1 73 "The displacement of the cylinder i n the vertical direction is denoted by " }{TEXT 276 1 "y" }{TEXT -1 18 ", the velocity by " }{TEXT 277 2 "yd" }{TEXT -1 25 " and the accel eration by " }{TEXT 278 3 "ydd" }{TEXT -1 58 ". The vertical displacem ent of the cylinder is denoted by " }{TEXT 279 1 "x" }{TEXT -1 18 ", t he velocity by " }{TEXT 280 2 "xd" }{TEXT -1 25 " and the acceleration by " }{TEXT 281 3 "xdd" }{TEXT -1 81 ". At last the rotation of the \+ cylinder with respect to its center is denoted by " }{TEXT 282 1 "f" } {TEXT -1 26 ", the rotational speed by " }{TEXT 283 1 "f" }{TEXT 289 3 "_d " }{TEXT -1 36 " and the rotational acceleration by " }{TEXT 284 1 "f" }{TEXT 290 3 "_dd" }{TEXT -1 127 ". The displacements, the v elocities and the accelerations are downward positve, the rotation is \+ positive counterclockwise. For " }{TEXT 286 3 "x=0" }{TEXT -1 2 ", " } {TEXT 287 3 "y=0" }{TEXT -1 5 " and " }{TEXT 285 1 "f" }{TEXT 288 2 "= 0" }{TEXT -1 75 " the spring should be free of stress. There should be no horizontal motion." }}{PARA 0 "" 0 "" {TEXT 324 68 "The exercise i s again to find the equation of motion for the system." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display([Fig_5_5()],scaling=constra ined,axes=none,title=\"Figure 5\");" }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 16 "Synthetic Method" }}{PARA 0 "" 0 "" {TEXT -1 467 "Before we con sider the equilibrium of the free bodies, we first consider the moment of inertia of the cylinder. The cylinder rotates around the connectio n point between cylinder and rope, because this point is fixed by the \+ rope (called point A). The rotation of the cylinder can be split for \+ small displacements into a rotation around the center C of the cylinde r and a translation in the vertical direction. This is pictured in the animations in fig. 6 and fig. 7. " }{TEXT 325 53 "(Click the picture \+ and press \"play\" in the menu bar.)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "display(Fig_5_6(),insequence=true,scaling=constrained ,axes=none,title=\"Figure 6\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "display(Fig_5_7(),insequence=true,scaling=constrained,axes=non e,title=\"Figure 7\");" }}}{PARA 0 "" 0 "" {TEXT -1 240 "In this examp le we consider the motion of the cylinder as a rotation around point C and a translation in the vertical direction, as fig. 8 shows. Then w e need the moment of inertia of the cylinder for the rotation around p oint C. From the " }{HYPERLNK 17 "dynamics" 2 "dynamics" "" }{TEXT -1 30 " package we get by use of the " }{HYPERLNK 17 "MI" 2 "MI" "" } {TEXT -1 9 " function" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "J : = MI('solidcylinder',M,r);" }}}{PARA 0 "" 0 "" {TEXT -1 75 "Now we con sider the equilibrium of the two free bodies as shown in Fig. 8. " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display([Fig_5_8()],scaling= constrained,axes=none,title=\"Figure 8\");" }}}{PARA 0 "" 0 "" {TEXT -1 92 "At first we write down the equilibrium of all forces in the ver tical direction for the cube." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "F_cube:=m*g-S2-m*xdd(t)=0;" "6#>%'F_cubeG/,(*&%\"mG\"\"\"%\"gGF)F)% #S2G!\"\"*&F(F)-%$xddG6#%\"tGF)F,\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 78 "The equilibrium of the forces in the vertical direction at the cyl inder yields" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "F_cylinder:=M* g+S1+S2-k*y(t)-M*ydd(t)=0;" "6#>%+F_cylinderG/,,*&%\"MG\"\"\"%\"gGF)F) %#S1GF)%#S2GF)*&%\"kGF)-%\"yG6#%\"tGF)!\"\"*&F(F)-%$yddG6#F2F)F3\"\"! " }}}{PARA 0 "" 0 "" {TEXT -1 86 "And the equilibrium of the moments a t the cylinder with respect to the center C yields" }}{EXCHG {PARA 0 " > " 0 "" {XPPEDIT 19 1 "M_cylinder_C:=S1*r-S2*r-J*phi_dd(t)=0;" "6#>%- M_cylinder_CG/,(*&%#S1G\"\"\"%\"rGF)F)*&%#S2GF)F*F)!\"\"*&%\"JGF)-%'ph i_ddG6#%\"tGF)F-\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 45 "Here the condit ion for the moment of inertia " }{TEXT 291 1 "J" }{TEXT -1 14 " is res pected." }}{PARA 0 "" 0 "" {TEXT -1 184 "Additionally there are some g eometrical relations between the displacements x(t) and y(t) and the r otation phi(t). In this example only small displacements are considere d, so that sin(" }{TEXT 293 1 "a" }{TEXT -1 1 ")" }{TEXT 296 4 " = a" }{TEXT -1 6 ", tan(" }{TEXT 294 1 "a" }{TEXT -1 1 ")" }{TEXT 295 3 " = " }{TEXT -1 10 "a and cos(" }{TEXT 292 1 "a" }{TEXT -1 2 ") " }{TEXT 297 3 "= 1" }{TEXT -1 34 ". To derive the relations between " }{TEXT 298 1 "x" }{TEXT -1 2 ", " }{TEXT 299 1 "y" }{TEXT -1 5 " and " } {TEXT 300 1 "f" }{TEXT -1 19 " we look to Fig. 9." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display([Fig_5_9()],scaling=constrained,axes= none,title=\"Figure 9\");" }}}{PARA 0 "" 0 "" {TEXT -1 17 "The relatio ns are" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "x(t) := 2*y(t):" "6# >-%\"xG6#%\"tG*&\"\"#\"\"\"-%\"yG6#F'F*" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "phi(t):=-x(t)/(2*r):" "6#>-%$phiG6#%\"tG,$*&-%\"xG6#F' \"\"\"*&\"\"#F-%\"rGF-!\"\"F1" }}}{PARA 0 "" 0 "" {TEXT -1 58 "Notice \+ especially the minus sign in the condition between " }{TEXT 301 1 "f" }{TEXT -1 5 " and " }{TEXT 302 1 "x" }{TEXT -1 1 "." }}{PARA 0 "" 0 " " {TEXT -1 69 "For the displacements and the rotation we write for the ir derivatives" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xd(t):=diff( x(t),t):" "6#>-%#xdG6#%\"tG-%%diffG6$-%\"xG6#F'F'" }}}{EXCHG {PARA 0 " > " 0 "" {XPPEDIT 19 1 "yd(t):=diff(y(t),t):" "6#>-%#ydG6#%\"tG-%%diff G6$-%\"yG6#F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "phi_d(t):= diff(phi(t),t):" "6#>-%&phi_dG6#%\"tG-%%diffG6$-%$phiG6#F'F'" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xdd(t):=diff(x(t),t,t):" "6#>- %$xddG6#%\"tG-%%diffG6%-%\"xG6#F'F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ydd(t):=diff(y(t),t,t):" "6#>-%$yddG6#%\"tG-%%diffG6%-% \"yG6#F'F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "phi_dd(t):=di ff(phi(t),t,t):" "6#>-%'phi_ddG6#%\"tG-%%diffG6%-%$phiG6#F'F'F'" }}} {PARA 0 "" 0 "" {TEXT -1 58 "From the equilibrium of the forces at the cube we isolate " }{TEXT 303 2 "S2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Sol1:=isolate(F_cube,S2);" }}}{PARA 0 "" 0 "" {TEXT -1 73 "We insert this result into the equilibrium of the forces at the cylinder " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Sol2:=subs(Sol 1,F_cylinder);" }}}{PARA 0 "" 0 "" {TEXT -1 16 "Here we isolate " } {TEXT 304 2 "S1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Sol3:=iso late(Sol2,S1);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "and substitute this an d " }{TEXT 305 4 "Sol1" }{TEXT -1 66 " into the equilibrium of moments at the cylinder. This yields the " }{TEXT 271 18 "equation of motion " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "eom_s:=subs(\{Sol3,Sol1 \},M_cylinder_C);" }}}{PARA 0 "" 0 "" {TEXT -1 21 "Some rewriting yiel ds" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "eom_s:=collect(simplif y(eom_s/r),\{y(t),diff(y(t),t,t),g\});" }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 15 "Analytic Method" }}{PARA 0 "" 0 "" {TEXT -1 107 "In this \+ system are no non-conservative forces active. We consider first the po tential energy of the system." }}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 16 "Potential Energy" }}{PARA 0 "" 0 "" {TEXT -1 43 "The potential energy of the spring is (see " }{HYPERLNK 17 "spring_energy" 2 "spring_energ y" "" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "V_sp ring:=spring_energy(k,y(t));" }}}{PARA 0 "" 0 "" {TEXT -1 68 "The pote ntial energy caused by the position of the cylinder is (see " } {HYPERLNK 17 "gravitational_energy" 2 "gravitational_energy" "" } {TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 34 "(notice the positive dir ection of " }{TEXT 306 1 "x" }{TEXT -1 5 " and " }{TEXT 307 1 "y" } {TEXT -1 12 " downwards!)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Vg_cylinder:=gravitational_energy(M,g,-y(t));" }}}{PARA 0 "" 0 "" {TEXT -1 40 "And for the cube the potential energy is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Vg_cube:=gravitational_energy(m,g,-x(t)); " }}}{PARA 0 "" 0 "" {TEXT -1 59 "Here and in the further calculation \+ the conditions between " }{TEXT 308 1 "x" }{TEXT -1 2 ", " }{TEXT 309 1 "y" }{TEXT -1 5 " and " }{TEXT 310 1 "f" }{TEXT -1 95 " are taken ov er from the synthetic method (see fig. 5).This effort is the same in b oth methods." }}{PARA 0 "" 0 "" {TEXT -1 29 "The total potential energ y is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "V:=V_spring+Vg_cylin der+Vg_cube;" }}}}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 14 "Kinetic Energy " }}{PARA 0 "" 0 "" {TEXT -1 219 "Now there are two masses. The cube m akes only a translation in the vertical direction. The motion of the c ylinder here should be considered as a rotation arround point A. The m oment of inertia for this rotation is (see " }{HYPERLNK 17 "section 4. 2.2" 1 "sec-04.mws" "Steiner" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "J_A:=J+M*r**2;" "6#>%$J_AG,&%\"JG\"\"\"*&%\"MGF'*$% \"rG\"\"#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "The velocities are" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xd(t):=diff(x(t),t);" "6#>-%#xdG6#%\"tG-%%diffG6$-%\"xG6#F'F'" } }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "yd(t):=diff(y(t),t);" "6#>-% #ydG6#%\"tG-%%diffG6$-%\"yG6#F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "phi_d(t):=diff(phi(t),t);" "6#>-%&phi_dG6#%\"tG-%%diffG 6$-%$phiG6#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 43 " So the kinetic energ y of the cube is (see " }{HYPERLNK 17 "kinetic_energy" 2 "kinetic_ener gy" "" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "T_c ube:=kinetic_energy(m,xd(t));" }}}{PARA 0 "" 0 "" {TEXT -1 69 "For the rotation of the cylinder the kinetic energy can be written by" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "T_cylinder:=kinetic_energy(J _A,phi_d(t));" }}}{PARA 0 "" 0 "" {TEXT -1 46 "with the conditions for the moment of inertia " }{TEXT 311 1 "J" }{TEXT -1 60 "_A and the rel ation between y and phi and their derivatives." }}{PARA 0 "" 0 "" {TEXT -1 27 "The total kinetic energy is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "T:=T_cube+T_cylinder;" }}}}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 25 "Use The Function LAGRANGE" }}{PARA 0 "" 0 "" {TEXT -1 48 "There are no non-conservative forces, so we have" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "F:=0:" "6#>%\"FG\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 53 "The equation of motion we get by use of the function " } {HYPERLNK 17 "LAGRANGE" 2 "LAGRANGE" "" }{TEXT -1 1 ":" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "eom_a:=LAGRANGE(T,V,[F],[y],t);" }} }{PARA 0 "" 0 "" {TEXT -1 33 "(Notice: For use of the function " } {HYPERLNK 17 "LAGRANGE" 2 "LAGRANGE" "" }{TEXT -1 1 " " }{TEXT 312 1 " F" }{TEXT -1 5 " and " }{TEXT 313 1 "y" }{TEXT -1 29 " have to be writ ten as sets.)" }}{PARA 0 "" 0 "" {TEXT -1 21 "Some rewriting yields" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "eom_s:=collect(simplify(eom _s),\{y(t),diff(y(t),t,t),g\});" }}}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 23 "Conclusion of Example 3" }}{PARA 0 "" 0 "" {TEXT -1 187 "In thi s example it is more efficient to use the analytic method, because all actions which have to be done are very schematic. This is especially \+ seen by the use of the prepared function " }{HYPERLNK 17 "LAGRANGE" 2 "LAGRANGE" "" }{TEXT -1 267 ". It is not necessary to write down the e quilibrium for the free bodies, and the complicated cutting of the con nections is avoided. Together with the conclusion of example one we ca n see that it depends on the particular problem to decide which method is more useful." }{TEXT 314 0 "" }{TEXT -1 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 93 "As the last example of this course we want to consider an example with two degrees of freedom" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 9 "Example 3" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "unassign( 'x','y','t','g','F','m'):" }}}{PARA 0 "" 0 "" {TEXT -1 34 "We consider the same system as in " }{HYPERLNK 17 "example 2" 1 "" "EX 2" }{TEXT -1 33 " but now with an additional mass " }{XPPEDIT 18 0 "me;" "6#%#me G" }{TEXT -1 25 " connected by the spring " }{XPPEDIT 18 0 "ke;" "6#%# keG" }{TEXT -1 40 " and driven by the time-dependent force " } {XPPEDIT 18 0 "F(t)" "6#-%\"FG6#%\"tG" }{TEXT -1 22 ", as shown in Fig . 10." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "display([Fig_5_10() ],scaling=constrained,axes=none,title=\"Figure 10\");" }}}{PARA 0 "" 0 "" {TEXT -1 154 "We leave out here the vertical guidances but we ass ume that there exists only vertical motions. In addition to the well k nown system we have now the mass " }{XPPEDIT 18 0 "me;" "6#%#meG" } {TEXT -1 57 ". Now we have the two independent Lagrangian coordinates \+ " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "y; " "6#%\"yG" }{TEXT -1 111 ". So we get two coupeld equations of motion for the system. We calculate the equations of motion by use of the " }{HYPERLNK 17 "LAGRANGE" 2 "LAGRANGE" "" }{TEXT -1 120 " function. The refore we have at first to calculate the kinetic energy and the potent ial of the system. The active force " }{XPPEDIT 18 0 "F(t)" "6#-%\"FG6 #%\"tG" }{TEXT -1 93 " will be considered as a non-conservative force. Here we don't consider the systhetic method." }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 14 "Kinetic Energy" }}{PARA 0 "" 0 "" {TEXT -1 143 "We def ine the state when all springs are stressless as the initial state of \+ the system. So we have the kinetic energy of the upper cuboid (see " } {HYPERLNK 17 "example 2" 1 "" "EX 2" }{TEXT -1 58 ") with use of the d efinition of the velocity of the cuboid" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xd(t) := diff(x(t),t):" "6#>-%#xdG6#%\"tG-%%diffG6$-%\" xG6#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 8 "given by" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "T_cuboid:=kinetic_energy(m,xd(t));" }}}{PARA 0 "" 0 "" {TEXT -1 23 "Tho motion of the mass " }{XPPEDIT 18 0 "me;" " 6#%#meG" }{TEXT -1 17 " has the velocity" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "yd(t) := diff(y(t),t):" "6#>-%#ydG6#%\"tG-%%diffG6$-%\" yG6#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 45 "So we get for the kinetic en ergy of the mass " }{XPPEDIT 18 0 "me;" "6#%#meG" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 31 "T_me:=kinetic_energy(me,yd(t));" }}}{PARA 0 " " 0 "" {TEXT -1 32 "The total kinetic energy is then" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "T:=T_cuboid+T_me;" }}}}{SECT 0 {PARA 5 " " 0 "" {TEXT -1 16 "Potential Energy" }}{PARA 0 "" 0 "" {TEXT -1 51 "T he potential energy of the upper springs are (see " }{HYPERLNK 17 "exa mple 2" 1 "" "EX 2" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "V1:=spring_energy(k1,x(t));" }}}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "V2:=sprin g_energy(k2,x(t));" }}}{PARA 0 "" 0 "" {TEXT -1 66 "The potential ener gy caused by the position of the upper cuboid is" }}{PARA 0 "" 0 "" {TEXT -1 34 "(notice the positive direction of " }{TEXT 315 1 "x" } {TEXT -1 12 " downwards!)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "Vg_cuboid:=gravitational_energy(m,g,-x(t));" }}}{PARA 0 "" 0 "" {TEXT -1 29 "The elongation of the spring " }{XPPEDIT 18 0 "ke;" "6#%# keG" }{TEXT -1 87 " is defined by the difference between the displacem ent of the both cuboids, which means" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Delta:=y(t)-x(t):" "6#>%&DeltaG,&-%\"yG6#%\"tG\"\"\"-% \"xG6#F)!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 32 "For the potential of th e spring " }{XPPEDIT 18 0 "ke;" "6#%#keG" }{TEXT -1 7 " we get" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Ve:=spring_energy(ke,Delta); " }}}{PARA 0 "" 0 "" {TEXT -1 22 "For the mass particle " }{XPPEDIT 18 0 "me;" "6#%#meG" }{TEXT -1 33 " we have the gravitational energy" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Vg_me:=gravitational_energ y(me,g,-y(t));" }}}{PARA 0 "" 0 "" {TEXT -1 211 "(Notice: For use of t he Lagrangian equation only the changing of the potential is of intere st. So it doesn't matter that we choose two different horizontals to f ormulate the potential of gravity for the bodies.)" }}{PARA 0 "" 0 "" {TEXT -1 29 "The total potential energy is" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 49 "V:=collect(V1+V2+Vg_cuboid+Ve+Vg_me,\{x(t),y(t)\}); " }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 25 "Use The Function LAGRANGE" }}{PARA 0 "" 0 "" {TEXT -1 47 "There only non-conservative force is th e force " }{XPPEDIT 18 0 "F(t)" "6#-%\"FG6#%\"tG" }{TEXT -1 235 " acti ng on the upper cuboid. To calculate the non-conservative forces with \+ respect to the Lagrangian coordinates we need in general to define a g lobal coordinate system. Here this is not necessary, but we will do it to show the method." }}{PARA 0 "" 0 "" {TEXT -1 21 "We define the glo bal " }{TEXT 316 6 "X-Y-Z-" }{TEXT -1 27 "coordinate system with the \+ " }{TEXT 317 1 "X" }{TEXT -1 39 "-axis in the horizontal direction, th e " }{TEXT 318 1 "Y" }{TEXT -1 40 "-axis orthogonal to the drawing and the " }{TEXT 319 1 "Z" }{TEXT -1 62 "-axis in the opposite direction \+ of the lagrangian coordinates " }{TEXT 320 1 "x" }{TEXT -1 4 " and" } {TEXT 321 2 " y" }{TEXT -1 43 ". So we get the equations of transforma tion" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "r1:=vector(3,[0,0,-x (t)]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "r2:=vector(3,[0,0 ,-y(t)]);" }}}{PARA 0 "" 0 "" {TEXT -1 163 "For the following calculat ion we need the derivative of these vectors with respect to the lagran gian coordinates. To get these we make some temporary substitutions" } }{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "sv := \{x(t) = lx, y(t) = ly \}:" "6#>%#svG<$/-%\"xG6#%\"tG%#lxG/-%\"yG6#F*%#lyG" }}{PARA 0 "> " 0 "" {XPPEDIT 19 1 "sr := \{lx = x(t), ly = y(t)\}:" "6#>%#srG<$/%#lxG-% \"xG6#%\"tG/%#lyG-%\"yG6#F+" }}}{PARA 0 "" 0 "" {TEXT -1 42 "Now we ca lculate the necessary derivatives" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "for i from 1 by 1 to 3 do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "rdiff11[i]:=subs(sr,diff(subs(sv,r1[i]),lx));" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "rdiff12[i]:=subs(sr,diff(subs(sv,r2 [i]),lx));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "rdiff21[i]:=subs(sr,d iff(subs(sv,r1[i]),ly));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "rdiff22 [i]:=subs(sr,diff(subs(sv,r2[i]),ly));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}}{PARA 0 "" 0 "" {TEXT -1 123 "The result of these deriv atives should be evident, so we needn't to talk about it in detail. We write the result as vectors" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "rd11:=vector(3,[rdiff11[1],rdiff11[2],rdiff11[3]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "rd12:=vector(3,[rdiff12[1],rdiff12[ 2],rdiff12[3]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "rd21:=v ector(3,[rdiff21[1],rdiff21[2],rdiff21[3]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "rd22:=vector(3,[rdiff22[1],rdiff22[2],rdiff22[3] ]);" }}}{PARA 0 "" 0 "" {TEXT -1 108 "The vectors of the resultant of \+ the nonconservative forces acting on both bodies with respect to the g lobal " }{TEXT 322 6 "X-Y-Z-" }{TEXT -1 22 " coordinate system are" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "F1:=vector(3,[0,0,F(t)]);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "F2:=vector(3,[0,0,0]);" }} }{PARA 0 "" 0 "" {TEXT -1 63 "Finaly we calculate the dotproduct and t he sum of the equation " }{XPPEDIT 18 0 "Q[k] = sum(F[i]*diff(r[i],q[k ]),i = 1 .. n)" "6#/&%\"QG6#%\"kG-%$sumG6$*&&%\"FG6#%\"iG\"\"\"-%%diff G6$&%\"rG6#F/&%\"qG6#F'F0/F/;F0%\"nG" }{TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Q[1]:=dotprod(F1,rd11)+dotprod(F2,rd12); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Q[2]:=dotprod(F1,rd21)+ dotprod(F2,rd22);" }}}{PARA 0 "" 0 "" {TEXT -1 61 "So we get at last t he vector of the nonconservative forces as" }}{EXCHG {PARA 0 "> " 0 " " {XPPEDIT 19 1 "Q := [Q[1], Q[2]];" "6#>%\"QG7$&F$6#\"\"\"&F$6#\"\"# " }}}{PARA 0 "" 0 "" {TEXT -1 92 "It can easily be seen that we can wr ite tis forces direct without this detailed calculation." }}{PARA 0 " " 0 "" {TEXT -1 40 "The set of the Lagrangian coordinates is" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xy := [x, y]:" "6#>%#xyG7$%\"x G%\"yG" }}}{PARA 0 "" 0 "" {TEXT -1 53 "The equation of motion we get \+ by use of the function " }{HYPERLNK 17 "LAGRANGE" 2 "LAGRANGE" "" } {TEXT -1 1 ":" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "eom:=LAGRAN GE(T,V,Q,xy,t);" }}}{PARA 0 "" 0 "" {TEXT -1 48 "So we get the two cou pled differential equations" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "EOM1:=eom[1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "EOM2:= eom[2];" }}}{PARA 0 "" 0 "" {TEXT -1 30 "We rearange the result and ge t" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "DGL1:=collect(EOM1,\{x( t),diff(x(t),t$2),y(t),diff(y(t),t$2)\});" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 62 "DGL2:=collect(EOM2,\{x(t),diff(x(t),t$2),y(t),diff( y(t),t$2)\});" }}}{PARA 0 "" 0 "" {TEXT -1 160 "For completeness we ca lculate the solution of this equations. Before doing this we make some assumptions for the initial state of the motion. We define that at " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 67 " the system starts \+ in it's initial state without motion, that means" }}{EXCHG {PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "IC1:=x(0)=0:" "6#>%$IC1G/-%\"xG6#\"\"!F)" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "IC2:=y(0)=0:" "6#>%$IC2G/-%\"y G6#\"\"!F)" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "IC3:=D(x)(0)=0: " "6#>%$IC3G/--%\"DG6#%\"xG6#\"\"!F," }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "IC4:=D(y)(0)=0:" "6#>%$IC4G/--%\"DG6#%\"yG6#\"\"!F," }} }{PARA 0 "" 0 "" {TEXT -1 49 "Finally we assume that the excitation is harmonic" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "F(t):=F0*sin(Omeg a*t):" "6#>-%\"FG6#%\"tG*&%#F0G\"\"\"-%$sinG6#*&%&OmegaGF*F'F*F*" }}} {PARA 0 "" 0 "" {TEXT -1 56 "and we use some concrete values for the s ystem constants" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=9.81:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "k1:=5.0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "k2:=5.0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "ke:=2.0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 " m:=1.0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "me:=0.2:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "F0:=0.01:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Omega:=5.0:" }}}{PARA 0 "" 0 "" {TEXT -1 47 "The system of the differential equations is now" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eval(DGL1);eval(DGL2);" }}}{PARA 0 "" 0 "" {TEXT -1 95 "To get the solution of this system of differential equati ons we use the adequate MAPLE funktion" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "LSG:=value(dsolve(\{eval(DGL1),eval(DGL2),IC1,IC2,IC3 ,IC4\},\{x(t),y(t)\})):" }}}{PARA 0 "" 0 "" {TEXT -1 100 "The result i s first very extensive, so we use some rearrangement by use of differe nt MAPLE funktions" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "xs(t): =simplify(evalf(combine(simplify(rhs(LSG[1])),trig)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "ys(t):=simplify(evalf(combine(simpl ify(rhs(LSG[2])),trig)));" }}}{PARA 0 "" 0 "" {TEXT -1 109 "Finally we see the time history of the motion of both bodies in Fig. 11 (green: \+ upper mass, blue: lower mass)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Tend:=10:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "P1:=plot( xs(t),t=0..Tend,color=green,thickness=3):" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 49 "P2:=plot(ys(t),t=0..Tend,color=blue,thickness=3):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "display(\{P1,P2\}, title= \"Figure 11\");" }}}{PARA 0 "" 0 "" {TEXT -1 67 "We see that the motio n is a vibration around the static equilibium." }}{PARA 0 "" 0 "" {TEXT -1 39 "Fig. 12 shows the motion of the system." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "display(Fig_5_11(xs(t),ys(t),eval(F(t)*5 0),Tend,t),insequence=true,scaling=constrained,axes=none, title=\"Figu re 12\");" }}}}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 152 "This is the end \+ of this course. Once you have understood these principles and examples , you have enough knowledge to solve simple dynamic problems. The " } {HYPERLNK 17 "dynamics" 2 "dynamics" "" }{TEXT -1 144 " package should give you some helpful functions for your own calculations. I wish yo u much fun in your future endeavors in classical mechanics." }}}} {MARK "0 0 0" 11 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }