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A body with finite mass and negligible dimensions is c alled a " }{TEXT 514 13 "mass particle" }{TEXT -1 13 " or simply a " } {TEXT 515 8 "particle" }{TEXT -1 4 ". A " }{TEXT 516 10 "rigid body" } {TEXT -1 267 " is a body with fixed distances among all of its points. Mass particles have negligible dimensions. This means that only trans lation and not rotation is defined for particles. The state of a mass \+ particle under motion is unambiguously defined by its position at time " }{TEXT 256 1 "t" }{TEXT -1 23 ". The position at time " }{TEXT 257 1 "t" }{TEXT -1 26 " is defined by the vector " }{TEXT 258 1 "r" } {TEXT -1 1 "(" }{TEXT 259 1 "t" }{TEXT -1 124 ") which starts at a fix ed point O of the inertial system of the particle. Accordingly, the po sition of the particle at time " }{TEXT 261 2 "t+" }{TEXT 260 1 "D" } {TEXT 262 1 "t" }{TEXT -1 33 " (call: \"Delta t\") is defined by " } {TEXT 263 1 "r" }{TEXT -1 1 "(" }{TEXT 265 2 "t+" }{TEXT 264 1 "D" } {TEXT 266 1 "t" }{TEXT -1 34 ") (see the animation in Fig. 1). " } {TEXT 391 74 "To see the animation, click on the diagram and hit \"pla y\" in the menu bar." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "display(Fig_2_1(),insequence=true,scaling =constrained,axes=none,title=\"Figure 1\");" }}}{PARA 0 "" 0 "" {TEXT -1 13 "The velocity " }{TEXT 267 1 "v" }{TEXT -1 1 "(" }{TEXT 268 1 "t " }{TEXT -1 15 ") is defined by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v(t):=diff(r(t),t):" "6#>-%\"vG6#%\"tG-%%diffG6$-%\"rG6#F'F'" }} }{PARA 0 "" 0 "" {TEXT -1 32 "Notice that in this expressions " } {TEXT 289 1 "r" }{TEXT -1 1 "(" }{TEXT 290 1 "t" }{TEXT -1 6 ") and " }{TEXT 292 1 "v" }{TEXT -1 1 "(" }{TEXT 291 1 "t" }{TEXT -1 14 ") are \+ vectors." }}{PARA 0 "" 0 "" {TEXT -1 25 "We can write this also as" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(r(t),t) = limit((r(t+Delt a_t)-r(t))/Delta_t,Delta_t = 0):" "6#/-%%diffG6$-%\"rG6#%\"tGF*-%&limi tG6$*&,&-F(6#,&F*\"\"\"%(Delta_tGF3F3-F(6#F*!\"\"F3F4F7/F4\"\"!" }}} {PARA 0 "" 0 "" {TEXT -1 33 "In the last equation, the vector " } {TEXT 278 1 "r" }{TEXT -1 1 "(" }{TEXT 274 2 "t+" }{TEXT 273 1 "D" } {TEXT 275 1 "t" }{TEXT -1 4 ") - " }{TEXT 277 1 "r" }{TEXT -1 1 "(" } {TEXT 276 1 "t" }{TEXT -1 87 ") points in the direction of the tangen t of the curve of the motion when the timestep " }{TEXT 269 1 "D" } {TEXT 270 1 "t" }{TEXT -1 52 " decreases and becomes infinitesimal. So the vector " }{TEXT 271 1 "v" }{TEXT -1 1 "(" }{TEXT 272 1 "t" } {TEXT -1 141 ") points in this direction, too. To see what happens in \+ this situation click the picture for Figure 1 above and press \"play\" in the menu bar." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "For two different times " }{TEXT 284 1 "t" }{TEXT -1 5 " \+ and " }{TEXT 286 2 "t+" }{TEXT 285 1 "D" }{TEXT 287 1 "t" }{TEXT -1 27 " we get for the velocities " }{TEXT 279 1 "v" }{TEXT -1 1 "(" } {TEXT 280 1 "t" }{TEXT -1 6 ") and " }{TEXT 288 1 "v" }{TEXT -1 1 "(" }{TEXT 282 2 "t+" }{TEXT 281 1 "D" }{TEXT 283 1 "t" }{TEXT -1 15 ") (s ee Fig. 2)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "display(Fig_2 _2(),scaling=constrained,axes=none, title=\"Figure 2\");" }}}{PARA 0 " " 0 "" {TEXT -1 30 "The acceleration is defined as" }}{EXCHG {PARA 0 " > " 0 "" {XPPEDIT 19 1 "a(t):=diff(v(t),t);" "6#>-%\"aG6#%\"tG-%%diffG 6$-%\"vG6#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 31 "Notice that in this ex presions " }{TEXT 293 1 "v" }{TEXT -1 1 "(" }{TEXT 294 1 "t" }{TEXT -1 6 ") and " }{TEXT 296 1 "a" }{TEXT -1 1 "(" }{TEXT 295 1 "t" } {TEXT -1 14 ") are vectors." }}{PARA 0 "" 0 "" {TEXT -1 20 "We can wri te this as" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(v(t),t) = l imit((v(t+Delta_t)-v(t))/Delta_t,Delta_t = 0):" "6#/-%%diffG6$-%\"vG6# %\"tGF*-%&limitG6$*&,&-F(6#,&F*\"\"\"%(Delta_tGF3F3-F(6#F*!\"\"F3F4F7/ F4\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 315 "A cartesian coordinat e system is a coordinate system in which the locations of points in sp ace are expressed by reference to three planes, called coordinate plan es, no two of which are parallel. We describe this system by three ort hogonal axes. The direction in space of every axis is described by the unit vectors " }{TEXT 297 1 "e" }{TEXT -1 3 "x, " }{TEXT 298 1 "e" } {TEXT -1 6 "y and " }{TEXT 299 1 "e" }{TEXT -1 1 "z" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "ex:=vector(3,[1,0,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "ey:=vector(3,[0,1,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "ez:=vector(3,[0,0,1]);" }}}{PARA 0 "" 0 " " {TEXT -1 19 "The position vector" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "r(t):=x(t)*ex+y(t)*ey+z(t)*ez;" }}}{PARA 0 "" 0 "" {TEXT -1 21 "with the coordinates " }{TEXT 300 1 "x" }{TEXT -1 1 "(" } {TEXT 304 1 "t" }{TEXT -1 1 ")" }{TEXT 305 3 ", y" }{TEXT -1 1 "(" } {TEXT 306 1 "t" }{TEXT -1 1 ")" }{TEXT 307 3 ", z" }{TEXT -1 1 "(" } {TEXT 308 1 "t" }{TEXT -1 49 ") describes the position of the particle at time " }{TEXT 301 1 "t" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 189 "The derivative of a variable with respect to time is usually d enoted by a dot above the variable. Here it is denoted by an appended \+ letter d. So we write for the coordinates of the velocity" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xd(t):=diff(x(t),t):" "6#>-%#xdG6#%\" tG-%%diffG6$-%\"xG6#F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "y d(t):=diff(y(t),t):" "6#>-%#ydG6#%\"tG-%%diffG6$-%\"yG6#F'F'" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "zd(t):=diff(z(t),t):" "6#>-%#z dG6#%\"tG-%%diffG6$-%\"zG6#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 27 "and f or the velocity vector" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v(t) :=diff(r(t),t);" "6#>-%\"vG6#%\"tG-%%diffG6$-%\"rG6#F'F'" }}}{PARA 0 " " 0 "" {TEXT -1 38 "Written in the form of a vector we get" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalm(v(t));" }}}{PARA 0 "" 0 "" {TEXT -1 33 "And for the acceleration we write" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "xdd(t):=diff(x(t),t$2):" "6#>-%$xddG6#%\"tG-%%diff G6$-%\"xG6#F'-%\"$G6$F'\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ydd(t):=diff(y(t),t$2):" "6#>-%$yddG6#%\"tG-%%diffG6$-%\"yG6#F'-%\" $G6$F'\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "zdd(t):=diff(z (t),t$2):" "6#>-%$zddG6#%\"tG-%%diffG6$-%\"zG6#F'-%\"$G6$F'\"\"#" }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "a(t):=diff(r(t),t$2);" "6#>-% \"aG6#%\"tG-%%diffG6$-%\"rG6#F'-%\"$G6$F'\"\"#" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 12 "evalm(a(t));" }}}{PARA 0 "" 0 "" {TEXT -1 22 " In general the vector " }{TEXT 303 1 "a" }{TEXT -1 1 "(" }{TEXT 302 1 "t" }{TEXT -1 117 ") does not have the direction of the tangent of the curve of the motion. We illustrate this in the following example." }} {SECT 0 {PARA 5 "" 0 "" {TEXT -1 38 "Example: Horizontal Throwing of a Ball" }}{PARA 0 "" 0 "" {TEXT -1 254 "Consider a man standing on a sm all hill. We state a coordinate system with the x-axis in the horizont al direction, the y-axis in the vertical and the z-axis orthogonal to \+ the picture plane. The origin of the system is at the head of the man \+ (see Fig. 3)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "display(Fig _2_3(10,0,0,0,0,15,0),scaling=constrained,axes=none, title=\"Figure 3 \");" }}}{PARA 0 "" 0 "" {TEXT -1 404 "Now the man throws a ball in th e horizontal direction, starting at the origin with velocity v0. Air r esistance is not taken into account. The velocity in the horizontal di rection then is constant and the acceleration is consequently 0 up to \+ the moment when the ball touches the ground. In the vertical direction , the acceleration is caused by gravity. There is no motion orthogonal to the picture plane ." }}{PARA 0 "" 0 "" {TEXT -1 63 "So we write fo r the accelerations in the x-, y- and z-direction" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ax(t):=0:" "6#>-%#axG6#%\"tG\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ay(t):=g:" "6#>-%#ayG6#%\"tG%\"gG" }} }{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "az(t) := 0:" "6#>-%#azG6#%\"t G\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 46 "where g is the acceleration of gravity = 9.81 " }{XPPEDIT 18 0 "m/s^2" "6#*&%\"mG\"\"\"*$%\"sG\"\"#! \"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 34 "We get for the ac celeration vector" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "a(t):=e valm(ax(t)*ex+ay(t)*ey+az(t)*ez);" }}}{PARA 0 "" 0 "" {TEXT -1 220 "At all times, the acceleration vector points in the vertical direction. \+ Because the velocity at the beginning points in the horizontal directi on we see already now that the directions of the two vectors are not t he same." }}{PARA 0 "" 0 "" {TEXT -1 33 "Integration with respect to t ime " }{TEXT 309 1 "t" }{TEXT -1 31 " yields for the velocity vector" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vx(t):=int(ax(t),t)+cx;" "6# >-%#vxG6#%\"tG,&-%$intG6$-%#axG6#F'F'\"\"\"%#cxGF/" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vy(t):=int(ay(t),t)+cy;" "6#>-%#vyG6#%\"tG,&- %$intG6$-%#ayG6#F'F'\"\"\"%#cyGF/" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "vz(t):=int(az(t),t)+cz;" "6#>-%#vzG6#%\"tG,&-%$intG6$-% #azG6#F'F'\"\"\"%#czGF/" }}}{PARA 0 "" 0 "" {TEXT -1 49 "From this int egration we get the three constants " }{TEXT 310 2 "cx" }{TEXT -1 2 ", " }{TEXT 311 2 "cy" }{TEXT -1 5 " and " }{TEXT 312 2 "cz" }{TEXT -1 86 ". We know the values of the components of the velocity in every di rection at the time " }{TEXT 323 3 "t=0" }{TEXT -1 41 ". These are the three initial conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ic_x:=subs(t=0,vx(t))=v0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ic_y:=subs(t=0,vy(t))=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ic_z:=subs(t=0,vz(t))=0;" }}}{PARA 0 "" 0 "" {TEXT -1 57 "Solv ing these conditions yields the integration constants" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "sol_1:=solve(\{ic_x,ic_y,ic_z\},\{cx,cy,c z\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "assign(sol_1);" }} }{PARA 0 "" 0 "" {TEXT -1 33 "So we get for the velocity vector" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "v(t):=evalm(vx(t)*ex+vy(t)*e y+vz(t)*ez);" }}}{PARA 0 "" 0 "" {TEXT -1 61 "One more integration yie lds the position of the ball at time " }{TEXT 316 1 "t" }{TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "x(t) := int(vx(t),t)+cxx; " "6#>-%\"xG6#%\"tG,&-%$intG6$-%#vxG6#F'F'\"\"\"%$cxxGF/" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "y(t) := int(vy(t),t)+cyy;" "6#>-%\"yG 6#%\"tG,&-%$intG6$-%#vyG6#F'F'\"\"\"%$cyyGF/" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "z(t) := int(vz(t),t)+czz;" "6#>-%\"zG6#%\"tG,&-%$i ntG6$-%#vzG6#F'F'\"\"\"%$czzGF/" }}}{PARA 0 "" 0 "" {TEXT -1 32 "Now w e have the three constants " }{TEXT 313 3 "cxx" }{TEXT -1 2 ", " } {TEXT 314 3 "cyy" }{TEXT -1 5 " and " }{TEXT 315 3 "czz" }{TEXT -1 161 ". We know the position of the ball at the beginning of the motion , because we defined this point as the origin. This information yields three initial conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ic_xx:=subs(t=0,x(t))=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ic_yy:=subs(t=0,y(t))=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ic_zz:=subs(t=0,z(t))=0;" }}}{PARA 0 "" 0 "" {TEXT -1 57 "Solv ing these conditions yields the integration constants" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "sol_2:=solve(\{ic_xx,ic_yy,ic_zz\},\{cxx, cyy,czz\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "assign(sol_2 );" }}}{PARA 0 "" 0 "" {TEXT -1 33 "So we get for the position vector " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "r(t):=evalm(x(t)*ex+y(t) *ey+z(t)*ez);" }}}{PARA 0 "" 0 "" {TEXT -1 95 "With this result the pr oblem is solved. But now we want to make some additional consideration s." }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }{TEXT 317 1 "h" }{TEXT -1 206 " be the vertical distance between the starting point of the ball \+ and the ground. Then we can calculate the time period the ball needs o n its way up to the moment when it touches the ground from the relatio n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eq:=eval(y(t))=h;" }}} {PARA 0 "" 0 "" {TEXT -1 53 "The solution of this equation yields the \+ desired time" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Tend:=solve( eq,t);" }}}{PARA 0 "" 0 "" {TEXT -1 203 "We get two solutions with the same absolute value but different signs. Only the positive solution m akes sense, so we use a little trick (note: here it is not clearly def ined whether g and h are positive)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Tend:=sqrt(Tend[1]**2);" }}}{PARA 0 "" 0 "" {TEXT -1 113 "At last we want to calculate the horizontal distance that the bal l travels. We get this by substituting the time " }{TEXT 318 4 "Tend" }{TEXT -1 23 " in the expression for " }{TEXT 319 1 "x" }{TEXT -1 1 "( " }{TEXT 322 1 "t" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "xend:=subs(t=Tend,eval(x(t)));" }}}{PARA 0 "" 0 "" {TEXT -1 65 "To show the result we choose some concrete values for the height " }{TEXT 320 1 "h" }{TEXT -1 18 " and the velocity " }{TEXT 321 2 "v0" }{TEXT -1 2 ". " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "h:=10:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "v0:=10:" }}} {PARA 0 "" 0 "" {TEXT -1 30 "The acceleration of gravity is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=9.81:" }}}{PARA 0 "" 0 "" {TEXT -1 144 "Fig. 4 shows the result. To see what happens click the picture and press \"play\" in the menu bar. (Notice that the man's arm actua lly moves =) )" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "display(F ig_2_3(h,v0,eval(x(t)),eval(y(t)),Tend,eval(xend),t),insequence=true,s caling=constrained,axes=none, title=\"Figure 4\");" }}}{PARA 0 "" 0 " " {TEXT -1 131 "In Fig 5 we see the velocity vector and the accelerati on vector together with the trajectory of the motion at different time steps." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "display(Fig_2_4( h,v0,eval(x(t)),eval(y(t)),Tend,eval(xend),t),insequence=false,scaling =constrained,axes=none, title=\"Figure 5\");" }}}{PARA 0 "" 0 "" {TEXT -1 108 "To try the example for different initial velocities and \+ different heights, change the values h and v0 above." }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 23 "2.1.2 Polar Coordinates" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Now we consider a coordinate system with rotate d initial vectors " }{TEXT 326 1 "e" }{TEXT -1 2 "r(" }{TEXT 342 1 "t " }{TEXT -1 6 ") and " }{TEXT 327 1 "e" }{TEXT 329 1 "f" }{TEXT -1 1 " (" }{TEXT 343 0 "" }{TEXT -1 0 "" }{TEXT 344 1 "t" }{TEXT -1 64 ") (in the figures, this is written as ephi) against the vectors " }{TEXT 324 1 "e" }{TEXT -1 6 "x and " }{TEXT 325 1 "e" }{TEXT -1 16 "y by the vector " }{TEXT 330 2 "f(" }{TEXT 341 1 "t" }{TEXT -1 45 ") (phi(t) i n the figures). The directions of " }{TEXT 331 1 "e" }{TEXT -1 6 "r an d " }{TEXT 332 1 "e" }{TEXT 333 1 "f" }{TEXT -1 77 " are changing. Se e the presentation in Fig. 6 to get the following relation." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "display(Fig_2_5(),scaling=co nstrained,axes=none, title=\"Figure 6\");" }}}{PARA 0 "" 0 "" {TEXT -1 66 "The relations between the different systems of initial vectors \+ are" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "er(t):=cos(phi(t))*ex+s in(phi(t))*ey:" "6#>-%#erG6#%\"tG,&*&-%$cosG6#-%$phiG6#F'\"\"\"%#exGF0 F0*&-%$sinG6#-F.6#F'F0%#eyGF0F0" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ephi(t):=-sin(phi(t))*ex+cos(phi(t))*ey:" "6#>-%%ephiG6#%\"tG,&* &-%$sinG6#-%$phiG6#F'\"\"\"%#exGF0!\"\"*&-%$cosG6#-F.6#F'F0%#eyGF0F0" }}}{PARA 0 "" 0 "" {TEXT -1 21 "Or written as vectors" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "evalm(er(t));" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 15 "evalm(ephi(t));" }}}{PARA 0 "" 0 "" {TEXT -1 20 "Th e initial vectors " }{TEXT 334 1 "e" }{TEXT -1 6 "x and " }{TEXT 335 1 "e" }{TEXT -1 90 "y are not dependent on time. So we get by the deri vative with respect to time for vectors " }{TEXT 336 1 "e" }{TEXT -1 6 "r and " }{TEXT 337 1 "e" }{TEXT 338 1 "f" }}{EXCHG {PARA 0 "> " 0 " " {XPPEDIT 19 1 "phi_d(t) := diff(phi(t),t):" "6#>-%&phi_dG6#%\"tG-%%d iffG6$-%$phiG6#F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "erd(t) :=phi_d(t)*(-sin(phi(t))*ex+cos(phi(t))*ey);" "6#>-%$erdG6#%\"tG*&-%&p hi_dG6#F'\"\"\",&*&-%$sinG6#-%$phiG6#F'F,%#exGF,!\"\"*&-%$cosG6#-F36#F 'F,%#eyGF,F,F," }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ephid(t):=- phi_d(t)*(cos(phi(t))*ex+sin(phi(t))*ey);" "6#>-%&ephidG6#%\"tG,$*&-%& phi_dG6#F'\"\"\",&*&-%$cosG6#-%$phiG6#F'F-%#exGF-F-*&-%$sinG6#-F46#F'F -%#eyGF-F-F-!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 21 "Or written as vecto rs" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalm(erd(t));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "evalm(ephid(t));" }}}{PARA 0 "" 0 "" {TEXT -1 21 "as the derivative of " }{TEXT 328 1 "f" }{TEXT -1 1 "(" }{TEXT 339 1 "t" }{TEXT -1 23 ") with respect to time " } {TEXT 340 1 "t" }{TEXT -1 37 " (spoken as: \"e r dot\" and ephid: \"e \+ " }{TEXT 345 1 "f" }{TEXT -1 7 " dot\")." }}{PARA 0 "" 0 "" {TEXT -1 113 "We can calculate this result by using Maple when we calculate the derivatives for every coordinate of the vectors" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 92 "erd(t):=vector(3,[diff(evalm(er(t))[1],t),diff (evalm(er(t))[2],t),diff(evalm(er(t))[3],t)]);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 100 "ephid(t):=vector(3,[diff(evalm(ephi(t))[1],t) ,diff(evalm(ephi(t))[2],t),diff(evalm(ephi(t))[3],t)]);" }}}{PARA 0 " " 0 "" {TEXT -1 126 "Because we consider only plane motions, there are no changes of the position in the direction orthogonal to the plane. \+ So the " }{TEXT 375 1 "z" }{TEXT -1 42 "-components of the velocity ve ctors are 0." }}{PARA 0 "" 0 "" {TEXT -1 76 "For clarity of the exposi tion that follows, we need to unassign the vectors " }{TEXT 363 1 "e" }{TEXT -1 2 "r(" }{TEXT 366 1 "t" }{TEXT -1 6 ") and " }{TEXT 364 1 "e " }{TEXT 365 1 "f" }{TEXT -1 1 "(" }{TEXT 367 1 "t" }{TEXT -1 24 ") an d their derivatives." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "unas sign('er(t)','ephi(t)','erd(t)','ephid(t)');" }}}{PARA 0 "" 0 "" {TEXT -1 28 "Comparison with the vectors " }{TEXT 346 1 "e" }{TEXT -1 2 "r(" }{TEXT 351 1 "t" }{TEXT -1 6 ") and " }{TEXT 347 1 "e" }{TEXT 348 1 "f" }{TEXT -1 1 "(" }{TEXT 352 1 "t" }{TEXT -1 21 ") yields the \+ relation" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "erd(t):=phi_d(t)*e phi(t);" "6#>-%$erdG6#%\"tG*&-%&phi_dG6#F'\"\"\"-%%ephiG6#F'F," }}} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ephid(t) :=-phi_d(t)*er(t);" " 6#>-%&ephidG6#%\"tG,$*&-%&phi_dG6#F'\"\"\"-%#erG6#F'F-!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 17 "for the velocity." }}{PARA 0 "" 0 "" {TEXT -1 188 "Now we consider a mass particle on a trajectory by use of polar c oordinates. We use the example with the man throwing a ball (see Fig. \+ 4). Fig. 7 shows the situation at an arbitrary time. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Tend := sqrt(2)*sqrt(1/g*h);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "display(Fig_2_6(10,10,eval(x(t)),e val(y(t)),Tend,eval(xend),t),insequence=false,scaling=constrained,axes =none, title =\"Figure 7\");" }}}{PARA 0 "" 0 "" {TEXT -1 52 "The posi tion of the ball is described by the vector " }{TEXT 349 1 "r" }{TEXT -1 1 "(" }{TEXT 350 1 "t" }{TEXT -1 2 ")." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "r(t):=R(t)*er(t);" "6#>-%\"rG6#%\"tG*&-%\"RG6#F'\"\"\"- %#erG6#F'F," }}}{PARA 0 "" 0 "" {TEXT -1 102 "with the distance R(t) b etween the origin and the current position of the ball and the initial vector " }{TEXT 353 1 "e" }{TEXT -1 2 "r(" }{TEXT 354 1 "t" }{TEXT -1 61 "). This vector is rotated from the fixed x-axes by the angle " }{TEXT 355 1 "f" }{TEXT -1 1 "(" }{TEXT 356 1 "t" }{TEXT -1 22 "). The initial vector " }{TEXT 357 1 "e" }{TEXT 358 1 "f" }{TEXT -1 1 "(" } {TEXT 359 1 "t" }{TEXT -1 27 ") is choosen orthogonal to " }{TEXT 360 1 "e" }{TEXT -1 2 "r(" }{TEXT 361 1 "t" }{TEXT -1 33 ") in the directi on of increasing " }{TEXT 362 3 "f. " }{TEXT -1 23 "We get for the vel ocity" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v(t):=diff(r(t),t);" "6#>-%\"vG6#%\"tG-%%diffG6$-%\"rG6#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 48 "and with the result above for the derivative of " }{TEXT 368 1 "e " }{TEXT -1 2 "r(" }{TEXT 369 1 "t" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 38 "v(t):=subs(diff(er(t),t)=erd(t),v(t));" }}} {PARA 0 "" 0 "" {TEXT -1 27 "For the acceleration we get" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "a(t):=diff(v(t),t);" }}}{PARA 0 "" 0 "" {TEXT -1 40 "and with the corresponding substitutions" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "a(t):=subs(\{diff(er(t),t)=erd(t),d iff(ephi(t),t)=ephid(t)\},a(t));" }}}{PARA 0 "" 0 "" {TEXT -1 14 "At l ast we get" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "a(t):=collect( a(t),\{er(t),ephi(t)\});" }}}{PARA 0 "" 0 "" {TEXT -1 88 "Now we colle ct the components of the velocity and the acceleration in the directio ns of " }{TEXT 370 1 "e" }{TEXT 371 1 "f" }{TEXT -1 1 "(" }{TEXT 372 1 "t" }{TEXT -1 6 ") and " }{TEXT 373 1 "e" }{TEXT -1 2 "r(" }{TEXT 374 1 "t" }{TEXT -1 2 ") " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "vr(t):=coeff(v(t),er(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "vphi(t):=coeff(v(t),ephi(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ar(t):=coeff(a(t),er(t));" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 29 "aphi(t):=coeff(a(t),ephi(t));" }}}{PARA 0 "" 0 "" {TEXT -1 36 "These relations are included in the " }{HYPERLNK 17 "dyna mics" 2 "dynamics" "" }{TEXT -1 15 " package. See " }{HYPERLNK 17 "po lar_vel" 2 "polar_vel" "" }{TEXT -1 2 ", " }{HYPERLNK 17 "polar_acc" 2 "polar_acc" "" }{TEXT -1 35 " to know how to use this functions." }} {SECT 0 {PARA 5 "" 0 "" {TEXT -1 53 "Example: Horizontal Throwing of a Ball (continuation)" }}{PARA 0 "" 0 "" {TEXT -1 81 "As an example we \+ consider again the man who throws a ball from the top of a hill." }} {PARA 0 "" 0 "" {TEXT -1 79 "But first we unassign the variables to de scribe the results in a general manner" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "unassign('h','v0','g');" }}}{PARA 0 "" 0 "" {TEXT -1 56 " Above we had calculated the position vector of the ball" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "r(t) := vector([v0*t, 1/2*g* t^2, 0]);" }}}{PARA 0 "" 0 "" {TEXT -1 66 "So the distance between the origin and the position of the ball is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "R(t):=sqrt((r(t)[1]**2+r(t)[2]**2+r(t)[3]**2));" "6#>-% \"RG6#%\"tG-%%sqrtG6#,(*$&-%\"rG6#F'6#\"\"\"\"\"#F2*$&-F/6#F'6#F3F3F2* $&-F/6#F'6#\"\"$F3F2" }}}{PARA 0 "" 0 "" {TEXT -1 30 "and for the angl e between the " }{TEXT 376 1 "x" }{TEXT -1 48 "-axis and the position \+ vector we get from Fig. 7" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "p hi(t):=arctan(eval(y(t))/eval(x(t)));" "6#>-%$phiG6#%\"tG-%'arctanG6#* &-%%evalG6#-%\"yG6#F'\"\"\"-F-6#-%\"xG6#F'!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 62 "Using the relations above we get for the velocity in the \+ form " }{XPPEDIT 18 0 "v_polar(t)=[vr(t),vphi(t)]" "6#/-%(v_polarG6#% \"tG7$-%#vrG6#F'-%%vphiG6#F'" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "v_polar(t):=polar_vel(R(t),phi(t),t);" }}}{PARA 0 "" 0 "" {TEXT -1 37 "and for the acceleration in the form " }{XPPEDIT 18 0 "a_polar( t)=[ar(t),aphi(t)]" "6#/-%(a_polarG6#%\"tG7$-%#arG6#F'-%%aphiG6#F'" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "a_polar(t):=polar_acc(R(t),p hi(t),t);" }}}{PARA 0 "" 0 "" {TEXT -1 240 "Of course this result is m uch more complicated than the result in cartesian coordinates. But thi s example shows how to calculate the velocity and the acceleration in \+ polar coordinates from the given position vector. And with the functio ns " }{HYPERLNK 17 "polar_vel" 2 "polar_vel" "" }{TEXT -1 5 " and " } {HYPERLNK 17 "polar_acc" 2 "polar_acc" "" }{TEXT -1 25 " the calculati on is easy." }}{PARA 0 "" 0 "" {TEXT -1 45 "With the same concrete val ues for the height " }{TEXT 377 1 "h" }{TEXT -1 18 " and the velocity \+ " }{TEXT 378 2 "v0" }{TEXT -1 9 " as above" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 6 "h:=10:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 " v0:=10:" }}}{PARA 0 "" 0 "" {TEXT -1 31 "and the acceleration of gravi ty" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=9.81:" }}}{PARA 0 " " 0 "" {TEXT -1 23 "we get for the velocity" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 27 "vr(t):=eval(v_polar(t)[1]);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 29 "vphi(t):=eval(v_polar(t)[2]);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "and for the acceleration" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 27 "ar(t):=eval(a_polar(t)[1]);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 29 "aphi(t):=eval(a_polar(t)[2]);" }}}{PARA 0 "" 0 "" {TEXT -1 26 "Fig. 8 shows the distance " }{TEXT 385 1 "R" }{TEXT -1 1 "(" }{TEXT 386 1 "t" }{TEXT -1 68 ") between the origin and the p osition of the ball, Fig. 9 the angle " }{TEXT 383 1 "f" }{TEXT -1 1 " (" }{TEXT 384 1 "t" }{TEXT -1 30 ") between the position vector " } {TEXT 387 1 "r" }{TEXT -1 1 "(" }{TEXT 388 1 "t" }{TEXT -1 10 ") and t he " }{TEXT 389 1 "x" }{TEXT -1 24 "-axis as a function of " }{TEXT 390 1 "t" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "Plot1p:=plot(R(t ),t=0..Tend, color=green, thickness=3, legend=\"R(t)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "Plot2p:=plot(phi(t),t=0..Tend, 0..P i/2, color=blue, thickness=3, legend=\"phi(t)\"):" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 36 "display(\{Plot1p\}, title=\"Figure 8\");" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "display(\{Plot2p\}, title= \"Figure 9\");" }}}{PARA 0 "" 0 "" {TEXT -1 88 "The result of the calc ulation is shown in Fig. 10 (velocity) and Fig. 11 (acceleration)." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "Plot1v:=plot(vr(t),t=0..Tend , color=green, thickness=3, legend=\"velocity in r-direction\"):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "Plot2v:=plot(vphi(t),t=0..Te nd, color=blue, thickness=3, legend=\"velocity in phi-direction\"):" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "display(\{Plot1v,Plot2v\}, title=\"Figure 10\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "P lot1a:=plot(ar(t),t=0..Tend, color=green, thickness=3, legend=\"accele ration in r-direction\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "Plot2a:=plot(aphi(t),t=0..Tend, color=blue, thickness=3, legend=\" acceleration in phi-direction\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "display(\{Plot1a,Plot2a\}, title=\"Figure 11\");" }}} {PARA 0 "" 0 "" {TEXT -1 32 "Remember that the origin of the " }{TEXT 379 1 "f" }{TEXT -1 1 "(" }{TEXT 380 1 "t" }{TEXT -1 4 ")-r(" }{TEXT 381 1 "t" }{TEXT -1 68 ")-coordinate system is at the starting point o f the motion. At time " }{TEXT 382 4 "Tend" }{TEXT -1 190 " the ball t ouches the ground, and the motion ends. You can see what happens when \+ the hill on which the man stands is higher or when he throws the ball \+ faster by changing the constants above." }}{PARA 0 "" 0 "" {TEXT -1 176 "The description of the motion in this example in polar coordinate s is not particularly useful, but it helps to understand the relations between cartesian and polar coordinates." }}}}{SECT 0 {PARA 3 "" 0 " " {TEXT -1 25 "2.1.3 Natural Coordinates" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "The last method for describing motion in this section is \+ to use " }{TEXT 517 19 "natural coordinates" }{TEXT -1 118 ". The traj ectory of the motion should be given in this case. The position of the partricle is given by the arc length " }{TEXT 392 1 "s" }{TEXT -1 30 ". Starting point is chosen at " }{TEXT 393 3 "s=0" }{TEXT -1 14 ". At the time " }{TEXT 394 1 "t" }{TEXT -1 33 " the position of the partic le is " }{TEXT 395 1 "s" }{TEXT -1 1 "(" }{TEXT 396 1 "t" }{TEXT -1 32 "). The tangential inital vector " }{TEXT 397 1 "e" }{TEXT -1 2 "t( " }{TEXT 398 1 "t" }{TEXT -1 51 ") points in the tangential direction \+ of increasing " }{TEXT 399 1 "s" }{TEXT -1 28 ". The normal initial ve ctor " }{TEXT 400 1 "e" }{TEXT -1 2 "n(" }{TEXT 401 1 "t" }{TEXT -1 111 ") points orthogonal to the tangential inital vector in the direct ion of the center of the circle (see Fig. 12)." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 71 "display(Fig_2_7()[1],scaling=constrained,axes= none, title=\"Figure 12\");" }}}{PARA 0 "" 0 "" {TEXT -1 56 "Now we co nsider two states of the motion, first at time " }{TEXT 402 1 "t" } {TEXT -1 20 " and second at time " }{TEXT 403 1 "t" }{TEXT -1 1 "+" } {TEXT 404 1 "D" }{TEXT 405 2 "t " }{TEXT -1 26 "(in the following figu res " }{TEXT 406 1 "D" }{TEXT -1 71 " is written \"Delta\"). The posit ions can be described by the arc length " }{TEXT 407 1 "s" }{TEXT -1 1 "(" }{TEXT 408 1 "t" }{TEXT -1 6 ") and " }{TEXT 409 1 "s" }{TEXT -1 1 "(" }{TEXT 431 2 "t+" }{TEXT 411 1 "D" }{TEXT -1 38 ") or as abov e by the position vectors " }{TEXT 410 1 "r" }{TEXT -1 1 "(" }{TEXT 416 1 "t" }{TEXT -1 7 ") and " }{TEXT 412 1 "r" }{TEXT -1 1 "(" } {TEXT 414 2 "t+" }{TEXT 413 1 "D" }{TEXT 415 1 "t" }{TEXT -1 60 "). Th e direction the particle is moving this time period is " }{TEXT 417 1 "D" }{TEXT 418 2 "s " }{TEXT -1 59 "(Delta_s in the figure). This situ ation is shown in Fig 13." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_2_7()[2],scaling=constrained,axes=none, title=\"Figure 13 \");" }}}{PARA 0 "" 0 "" {TEXT -1 133 "The velocity points in the dire ction of the trajectory of the motion, as was shown above. That means \+ that the velocity at every time " }{TEXT 419 1 "t" }{TEXT -1 28 " poin ts in the direction of " }{TEXT 420 1 "e" }{TEXT -1 2 "t(" }{TEXT 421 1 "t" }{TEXT -1 43 ") and has no component in the direction of " } {TEXT 422 1 "e" }{TEXT -1 2 "n(" }{TEXT 423 1 "t" }{TEXT -1 26 "). Wit h the derivative of " }{TEXT 450 1 "s" }{TEXT -1 22 " with respect to \+ time " }{TEXT 449 1 "t" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "sd(t ):=diff(s(t),t):" "6#>-%#sdG6#%\"tG-%%diffG6$-%\"sG6#F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 23 "we get for the velocity" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v(t):=sd(t)*et(t);" "6#>-%\"vG6#%\"tG*&-%#sdG6#F'\" \"\"-%#etG6#F'F," }}}{PARA 0 "" 0 "" {TEXT -1 125 "Next we want to der ive the acceleration of the motion of the particle. The derivative of \+ the velocity with respect to time is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "a(t):=diff(v(t),t);" "6#>-%\"aG6#%\"tG-%%diffG6$-%\"vG6 #F'F'" }}}{PARA 0 "" 0 "" {TEXT -1 80 "The question is at this point w hat happens after a change of the initial vector " }{TEXT 451 1 "e" } {TEXT -1 2 "t(" }{TEXT 452 1 "t" }{TEXT -1 43 ") in time. In Fig. 14 t he velocity vectors " }{TEXT 427 1 "v" }{TEXT -1 1 "(" }{TEXT 428 1 "t " }{TEXT -1 6 ") and " }{TEXT 429 1 "v" }{TEXT -1 2 "( " }{TEXT 424 1 "t" }{TEXT -1 1 "+" }{TEXT 425 1 "D" }{TEXT 426 1 "t" }{TEXT -1 12 ") \+ are shown." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_2_ 7()[3],scaling=constrained,axes=none, title=\"Figure 14\");" }}}{PARA 0 "" 0 "" {TEXT -1 77 "The radius of curvature for both considered tim e points are drawn in Fig. 15." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_2_7()[4],scaling=constrained,axes=none, title=\"Figu re 15\");" }}}{PARA 0 "" 0 "" {TEXT -1 47 "We relocate the vector of t he velocity at time " }{TEXT 430 1 "t" }{TEXT -1 23 " parallel to the \+ point " }{TEXT 432 1 "s" }{TEXT -1 1 "(" }{TEXT 434 2 "t+" }{TEXT 433 1 "D" }{TEXT -1 11 ") (Fig. 16)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_2_7()[5],scaling=constrained,axes=none, title=\"Fi gure 16\");" }}}{PARA 0 "" 0 "" {TEXT -1 44 "Next we replace the vecto rs of the velocity " }{TEXT 446 1 "v" }{TEXT -1 1 "(" }{TEXT 445 1 "t " }{TEXT -1 7 ") and " }{TEXT 441 1 "v" }{TEXT -1 1 "(" }{TEXT 443 2 "t+" }{TEXT 442 1 "D" }{TEXT 444 1 "t" }{TEXT -1 26 ") by the initial \+ vectors " }{TEXT 435 1 "e" }{TEXT -1 2 "t(" }{TEXT 440 1 "t" }{TEXT -1 7 ") and " }{TEXT 436 1 "e" }{TEXT -1 2 "t(" }{TEXT 438 2 "t+" } {TEXT 437 1 "D" }{TEXT 439 1 "t" }{TEXT -1 46 "). Additionally we incl ude the initial vector " }{TEXT 447 1 "e" }{TEXT -1 2 "n(" }{TEXT 448 1 "t" }{TEXT -1 16 ") (see Fig. 17)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_2_7()[6],scaling=constrained,axes=none, t itle=\"Figure 17\");" }}}{PARA 0 "" 0 "" {TEXT -1 68 "The changing of \+ the tangential initial vector while the motion from " }{TEXT 456 1 "s " }{TEXT -1 1 "(" }{TEXT 457 1 "t" }{TEXT -1 5 ") to " }{TEXT 458 1 "s " }{TEXT -1 1 "(" }{TEXT 460 2 "t+" }{TEXT 459 1 "D" }{TEXT 461 1 "t" }{TEXT -1 15 ") is given by " }{TEXT 467 1 "e" }{TEXT -1 2 "t(" } {TEXT 454 2 "t+" }{TEXT 453 1 "D" }{TEXT 455 1 "t" }{TEXT -1 2 ")-" } {TEXT 462 1 "e" }{TEXT -1 2 "t(" }{TEXT 463 1 "t" }{TEXT -1 16 ") (see Fig. 18)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(Fig_2_ 7()[7],scaling=constrained,axes=none, title=\"Figure 18\");" }}}{PARA 0 "" 0 "" {TEXT -1 34 "Now we consider the situation for " }{TEXT 464 1 "D" }{TEXT 465 7 "t -> 0." }{TEXT -1 118 " The triangles (in Fig. 19 painted cyan and red) are similar (that means that the angles in both triangles are equal)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "di splay(Fig_2_7()[8],scaling=constrained,axes=none, title=\"Figure 19\") ;" }}}{PARA 0 "" 0 "" {TEXT -1 26 "So we can see the relation" }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "REL_1 := abs(et(t+Delta_t)-et( t))/abs(et(t)) = Delta_s/rho(t):" "6#>%&REL_1G/*&-%$absG6#,&-%#etG6#,& %\"tG\"\"\"%(Delta_tGF0F0-F,6#F/!\"\"F0-F(6#-F,6#F/F4*&%(Delta_sGF0-%$ rhoG6#F/F4" }}}{PARA 0 "" 0 "" {TEXT -1 30 "Notice that the angle betw een " }{TEXT 466 1 "e" }{TEXT -1 2 "t(" }{TEXT 472 1 "t" }{TEXT -1 7 " ) and " }{TEXT 468 1 "e" }{TEXT -1 2 "t(" }{TEXT 470 2 "t+" }{TEXT 469 1 "D" }{TEXT 471 1 "t" }{TEXT -1 14 ") and between " }{TEXT 473 1 "r" }{TEXT -1 1 "(" }{TEXT 475 2 "t+" }{TEXT 474 1 "D" }{TEXT 476 1 "t " }{TEXT -1 6 ") and " }{TEXT 477 2 " r" }{TEXT -1 1 "(" }{TEXT 479 2 "t+" }{TEXT 478 1 "D" }{TEXT 480 1 "t" }{TEXT -1 20 ") for infinitesim al " }{TEXT 481 1 "D" }{TEXT 482 1 "t" }{TEXT -1 32 " is infinitesimal , too. The arc " }{TEXT 483 1 "D" }{TEXT 484 2 "s " }{TEXT -1 31 "is t hen approximately straight." }}{PARA 0 "" 0 "" {TEXT -1 68 "The absolu te value of all initial vectors, especially the vector et(" }{TEXT 485 1 "t" }{TEXT -1 7 ") is 1." }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "REL_2:=abs(et(t))=1:" "6#>%&REL_2G/-%$absG6#-%#etG6#%\"tG\"\"\" " }}}{PARA 0 "" 0 "" {TEXT -1 46 "The changing of the tangential initi al vector " }{TEXT 486 1 "e" }{TEXT -1 2 "t(" }{TEXT 487 1 "t" }{TEXT -1 15 ") is defined by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff (et(t),t) = limit((et(t+Delta_t)-et(t))/Delta_t,Delta_t = 0):" "6#/-%% diffG6$-%#etG6#%\"tGF*-%&limitG6$*&,&-F(6#,&F*\"\"\"%(Delta_tGF3F3-F(6 #F*!\"\"F3F4F7/F4\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 4 "For " }{TEXT 490 1 "D" }{TEXT 491 6 "t -> 0" }{TEXT -1 18 " the direction of " } {TEXT 497 1 "e" }{TEXT -1 2 "t(" }{TEXT 493 2 "t+" }{TEXT 492 1 "D" } {TEXT 494 1 "t" }{TEXT -1 2 ")-" }{TEXT 495 1 "e" }{TEXT -1 2 "t(" } {TEXT 496 1 "t" }{TEXT -1 22 ") is the direction of " }{TEXT 498 1 "e " }{TEXT -1 2 "n(" }{TEXT 499 1 "t" }{TEXT -1 13 "). So we get " }} {EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(et(t),t) = limit(abs(et(t +Delta_t)-et(t))/Delta_t,Delta_t = 0)*en(t):" "6#/-%%diffG6$-%#etG6#% \"tGF**&-%&limitG6$*&-%$absG6#,&-F(6#,&F*\"\"\"%(Delta_tGF7F7-F(6#F*! \"\"F7F8F;/F8\"\"!F7-%#enG6#F*F7" }}}{PARA 0 "" 0 "" {TEXT -1 28 "Toge ther with the relations " }{TEXT 489 5 "REL_1" }{TEXT -1 5 " and " } {TEXT 488 5 "REL_2" }{TEXT -1 7 " we get" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(et(t),t) = limit(Delta_s/Delta_t/rho(t),Delta_t = \+ 0)*en(t):" "6#/-%%diffG6$-%#etG6#%\"tGF**&-%&limitG6$*(%(Delta_sG\"\" \"%(Delta_tG!\"\"-%$rhoG6#F*F3/F2\"\"!F1-%#enG6#F*F1" }}}{PARA 0 "" 0 "" {TEXT -1 8 "And with" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "lim it(Delta_s/Delta_t,Delta_t=0) = diff(s(t),t):" "6#/-%&limitG6$*&%(Delt a_sG\"\"\"%(Delta_tG!\"\"/F*\"\"!-%%diffG6$-%\"sG6#%\"tGF4" }}}{PARA 0 "" 0 "" {TEXT -1 7 "we get " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "RES:=diff(et(t),t) = diff(s(t),t)/rho(t)*en(t);" }}}{PARA 0 "" 0 "" {TEXT -1 6 "Using " }{TEXT 500 3 "RES" }{TEXT -1 38 " we obtain f or the acceleration vector" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "a(t):=subs(RES,a(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "The \+ relations for the velocity and the acceleration vector are included in the " }{HYPERLNK 17 "dynamics" 2 "dynamics" "" }{TEXT -1 15 " package . See " }{HYPERLNK 17 "natural_vel" 2 "natural_vel" "" }{TEXT -1 2 ", " }{HYPERLNK 17 "natural_acc" 2 "natural_acc" "" }{TEXT -1 35 " to se e how to use these functions." }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 50 "Example: Horizontal Throw of a Ball (continuation)" }}{PARA 0 "" 0 " " {TEXT -1 61 "As an example we consider again the man who throws the ball." }}{PARA 0 "" 0 "" {TEXT -1 75 "First we unassign the variables to describe the results in a general manner" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "unassign('h','v0','g');" }}}{PARA 0 "" 0 "" {TEXT -1 60 " Above we had calculated for the position vector of the ball" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "r(t) := vector([v0*t, 1/2*g *t^2, 0]);" }}}{PARA 0 "" 0 "" {TEXT -1 103 "The position of the ball \+ is in natural coordinates described by the arc length. For a given fun ction f(" }{TEXT 501 1 "x" }{TEXT -1 36 ") the arc length between the \+ points " }{TEXT 502 2 "s1" }{TEXT -1 5 " and " }{TEXT 503 2 "s2" } {TEXT -1 12 " is given by" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "s = int(sq rt(1+diff(f(x),x)^2),x = s1 .. s2);" "6#/%\"sG-%$intG6$-%%sqrtG6#,&\" \"\"F,*$-%%diffG6$-%\"fG6#%\"xGF4\"\"#F,/F4;%#s1G%#s2G" }}{PARA 0 "" 0 "" {TEXT -1 118 "Additionally we need the radius of curvature of the trajectory for every position of the ball and so for every time t." } }{PARA 0 "" 0 "" {TEXT -1 16 "This is given by" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "rho=sqrt((1+diff(f(x),x)**2)**3)/diff(f(x),x$2):" "6#/% $rhoG*&-%%sqrtG6#*$,&\"\"\"F+*$-%%diffG6$-%\"fG6#%\"xGF3\"\"#F+\"\"$F+ -F.6$-F16#F3-%\"$G6$F3F4!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 76 "In the e xample we know the function of the trajectory as a function of time." }}{PARA 0 "" 0 "" {TEXT -1 4 "With" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "x(t):=r(t)[1];" }}}{PARA 0 "" 0 "" {TEXT -1 3 "and" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "y(t):=r(t)[2];" }}}{PARA 0 "" 0 "" {TEXT -1 7 "we get " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "y(x):=subs(t=solve(x(t)=x,t),y(t));" }}}{PARA 0 "" 0 "" {TEXT -1 29 "Remember that the coordinate " }{TEXT 504 2 "y " }{TEXT -1 17 "poi nts downwards." }}{PARA 0 "" 0 "" {TEXT -1 43 "For the arc length of t he trajectory we get" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "s(xi) \+ := expand(int(sqrt(1+diff(y(x),x)^2),x = 0 .. xi));" "6#>-%\"sG6#%#xiG -%'expandG6#-%$intG6$-%%sqrtG6#,&\"\"\"F2*$-%%diffG6$-%\"yG6#%\"xGF:\" \"#F2/F:;\"\"!F'" }}}{PARA 0 "" 0 "" {TEXT -1 51 "And the relation for the radius of curvature yields" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rho(x) := sqrt((1+diff(y(x),x)^2)^3)/diff(y(x),x$2);" "6#>-%$rho G6#%\"xG*&-%%sqrtG6#*$,&\"\"\"F.*$-%%diffG6$-%\"yG6#F'F'\"\"#F.\"\"$F. -F16$-F46#F'-%\"$G6$F'F6!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 28 "Now we \+ replace the variable " }{TEXT 505 1 "x" }{TEXT -1 3 "by " }{TEXT 507 1 "x" }{TEXT -1 1 "(" }{TEXT 506 1 "t" }{TEXT -1 1 ")" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "s(t):=subs(xi=x(t),s(xi));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "rho(t):=subs(x=x(t),rho(x));" }}} {PARA 0 "" 0 "" {TEXT -1 53 "We get for the velocity vector in natural coordinates" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "v_natural(t) :=natural_vel(s(t),t);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "and for the ac celeration" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "a_natural(t):= natural_acc(s(t),rho(t),t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "W e don't look whether this expressions can be simplified but we use the same values as above for the example" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "h:=10:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "v0 :=10:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=9.81:" }}}{PARA 0 "" 0 "" {TEXT -1 42 "With this values we get for the trajectory" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "s(t):=eval(s(t));" }}}{PARA 0 "" 0 "" {TEXT -1 17 "and the curvature" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "rho(t):=eval(rho(t));" }}}{PARA 0 "" 0 "" {TEXT -1 28 "Then we get for the velocity" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "vt(t):=simplify(eval(v_natural(t)[1]));" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 29 "vn(t):=eval(v_natural(t)[2]);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "and for the acceleration" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 39 "at(t):=simplify(eval(a_natural(t)[1]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "an(t):=simplify(eval(a_natural(t)[2 ]));" }}}{PARA 0 "" 0 "" {TEXT -1 29 "Fig. 20 shows the trajectory " } {TEXT 511 1 "s" }{TEXT -1 1 "(" }{TEXT 512 1 "t" }{TEXT -1 59 ") of th e motion of the ball, Fig.9 the radius of curvature " }{TEXT 509 1 "r " }{TEXT -1 1 "(" }{TEXT 510 1 "t" }{TEXT -1 28 ") in dependence of th e time " }{TEXT 513 1 "t" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 " Plot1n:=plot(s(t),t=0..Tend, color=green, thickness=3, legend=\"s(t)\" ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "Plot2n:=plot(rho(t),t =0..Tend, color=blue, thickness=3, legend=\"rho(t)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "display(\{Plot1n\}, title=\"Figure \+ 20\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "display(\{Plot2n \}, title=\"Figure 21\");" }}}{PARA 0 "" 0 "" {TEXT -1 89 "This result of the calculation is shown in Fig. 22 (velocity) and Fig. 23 (accele ration)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "Plot1nv:=plot(vt (t),t=0..Tend, color=green, thickness=3, legend=\"velocity in t-direct ion\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "Plot2nv:=plot(vn (t),t=0..Tend, color=blue, thickness=3, legend=\"velocity in n-directi on\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "display(\{Plot1nv ,Plot2nv\}, title=\"Figure 22\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "Plot1na:=plot(at(t),t=0..Tend, color=green, thickness =3, legend=\"acceleration in t-direction\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "Plot2na:=plot(an(t),t=0..Tend, color=blue, thick ness=3, legend=\"acceleration in n-direction\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "display(\{Plot1na,Plot2na\}, title=\"Figure 2 3\");" }}}{PARA 0 "" 0 "" {TEXT -1 8 "At time " }{TEXT 508 4 "Tend" } {TEXT -1 189 " the ball touches the ground and the motion ends. You ca n see what happens when the hill on which the man stands is higher or \+ when he throws the ball faster by changing the constants above." }} {PARA 0 "" 0 "" {TEXT -1 316 "As with the description of the motion in this example in polar coordinates, it makes no sense in general to d escribe the solution in natural coordinates, but it helps to understan d the relations between cartesian and natural coordinates. Natural coo rdinates are useful if the trajectory of the motion is prescribed." }} }}{PARA 3 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 15 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }