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}}}{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/m6 dynfig\",libname:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "with(l inalg):with(plots):with(plottools):with(dynamics);with(figures_chapter _4);" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 22 "4.2 Plane Rigid Bodies " }}{PARA 0 "" 0 "" {TEXT -1 370 "In this section we derive some geome trical properties of the apportionment of the mass of rigid bodies. Th ese properties don't depend on the motion of the body but they may aff ect it. Here we consider a special kind of body. We consider bodies wi th a symmetry plane with respect to their mass distribution, i.e., the y are cylindrical (or prismatic). We will call these " }{TEXT 256 19 "planar rigid bodies" }{TEXT -1 72 ". In this special case it is enoug h to consider only the symmetry plane." }}{SECT 0 {PARA 5 "" 0 "4.2.1 " {TEXT -1 49 "4.2.1 Kinematics of Planar Motion of Rigid Bodies" }} {PARA 0 "" 0 "" {TEXT 367 8 "Velocity" }}{PARA 0 "" 0 "" {TEXT -1 68 " Now we consider a rigid body that is moving with respect to a fixed " }{TEXT 284 7 "(x,y,z)" }{TEXT -1 30 "-reference system with origin " } {TEXT 285 1 "O" }{TEXT -1 123 ". We talk about planar motions when all points of the considered body are moving in parallel planes. Usually \+ we define the " }{TEXT 286 1 "z" }{TEXT -1 76 "-axis orthogonal to thi s plane. Additionally we define a local body - fixed " }{TEXT 287 7 "( X,Y,Z)" }{TEXT -1 24 "-system, again with the " }{TEXT 288 1 "Z" } {TEXT -1 48 "-axis orthogonal to the plane of motion. So the " }{TEXT 289 1 "Z" }{TEXT -1 14 "-axis and the " }{TEXT 290 1 "z" }{TEXT -1 83 "-axis are always parallel. The origin of the moving system is the bod y-fixed point " }{TEXT 291 1 "P" }{TEXT -1 33 ". In Fig. 8 we see the \+ situation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "display(Fig_4_ 5(),scaling=constrained,axes=none,title=\"Figure 8\");" }}}{PARA 0 "" 0 "" {TEXT -1 56 "The position of the rigid body is defined by the vec tor " }{TEXT 292 2 "rP" }{TEXT -1 9 " between " }{TEXT 293 1 "O" } {TEXT -1 5 " and " }{TEXT 294 2 "P " }{TEXT -1 15 "and the angle " } {XPPEDIT 18 0 "phi" "6#%$phiG" }{TEXT -1 20 " between the global " } {TEXT 295 1 "x" }{TEXT -1 20 "-axis and the local " }{TEXT 296 1 "X" } {TEXT -1 102 "-axis. (For clarity this angle is not shown in the figur e above, but it should be clear what we mean)." }}{PARA 0 "" 0 "" {TEXT -1 48 "The velocity of the body is given by the vector " }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "vP=diff(rP,t)" "6#/%#vPG-%%diffG6$%#r PG%\"tG" }}{PARA 0 "" 0 "" {TEXT -1 13 "of the point " }{TEXT 297 1 "P " }{TEXT -1 75 " with respect to the inertial system and the vector of the angular velocity" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "omega=diff(p hi,t)*eZ" "6#/%&omegaG*&-%%diffG6$%$phiG%\"tG\"\"\"%#eZGF+" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "As in " }{HYPERLNK 17 "section 2" 1 "sec-02.mws" "" }{TEXT -1 92 ", when we described the motion in rela tive coordinates, we consider now an additional point " }{TEXT 298 1 " Q" }{TEXT -1 271 ". But now the point Q is fixed on the body and so it is fixed in the local coordinate system. Before we continue we repeat some details about the formulation for motions in relative coordinate s. As in the prior section, we define the unit vectors in the directio n of the " }{TEXT 299 2 "x-" }{TEXT -1 9 " and the " }{TEXT 300 2 "y- " }{TEXT -1 7 "axis as" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ex:= vector(3,[1,0,0]);" "6#>%#exG-%'vectorG6$\"\"$7%\"\"\"\"\"!F+" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ey:=vector(3,[0,1,0]);" "6#>%# eyG-%'vectorG6$\"\"$7%\"\"!\"\"\"F*" }}}{PARA 0 "" 0 "" {TEXT -1 41 "T he unit vectors in the direction of the " }{TEXT 301 1 "X" }{TEXT -1 10 "- and the " }{TEXT 302 2 "Y-" }{TEXT -1 40 "axis of the moving sys tem are defined by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eX:=vect or(3,[eXx,eXy,0]);" "6#>%#eXG-%'vectorG6$\"\"$7%%$eXxG%$eXyG\"\"!" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eY:=vector(3,[eYx,eYy,0]);" "6 #>%#eYG-%'vectorG6$\"\"$7%%$eYxG%$eYyG\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 71 "As before we desribe the derivative with respect to time by app ending \"" }{TEXT 306 2 "_d" }{TEXT -1 43 "\" to the variable and sayi ng for example: \"" }{TEXT 307 1 "r" }{TEXT -1 6 " dot\"." }}{PARA 0 " " 0 "" {TEXT -1 88 "The current position and the motion of the conside red point Q is described by the vector" }{TEXT 303 3 " rQ" }{TEXT -1 101 ", which is affected by the current position of the moving coordin ate system, described by the vector " }{TEXT 304 2 "rP" }{TEXT -1 118 ", and the relative position of the point Q with respect to the origin P of the moving system, described by the vector " }{TEXT 305 1 "r" } {TEXT -1 7 " (rho)." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rP := x P*ex+yP*ey:" "6#>%#rPG,&*&%#xPG\"\"\"%#exGF(F(*&%#yPGF(%#eyGF(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalm(rP);" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rho := Xrho*eX+Yrho*eY:" "6#>%$rhoG,&*&%%Xr hoG\"\"\"%#eXGF(F(*&%%YrhoGF(%#eYGF(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalm(rho);" }}}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Xrho*e Xx;" "6#*&%%XrhoG\"\"\"%$eXxGF%" }{TEXT -1 52 " is the x-component of \+ the projection of the vector " }{XPPEDIT 18 0 "rho" "6#%$rhoG" }{TEXT -1 48 " on the X-axis of the moving coordinate system, " }{XPPEDIT 18 0 "Yrho*eYx;" "6#*&%%YrhoG\"\"\"%$eYxGF%" }{TEXT -1 53 " is the x-com ponent of the projection of the vector " }{XPPEDIT 18 0 "rho" "6#%$rho G" }{TEXT -1 60 " on the Y-axis of the moving coordinate system. Accor dingly " }{XPPEDIT 18 0 "Xrho*eXy;" "6#*&%%XrhoG\"\"\"%$eXyGF%" } {TEXT -1 52 " is the y-component of the projection of the vector " } {XPPEDIT 18 0 "rho" "6#%$rhoG" }{TEXT -1 48 " on the X-axis of the mov ing coordinate system, " }{XPPEDIT 18 0 "Yrho*eYy;" "6#*&%%YrhoG\"\"\" %$eYyGF%" }{TEXT -1 53 " is the y-component of the projection of the \+ vector " }{XPPEDIT 18 0 "rho" "6#%$rhoG" }{TEXT -1 48 " on the Y-axis \+ of the moving coordinate system. " }}{PARA 0 "" 0 "" {TEXT -1 12 "Agai n as in " }{HYPERLNK 17 "section 2" 1 "sec-02.mws" "" }{TEXT -1 26 " w e make the substitutions" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "su bs1:=Xrho*eXx=XQx:" "6#>%&subs1G/*&%%XrhoG\"\"\"%$eXxGF(%$XQxG" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs2:=Yrho*eYx=YQx:" "6#>%&su bs2G/*&%%YrhoG\"\"\"%$eYxGF(%$YQxG" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs3:=Xrho*eXy=XQy:" "6#>%&subs3G/*&%%XrhoG\"\"\"%$eXy GF(%$XQyG" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs4:=Yrho*eYy= YQy:" "6#>%&subs4G/*&%%YrhoG\"\"\"%$eYyGF(%$YQyG" }}}{PARA 0 "" 0 "" {TEXT -1 24 "For the vector rQ we get" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rQ := rP+rho;" "6#>%#rQG,&%#rPG\"\"\"%$rhoGF'" }}} {PARA 0 "" 0 "" {TEXT -1 22 "This yields by use of " }{TEXT 309 5 "sub s1" }{TEXT -1 4 " to " }{TEXT 308 5 "subs4" }{TEXT -1 29 " (we use the appended letter " }{TEXT 310 1 "s" }{TEXT -1 61 " to distinguish the \+ expression with and without substutution)" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rQs:=subs(\{subs1,subs2,subs3,subs4\},evalm(rQ));" "6#> %$rQsG-%%subsG6$<&%&subs1G%&subs2G%&subs3G%&subs4G-%&evalmG6#%#rQG" }} }{PARA 0 "" 0 "" {TEXT -1 20 "The rotation of the " }{TEXT 337 7 "(X,Y ,Z)" }{TEXT -1 34 "-system is described by the vector" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Omega:=vector(3,[0,0,omega]);" "6#>%&OmegaG -%'vectorG6$\"\"$7%\"\"!F*%&omegaG" }}}{PARA 0 "" 0 "" {TEXT -1 20 "Fo r the velocity of " }{TEXT 312 2 "Q " }{TEXT -1 27 "we go three compon enets in " }{HYPERLNK 17 "section 2" 1 "sec-02.mws" "" }{TEXT -1 47 ". At first there was the velocity of the point " }{TEXT 311 1 "P" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vP:=vector(3,[xP_d,yP_d,0]);" "6#>%#vPG-%'vectorG6$\"\"$7%%%xP_dG%%yP_dG\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 46 "Next there is the changing of the position of " }{TEXT 313 1 "Q" }{TEXT -1 44 " inside the relative system. In contrast to " }{HYPERLNK 17 "section 2" 1 "sec-02.mws" "" }{TEXT -1 57 " there is in the current situation no relative motion of " }{TEXT 314 1 "Q" } {TEXT -1 14 " and so we get" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vrel := vector(3,[0, 0, 0]);" "6#>%%vrelG-%'vectorG6$\"\"$7%\"\"!F*F* " }}}{PARA 0 "" 0 "" {TEXT -1 104 "At last we have the part of the vel ocity which is caused by the rotation of the moving coordinate system " }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vrot:=crossprod(Omega,rho) ;" "6#>%%vrotG-%*crossprodG6$%&OmegaG%$rhoG" }}}{PARA 0 "" 0 "" {TEXT -1 38 "or with use of the substitutions above" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vrots:=subs(\{subs1,subs2,subs3,subs4\},evalm(vrot ));" "6#>%&vrotsG-%%subsG6$<&%&subs1G%&subs2G%&subs3G%&subs4G-%&evalmG 6#%%vrotG" }}}{PARA 0 "" 0 "" {TEXT -1 14 "For the point " }{TEXT 335 1 "Q" }{TEXT -1 26 " we get the total velocity" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vQ := vP+vrel+vrots;" "6#>%#vQG,(%#vPG\"\"\"%%vrel GF'%&vrotsGF'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalm(vQ);" "6#-%&evalmG6#%#vQG" }}}{PARA 0 "" 0 "" {TEXT -1 24 "or written as com ponents" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vx:=evalm(vQ)[1];" "6#>%#vxG&-%&evalmG6#%#vQG6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vy:=evalm(vQ)[2];" "6#>%#vyG&-%&evalmG6#%#vQG6#\"\"#" } }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vz:=evalm(vQ)[3];" "6#>%#vzG &-%&evalmG6#%#vQG6#\"\"$" }}}{PARA 0 "" 0 "" {TEXT -1 104 "We see that this consideration is a special case of the description of motions in relative coordinates. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 196 "Next we consider the same situation as before. But \+ now we choose another point inside the rigid body as the origin of the local coordinate system. As shown in Fig. 9 we define an additional p oint " }{TEXT 315 2 "P\264" }{TEXT -1 45 " inside the rigid body. The \+ question is now: " }{TEXT 349 69 "Is the motion of the system dependen t on the chosen reference points?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "display(Fig_4_6(),scaling=constrained,axes=none,title =\"Figure 9\");" }}}{PARA 0 "" 0 "" {TEXT -1 16 "The position of " } {TEXT 316 2 "P\264" }{TEXT -1 68 " with respect to the fixed coordinat e system is given by the vector " }{TEXT 317 3 "rP\264" }{TEXT -1 19 " , the position of " }{TEXT 318 1 "Q" }{TEXT -1 17 " with respect to \+ " }{TEXT 319 2 "P\264" }{TEXT -1 24 " is given by the vector " } {XPPEDIT 18 0 "rho\264" "6#%%rho|_vG" }{TEXT -1 26 ". At last the posi tion of " }{TEXT 320 2 "P\264" }{TEXT -1 17 " with respect to " } {TEXT 321 1 "P" }{TEXT -1 29 " is described by the vector " }{TEXT 322 4 "rPP\264" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 68 "With th e relations above we want to formulate the velocity of point " }{TEXT 323 1 "Q" }{TEXT -1 48 " when we describe the situation by use of poin t " }{TEXT 324 2 "P\264" }{TEXT -1 18 " instead of point " }{TEXT 325 1 "P" }{TEXT -1 108 ". At first we have to make some preparatory work, which is the same as we done above but with new variables." }}{PARA 0 "" 0 "" {TEXT -1 57 "The unit vectors of the local coordinate system in point " }{TEXT 331 2 "P\264" }{TEXT -1 15 " are defined by" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`eX\264` := vector(3,[`eXx\264 `, `eXy\264`, 0]);" "6#>%$eX|_vG-%'vectorG6$\"\"$7%%%eXx|_vG%%eXy|_vG \"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`eY\264` := vector(3 ,[`eYx\264`, `eYy\264`, 0]);" "6#>%$eY|_vG-%'vectorG6$\"\"$7%%%eYx|_vG %%eYy|_vG\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 129 "The current position \+ of the point Q is affected by the current position of the moving coord inate system, described by the vector " }{TEXT 326 3 "rP\264" }{TEXT -1 119 ", and the relative position of the point Q with respect to the origin P\264 of the moving system, described by the vector " }{TEXT 332 4 "rho\264" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`rP\264` := `xP\264`*ex+`yP\264`*ey:" "6#>%$rP|_vG,&*&%$xP|_vG\" \"\"%#exGF(F(*&%$yP|_vGF(%#eyGF(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalm(rP\264);" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`rho\264` := `Xrho\264`*`eX\264`+`Yrho\264`*`eY\264`:" "6#>%%rho |_vG,&*&%&Xrho|_vG\"\"\"%$eX|_vGF(F(*&%&Yrho|_vGF(%$eY|_vGF(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalm(rho\264);" }}}{PARA 0 "" 0 "" {TEXT -1 16 "The position of " }{TEXT 343 2 "P\264" }{TEXT -1 17 " with respect to " }{TEXT 344 1 "P" }{TEXT -1 18 " can be written \+ as" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "rPP\264:=rho-rho\264: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalm(rPP\264);" }}} {PARA 0 "" 0 "" {TEXT -1 32 "Again we make some substitutions" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`subs1\264` := `Xrho\264`*`eXx \264` = `XQx\264`:" "6#>%'subs1|_vG/*&%&Xrho|_vG\"\"\"%%eXx|_vGF(%%XQx |_vG" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`subs2\264` := `Yrho \264`*`eYx\264` = `YQx\264`:" "6#>%'subs2|_vG/*&%&Yrho|_vG\"\"\"%%eYx| _vGF(%%YQx|_vG" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`subs3\264` := `Xrho\264`*`eXy\264` = `XQy\264`:" "6#>%'subs3|_vG/*&%&Xrho|_vG\" \"\"%%eXy|_vGF(%%XQy|_vG" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "` subs4\264` := `Yrho\264`*`eYy\264` = `YQy\264`:" "6#>%'subs4|_vG/*&%&Y rho|_vG\"\"\"%%eYy|_vGF(%%YQy|_vG" }}}{PARA 0 "" 0 "" {TEXT -1 29 "And for the vector rQ\264 we get" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`rQ\264` := `rP\264`+`rho\264`;" "6#>%$rQ|_vG,&%$rP|_vG\"\"\"%%rho| _vGF'" }}}{PARA 0 "" 0 "" {TEXT -1 24 "This yields by using of " } {TEXT 328 5 "subs1" }{TEXT -1 4 " to " }{TEXT 327 5 "subs4" }{TEXT -1 29 " (we use the appended letter " }{TEXT 329 1 "s" }{TEXT -1 61 " to \+ distinguish the expression with and without substutution)" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`rQs\264` := subs(\{`subs1\264`, `sub s2\264`, `subs3\264`, `subs4\264`\},evalm(`rQ\264`));" "6#>%%rQs|_vG-% %subsG6$<&%'subs1|_vG%'subs2|_vG%'subs3|_vG%'subs4|_vG-%&evalmG6#%$rQ| _vG" }}}{PARA 0 "" 0 "" {TEXT -1 21 "The rotation of the (" }{TEXT 336 10 "X\264,Y\264,Z\264)-" }{TEXT -1 66 "system with respect to the \+ fixed system is described by the vector" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`Omega\264` := vector(3,[0, 0, `omega\264`]);" "6#>%'Om ega|_vG-%'vectorG6$\"\"$7%\"\"!F*%'omega|_vG" }}}{PARA 0 "" 0 "" {TEXT -1 26 "The velocity of the point " }{TEXT 330 2 "P\264" }{TEXT -1 3 " is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`vP\264` := vecto r(3,[`xP\264_d`, `yP\264_d`, 0]);" "6#>%$vP|_vG-%'vectorG6$\"\"$7%%&xP |_v_dG%&yP|_v_dG\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 77 "The component o f the velocity which is caused by the rotation of the moving (" } {TEXT 333 10 "X\264,Y\264,Z\264)-" }{TEXT -1 20 "coordinate system is " }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`vrot\264` := crossprod(`O mega\264`,`rho\264`);" "6#>%&vrot|_vG-%*crossprodG6$%'Omega|_vG%%rho|_ vG" }}}{PARA 0 "" 0 "" {TEXT -1 38 "or with use of the substitutions a bove" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`vrots\264` := subs(\{ `subs1\264`, `subs2\264`, `subs3\264`, `subs4\264`\},evalm(`vrot\264`) );" "6#>%'vrots|_vG-%%subsG6$<&%'subs1|_vG%'subs2|_vG%'subs3|_vG%'subs 4|_vG-%&evalmG6#%&vrot|_vG" }}}{PARA 0 "" 0 "" {TEXT -1 22 "The total \+ velocity of " }{TEXT 334 1 "Q" }{TEXT -1 4 " is " }}{EXCHG {PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "`vQ\264` := `vP\264`+`vrots\264`;" "6#>%$vQ|_vG, &%$vP|_vG\"\"\"%'vrots|_vGF'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalm(`vQ\264`);" "6#-%&evalmG6#%$vQ|_vG" }}}{PARA 0 "" 0 "" {TEXT -1 24 "or written as components" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`vx\264` := evalm(`vQ\264`)[1];" "6#>%$vx|_vG&-%&evalmG6#%$vQ|_v G6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`vy\264` := eval m(`vQ\264`)[2];" "6#>%$vy|_vG&-%&evalmG6#%$vQ|_vG6#\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "`vz\264` := evalm(`vQ\264`)[3];" "6#> %$vz|_vG&-%&evalmG6#%$vQ|_vG6#\"\"$" }}}{PARA 0 "" 0 "" {TEXT -1 16 "W ith respect to " }{TEXT 338 1 "P" }{TEXT -1 24 " we get the velocity o f " }{TEXT 339 2 "P\264" }{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "vP\264:=vP+crossprod(Omega\264,rPP\264);" }}}{PARA 0 "" 0 "" {TEXT -1 32 "This yields for the velocity of " }{TEXT 340 1 "Q " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalm(vQ\264);" }}} {PARA 0 "" 0 "" {TEXT -1 44 "We compare this result with the velocity \+ of " }{TEXT 341 1 "Q" }{TEXT -1 49 " by use of the description with re spect to point " }{TEXT 342 1 "P" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalm(vQ);" }}}{PARA 0 "" 0 "" {TEXT -1 119 "Of course all com ponents of the velocity must be the same, independent of the reference point, so we have the equations" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eq1:=evalm(vQ)[1]=evalm(vQ\264)[1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eq2:=evalm(vQ)[2]=evalm(vQ\264)[2];" }}}{PARA 0 " " 0 "" {TEXT -1 8 "for the " }{TEXT 345 1 "x" }{TEXT -1 9 " and the " }{TEXT 346 1 "y" }{TEXT -1 84 " component. The solution of both equati on yields with use of the substitutions above" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "solve(\{subs(\{`subs1\264`, `subs2\264`, `subs3 \264`, `subs4\264`,subs1, subs2, subs3, subs4\},eq1)\},\{omega\264\}); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "solve(\{subs(\{`subs1 \264`, `subs2\264`, `subs3\264`, `subs4\264`,subs1, subs2, subs3, subs 4\},eq2)\},\{omega\264\});" }}}{PARA 0 "" 0 "" {TEXT -1 20 "That means that the " }{TEXT 350 90 "rotational velocity of a rigid body is inde pendent of the position of the reference points" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "Next we c onsider the question: " }{TEXT 347 17 "Is there a point " }{TEXT 355 1 "M" }{TEXT 356 9 " which ha" }{TEXT 351 0 "" }{TEXT 352 16 "s zero v elocity?" }{TEXT 437 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "When we know the velocity of point " }{TEXT 353 1 "P" }{TEXT -1 36 " we know that the velocity of point " }{TEXT 354 2 "M " }{TEXT -1 13 "is defined by" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "v[M] = v[P]+cros sprod(Omega,rho[M]);" "6#/&%\"vG6#%\"MG,&&F%6#%\"PG\"\"\"-%*crossprodG 6$%&OmegaG&%$rhoG6#F'F," }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "When we postulate that the velocity of point " }{TEXT 357 1 "M" } {TEXT -1 4 " is " }{XPPEDIT 18 0 "v[M]=0" "6#/&%\"vG6#%\"MG\"\"!" } {TEXT -1 13 " we can write" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "v[P]=-c rossprod(Omega,rho[M])" "6#/&%\"vG6#%\"PG,$-%*crossprodG6$%&OmegaG&%$r hoG6#%\"MG!\"\"" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "When we supposed that " }{XPPEDIT 18 0 "v[P]" "6#&%\"vG6#%\"PG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 16 " are known th en " }{XPPEDIT 18 0 "rho[M]" "6#&%$rhoG6#%\"MG" }{TEXT -1 30 " is defi ned by this relation. " }{XPPEDIT 18 0 "rho[M]" "6#&%$rhoG6#%\"MG" } {TEXT -1 26 " is the vector from point " }{TEXT 358 1 "P" }{TEXT -1 4 " to " }{TEXT 359 1 "M" }{TEXT -1 49 ". In case we know the velocity o f a second point " }{TEXT 360 2 "P\264" }{TEXT -1 14 " we get point " }{TEXT 361 1 "M" }{TEXT -1 72 " from the intersection of both orthogon al lines to the velocity vectors " }{XPPEDIT 18 0 "v[P]" "6#&%\"vG6#% \"PG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "v[P\264]" "6#&%\"vG6#%#P|_vG " }{TEXT -1 21 " as shown in Fig. 10." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "display(Fig_4_7(),scaling=constrained,axes=none,title =\"Figure 10\");" }}}{PARA 0 "" 0 "" {TEXT -1 60 "Otherwise we get the velocity of every point by the relation" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "v[P]=crossprod(Omega,R)" "6#/&%\"vG6#%\"PG-%*crossprodG 6$%&OmegaG%\"RG" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{TEXT 362 1 "R" }{TEXT -1 20 " is the vector from " }{TEXT 363 1 "M" }{TEXT -1 14 " to the point " }{TEXT 364 2 "P." }{TEXT -1 77 " Every g eneral motion can momentarily be considered as a rotation around the \+ " }{TEXT 365 28 "momentary center of rotation" }{TEXT -1 165 ". The wo rd \"momentary\" shows that the center of rotation in general changes \+ its position during the motion. This fact is shown on the basis of the following example." }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 22 "Example: M an on Ladder" }}{PARA 0 "" 0 "" {TEXT -1 232 "Let's consider the follo wing situation: A man climbs a ladder. When he reaches the top point, \+ the one end of the ladder slips off. Don't worry, nothing happens to t he man, as you can see in Fig. 11, where the situation is animated. " }{TEXT 348 59 "(Click again the picture and press \"play\" in the menu bar.)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "display(Fig_4_8(), insequence=true,scaling=constrained,axes=none,title=\"Figure 11\");" } }}{PARA 0 "" 0 "" {TEXT -1 131 "The motion of the left part of the lad der is clear, we have a rotation around the left support of the ladder . This point is fixed. " }}{PARA 0 "" 0 "" {TEXT -1 267 "The motion of the right part of the ladder is a combination of a translation and a \+ rotation. As described above we can formulate every momentary situatio n during the motion as a rotation around a momentary center of rotatio n. We get this momentary center of rotation " }{TEXT 366 1 "M" }{TEXT -1 12 " as follows." }}{PARA 0 "" 0 "" {TEXT -1 455 "Because the motio n of the left part of the ladder is known as a rotation around the lef t support we know the motion of the joint between both parts of the la dder. The velocity of this joint points orthogonal to the vector from \+ the left support to the joint. Because this joint also belongs to the \+ right part of the ladder, the momentary center of the rotation of this part of the ladder must lie on the line orthogonal to the velocity ve ctor of the joint." }}{PARA 0 "" 0 "" {TEXT -1 360 "Second, we know th at the right support of the ladder slips on the ground, which in this \+ example should be the horizontal. The velocity vector of the right sup port points in the horizontal direction, too. The momentary center of \+ rotation lies again on the orthogonal line to the velocity vector, whi ch means in this case on a vertical line at the right support." }} {PARA 0 "" 0 "" {TEXT -1 250 "Now we know the position of the momentar y center of rotation of the right part of the ladder: it lies on the i ntersection of both lines. In Fig. 12 these two lines are drawn dotted in red. The momentary center of rotation is marked by the red point. \+ " }{TEXT 438 137 "When you click on the figure and press \"play\" on t he menu bar you see the trajectory of the position of the momentary ce nter of rotation." }{TEXT -1 112 " It is a circular arc around the lef t support of the ladder with the radius the double the length of the l adder." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "display(Fig_4_9(), insequence=true,scaling=constrained,axes=none,title=\"Figure 12\");" } }}{PARA 0 "" 0 "" {TEXT -1 294 "Finally it should be said that the mot ion of the man is not a rotation but a translational motion on a circu lar arc. For completeness it should also be said that a translational \+ motion can be considered as a rotation with infinite distance between \+ the momentary center of rotation and the body." }}}{PARA 0 "" 0 "" {TEXT 368 12 "Acceleration" }}{PARA 0 "" 0 "" {TEXT -1 63 "Next we cal culate the acceleration of an arbitrary fixed point " }{TEXT 369 1 "Q " }{TEXT -1 84 " of a rigid body with respect to the inertial system. \+ We know about the velocity of " }{TEXT 370 1 "Q" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "vQ := vP+crossprod(Omega,rho):" "6#>%#vQG,&%#vPG\"\"\"- %*crossprodG6$%&OmegaG%$rhoGF'" }}{PARA 0 "" 0 "" {TEXT -1 12 "Notice \+ that " }{XPPEDIT 18 0 "vQ, vP, aQ, aP" "6&%#vQG%#vPG%#aQG%#aPG" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 13 " \+ are vectors." }}{PARA 0 "" 0 "" {TEXT -1 43 "Differentiating with resp ect to time yields" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "aQ:=diff(vQ,t): " "6#>%#aQG-%%diffG6$%#vQG%\"tG" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "aQ :=aP+crossprod(diff(Omega,t),rho)+crossprod(Omega,diff(rho,t));" "6#>% #aQG,(%#aPG\"\"\"-%*crossprodG6$-%%diffG6$%&OmegaG%\"tG%$rhoGF'-F)6$F. -F,6$F0F/F'" }}{PARA 0 "" 0 "" {TEXT -1 21 "As shown above it is " } {XPPEDIT 18 0 "diff(rho,t)=crossprod(Omega,rho)" "6#/-%%diffG6$%$rhoG% \"tG-%*crossprodG6$%&OmegaGF'" }{TEXT -1 7 " and so" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "aQ:=aP+crossprod(diff(Omega,t),rho)+crossprod(Omega, crossprod(Omega,rho));" "6#>%#aQG,(%#aPG\"\"\"-%*crossprodG6$-%%diffG6 $%&OmegaG%\"tG%$rhoGF'-F)6$F.-F)6$F.F0F'" }}{PARA 0 "" 0 "" {TEXT -1 11 "or because " }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "Omega = diff(phi,t )*eZ;" "6#/%&OmegaG*&-%%diffG6$%$phiG%\"tG\"\"\"%#eZGF+" }}{PARA 0 "" 0 "" {TEXT -1 6 "and so" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "aQ := cros sprod(Omega,crossprod(Omega,rho)) = -diff(phi,t)^2*rho;" "6#>%#aQG/-%* crossprodG6$%&OmegaG-F'6$F)%$rhoG,$*&-%%diffG6$%$phiG%\"tG\"\"#F,\"\" \"!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 12 "we can write" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "aQ:=aP+crossprod(diff(Omega,t),rho)-(diff(phi,t))* *2*rho;" "6#>%#aQG,(%#aPG\"\"\"-%*crossprodG6$-%%diffG6$%&OmegaG%\"tG% $rhoGF'*&-F,6$%$phiGF/\"\"#F0F'!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 10 "R emember: " }{TEXT 375 2 "eZ" }{TEXT -1 23 " is the unit vector in " } {TEXT 376 1 "Z" }{TEXT -1 11 "-direction." }}{PARA 0 "" 0 "" {TEXT -1 124 "This formulations are not exactly as they are to be used in MAPLE , but for better presentation we make some simplifications." }}{PARA 0 "" 0 "" {TEXT -1 26 "The total acceleration of " }{TEXT 371 1 "Q" } {TEXT -1 54 " so has three components: first the velocity of point " } {TEXT 372 1 "P" }{TEXT -1 26 ", the radial acceleration " }{XPPEDIT 18 0 "-diff(phi,t)^2*rho" "6#,$*&-%%diffG6$%$phiG%\"tG\"\"#%$rhoG\"\" \"!\"\"" }{TEXT -1 33 " and the tangential acceleration " }{XPPEDIT 18 0 "crossprod(diff(Omega,t),rho)" "6#-%*crossprodG6$-%%diffG6$%&Omeg aG%\"tG%$rhoG" }{TEXT -1 55 ". The last two parts result from the rota tion of point " }{TEXT 373 1 "Q" }{TEXT -1 14 " around point " }{TEXT 374 1 "P" }{TEXT -1 9 ". We say " }{XPPEDIT 18 0 "diff(Omega,t)= diff( phi,t$2)*eZ" "6#/-%%diffG6$%&OmegaG%\"tG*&-F%6$%$phiG-%\"$G6$F(\"\"#\" \"\"%#eZGF1" }{TEXT -1 77 " is the angular acceleration of the body wi th respect to the inertial system." }}{PARA 0 "" 0 "" {TEXT -1 105 "As the relations for the velocity this results are analogous to the rela tions for the relative motion in " }{HYPERLNK 17 "section 2" 1 "sec-02 .mws" "" }{TEXT -1 48 ", but without any relative motion between point " }{TEXT 377 1 "Q" }{TEXT -1 26 " and the local body-fixed " }{TEXT 378 7 "(X,Y,Z)" }{TEXT -1 9 "-system. " }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 24 "4.2.1 The Center of Mass" }}{PARA 0 "" 0 "" {TEXT -1 28 " Here it is assumed that the " }{TEXT 257 17 "center of gravity" } {TEXT -1 192 " is known from the statics. The center of mass is define d equivalently, but we constitute the formulation with respect to infi nite mass particles of the body instead of infinite partial areas." }} {PARA 0 "" 0 "" {TEXT -1 72 "Consider the planar rigid body shown in F ig. 13. We choose an arbitrary " }{TEXT 259 8 "(X,Y,Z)-" }{TEXT -1 30 "coordinate system with origin " }{TEXT 258 1 "P" }{TEXT -1 9 " and th e " }{TEXT 260 1 "Z" }{TEXT -1 54 "-axis perpendicular to the symmetry plane of the body." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "displ ay(Fig_4_10(),scaling=constrained,axes=none,title=\"Figure 13\");" }}} {PARA 0 "" 0 "" {TEXT -1 32 "The center of mass is defined by" }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "rho[S] := int(rho,m)/int(1,m):" "6#>& %$rhoG6#%\"SG*&-%$intG6$F%%\"mG\"\"\"-F*6$F-F,!\"\"" }{TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 6 "Where " }{XPPEDIT 18 0 "int(1,m) = m;" "6# /-%$intG6$\"\"\"%\"mGF(" }{TEXT -1 33 " is the total mass of the body. \"" }{XPPEDIT 18 0 "rho" "6#%$rhoG" }{TEXT -1 120 "\" is written \"rh o\" in the figure. The integrals in the equation are taken over the to tal mass of the body. In case that " }{XPPEDIT 18 0 "P=S" "6#/%\"PG%\" SG" }{TEXT -1 7 " it is " }{XPPEDIT 18 0 "rho[S]=0" "6#/&%$rhoG6#%\"SG \"\"!" }{TEXT -1 8 " and so " }{XPPEDIT 18 0 "int(rho,m) = 0;" "6#/-%$ intG6$%$rhoG%\"mG\"\"!" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 113 "When the mass of the body is uniform over the area the center of \+ mass is the same point as the center of gravity." }}}{SECT 0 {PARA 5 " " 0 "" {TEXT -1 23 "4.2.2 Moment of Inertia" }}{PARA 0 "" 0 "" {TEXT -1 51 "The moment of inertia of a rigid body is defined by" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "J[P] := int(rho*rho,m) = int(X^2+Y^2,m);" " 6#>&%\"JG6#%\"PG/-%$intG6$*&%$rhoG\"\"\"F-F.%\"mG-F*6$,&*$%\"XG\"\"#F. *$%\"YGF5F.F/" }}{PARA 0 "" 0 "" {TEXT -1 50 "with the integral over t he total mass of the body." }}{PARA 0 "" 0 "" {TEXT -1 61 "In the spec ial case of prismatic bodies with constant height " }{TEXT 261 1 "h" } {TEXT -1 13 " and density " }{XPPEDIT 18 0 "gamma" "6#%&gammaG" } {TEXT -1 7 " we get" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "dm=gamma*h*dA " "6#/%#dmG*(%&gammaG\"\"\"%\"hGF'%#dAGF'" }{TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 5 "with " }{XPPEDIT 18 0 "dA" "6#%#dAG" }{TEXT -1 34 " the cross section of the element " }{XPPEDIT 18 0 "dm" "6#%#dmG" } {TEXT -1 8 " in the " }{TEXT 262 5 "(X,Y)" }{TEXT -1 7 "-plane." }} {PARA 0 "" 0 "" {TEXT -1 22 "So we get the relation" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "J[P]=gamma*h*I[P,polar]" "6#/&%\"JG6#%\"PG*(%&gammaG \"\"\"%\"hGF*&%\"IG6$F'%&polarGF*" }}{PARA 0 "" 0 "" {TEXT -1 9 "with \+ the " }{TEXT 263 27 "polar second moment of area" }{TEXT -1 1 " " } {XPPEDIT 18 0 "I[P,polar]" "6#&%\"IG6$%\"PG%&polarG" }{TEXT -1 33 " wh at is known from the subject \"" }{TEXT 264 21 "Strength of materials " }{TEXT -1 2 "\"." }}{SECT 0 {PARA 5 "" 0 "Steiner" {TEXT -1 18 "Theo rem of Steiner" }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }{XPPEDIT 18 0 "J[ S]" "6#&%\"JG6#%\"SG" }{TEXT -1 67 " be the known moment of inertia wi th respect to the center of mass " }{TEXT 265 1 "S" }{TEXT -1 93 ". We are now asking for the moment of inertia of the body with respect to \+ an arbitrary point " }{TEXT 266 1 "P" }{TEXT -1 36 ". The situation is shown in Fig. 14." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "displa y(Fig_4_11(),scaling=constrained,axes=none,title=\"Figure 14\");" }}} {PARA 0 "" 0 "" {TEXT -1 36 "From the figure we get the relations" }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "X = x-a:" "6#/%\"XG,&%\"xG\"\"\"%\"aG !\"\"" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "Y = y-b:" "6#/%\"YG,&%\"yG\" \"\"%\"bG!\"\"" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "a^2+b^2 = r^2:" "6# /,&*$%\"aG\"\"#\"\"\"*$%\"bGF'F(*$%\"rGF'" }}{PARA 0 "" 0 "" {TEXT -1 76 "The integrals are over the entire surface. From the definition abo ve we know" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "J[P]=int(X**2+Y**2,m); " "6#/&%\"JG6#%\"PG-%$intG6$,&*$%\"XG\"\"#\"\"\"*$%\"YGF.F/%\"mG" }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "J[P] := int(x^2+y^2,m)+(a^2+b^2)*int( 1,m)-2*a*int(x,m)-2*b*int(y,m):" "6#>&%\"JG6#%\"PG,*-%$intG6$,&*$%\"xG \"\"#\"\"\"*$%\"yGF/F0%\"mGF0*&,&*$%\"aGF/F0*$%\"bGF/F0F0-F*6$F0F3F0F0 *(F/F0F7F0-F*6$F.F3F0!\"\"*(F/F0F9F0-F*6$F2F3F0F?" }}{PARA 0 "" 0 "" {TEXT -1 19 "The last two parts " }{XPPEDIT 18 0 "2*a*int(x,m)" "6#*( \"\"#\"\"\"%\"aGF%-%$intG6$%\"xG%\"mGF%" }{TEXT -1 5 " and " } {XPPEDIT 18 0 "2*b*int(y,m)" "6#*(\"\"#\"\"\"%\"bGF%-%$intG6$%\"yG%\"m GF%" }{TEXT -1 18 " are zero because " }{TEXT 267 1 "x" }{TEXT -1 5 " \+ and " }{TEXT 268 1 "y" }{TEXT -1 73 " are the coordinates with respect to the center of mass. Then we get the " }{TEXT 269 18 "theorem of St einer" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "J[P]=J[S]+m*r**2" "6#/&%\"JG 6#%\"PG,&&F%6#%\"SG\"\"\"*&%\"mGF,*$%\"rG\"\"#F,F," }}{PARA 0 "" 0 "" {TEXT -1 172 "Notice that the moment of inertia with respect to the ce nter of mass has always the smallest value in comparison with the mome nt of inertia with respect to any other point." }}{PARA 0 "" 0 "" {TEXT -1 193 "In the following examples we deduce the moments of inert ia of some special bodies. The results of these examples and the momen ts of inertia for some more plane rigid bodies are included in the " } {HYPERLNK 17 "dynamics" 2 "dynamics" "" }{TEXT -1 14 " package. See " }{HYPERLNK 17 "MI" 2 "MI" "" }{TEXT -1 22 " for more information." }}} {SECT 0 {PARA 5 "" 0 "" {TEXT -1 20 "Example: Slender Rod" }}{PARA 0 " " 0 "" {TEXT -1 110 "We consider the rod shown in Fig. 15. The dimens ions of the rod orthogonal to the X-axis shall be negligible." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display(Fig_4_12(),scaling=c onstrained,axes=none,title=\"Figure 15\");" }}}{PARA 0 "" 0 "" {TEXT -1 85 "We are at first interested in the moment of inertia of the rod \+ with respect to point " }{TEXT 270 1 "P" }{TEXT -1 45 ". Therefore we \+ consider a small mass element " }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "dm \+ = m/L*dX;" "6#/%#dmG*(%\"mG\"\"\"%\"LG!\"\"%#dXGF'" }}{PARA 0 "" 0 "" {TEXT -1 5 "Here " }{TEXT 271 1 "m" }{TEXT -1 34 " is the total mass o f the rod and " }{TEXT 272 1 "L" }{TEXT -1 36 " is its length. Substit ute this into" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "J[P]:=int(X**2,m)" " 6#>&%\"JG6#%\"PG-%$intG6$*$%\"XG\"\"#%\"mG" }{TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 6 "yields" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "J [P]:=int(X**2*m/L,X=0..L);" "6#>&%\"JG6#%\"PG-%$intG6$*(%\"XG\"\"#%\"m G\"\"\"%\"LG!\"\"/F,;\"\"!F0" }}}{PARA 0 "" 0 "" {TEXT -1 32 "The dist ance between the points " }{TEXT 273 1 "P" }{TEXT -1 5 " and " }{TEXT 274 1 "S" }{TEXT -1 4 " is " }{XPPEDIT 18 0 "L/2" "6#*&%\"LG\"\"\"\"\" #!\"\"" }{TEXT -1 97 ". For the moment of inertia with respect to the \+ center of mass we get from the theorem of Steiner" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "J[S]:=J[P]-m*(L/2)**2;" "6#>&%\"JG6#%\"SG,&&F%6 #%\"PG\"\"\"*&%\"mGF,*$*&%\"LGF,\"\"#!\"\"F2F,F3" }}}{PARA 0 "" 0 "" {TEXT -1 141 "Because we consider only planar systems, the depth of th e rod has no influence on the result. So this relation is also valid f or thin plates." }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 27 "Example: Semi -Circular Disk" }}{PARA 0 "" 0 "" {TEXT -1 75 "Next we consider the mo ment of inertia of a semi-circular disk with radius " }{TEXT 275 1 "R " }{TEXT -1 12 ", thickness " }{TEXT 278 1 "t" }{TEXT -1 10 " and mass " }{TEXT 276 1 "m" }{TEXT -1 21 " as shown in Fig. 16." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display(Fig_4_13(),scaling=constrai ned,axes=none,title=\"Figure 16\");" }}}{PARA 0 "" 0 "" {TEXT -1 80 "T he distance between the center of mass and the center of the circle is given by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "s:=4*R/3/Pi:" "6# >%\"sG**\"\"%\"\"\"%\"RGF'\"\"$!\"\"%#PiGF*" }}}{PARA 0 "" 0 "" {TEXT -1 38 "For the calculations consider Fig. 17." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display(Fig_4_14(),scaling=constrained,axes=none ,title=\"Figure 17\");" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 277 0 "" }{TEXT -1 44 "The moment of inertia with respect to point " }{TEXT 279 1 "M" }{TEXT -1 14 " is defined by" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "J[M] := int(x^2+y^2,m) = int(r^2,m);" "6#>&%\"JG6#%\"MG/-%$intG6 $,&*$%\"xG\"\"#\"\"\"*$%\"yGF/F0%\"mG-F*6$*$%\"rGF/F3" }}{PARA 0 "" 0 "" {TEXT -1 143 "The thickness of the disk has to be constant, the mat erial homogenous, so we can write for the infinite mass element which \+ is marked in Fig. 17" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "dm := t*rho*r *dr*dphi;" "6#>%#dmG*,%\"tG\"\"\"%$rhoGF'%\"rGF'%#drGF'%%dphiGF'" }} {PARA 0 "" 0 "" {TEXT -1 29 "with the infinitesimal angle " }{TEXT 280 8 "dphi (=d" }{TEXT 282 1 "f" }{TEXT 283 1 ")" }{TEXT 281 1 "." }} {PARA 0 "" 0 "" {TEXT -1 30 "The density of the material is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rho:=m/t/(R**2*Pi/2):" "6#>%$rhoG*(% \"mG\"\"\"%\"tG!\"\"*(%\"RG\"\"#%#PiGF'F,F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 90 "The integral in the definition of the moment of inertia i s now divided into two integrals:" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "J[M]:=int(int(r**2*t*rho*r,r=0..R),phi=0..Pi);" "6#>&% \"JG6#%\"MG-%$intG6$-F)6$**%\"rG\"\"#%\"tG\"\"\"%$rhoGF1F.F1/F.;\"\"!% \"RG/%$phiG;F5%#PiG" }}}{PARA 0 "" 0 "" {TEXT -1 108 "We get the momen t of inertia of the disk with respect to the center of mass by use of \+ the theorem of Steiner" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "J[S] :=J[M]+m*s**2;" "6#>&%\"JG6#%\"SG,&&F%6#%\"MG\"\"\"*&%\"mGF,*$%\"sG\" \"#F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "collect(J[S],\{m ,R\})=evalf(J[S]);" }}}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 42 "4.2.3 Th e Angular Momentum of Rigid Bodies" }}{PARA 0 "" 0 "" {TEXT -1 51 "We \+ continue now the considerations form subsection " }{HYPERLNK 17 "4.1.2 " 1 "sec-04.mws" "4.1.2" }{TEXT -1 134 " (The Angular Momentum for Sys tems of Mass Particles). One of the final results was the axiom of the angular momentum for rigid bodies" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "Lo = int(crossprod(r,rd),m);" "6#/%#LoG-%$intG6$-%*crossprodG6$%\"rG% #rdG%\"mG" }}{PARA 0 "" 0 "" {TEXT -1 5 "with " }{XPPEDIT 18 0 "rd = d iff(r,t);" "6#/%#rdG-%%diffG6$%\"rG%\"tG" }{TEXT -1 19 " the derivativ e of " }{XPPEDIT 18 0 "r;" "6#%\"rG" }{TEXT -1 22 " with respect to ti me." }}{PARA 0 "" 0 "" {TEXT -1 192 "To calculate the angular momentum we need the position and the velocity of every mass particle of the r igid body. We know from above that the motion of a rigid body is known when the velocity " }{XPPEDIT 18 0 "v[P]" "6#&%\"vG6#%\"PG" }{TEXT -1 23 " of an arbitrary point " }{XPPEDIT 18 0 "P" "6#%\"PG" }{TEXT -1 26 " and the angular velocity " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG " }{TEXT -1 46 " is given. For an arbitrary mass particle then" }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "r:=r[P]+rho" "6#>%\"rG,&&F$6#%\"PG\" \"\"%$rhoGF)" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "diff(r,t):=v=v[P]+cro ssprod(Omega,rho):" "6#>-%%diffG6$%\"rG%\"tG/%\"vG,&&F*6#%\"PG\"\"\"-% *crossprodG6$%&OmegaG%$rhoGF/" }}{PARA 0 "" 0 "" {TEXT -1 38 "is valid . See Fig. 18 for the details." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display(Fig_4_15(),scaling=constrained,axes=none,title=\"Figure \+ 18\");" }}}{PARA 0 "" 0 "" {TEXT -1 61 "The only value which denotes t he mass particle is the vector " }{XPPEDIT 18 0 "rho" "6#%$rhoG" } {TEXT -1 74 ", which specifies the position of the mass particle with \+ respect to point " }{TEXT 379 1 "P" }{TEXT -1 17 ". So we can write" } }{PARA 257 "" 0 "" {XPPEDIT 18 0 "Lo = int(crossprod(r[P]+rho,v[P]+cro ssprod(Omega,rho)),m);" "6#/%#LoG-%$intG6$-%*crossprodG6$,&&%\"rG6#%\" PG\"\"\"%$rhoGF0,&&%\"vG6#F/F0-F)6$%&OmegaGF1F0%\"mG" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "Lo = crossprod(r[P],v[P])*int(1,m)+crossprod(r[P], crossprod(Omega,int(rho,m)))+crossprod(int(rho,m),v[P])+int(crossprod( rho,crossprod(Omega,rho)),m):" "6#/%#LoG,**&-%*crossprodG6$&%\"rG6#%\" PG&%\"vG6#F-\"\"\"-%$intG6$F1%\"mGF1F1-F(6$&F+6#F--F(6$%&OmegaG-F36$%$ rhoGF5F1-F(6$-F36$F?F5&F/6#F-F1-F36$-F(6$F?-F(6$F " 0 "" {MPLTEXT 1 0 14 " unassign('J'):" }}}{PARA 0 "" 0 "" {TEXT -1 127 "We consider the situa tion shown in Fig. 19. There are two rigid bodies, a cuboid and a cyli nder, both on a slope with gradient " }{XPPEDIT 18 0 "alpha" "6#%&alph aG" }{TEXT -1 33 ". Both bodies have the same mass " }{TEXT 394 1 "m" }{TEXT -1 32 ". The radius of the cylinder is " }{TEXT 395 1 "R" } {TEXT -1 43 ", the length of the edges of the cuboid is " }{XPPEDIT 18 0 "2*R" "6#*&\"\"#\"\"\"%\"RGF%" }{TEXT 396 1 "." }{TEXT -1 28 " In the initial position at " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" } {TEXT -1 36 " the distance between the bodies is " }{TEXT 397 1 "d" } {TEXT -1 148 ". At last we assume that there is always adherence betwe en the cylinder and the ground, and there is no friction between the c uboid and the ground (" }{XPPEDIT 18 0 "mu=0" "6#/%#muG\"\"!" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 21 "The question is now: " } {TEXT 439 58 "Will the distance between the bodies increase or decreas e?" }{TEXT -1 1 " " }{TEXT 440 66 "And when it decreases: After what t ime do the bodies have contact?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_4_16()[1],scaling=constrained,axes=none,title=\"Fi gure 19\");" }}}{PARA 0 "" 0 "" {TEXT -1 258 "First we consider the mo tion of the cylinder. We choose the coordinate system as shown in Fig. 20. In this example we have a general plane motion. There is no rotat ion around a fixed axis but the momentary pole of the rotation is chan ging during the motion. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 " display(Fig_4_16()[2],scaling=constrained,axes=none,title=\"Figure 20 \");" }}}{PARA 0 "" 0 "" {TEXT -1 73 "We consider now the three equati ons of the equilibrium for the forces in " }{TEXT 398 1 "x" }{TEXT -1 6 "- and " }{TEXT 399 1 "y" }{TEXT -1 52 "-direction and the moment ar ound the center of mass " }{TEXT 400 1 "S" }{TEXT -1 101 ". Therefore \+ we cut the cylinder free and consider all active and passive forces, a s shown in Fig. 21." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "displ ay(Fig_4_16()[3],scaling=constrained,axes=none,title=\"Figure 21\");" }}}{PARA 0 "" 0 "" {TEXT -1 38 "In Fig. 21 we have the inertial forces " }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Tx:=m*diff(x(t),t$2):" "6# >%#TxG*&%\"mG\"\"\"-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F.\"\"#F'" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Ty := m*diff(y(t),`$`(t,2)):" "6#>%#TyG*&%\"mG\"\"\"-%%diffG6$-%\"yG6#%\"tG-%\"$G6$F.\"\"#F'" }}} {PARA 0 "" 0 "" {TEXT -1 184 "We assume that the acceleration is measu red in the same direction as the displacements. So the inertial forces are assumed to point against the coordinate axis when they are positi ve. " }}{PARA 0 "" 0 "" {TEXT -1 65 "The inertia moment (Do not confus e with the moment of inertia) is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Mz := J[S]*diff(phi(t),t$2):" "6#>%#MzG*&&%\"JG6#%\"SG\"\"\"-%%d iffG6$-%$phiG6#%\"tG-%\"$G6$F1\"\"#F*" }}}{PARA 0 "" 0 "" {TEXT -1 37 "Additionally we have the weight force" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "G:=m*g:" "6#>%\"GG*&%\"mG\"\"\"%\"gGF'" }}}{PARA 0 "" 0 "" {TEXT -1 33 "with the acceleration of gravity " }{TEXT 401 1 "g" }{TEXT -1 24 " and the contact forces " }{TEXT 402 1 "N" }{TEXT -1 5 " and " }{TEXT 403 1 "H" }{TEXT -1 64 ". So we get from Fig. 21 the thr ee equations for the equilibrium" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ_Fx:=Tx-G*sin(alpha)+H=0;" "6#>%&EQ_FxG/,(%#TxG\"\"\"*&%\"GGF( -%$sinG6#%&alphaGF(!\"\"%\"HGF(\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ_Fy:=Ty+G*cos(alpha)-N=0;" "6#>%&EQ_FyG/,(%#TyG\"\"\" *&%\"GGF(-%$cosG6#%&alphaGF(F(%\"NG!\"\"\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ_MS:=Mz-H*R=0;" "6#>%&EQ_MSG/,&%#MzG\"\"\"*&%\"H GF(%\"RGF(!\"\"\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 107 "Further we have some additive geometrical conditions. We know that the distance betwe en the center of mass " }{TEXT 404 1 "S" }{TEXT -1 19 " and the ground is " }{TEXT 405 3 "R. " }{TEXT -1 10 "This means" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "y(t):=0;" }}}{PARA 0 "" 0 "" {TEXT -1 31 "and \+ so the acceleration in the " }{TEXT 406 1 "y" }{TEXT -1 13 "-direction is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(y(t),`$`(t,2));" " 6#-%%diffG6$-%\"yG6#%\"tG-%\"$G6$F)\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 66 "From the condition that the cylinder rolls without sliding, we get " }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "phi(t):=x(t)/R:" "6#>-%$ph iG6#%\"tG*&-%\"xG6#F'\"\"\"%\"RG!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 6 " and so" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(phi(t),t$2);" " 6#-%%diffG6$-%$phiG6#%\"tG-%\"$G6$F)\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 41 "This yields for the equilibrium equations" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 6 "EQ_Fx;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "EQ_Fy;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "EQ_MS;" }}} {PARA 0 "" 0 "" {TEXT -1 69 "From the equation EQ_Fy we get the inform ation that the normal force " }{TEXT 407 1 "N" }{TEXT -1 108 " is inde pendent from the motion. It has the same value as in the case of stati cal considerations.We get for " }{TEXT 408 1 "N" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 18 "N:=solve(EQ_Fy,N);" }}}{PARA 0 "" 0 "" {TEXT -1 60 "From the equation EQ_MS we get the tangential contact force " } {TEXT 409 1 "H" }{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "H:=solve(EQ_MS,H);" }}}{PARA 0 "" 0 "" {TEXT -1 61 "This force i s dependent on the motion, as we see in the term " }{XPPEDIT 18 0 "dif f(x(t),t$2)" "6#-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F)\"\"#" }{TEXT -1 17 " in the relation." }}{PARA 0 "" 0 "" {TEXT -1 18 "From the function " }{HYPERLNK 17 "MI" 2 "MI" "" }{TEXT -1 10 " from the " }{HYPERLNK 17 " dynamics" 2 "dynamics" "" }{TEXT -1 37 " package we get the moment of \+ inertia" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "J[S] := MI('solid cylinder',m,R);" }}}{PARA 0 "" 0 "" {TEXT -1 71 "At last we get from t he equation EQ_Fx the acceleration of the cylinder" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 43 "a_cylinder(t):=solve(EQ_Fx,diff(x(t),t$2)); " }}}{PARA 0 "" 0 "" {TEXT -1 86 "Next we consider the situation of th e cuboid. Therefore we reset some variables first." }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 51 "unassign('N','Tx','Ty','G','EQ_Fx','EQ_Fy',' y(t)'):" }}}{PARA 0 "" 0 "" {TEXT -1 48 "The situation of the cuboid i s shown in Fig. 22." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "displ ay(Fig_4_16()[4],scaling=constrained,axes=none,title=\"Figure 22\");" }}}{PARA 0 "" 0 "" {TEXT -1 189 "Again we consider first the equations of the equilibrium, whereby we don't need to regard any balance of th e moments, because there is no rotation. We consider the cut free body in Fig. 23." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_ 4_16()[5],scaling=constrained,axes=none,title=\"Figure 23\");" }}} {PARA 0 "" 0 "" {TEXT -1 40 "Because the mass of the cuboid is again \+ " }{TEXT 410 1 "m" }{TEXT -1 130 " and we use the same coordinates to \+ describe the situation as before we get the same inertia forces in Fig . 23 as for the cylinder" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Tx :=m*diff(x(t),t$2):" "6#>%#TxG*&%\"mG\"\"\"-%%diffG6$-%\"xG6#%\"tG-%\" $G6$F.\"\"#F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Ty := m*diff (y(t),`$`(t,2)):" "6#>%#TyG*&%\"mG\"\"\"-%%diffG6$-%\"yG6#%\"tG-%\"$G6 $F.\"\"#F'" }}}{PARA 0 "" 0 "" {TEXT -1 29 "And the weight force is ag ain" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "G:=m*g:" "6#>%\"GG*&%\" mG\"\"\"%\"gGF'" }}}{PARA 0 "" 0 "" {TEXT -1 36 "We get now the equili brium equations" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ_Fx := Tx -G*sin(alpha) = 0;" "6#>%&EQ_FxG/,&%#TxG\"\"\"*&%\"GGF(-%$sinG6#%&alph aGF(!\"\"\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ_Fy:=Ty+G *cos(alpha)-N=0;" "6#>%&EQ_FyG/,(%#TyG\"\"\"*&%\"GGF(-%$cosG6#%&alphaG F(F(%\"NG!\"\"\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 65 "Further we have s ame geometrical condition for the motion in the " }{TEXT 415 1 "y" } {TEXT -1 76 "-directuion as before, that is that the distance between \+ the center of mass " }{TEXT 411 1 "S" }{TEXT -1 19 " and the ground is " }{TEXT 412 3 "R. " }{TEXT -1 10 "This means" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "y(t):=0;" }}}{PARA 0 "" 0 "" {TEXT -1 27 "and so \+ the acceleration is " }{TEXT 413 1 "y" }{TEXT -1 13 "-direction is" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(y(t),`$`(t,2));" "6#-%%di ffG6$-%\"yG6#%\"tG-%\"$G6$F)\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 41 "Thi s yields for the equilibrium equations" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "EQ_Fx;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "EQ _Fy;" }}}{PARA 0 "" 0 "" {TEXT -1 48 "From the equation EQ_Fy we get t he normal force " }{TEXT 414 1 "N" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "N:=solve(EQ_Fy,N);" }}}{PARA 0 "" 0 "" {TEXT -1 63 "T his is the same result as that for the motion of the cylinder." }} {PARA 0 "" 0 "" {TEXT -1 65 "And from the equation EQ_Fx we get the ac celeration of the cuboid" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " a_cuboid(t):=solve(EQ_Fx,diff(x(t),t$2));" }}}{PARA 0 "" 0 "" {TEXT -1 48 "Comparison with the acceleration of the cylinder" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "a_cylinder(t);" }}}{PARA 0 "" 0 "" {TEXT -1 127 "shows that the accelerations are different. The rolling \+ cylinder is slower than the cuboid. So we need to answer the question, " }{TEXT 441 32 "when do the bodies make contact?" }{TEXT -1 151 ". T o answer this question we have to calculate the displacement as a func tion of time for both bodies. To get this we integrate the acceleratio n twice." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "v_cylinder(t):=i nt(a_cylinder(t),t)+C1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 " s_cylinder(t):=int(v_cylinder(t),t)+C2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "v_cuboid(t):=int(a_cuboid(t),t)+D1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "s_cuboid(t):=int(v_cuboid(t),t)+D2;" }}} {PARA 0 "" 0 "" {TEXT -1 27 "Here we use the coordinate " }{TEXT 417 10 "s_cylinder" }{TEXT -1 5 " and " }{TEXT 416 8 "s_cuboid" }{TEXT -1 324 " to describe the displacement and assume that both coordinates st art in both bodies' center of mass at the beginning. From the integrat ion we get two constants for the motion of each body. To compute these we need four inital conditions. We assume that both bodies are at res t at the beginning of the time period. That means" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "IC1:=subs(t=0,v_cylinder(t))=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "IC2:=subs(t=0,s_cylinder(t))=0;" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "IC3:=subs(t=0,v_cuboid(t))= 0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "IC4:=subs(t=0,s_cuboi d(t))=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "sol:=solve(\{IC 1,IC2,IC3,IC4\},\{C1,C2,D1,D2\});" }}}{PARA 0 "" 0 "" {TEXT -1 46 "So \+ we get for the displacement of the cylinder" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 34 "s_cyl(t):=subs(sol,s_cylinder(t));" }}}{PARA 0 "" 0 "" {TEXT -1 18 "and for the cuboid" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "s_cub(t):=subs(sol,s_cuboid(t));" }}}{PARA 0 "" 0 "" {TEXT -1 80 "We must compute, after what time is the displacement of t he cuboid by the value " }{TEXT 418 1 "d" }{TEXT -1 96 " (initial dist ance between both bodies) larger than the displacement of the cylinder ? This means" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Cond:=s_cub( t)=s_cyl(t)+d;" }}}{PARA 0 "" 0 "" {TEXT -1 40 "And the solution of th is equation yields" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "T:=sol ve(Cond,t);" }}}{PARA 0 "" 0 "" {TEXT -1 83 "We know that time has to \+ be positive, so we need only consider the positive result." }}{PARA 0 "" 0 "" {TEXT -1 40 "At last we consider the special case of " } {XPPEDIT 18 0 "alpha=30" "6#/%&alphaG\"#I" }{TEXT -1 14 "\260, that me ans " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "alpha := Pi/180*30; " }}}{PARA 0 "" 0 "" {TEXT -1 38 "and we assume for the initial distan ce" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "d:=10:" }}}{PARA 0 "" 0 "" {TEXT -1 39 "For the acceleration of gravity we know" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=9.81:" }}}{PARA 0 "" 0 "" {TEXT -1 23 "and so we get the value" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Time:=evalf(abs(T[1]));" }}}{PARA 0 "" 0 "" {TEXT -1 116 "We see that the motion of both bodies and in the end the time of the contact don't depend on the mass of the bodies." }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 47 "4.2.5 Balance of Energy, Conservation of Energy" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "unassign('J'):" }}}{PARA 0 " " 0 "" {TEXT -1 11 "In section " }{HYPERLNK 17 "3.2" 1 "sec-03.mws" "3 .2" }{TEXT -1 204 " we had considered the balance of energy and the co nservation of energy of mass particles. Now we continue this for rigid bodies. Fig. 24 shows an arbitrary rigid body with the notation which is used now." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display(Fig _4_17(),scaling=constrained,axes=none,title=\"Figure 24\");" }}}{PARA 0 "" 0 "" {TEXT -1 204 "We describe the general motion of the body by \+ the translation of the center of mass and the rotation of the body aro und its center of mass. The velocity of the center of mass is describe d by the variable " }{XPPEDIT 18 0 "vV[S];" "6#&%#vVG6#%\"SG" }{TEXT -1 204 " (vS in the figure). To avoid confusion we use in this section the notation vV to describe the velocity vectors and v to describe th e scalar value of the velocity. The rotation is described by the vecto r " }{XPPEDIT 18 0 "Omega=omega*eZ" "6#/%&OmegaG*&%&omegaG\"\"\"%#eZGF '" }{TEXT -1 22 " with the unit vector " }{XPPEDIT 18 0 "eZ" "6#%#eZG " }{TEXT -1 18 " along the Z-axis." }}{PARA 0 "" 0 "" {TEXT -1 105 "Fo r an arbitrary mass element of the rigid body (in the figure yellow co lered) we get the velocity vector" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 " vV = vV[S]+crossprod(Omega,rho);" "6#/%#vVG,&&F$6#%\"SG\"\"\"-%*crossp rodG6$%&OmegaG%$rhoGF)" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "o r with " }{XPPEDIT 18 0 "crossprod(Omega,rho) = vV[Rot];" "6#/-%*cross prodG6$%&OmegaG%$rhoG&%#vVG6#%$RotG" }{TEXT -1 13 " we can write" }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "vV = vV[S]+vV[Rot];" "6#/%#vVG,&&F$6# %\"SG\"\"\"&F$6#%$RotGF)" }}{PARA 0 "" 0 "" {TEXT -1 90 "Considering t he mass element as a mass particle yields the kinetic energy (see subs ection " }{HYPERLNK 17 "3.2" 1 "sec-03.mws" "3.2" }{TEXT -1 1 ")" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T:=kinetic_energy(v,dm);" "6#> %\"TG-%/kinetic_energyG6$%\"vG%#dmG" }}}{PARA 0 "" 0 "" {TEXT -1 50 "F or the total rigid body we get the kinetic energy" }}{PARA 257 "" 0 " " {XPPEDIT 18 0 "T := int(v^2,m)/2;" "6#>%\"TG*&-%$intG6$*$%\"vG\"\"#% \"mG\"\"\"F+!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 38 "The square of the ve locity is given by" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "v^2 = dotprod(v V,vV);" "6#/*$%\"vG\"\"#-%(dotprodG6$%#vVGF*" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "v^2 = dotprod(vV[S],vV[S])+2*dotprod(vV[S],crossprod(Om ega,rho))+dotprod(crossprod(Omega,rho),crossprod(Omega,rho));" "6#/*$% \"vG\"\"#,(-%(dotprodG6$&%#vVG6#%\"SG&F,6#F.\"\"\"*&F&F1-F)6$&F,6#F.-% *crossprodG6$%&OmegaG%$rhoGF1F1-F)6$-F86$F:F;-F86$F:F;F1" }}{PARA 0 " " 0 "" {TEXT -1 20 "We use the relations" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "Omega=omega*eZ" "6#/%&OmegaG*&%&omegaG\"\"\"%#eZGF'" }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "rho=X*eX+Y*eY" "6#/%$rhoG,&*&%\"XG\" \"\"%#eXGF(F(*&%\"YGF(%#eYGF(F(" }}{PARA 0 "" 0 "" {TEXT -1 58 "with t he unit vectors of the body fixed coordinate system " }{XPPEDIT 18 0 " eX" "6#%#eXG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "eY" "6#%#eYG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "eZ" "6#%#eZG" }{TEXT -1 9 " and thus" }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "crossprod(Omega,rho)=omega*(X*eX+Y*eY )" "6#/-%*crossprodG6$%&OmegaG%$rhoG*&%&omegaG\"\"\",&*&%\"XGF+%#eXGF+ F+*&%\"YGF+%#eYGF+F+F+" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "dotprod(cro ssprod(Omega,rho),crossprod(Omega,rho))=omega**2*(X**2+Y**2)" "6#/-%(d otprodG6$-%*crossprodG6$%&OmegaG%$rhoG-F(6$F*F+*&%&omegaG\"\"#,&*$%\"X GF0\"\"\"*$%\"YGF0F4F4" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "So we get from the description with use of vectors to the scalar nota tion. Additionally we use the known relations" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "int(1,m) = m,int(X,m) = 0,int(Y,m) = 0,int(X^2+y^2,m) = J[S];" "6&/-%$intG6$\"\"\"%\"mGF(/-F%6$%\"XGF(\"\"!/-F%6$%\"YGF(F-/-F %6$,&*$F,\"\"#F'*$%\"yGF7F'F(&%\"JG6#%\"SG" }}{PARA 0 "" 0 "" {TEXT -1 34 "Thus we get for the kinetic energy" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "T:=1/2*m*v[S]**2+1/2*J[S]*omega**2" "6#>%\"TG,&**\"\"\" F'\"\"#!\"\"%\"mGF'&%\"vG6#%\"SGF(F'**F'F'F(F)&%\"JG6#F.F'%&omegaGF(F' " }}{PARA 0 "" 0 "" {TEXT -1 63 "The kinetic energy of a rigid body is seperated into two parts:" }}{PARA 0 "" 0 "" {TEXT -1 52 "The kinetic energy of the translation with velocity " }{XPPEDIT 18 0 "v[S];" "6#& %\"vG6#%\"SG" }{TEXT -1 1 " " }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "T[Tra ns] := 1/2*m*v[S]^2;" "6#>&%\"TG6#%&TransG**\"\"\"F)\"\"#!\"\"%\"mGF)& %\"vG6#%\"SGF*" }}{PARA 0 "" 0 "" {TEXT -1 115 "and the kinetic energy of the rotational motion of the body around its center of mass with t he rotational velocity " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "T[Rot] := 1*J[S]*omega^2/2;" "6#>&%\"TG6#%$ RotG**\"\"\"F)&%\"JG6#%\"SGF)%&omegaG\"\"#F/!\"\"" }{TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 24 "We can use the function " }{HYPERLNK 17 " kinetic_energy" 2 "kinetic_energy" "" }{TEXT -1 18 " from the package \+ " }{HYPERLNK 17 "dynamics" 2 "dynamics" "" }{TEXT -1 130 " to calculat e both parts, the translational and the rotational kinetic energy, bec ause the format of both expressions is the same." }}{EXCHG {PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "T[Trans] := kinetic_energy(m,v[S]);" "6#>&%\"TG6 #%&TransG-%/kinetic_energyG6$%\"mG&%\"vG6#%\"SG" }}}{EXCHG {PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "T[Rot] := kinetic_energy(J[S],omega);" "6#>&%\"T G6#%$RotG-%/kinetic_energyG6$&%\"JG6#%\"SG%&omegaG" }}}{PARA 0 "" 0 " " {TEXT -1 110 "In the case that we consider a motion of a rigid body \+ in form of a rotation around a fixed axis through point " }{TEXT 419 1 "P" }{TEXT -1 84 " it is convenient to describe the motion with resp ect to this axis. With the vector " }{XPPEDIT 18 0 "rho[P]" "6#&%$rhoG 6#%\"PG" }{TEXT -1 6 " from " }{TEXT 421 1 "S" }{TEXT -1 4 " to " } {TEXT 420 1 "P" }{TEXT -1 14 " we know from " }{HYPERLNK 17 "4.2.1" 1 "" "4.2.1" }{TEXT -1 35 " the velocity of the center of mass" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "V[S]=crossprod(Omega,rho[P])" "6#/&%\"VG6#% \"SG-%*crossprodG6$%&OmegaG&%$rhoG6#%\"PG" }}{PARA 0 "" 0 "" {TEXT -1 8 "and thus" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "v[S]**2=omega**2*rho[P ]**2" "6#/*$&%\"vG6#%\"SG\"\"#*&%&omegaGF)&%$rhoG6#%\"PGF)" }}{PARA 0 "" 0 "" {TEXT -1 29 "For the kinetic energy we get" }}{PARA 257 "" 0 " " {XPPEDIT 18 0 "T=(J[S]+m*rho[P]**2)*omega**2/2" "6#/%\"TG*(,&&%\"JG6 #%\"SG\"\"\"*&%\"mGF+*$&%$rhoG6#%\"PG\"\"#F+F+F+*$%&omegaGF3F+F3!\"\" " }}{PARA 0 "" 0 "" {TEXT -1 18 "and according the " }{HYPERLNK 17 "ax iom of Steiner" 1 "" "Steiner" }{TEXT -1 11 " this means" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T=kinetic_energy(J[P],omega);" "6#/% \"TG-%/kinetic_energyG6$&%\"JG6#%\"PG%&omegaG" }}}{PARA 0 "" 0 "" {TEXT -1 49 "As we see we can use again our function from the " } {HYPERLNK 17 "dynamics" 2 "dynamics" "" }{TEXT -1 29 "-package for the calculation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 118 "Next we consider the balance of energy of the rigid body . Here we regard the rigid body as a system of mass particles " } {TEXT 422 2 "dm" }{TEXT -1 50 ". We know that the passive forces do no work (see " }{HYPERLNK 17 "3.2" 1 "sec-03.mws" "3.2" }{TEXT -1 24 "). Further we know from " }{HYPERLNK 17 "4.1.1" 1 "" "4.1.1" }{TEXT -1 253 " that the inner forces are equal by pairs and point against each \+ other. At last there is no relative motion between the mass particles, by definition of a rigid body. The balance of energy for the rigid bo dy is given by the summation of the work of all " }{TEXT 423 1 "n" } {TEXT -1 146 " bodies acting forces on the one side of the equation, a nd the integration of the kinetic energy over the total mass of the bo dy on the other side" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "sum(int(F[i], r[i]=r[i1]..r[i2]),i=1..n)=int(v[2]**2,m)/2-int(v[1]**2,m)/2" "6#/-%$s umG6$-%$intG6$&%\"FG6#%\"iG/&%\"rG6#F-;&F06#%#i1G&F06#%#i2G/F-;\"\"\"% \"nG,&*&-F(6$*$&%\"vG6#\"\"#FE%\"mGF;FE!\"\"F;*&-F(6$*$&FC6#F;FEFFF;FE FGFG" }}{PARA 0 "" 0 "" {TEXT -1 12 "The indices " }{TEXT 424 1 "1" } {TEXT -1 4 " and" }{TEXT 425 0 "" }{TEXT -1 1 " " }{TEXT 426 1 "2" } {TEXT -1 35 " denote here two different states. " }}{PARA 0 "" 0 "" {TEXT -1 78 "For the right side of the equation we can write according to the results above" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "T[2,Trans]+T [2,Rot]-T[1,Trans]-T[1,Rot];" "6#,*&%\"TG6$\"\"#%&TransG\"\"\"&F%6$F'% $RotGF)&F%6$F)F(!\"\"&F%6$F)F,F/" }}{PARA 0 "" 0 "" {TEXT -1 80 "The l eft side of the equation is the work of all active forces on the path \+ from " }{TEXT 427 1 "1" }{TEXT -1 4 " to " }{TEXT 428 1 "2" }{TEXT -1 35 ". It will be rearranged as follows:" }}{PARA 0 "" 0 "" {TEXT -1 4 "With" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "dr[i] = vV[i]*dt,vV[i] = vV[ S]+crossprod(Omega,rho[i]);" "6$/&%#drG6#%\"iG*&&%#vVG6#F'\"\"\"%#dtGF ,/&F*6#F',&&F*6#%\"SGF,-%*crossprodG6$%&OmegaG&%$rhoG6#F'F," }}{PARA 0 "" 0 "" {TEXT -1 6 "we get" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "sum(i nt(F[i]*vV[S],t = t[1] .. t[2])+int(dotprod(F[i],crossprod(Omega,rho[i ])),t = t[1] .. t[2]),i = 1 .. n) = sum(int(F[i],r[S] = r[S1] .. r[S2] )+int(dotprod(crossprod(rho[i],F[i]),Omega),t = t[1] .. t[2]),i = 1 .. n);" "6#/-%$sumG6$,&-%$intG6$*&&%\"FG6#%\"iG\"\"\"&%#vVG6#%\"SGF0/%\" tG;&F66#F0&F66#\"\"#F0-F)6$-%(dotprodG6$&F-6#F/-%*crossprodG6$%&OmegaG &%$rhoG6#F//F6;&F66#F0&F66#F " 0 "" {MPLTEXT 1 0 65 "unassign('H','R','N','L',' s','g','alpha','W(s)','omega','E','v'):" }}}{PARA 0 "" 0 "" {TEXT -1 66 "We consider a system with a cuboid and a cylinder, each with mass \+ " }{TEXT 430 1 "m" }{TEXT -1 66 " and connected by a rope. The cylinde r lies on a slope with angle " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" } {TEXT -1 305 ". Gravity causes the system to move. Have a look at Fig. 25, where the initial situation is shown. Here the velocity of all pa rts of the system is zero. Assume that the first part of the surface w here the cuboid lies has no friction (green color) and on the second p art there is friction with coefficient " }{XPPEDIT 18 0 "mu" "6#%#muG " }{TEXT -1 265 " between ground and cuboid (brown color). Between the cylinder and the ground always exist adherence. Additionally we assum e that the rope never sags and the displacement of the cuboid is the s ame as the displacement of the cylinder. So we use only one coordinate , " }{XPPEDIT 18 0 "s" "6#%\"sG" }{TEXT -1 38 ", to describe the momen tary situation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(F ig_4_18()[1],scaling=constrained,axes=none,title=\"Figure 25\");" }}} {PARA 0 "" 0 "" {TEXT -1 166 "It is important to see what happens whil e the cuboid passes the transition from the part without friction to t he part with friction, That means the situation between " }{XPPEDIT 18 0 "s=L1" "6#/%\"sG%#L1G" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "s=L1+2 *R" "6#/%\"sG,&%#L1G\"\"\"*&\"\"#F'%\"RGF'F'" }{TEXT -1 53 ". These si tuations are shown in the Fig.'s 26 and 27." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 71 "display(Fig_4_18()[2],scaling=constrained,axes=none ,title=\"Figure 26\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "d isplay(Fig_4_18()[3],scaling=constrained,axes=none,title=\"Figure 27\" );" }}}{PARA 0 "" 0 "" {TEXT -1 267 "Here we think it should be a good approximation of reality to assume that the friction force increases \+ linearly from zero up to the maximum value. Further we assume that the friction force is not dependent on the velocity but solely on the law for the friction force " }{XPPEDIT 18 0 "FR" "6#%#FRG" }{TEXT -1 10 " , given by" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "FR:=mu*N:" "6#>%#FRG*&% #muG\"\"\"%\"NGF'" }}{PARA 0 "" 0 "" {TEXT -1 21 "with the normal forc e" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "N:=m*g:" "6#>%\"NG*&%\"mG\"\"\"% \"gGF'" }}{PARA 0 "" 0 "" {TEXT -1 78 "We can write the friction force as a function of the position of the cuboid by" }}{EXCHG {PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "FR(s) := piecewise(s < L1,0,L1 <= s and s < L1+2 *R,mu*m*g/(2*R)*(s-L1),L1+2*R <= s,mu*m*g);" "6#>-%#FRG6#%\"sG-%*piece wiseG6(2F'%#L1G\"\"!31F,F'2F',&F,\"\"\"*&\"\"#F2%\"RGF2F2*,%#muGF2%\"m GF2%\"gGF2*&F4F2F5F2!\"\",&F'F2F,F;F21,&F,F2*&F4F2F5F2F2F'*(F7F2F8F2F9 F2" }}}{PARA 0 "" 0 "" {TEXT -1 86 "We will see later in this example \+ by use of some concrete values that this is correct." }}{PARA 0 "" 0 " " {TEXT -1 187 "The end of the motion is reached when the friction for ce has dissipated all energy or when the cylinder runs against the wal l at the end of the slope as shown in Fig. 28. That means that " } {XPPEDIT 18 0 "s=L3" "6#/%\"sG%#L3G" }{TEXT -1 12 " is reached." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_4_18()[4],scalin g=constrained,axes=none,title=\"Figure 28\");" }}}{PARA 0 "" 0 "" {TEXT -1 390 "To formulate the expressions for the energy of the syste m in the different states we need to define a reference horizontal. Th e altitude of the cuboid doesn't change during the motion, so we don't need to consider it. It seems to be useful to define the position of \+ the cylinder in the initial state as the reference horizontal. The alt itude of the cylinder as a function of the coordinate " }{TEXT 431 1 " s" }{TEXT -1 9 " given by" }}{EXCHG {PARA 258 "> " 0 "" {XPPEDIT 19 1 "H(s) := -s*sin(alpha):" "6#>-%\"HG6#%\"sG,$*&F'\"\"\"-%$sinG6#%&alpha GF*!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 387 "Next we need some geometric al conditions about the motion of the cylinder and the cuboid. The cub oid's motion is a pure translation in the horizontal direction. The mo tion of the cylinder can be desribed as a translation of its center of mass along the slope and an additional rotation of the cylinder aroun d the center of mass. Here the rotation of the cylinder is given by th e relation" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "phi(s) = s/R:" "6#/-%$p hiG6#%\"sG*&F'\"\"\"%\"RG!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 87 "This yi elds for the translational and rotational velocity of the cylinder the condition" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "omega = diff(phi(s),t): " "6#/%&omegaG-%%diffG6$-%$phiG6#%\"sG%\"tG" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "omega=v_cylinder/R:" "6#/%&omegaG*&%+v_cylinderG\"\"\"% \"RG!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 197 "(Remark: We can also consid er the motion of the cylinder as a pure momentary rotation around the \+ contact point between the cylinder and the ground. We come back to thi s subject in a later example.)" }}{PARA 0 "" 0 "" {TEXT -1 49 "For the moment of inertia of the cylinder we get " }}{EXCHG {PARA 0 "> " 0 " " {XPPEDIT 19 1 "J := MI('solidcylinder',m,R);" "6#>%\"JG-%#MIG6%.%.so lidcylinderG%\"mG%\"RG" }}}{PARA 0 "" 0 "" {TEXT -1 68 "Now we conside r the states which are described in the figures above." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 5 "" 0 "" {TEXT -1 17 "State 1 (Fig. 25)" } }{PARA 0 "" 0 "" {TEXT -1 110 "At this time the velocity of all parts \+ of the system is zero, as defined above. So the cuboid has the velocit y" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v[1] := 0:" "6#>&%\"vG6# \"\"\"\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 43 "and the rotational speed \+ of the cylinder is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "omega[1] := 0:" "6#>&%&omegaG6#\"\"\"\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 57 "So we get for the kinetic energy of the cuboid at state 1" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cuboid[1] := kinetic_energy(m,v[1]) ;" "6#>&%)T_cuboidG6#\"\"\"-%/kinetic_energyG6$%\"mG&%\"vG6#F'" }}} {PARA 0 "" 0 "" {TEXT -1 42 "and for the kinetic energy of the cylinde r" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cylinder[1] := kinetic_ energy(m,v[1])+kinetic_energy(J,omega[1]);" "6#>&%+T_cylinderG6#\"\"\" ,&-%/kinetic_energyG6$%\"mG&%\"vG6#F'F'-F*6$%\"JG&%&omegaG6#F'F'" }}} {PARA 0 "" 0 "" {TEXT -1 132 "Because we have choosen the reference ho rizont in this position the height of the cylinder over this horzontal has to be set to zero" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "h[1] := 0:" "6#>&%\"hG6#\"\"\"\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 46 "The p otential of the system at state 1 is then" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "V[1] := gravitational_energy(m,g,h[1]);" "6#>&%\"VG6#\" \"\"-%5gravitational_energyG6%%\"mG%\"gG&%\"hG6#F'" }}}{PARA 0 "" 0 " " {TEXT -1 96 "The potential of the cuboid doesn't change during the m otion and so it needn't to be considered." }}{PARA 0 "" 0 "" {TEXT -1 48 "The total energy of the System at state 1 is now" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "E[1]:=T_cuboid[1]+T_cylinder[1]+V[1];" "6#> &%\"EG6#\"\"\",(&%)T_cuboidG6#F'F'&%+T_cylinderG6#F'F'&%\"VG6#F'F'" }} }{PARA 5 "" 0 "" {TEXT -1 17 "State 2 (Fig. 26)" }}{PARA 0 "" 0 "" {TEXT -1 12 "Now we have " }{XPPEDIT 18 0 "s = L1;" "6#/%\"sG%#L1G" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "state2:=s=L1;" "6#>%'state2G/% \"sG%#L1G" }}}{PARA 0 "" 0 "" {TEXT -1 13 "The velocity " }{XPPEDIT 18 0 "v[2];" "6#&%\"vG6#\"\"#" }{TEXT -1 79 " of the cuboid and the cy linder must be the same, because the rope doesn't sag." }}{PARA 0 "" 0 "" {TEXT -1 80 "We get the rotational speed of the cylinder from the geometrical condition above" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "omega[2] := v[2]/R;" "6#>&%&omegaG6#\"\"#*&&%\"vG6#F'\"\"\"%\"RG!\" \"" }}}{PARA 0 "" 0 "" {TEXT -1 54 "We get for the kinetic energy of t he cuboid at state 2" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cubo id[2] := kinetic_energy(m,v[2]);" "6#>&%)T_cuboidG6#\"\"#-%/kinetic_en ergyG6$%\"mG&%\"vG6#F'" }}}{PARA 0 "" 0 "" {TEXT -1 42 "and for the ki netic energy of the cylinder" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_translation_cylinder[2] := kinetic_energy(m,v[2]);" "6#>&%7T_tran slation_cylinderG6#\"\"#-%/kinetic_energyG6$%\"mG&%\"vG6#F'" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_rotation_cylinder[2] := kine tic_energy(J,omega[2]);" "6#>&%4T_rotation_cylinderG6#\"\"#-%/kinetic_ energyG6$%\"JG&%&omegaG6#F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cylinder[2] := T_translation_cylinder[2]+T_rotation_cylinder[2]; " "6#>&%+T_cylinderG6#\"\"#,&&%7T_translation_cylinderG6#F'\"\"\"&%4T_ rotation_cylinderG6#F'F," }}}{PARA 0 "" 0 "" {TEXT -1 123 "The altitud e of the cuboid doesn't change, but the changing of the height of the \+ cylinder is given by use of the condition " }{TEXT 432 6 "state2" } {TEXT -1 3 " by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "h[2] := sub s(state2,H(s));" "6#>&%\"hG6#\"\"#-%%subsG6$%'state2G-%\"HG6#%\"sG" }} }{PARA 0 "" 0 "" {TEXT -1 41 "The potential of the system at state 2 i s" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "V[2] := gravitational_ene rgy(m,g,h[2]);" "6#>&%\"VG6#\"\"#-%5gravitational_energyG6%%\"mG%\"gG& %\"hG6#F'" }}}{PARA 0 "" 0 "" {TEXT -1 44 "The total energy of the sys tem at state 2 is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "E[2] := T _cuboid[2]+T_cylinder[2]+V[2];" "6#>&%\"EG6#\"\"#,(&%)T_cuboidG6#F'\" \"\"&%+T_cylinderG6#F'F,&%\"VG6#F'F," }}}{PARA 0 "" 0 "" {TEXT -1 58 " The only unknown value in this expression is the velocity " }{XPPEDIT 18 0 "v_cuboid[2]" "6#&%)v_cuboidG6#\"\"#" }{TEXT -1 53 ". We can calc ulate it by use of the balance of energy" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eq1:=E[2]=E[1];" "6#>%$eq1G/&%\"EG6#\"\"#&F'6#\"\"\"" } }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "sol1 := solve(eq1,v[2]);" "6 #>%%sol1G-%&solveG6$%$eq1G&%\"vG6#\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 96 "Because we know that only the positive value can be the right one, we use again the little trick" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v[2] := sqrt(sol1[1]^2);" "6#>&%\"vG6#\"\"#-%%sqrtG6#*$&%%sol1G6 #\"\"\"F'" }}}{PARA 0 "" 0 "" {TEXT -1 44 "The angular velocity of the cylinder is then" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "omega[2]; " "6#&%&omegaG6#\"\"#" }}}{PARA 5 "" 0 "" {TEXT -1 17 "State 3 (Fig. 2 7)" }}{PARA 0 "" 0 "" {TEXT -1 12 "Now we have " }{XPPEDIT 18 0 "s = L 1+2*R;" "6#/%\"sG,&%#L1G\"\"\"*&\"\"#F'%\"RGF'F'" }}{EXCHG {PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "state3:=s=L1+2*R;" "6#>%'state3G/%\"sG,&%#L1G\" \"\"*&\"\"#F)%\"RGF)F)" }}}{PARA 0 "" 0 "" {TEXT -1 6 "Again " } {XPPEDIT 18 0 "v[3];" "6#&%\"vG6#\"\"$" }{TEXT -1 173 " is unknown. An d again we know that the cuboid and the cylinder must have the same ve locity. We get the rotational speed of the cylinder from the geometric al condition above" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "omega[3] := v[3]/R;" "6#>&%&omegaG6#\"\"$*&&%\"vG6#F'\"\"\"%\"RG!\"\"" }}} {PARA 0 "" 0 "" {TEXT -1 46 "The kinetic energy of the cuboid at state 3 is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cuboid[3] := kineti c_energy(m,v[3]);" "6#>&%)T_cuboidG6#\"\"$-%/kinetic_energyG6$%\"mG&% \"vG6#F'" }}}{PARA 0 "" 0 "" {TEXT -1 45 "and for the kinetic energy o f the cylinder is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_transla tion_cylinder[3] := kinetic_energy(m,v[3]);" "6#>&%7T_translation_cyli nderG6#\"\"$-%/kinetic_energyG6$%\"mG&%\"vG6#F'" }}}{EXCHG {PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "T_rotation_cylinder[3] := kinetic_energy(J,omega [3]);" "6#>&%4T_rotation_cylinderG6#\"\"$-%/kinetic_energyG6$%\"JG&%&o megaG6#F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cylinder[3] := T_translation_cylinder[3]+T_rotation_cylinder[3];" "6#>&%+T_cylinderG 6#\"\"$,&&%7T_translation_cylinderG6#F'\"\"\"&%4T_rotation_cylinderG6# F'F," }}}{PARA 0 "" 0 "" {TEXT -1 76 "The changing of the height of th e cylinder is given by use of the condition " }{TEXT 433 6 "state3" } {TEXT -1 3 " by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "h[3] := sub s(state3,H(s));" "6#>&%\"hG6#\"\"$-%%subsG6$%'state3G-%\"HG6#%\"sG" }} }{PARA 0 "" 0 "" {TEXT -1 41 "The potential of the system at state 3 i s" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "V[3] := gravitational_ene rgy(m,g,h[3]);" "6#>&%\"VG6#\"\"$-%5gravitational_energyG6%%\"mG%\"gG& %\"hG6#F'" }}}{PARA 0 "" 0 "" {TEXT -1 44 "The total energy of the sys tem at state 3 is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "E[3] := T _cuboid[3]+T_cylinder[3]+V[3];" "6#>&%\"EG6#\"\"$,(&%)T_cuboidG6#F'\" \"\"&%+T_cylinderG6#F'F,&%\"VG6#F'F," }}}{PARA 0 "" 0 "" {TEXT -1 196 "Now we have to consider the energy which is dissipated by the frictio n on the way from state 2 to state 3. Above we have declared that the \+ friction force increases linearly with the increasing of " }{TEXT 434 1 "s" }{TEXT -1 22 " in the range between " }{TEXT 435 2 "L1" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "L1+R" "6#,&%#L1G\"\"\"%\"RGF%" }{TEXT -1 36 ". So we have in the considered range" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "FR[2 .. 3](s) := mu*m*g*(s-L1)/(2*R):" "6#>-&%#FRG6#;\" \"#\"\"$6#%\"sG*,%#muG\"\"\"%\"mGF/%\"gGF/,&F,F/%#L1G!\"\"F/*&F)F/%\"R GF/F4" }}}{PARA 0 "" 0 "" {TEXT -1 57 "The energy which is dissipated \+ on this path is defined by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 " W[2 .. 3] := -int(FR[2 .. 3](s),s = L1 .. L1+2*R);" "6#>&%\"WG6#;\"\"# \"\"$,$-%$intG6$-&%#FRG6#;F(F)6#%\"sG/F4;%#L1G,&F7\"\"\"*&F(F9%\"RGF9F 9!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 101 "Notice that the friction forc e points against the direction of the motion, so we get a negative wor k." }}{PARA 0 "" 0 "" {TEXT -1 21 "Simplification yields" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "W[2..3]:=simplify(W[2..3]);" "6#>&%\" WG6#;\"\"#\"\"$-%)simplifyG6#&F%6#;F(F)" }}}{PARA 0 "" 0 "" {TEXT -1 28 "The balance of energy is now" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eq2 := E[3] = E[2]+W[2 .. 3];" "6#>%$eq2G/&%\"EG6#\"\"$,&&F'6#\" \"#\"\"\"&%\"WG6#;F-F)F." }}}{PARA 0 "" 0 "" {TEXT -1 49 "From this eq uation we get the velocity at state 3" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "sol2 := solve(eq2,v[3]);" "6#>%%sol2G-%&solveG6$%$eq2G& %\"vG6#\"\"$" }}}{PARA 0 "" 0 "" {TEXT -1 142 "By our specification of the situation, the velocity at state 3 cannot be negative. So only on e solution can be the right one. Our trick yields" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v[3]:=sqrt(sol2[1]**2);" "6#>&%\"vG6#\"\"$-%%sq rtG6#*$&%%sol2G6#\"\"\"\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 44 "The angu lar velocity of the cylinder is then" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "omega[3];" "6#&%&omegaG6#\"\"$" }}}{PARA 0 "" 0 "" {TEXT -1 45 "Here is an additional interesting question : " }{TEXT 442 49 "How can we know whether the bodies reach state 3?" }{TEXT -1 1 " " }{TEXT 443 106 "Perhaps the friction is strong enough to dissipa te the total energy of the system before state 3 is reched" }{TEXT -1 73 ". The limit situation is given when the velocity at state 3 is jus t zero." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "C1:=v[3]=0;" "6#>%# C1G/&%\"vG6#\"\"$\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 46 "What is the fr iction coefficient in this case?" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "mu[limit1] := solve(C1,mu);" "6#>&%#muG6#%'limit1G-%&solveG6$%#C 1GF%" }}}{PARA 0 "" 0 "" {TEXT -1 20 "Now we can say: For " }{XPPEDIT 18 0 "mu <= mu[grenz1];" "6#1%#muG&F$6#%'grenz1G" }{TEXT -1 106 " the \+ cuboid reaches state 3. For the next step of this calculation we assum e that this condition is valid." }}{PARA 5 "" 0 "" {TEXT -1 17 "State \+ 4 (Fig. 28)" }}{PARA 0 "" 0 "" {TEXT -1 12 "Now we have " }{XPPEDIT 18 0 "s = L1+L2;" "6#/%\"sG,&%#L1G\"\"\"%#L2GF'" }}{EXCHG {PARA 0 "> \+ " 0 "" {XPPEDIT 19 1 "state4 := s = L1+L2;" "6#>%'state4G/%\"sG,&%#L1G \"\"\"%#L2GF)" }}}{PARA 0 "" 0 "" {TEXT -1 39 "The rotational speed of the cylinder is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "omega[4] : = v[4]/R;" "6#>&%&omegaG6#\"\"%*&&%\"vG6#F'\"\"\"%\"RG!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 46 "The kinetic energy of the cuboid at state 4 is " }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cuboid[4] := kinetic_ene rgy(m,v[4]);" "6#>&%)T_cuboidG6#\"\"%-%/kinetic_energyG6$%\"mG&%\"vG6# F'" }}}{PARA 0 "" 0 "" {TEXT -1 45 "and for the kinetic energy of the \+ cylinder is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_translation_c ylinder[4] := kinetic_energy(m,v[4]);" "6#>&%7T_translation_cylinderG6 #\"\"%-%/kinetic_energyG6$%\"mG&%\"vG6#F'" }}}{EXCHG {PARA 0 "> " 0 " " {XPPEDIT 19 1 "T_rotation_cylinder[4] := kinetic_energy(J,omega[4]); " "6#>&%4T_rotation_cylinderG6#\"\"%-%/kinetic_energyG6$%\"JG&%&omegaG 6#F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cylinder[4] := T_tr anslation_cylinder[4]+T_rotation_cylinder[4];" "6#>&%+T_cylinderG6#\" \"%,&&%7T_translation_cylinderG6#F'\"\"\"&%4T_rotation_cylinderG6#F'F, " }}}{PARA 0 "" 0 "" {TEXT -1 45 "The changing of the height of the cy linder is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "h[4] := subs(stat e4,H(s));" "6#>&%\"hG6#\"\"%-%%subsG6$%'state4G-%\"HG6#%\"sG" }}} {PARA 0 "" 0 "" {TEXT -1 41 "The potential of the system at state 4 is " }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "V[4] := gravitational_ener gy(m,g,h[4]);" "6#>&%\"VG6#\"\"%-%5gravitational_energyG6%%\"mG%\"gG&% \"hG6#F'" }}}{PARA 0 "" 0 "" {TEXT -1 44 "The total energy of the syst em at state 4 is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "E[4] := T_ cuboid[4]+T_cylinder[4]+V[4];" "6#>&%\"EG6#\"\"%,(&%)T_cuboidG6#F'\"\" \"&%+T_cylinderG6#F'F,&%\"VG6#F'F," }}}{PARA 0 "" 0 "" {TEXT -1 132 "C onsider again the energy which is dissipated by the friction on the wa y from state 3 to state 4. Now the friction force is constant" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "FR[3 .. 4](s) := mu*m*g:" "6#> -&%#FRG6#;\"\"$\"\"%6#%\"sG*(%#muG\"\"\"%\"mGF/%\"gGF/" }}}{PARA 0 "" 0 "" {TEXT -1 56 "The energy which is dissipated on this way is define d by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "W[3 .. 4] := -int(FR[3 .. 4](s),s = L1+2*R .. L1+L2);" "6#>&%\"WG6#;\"\"$\"\"%,$-%$intG6$-&% #FRG6#;F(F)6#%\"sG/F4;,&%#L1G\"\"\"*&\"\"#F9%\"RGF9F9,&F8F9%#L2GF9!\" \"" }}}{PARA 0 "" 0 "" {TEXT -1 28 "The balance of energy is now" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eq3 := E[4] = E[3]+W[3 .. 4]; " "6#>%$eq3G/&%\"EG6#\"\"%,&&F'6#\"\"$\"\"\"&%\"WG6#;F-F)F." }}}{PARA 0 "" 0 "" {TEXT -1 38 "and for the velocity at state 4 we get" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "sol3 := solve(eq3,v[4]);" "6#> %%sol3G-%&solveG6$%$eq3G&%\"vG6#\"\"%" }}}{PARA 0 "" 0 "" {TEXT -1 71 "Again only the positive value of the velocity makes sense and so we g et" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v[4] := simplify(sqrt(so l3[1]^2));" "6#>&%\"vG6#\"\"%-%)simplifyG6#-%%sqrtG6#*$&%%sol3G6#\"\" \"\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 39 "The angular velocity of the c ylinder is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "omega[4];" "6#&% &omegaG6#\"\"%" }}}{PARA 0 "" 0 "" {TEXT 444 50 "How can we know wheth er the bodies reach state 4? " }{TEXT -1 71 "The limit situation is gi ven when the velocity at state 4 is just zero." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "C2:=v[4]=0;" "6#>%#C2G/&%\"vG6#\"\"%\"\"!" }}} {PARA 0 "" 0 "" {TEXT -1 46 "What is the friction coefficient in this \+ case?" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "mu[limit2] := solve(C 2,mu);" "6#>&%#muG6#%'limit2G-%&solveG6$%#C2GF%" }}}{PARA 0 "" 0 "" {TEXT -1 23 "We get the result: For " }{XPPEDIT 18 0 "mu <= mu[grenz2] ;" "6#1%#muG&F$6#%'grenz2G" }{TEXT -1 28 " the cuboid reaches state 4. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "For c ompleteness we consider some concrete " }{TEXT 436 16 "numerical value s" }{TEXT -1 1 ":" }}{EXCHG {PARA 0 "> " 0 "numval" {MPLTEXT 1 0 51 "S U:=\{mu=0.3,m=1,R=1,L1=2,g=9.81,alpha=Pi/12,L2=10\}:" }}}{PARA 0 "" 0 "" {TEXT -1 40 "This values yield for the friction force" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "FRk:=subs(SU,FR(s));" }}}{PARA 0 " " 0 "" {TEXT -1 45 "Fig. 29 shows the curve of the friction force" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "plot(FRk,s=0..subs(SU,L1+L2) , thickness=3, title=\"Figure 29\");" }}}{PARA 0 "" 0 "" {TEXT -1 49 " The limit values for the friction coefficient are" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalf(subs(SU,mu[limit1]));" "6#-%&evalfG6#-%%s ubsG6$%#SUG&%#muG6#%'limit1G" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalf(subs(SU,mu[limit2]));" "6#-%&evalfG6#-%%subsG6$%#SUG&%#muG6#% 'limit2G" }}}{PARA 0 "" 0 "" {TEXT -1 188 "We see that for the chosen \+ values the limit friction coefficients are higher than the current val ue, so the system moves from state 1 up to state 4. The velocity is in the different states" }}{PARA 5 "" 0 "" {TEXT -1 17 "State 1 (Fig. 25 )" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs(SU,v[1]);" "6#-%%sub sG6$%#SUG&%\"vG6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "su bs(SU,omega[1]);" "6#-%%subsG6$%#SUG&%&omegaG6#\"\"\"" }}}{PARA 5 "" 0 "" {TEXT -1 17 "State 2 (Fig. 26)" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalf(subs(SU,v[2]));" "6#-%&evalfG6#-%%subsG6$%#SUG&% \"vG6#\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalf(subs(SU, omega[2]));" "6#-%&evalfG6#-%%subsG6$%#SUG&%&omegaG6#\"\"#" }}}{PARA 5 "" 0 "" {TEXT -1 17 "State 3 (Fig. 27)" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalf(subs(SU,v[3]));" "6#-%&evalfG6#-%%subsG6$%#SUG&% \"vG6#\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalf(subs(SU, omega[3]));" "6#-%&evalfG6#-%%subsG6$%#SUG&%&omegaG6#\"\"$" }}}{PARA 5 "" 0 "" {TEXT -1 17 "State 4 (Fig. 28)" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalf(subs(SU,v[4]));" "6#-%&evalfG6#-%%subsG6$%#SUG&% \"vG6#\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "evalf(subs(SU, omega[4]));" "6#-%&evalfG6#-%%subsG6$%#SUG&%&omegaG6#\"\"%" }}}{PARA 0 "" 0 "" {TEXT -1 30 "Notice that, as in subsection " }{HYPERLNK 17 " 3.2" 1 "sec-03.mws" "3.2" }{TEXT -1 42 ", time appears nowhere in thes e relations." }}{PARA 0 "" 0 "" {TEXT -1 47 "At last we want to consid er an arbitrary state " }{XPPEDIT 18 0 "s=x" "6#/%\"sG%\"xG" }{TEXT -1 73 " by use of the above defined numerical values. The friction for ce is then" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "FRk;" }}}{PARA 0 "" 0 "" {TEXT -1 67 "For the work which is done from the friction fo rce from state 1 to " }{XPPEDIT 18 0 "x=s" "6#/%\"xG%\"sG" }{TEXT -1 3 " is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Ws(x):=-int(FRk,s= 0..x);" }}}{PARA 0 "" 0 "" {TEXT -1 39 "The kinetic energy of the cubo id is now" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cub(x) := evalf (subs(SU,kinetic_energy(m,vs(x))));" "6#>-%&T_cubG6#%\"xG-%&evalfG6#-% %subsG6$%#SUG-%/kinetic_energyG6$%\"mG-%#vsG6#F'" }}}{PARA 0 "" 0 "" {TEXT -1 45 "and for the kinetic energy of the cylinder is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_translation_cyl(x) := evalf(subs(SU ,kinetic_energy(m,vs(x))));" "6#>-%2T_translation_cylG6#%\"xG-%&evalfG 6#-%%subsG6$%#SUG-%/kinetic_energyG6$%\"mG-%#vsG6#F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_rotation_cyl(x) := evalf(subs(SU,kinetic_ energy(J,vs(x)/R)));" "6#>-%/T_rotation_cylG6#%\"xG-%&evalfG6#-%%subsG 6$%#SUG-%/kinetic_energyG6$%\"JG*&-%#vsG6#F'\"\"\"%\"RG!\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "T_cyl(x) := T_translation_cyl( x)+T_rotation_cyl(x);" "6#>-%&T_cylG6#%\"xG,&-%2T_translation_cylG6#F' \"\"\"-%/T_rotation_cylG6#F'F," }}}{PARA 0 "" 0 "" {TEXT -1 45 "The ch anging of the height of the cylinder is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "hs(x) := evalf(subs(SU,subs(s = x,H(s))));" "6#>-%#hsG6 #%\"xG-%&evalfG6#-%%subsG6$%#SUG-F,6$/%\"sGF'-%\"HG6#F2" }}}{PARA 0 " " 0 "" {TEXT -1 47 "The potential of the system is in this position" } }{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Vs(x) := evalf(subs(SU,gravit ational_energy(m,g,hs(x))));" "6#>-%#VsG6#%\"xG-%&evalfG6#-%%subsG6$%# SUG-%5gravitational_energyG6%%\"mG%\"gG-%#hsG6#F'" }}}{PARA 0 "" 0 "" {TEXT -1 45 "And for the total energy of the system we get" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Es(x) := T_cub(x)+T_cyl(x)+Vs(x);" "6 #>-%#EsG6#%\"xG,(-%&T_cubG6#F'\"\"\"-%&T_cylG6#F'F,-%#VsG6#F'F," }}} {PARA 0 "" 0 "" {TEXT -1 37 "The balance of energy is now given by" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eqs := Es(x) = E[1]+Ws(x);" "6 #>%$eqsG/-%#EsG6#%\"xG,&&%\"EG6#\"\"\"F.-%#WsG6#F)F." }}}{PARA 0 "" 0 "" {TEXT -1 27 "and for the velocity we get" }}{EXCHG {PARA 0 "> " 0 " " {XPPEDIT 19 1 "sols := solve(eqs,vs(x));" "6#>%%solsG-%&solveG6$%$eq sG-%#vsG6#%\"xG" }}}{PARA 0 "" 0 "" {TEXT -1 71 "Again only the positi ve value of the velocity makes sense and so we get" }}{EXCHG {PARA 0 " > " 0 "" {XPPEDIT 19 1 "vv(x) := simplify(sqrt(sols[1]^2));" "6#>-%#vv G6#%\"xG-%)simplifyG6#-%%sqrtG6#*$&%%solsG6#\"\"\"\"\"#" }}}{PARA 0 " " 0 "" {TEXT -1 80 "Fig. 30 shows at last the velocity of the bodies i n dependence of their position" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot(vv(x),x=1..subs(SU,L1+L2), thickness=3, title=\"Figure 30\" );" }}}{PARA 0 "" 0 "" {TEXT -1 46 "You can test what happens when you change the " }{HYPERLNK 17 "numerical values" 1 "" "numval" }{TEXT -1 45 ". But consider whether the result makes sense" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 11 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }