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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 30 "Classical Mechanics with \+
Maple" }}{PARA 256 "" 0 "" {TEXT -1 38 "Section 4.1: Systems of Mass P
articles" }}{PARA 19 "" 0 "" {TEXT -1 41 "Dr. Harald Kammerer\nmaple@j
ademountain.de" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 14 "Initialisation
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "libname:=\"C:/mylib/m6dynlib\",\"C:
/mylib/m6dynfig\",libname:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
80 "with(linalg):with(plots):with(plottools):with(dynamics);with(figur
es_chapter_4);" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 29 "4.1 Systems o
f Mass Particles" }}{PARA 0 "" 0 "" {TEXT -1 105 "Later in this sectio
n we consider the motions of rigid bodies. To transition from the mass
 particles in  " }{HYPERLNK 17 "section 2" 1 "sec-02.mws" "" }{TEXT 
-1 5 " and " }{HYPERLNK 17 "section 3" 1 "sec-03.mws" "" }{TEXT -1 
650 " we consider now systems of mass particles. We can imagine every \+
real rigid body as a system of mass particles which are held together \+
by inner forces. In the same manner we can consider the system of the \+
earth and the moon as a system of two mass particles that are hold tog
ether by gravitational forces. In relation to the distance between the
m their own dimensions are approximately negligible. The following cal
culations are valid for all kinds of systems of mass particles, especi
ally when the mass particles are close together in the case of rigid b
odies. So the following considerations are very useful for later calcu
lations for rigid bodies." }}{SECT 0 {PARA 5 "" 0 "4.1.1" {TEXT -1 24 
"4.1.1 The Center of Mass" }}{PARA 0 "" 0 "" {TEXT -1 119 "We consider
 a group of some mass particles, which are held together by rigid or e
lastic boundaries, as shown in Fig. 1." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 69 "display(Fig_4_1()[1],scaling=constrained,axes=none,ti
tle=\"Figure 1\");" }}}{PARA 0 "" 0 "" {TEXT -1 113 "When we cut the b
oundaries we have to add the corresponding forces between the mass par
ticles, as shown in Fig. 2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
69 "display(Fig_4_1()[2],scaling=constrained,axes=none,title=\"Figure \+
2\");" }}}{PARA 0 "" 0 "" {TEXT -1 47 "Now we consider two neighboring
 mass particles " }{XPPEDIT 18 0 "m[i];" "6#&%\"mG6#%\"iG" }{TEXT -1 
5 " and " }{XPPEDIT 18 0 "m[j];" "6#&%\"mG6#%\"jG" }{TEXT -1 33 ". Acc
ording to Newton's Third Law" }{TEXT 256 2 ", " }{XPPEDIT 18 0 "F[j,i]
;" "6#&%\"FG6$%\"jG%\"iG" }{XPPEDIT 18 0 "m[j];" "6#&%\"mG6#%\"jG" }
{XPPEDIT 18 0 "m[i];" "6#&%\"mG6#%\"iG" }{XPPEDIT 18 0 "F[i,j]" "6#&%
\"FG6$%\"iG%\"jG" }{XPPEDIT 18 0 "m[i]" "6#&%\"mG6#%\"iG" }{XPPEDIT 
18 0 "m[j]" "6#&%\"mG6#%\"jG" }{TEXT -1 1 " " }{XPPEDIT 18 0 "F[j,i]+F
[i,j] = 0;" "6#/,&&%\"FG6$%\"jG%\"iG\"\"\"&F&6$F)F(F*\"\"!" }{TEXT -1 
13 " (see Fig. 3)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "display
(Fig_4_1()[3],scaling=constrained,axes=none,title=\"Figure 3\");" }}}
{PARA 0 "" 0 "" {TEXT -1 23 "Additionally we define " }{XPPEDIT 18 0 "
F[i,i] = 0;" "6#/&%\"FG6$%\"iGF'\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 
"" {TEXT -1 45 "We consider now the freely cut mass particle " }
{XPPEDIT 18 0 "m[i];" "6#&%\"mG6#%\"iG" }{TEXT -1 53 ". The resultant \+
of all acting active forces on it is " }{XPPEDIT 18 0 "F[i];" "6#&%\"F
G6#%\"iG" }{TEXT -1 55 ". The position vector of the mass particle is \+
given by " }{XPPEDIT 18 0 "r[i];" "6#&%\"rG6#%\"iG" }{TEXT -1 34 " and
 its acceleration is given by " }{XPPEDIT 18 0 "r_dd[i]" "6#&%%r_ddG6#
%\"iG" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "r_dd[i] := diff(r[i],`$`(t
,2));" "6#>&%%r_ddG6#%\"iG-%%diffG6$&%\"rG6#F'-%\"$G6$%\"tG\"\"#" }
{TEXT -1 33 ". From Newton's Second Law we get" }}{PARA 257 "" 0 "" 
{XPPEDIT 18 0 "m[i]*r_dd[i] = F[i]+sum(F[j,i],j = 1 .. n):" "6#/*&&%\"
mG6#%\"iG\"\"\"&%%r_ddG6#F(F),&&%\"FG6#F(F)-%$sumG6$&F/6$%\"jGF(/F6;F)
%\"nGF)" }}{PARA 0 "" 0 "" {TEXT -1 43 "When we consider all mass part
icles we get " }{TEXT 257 1 "n" }{TEXT -1 48 " such equations. We suma
rize all of them and get" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "sum(m[i]*
r_dd[i],i=1..n)=sum(F[i],i=1..n)+sum(sum(F[j,i],j=1..n),i=1..n);" "6#/
-%$sumG6$*&&%\"mG6#%\"iG\"\"\"&%%r_ddG6#F+F,/F+;F,%\"nG,&-F%6$&%\"FG6#
F+/F+;F,F2F,-F%6$-F%6$&F76$%\"jGF+/FA;F,F2/F+;F,F2F," }}{PARA 0 "" 0 "
" {TEXT -1 77 "We know from the statics about the center of gravity of
 a system of particles" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "r[s]=sum(m[
i]*r[i],i=1..n)/sum(m[i],i=1..n):" "6#/&%\"rG6#%\"sG*&-%$sumG6$*&&%\"m
G6#%\"iG\"\"\"&F%6#F0F1/F0;F1%\"nGF1-F*6$&F.6#F0/F0;F1F6!\"\"" }}
{PARA 0 "" 0 "" {TEXT -1 34 "with the total mass of the system " }
{XPPEDIT 18 0 "sum(m[i],i=1..n)=m[ges]" "6#/-%$sumG6$&%\"mG6#%\"iG/F*;
\"\"\"%\"nG&F(6#%$gesG" }{TEXT -1 50 " and the position vector of the \+
center of gravity " }{XPPEDIT 18 0 "r[s]" "6#&%\"rG6#%\"sG" }{TEXT -1 
58 ". The acceleration of the center of gravity is defined by " }
{XPPEDIT 18 0 "r_dd[s]=diff(r[s],t$2)" "6#/&%%r_ddG6#%\"sG-%%diffG6$&%
\"rG6#F'-%\"$G6$%\"tG\"\"#" }{TEXT -1 53 ". This yields for the left s
ide of the equation above" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "sum(m[i]
*r_dd[i],i=1..n)=diff(sum(m[i]*r[i],i=1..n),t$2);" "6#/-%$sumG6$*&&%\"
mG6#%\"iG\"\"\"&%%r_ddG6#F+F,/F+;F,%\"nG-%%diffG6$-F%6$*&&F)6#F+F,&%\"
rG6#F+F,/F+;F,F2-%\"$G6$%\"tG\"\"#" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 
"sum(m[i]*r_dd[i],i = 1 .. n) = m[ges]*r_dd[s]" "6#/-%$sumG6$*&&%\"mG6
#%\"iG\"\"\"&%%r_ddG6#F+F,/F+;F,%\"nG*&&F)6#%$gesGF,&F.6#%\"sGF," }}
{PARA 0 "" 0 "" {TEXT -1 43 "On the right side of the equation we inse
rt" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "sum(F[i],i=1..n)=F[res]" "6#/-%
$sumG6$&%\"FG6#%\"iG/F*;\"\"\"%\"nG&F(6#%$resG" }}{PARA 0 "" 0 "" 
{TEXT -1 6 "where " }{XPPEDIT 18 0 "F[res]" "6#&%\"FG6#%$resG" }{TEXT 
-1 83 " is the resultant of all active forces that are acting from the
 outside on all the " }{TEXT 258 1 "n" }{TEXT -1 16 " mass particles.
" }}{PARA 0 "" 0 "" {TEXT -1 135 "At last we have the double sum on th
e right side of the equation. This double sum is zero, because all inn
er forces are annul by pairs." }}{PARA 0 "" 0 "" {TEXT -1 14 "So we ge
t the " }{TEXT 259 36 "first axiom of the center of gravity" }{TEXT 
-1 3 " as" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "m[ges]*r_dd[s]=F[res]" "
6#/*&&%\"mG6#%$gesG\"\"\"&%%r_ddG6#%\"sGF)&%\"FG6#%$resG" }}{PARA 0 "
" 0 "" {TEXT -1 195 "This means that we can consider the motion of the
 center of gravity of a system of mass particles as the motion of a ma
ss particle with the mass equal to the sum of all the single mass part
icles." }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 37 "Example: Motion of A Ca
r With Trailer" }}{PARA 0 "" 0 "" {TEXT -1 174 "Now we consider the mo
tion of a car with a trailer. Of course we consider a simplified model
. Both the car and the trailer are considered as mass particles. The c
ar has mass " }{TEXT 260 2 "m1" }{TEXT -1 18 ", the trailer mass" }
{TEXT 261 3 " m2" }{TEXT -1 157 ". Both are moving without friction on
 the road. The interconnection between car and trailer is elastic and \+
described by an elastic spring with the stiffness " }{TEXT 262 1 "k" }
{TEXT -1 35 ". In Fig. 4 the situation is shown." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 69 "display(Fig_4_2()[1],scaling=constrained,axes=
none,title=\"Figure 4\");" }}}{PARA 0 "" 0 "" {TEXT -1 54 "The distanc
e between car and trailer is defined to be " }{TEXT 263 1 "L" }{TEXT 
-1 56 " when the spring is free of stress The Position of mass " }
{TEXT 264 2 "m1" }{TEXT -1 17 " is described by " }{TEXT 265 2 "x1" }
{TEXT -1 22 ", the position of mass" }{TEXT 266 3 " m2" }{TEXT -1 3 " \+
by" }{TEXT 267 3 " x2" }{TEXT -1 56 ". The position of the center of g
ravity is described by " }{TEXT 269 3 "xs." }{TEXT -1 33 " The acceler
ations are given by  " }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "a1=diff(x1,t
$2):" "6#/%#a1G-%%diffG6$%#x1G-%\"$G6$%\"tG\"\"#" }}{PARA 257 "" 0 "" 
{XPPEDIT 18 0 "a2=diff(x2,t$2):" "6#/%#a2G-%%diffG6$%#x2G-%\"$G6$%\"tG
\"\"#" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 257 "
" 0 "" {XPPEDIT 18 0 "as = diff(xs,`$`(t,2)):" "6#/%#asG-%%diffG6$%#xs
G-%\"$G6$%\"tG\"\"#" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 219 "Fi
rst we use the axiom above to describe the motion of the center of gra
vity of the total system. Because there are no active forces in the ho
rizontal direction we get for the equation of motion of the center of \+
gravity" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eoms:=(m1+m2)*as=0:
" "6#>%%eomsG/*&,&%#m1G\"\"\"%#m2GF)F)%#asGF)\"\"!" }}}{PARA 0 "" 0 "
" {TEXT -1 338 "Dependent on the initial conditions is the center of g
ravity of the system moving with constant velocity. The inner force, h
ere the spring force, has no influence on the motion of the center of \+
gravity. We only need to consider it when we want to describe the moti
on of the separate mass particles. Cutting both masses free yields Fig
. 5." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "display(Fig_4_2()[2]
,scaling=constrained,axes=none,title=\"Figure 5\");" }}}{PARA 0 "" 0 "
" {TEXT -1 36 "Here the spring force is denoted by " }{TEXT 268 1 "F" 
}{TEXT -1 39 ". So we get the two equations of motion" }}{EXCHG {PARA 
0 "> " 0 "" {XPPEDIT 19 1 "eom1:=m1*a1+F=0:" "6#>%%eom1G/,&*&%#m1G\"\"
\"%#a1GF)F)%\"FGF)\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eo
m2:=m2*a2-F=0:" "6#>%%eom2G/,&*&%#m2G\"\"\"%#a2GF)F)%\"FG!\"\"\"\"!" }
}}{PARA 0 "" 0 "" {TEXT -1 102 "In the situation that the spring is un
stressed the distance between both mass particles is defined as " }
{TEXT 270 1 "L" }{TEXT -1 36 ". The relative displacement between " }
{TEXT 271 3 "m1 " }{TEXT -1 4 "and " }{TEXT 272 2 "m2" }{TEXT -1 12 " \+
is given by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs1:=u=x1-x2-
L:" "6#>%&subs1G/%\"uG,(%#x1G\"\"\"%#x2G!\"\"%\"LGF+" }}}{PARA 0 "" 0 
"" {TEXT -1 41 "and the force law for the spring force is" }}{EXCHG 
{PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs2 := F = k*u;" "6#>%&subs2G/%\"FG
*&%\"kG\"\"\"%\"uGF)" }}}{PARA 0 "" 0 "" {TEXT -1 42 "We multiply both
 equations of motion with " }{TEXT 273 2 "m2" }{TEXT -1 19 " respectiv
ely with " }{TEXT 274 2 "m1" }{TEXT -1 50 " and calculate the differen
ce of both. This yields" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "e
omr:=eom1*m2-eom2*m1;" }}}{PARA 0 "" 0 "" {TEXT -1 82 "After some rear
rangement and substituting the relation for the spring force we get" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "eomr1:=subs(subs2,sort(sort
(expand(eomr/m1/m2),a1),a1));" }}}{PARA 0 "" 0 "" {TEXT -1 5 "From " }
{TEXT 275 5 "subs1" }{TEXT -1 62 " we get for the relative acceleratio
n between car and trailer " }{TEXT 276 3 "udd" }}{EXCHG {PARA 0 "> " 
0 "" {XPPEDIT 19 1 "subs3 := a1-a2 = udd:" "6#>%&subs3G/,&%#a1G\"\"\"%
#a2G!\"\"%$uddG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "eomr2:=c
ollect(expand(algsubs(subs3,eomr1)),u);" }}}{PARA 0 "" 0 "" {TEXT -1 
89 "This equation is a differential equation of degree 2 with respect \+
to the relative motion " }{TEXT 277 1 "u" }{TEXT -1 377 " between both
 mass particles. At this point, we don't want to go into detail about \+
the solution of this kind of differential equation. But notice that th
is is a very important kind of differential equation which describes t
he nature of vibrations. Whithout special indication we had considered
 the linearized equation of motion of the mathematical pendulum and it
s solution in " }{HYPERLNK 17 "section 3" 1 "sec-03.mws" "" }{TEXT -1 
180 ". Because we use MAPLE for this course we can complete these exam
ple without supposing that everyone has the necessary mathematical bac
kround to understand this solution in detail." }}{PARA 0 "" 0 "" 
{TEXT -1 41 "To solve the equation above we substitute" }}{EXCHG 
{PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs4:=udd=diff(u(t),t$2):" "6#>%&sub
s4G/%$uddG-%%diffG6$-%\"uG6#%\"tG-%\"$G6$F-\"\"#" }}}{PARA 0 "" 0 "" 
{TEXT -1 3 "and" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs5:=u=u(
t):" "6#>%&subs5G/%\"uG-F&6#%\"tG" }}}{PARA 0 "" 0 "" {TEXT -1 19 "Fin
ally we use the " }{TEXT 285 25 "natural angular frequency" }}{EXCHG 
{PARA 0 "> " 0 "" {XPPEDIT 19 1 "subs6 := coeff(lhs(eomr2),u) = omega^
2;" "6#>%&subs6G/-%&coeffG6$-%$lhsG6#%&eomr2G%\"uG*$%&omegaG\"\"#" }}}
{PARA 0 "" 0 "" {TEXT -1 17 "Then we get from " }{TEXT 278 5 "eomr2" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "eomr3:=subs(\{subs4,subs5,s
ubs6\},eomr2);" }}}{PARA 0 "" 0 "" {TEXT -1 137 "Additionally we presu
me that we know the initial contitions of the motion. Let the relative
 displacement between car and trailer at time " }{TEXT 279 3 "t=0" }
{TEXT -1 4 " be " }{TEXT 280 2 "u0" }{TEXT -1 40 " and the relative ve
locity at this time " }{TEXT 281 3 "ud0" }{TEXT -1 15 ". Then we write
" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ic1:=u(0)=u0:" "6#>%$ic1G/
-%\"uG6#\"\"!%#u0G" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "ic2:=D(
u)(0)=ud0:" "6#>%$ic2G/--%\"DG6#%\"uG6#\"\"!%$ud0G" }}}{PARA 0 "" 0 "
" {TEXT -1 187 "The next step is done by the appropriate MAPLE-command
 and we don't look at this process in detail. Please see the MAPLE Hel
p for details about the use of the operator D and the function " }
{TEXT 282 7 "dsolve." }}{PARA 0 "" 0 "" {TEXT -1 84 "We get the soluti
on for the relative motion by considerating the initial conditions " }
{TEXT 283 3 "ic1" }{TEXT -1 5 " and " }{TEXT 284 3 "ic2" }{TEXT -1 1 "
:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "sol:=dsolve(\{eomr3,ic1
,ic2\},u(t));" }}}{PARA 0 "" 0 "" {TEXT -1 182 "We see that the motion
 between car and trailer is a vibration. In reality we have this pheno
menon for example in trains. The use of dampers reduces the vibrations
 between the wagons." }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 24 "The Case
 of Rigid Bodies" }}{PARA 0 "" 0 "" {TEXT -1 91 "When a system has con
stant distances between all mass particles, we denote the system as a \+
" }{TEXT 286 10 "rigid body" }{TEXT -1 268 ". This means that the axio
m of the center of gravity is also valid for rigid bodies. In such rig
id bodies the center of gravity is fixed. Now we see that we made in t
he previous considerations no error, because we considered always the \+
motion of the center of gravity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 35 "In general of course the resultant " }
{TEXT 287 5 "Fres " }{TEXT -1 66 "of all active forces doesn't point t
hrough the center of gravity (" }{TEXT 288 3 "cog" }{TEXT -1 163 "). B
ut we can replace every resultant which doesn't point through the cent
er of gravity by a parallel force trough the center of gravity and an \+
appropriate moment " }{TEXT 289 4 "Mres" }{TEXT -1 20 ", as shown in F
ig. 6" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "display(Fig_4_3(),s
caling=constrained,axes=none,title=\"Figure 6\");" }}}{PARA 0 "" 0 "" 
{TEXT -1 69 "The next section deals with the effect of this moment on \+
rigid bodies" }}}}{SECT 0 {PARA 5 "" 0 "4.1.2" {TEXT -1 55 "4.1.2 The \+
Angular Momentum of Systems of Mass Particles" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 14 "unassign('F'):" }}}{PARA 0 "" 0 "" {TEXT -1 3 "I
n " }{HYPERLNK 17 "section 3.3" 1 "sec-03.mws" "" }{TEXT -1 234 " we c
onsidered the linear momentum. Now we consider the case that there act
s not only a resultant force on the center of gravity of a mass partic
le or a system of those but also a resultant moment (rotation), as sho
wn in Fig. 6 above." }}{PARA 0 "" 0 "" {TEXT -1 70 "At first we have a
 look at the effect of a moment and the motion of a " }{TEXT 302 20 "s
ingle mass particle" }{TEXT -1 37 ". In Fig. 7 we see the mass particl
e " }{TEXT 290 1 "m" }{TEXT -1 26 " with its position vector " }{TEXT 
292 1 "r" }{TEXT -1 49 " with respect to the inertial system with orig
in " }{TEXT 291 1 "O" }{TEXT -1 224 ". We consider here the special ca
se of planar motion. We can do this by choosing the coordinate system \+
accordingly. This makes the representation concise. But the following \+
calculations are also valid for special situations." }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 66 "display(Fig_4_4(),scaling=constrained,axes=
none,title=\"Figure 7\");" }}}{PARA 0 "" 0 "" {TEXT -1 39 "The positio
n is described by the vector" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 
1 "r:=vector(3,[rx,ry,0]);" "6#>%\"rG-%'vectorG6$\"\"$7%%#rxG%#ryG\"\"
!" }}}{PARA 0 "" 0 "" {TEXT -1 10 "The force " }{TEXT 306 1 "F" }
{TEXT -1 27 " acts on the mass particle " }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "F:=vector(3,[Fx,Fy,0]);" "6#>%\"FG-%'vectorG6$\"\"$7%%#
FxG%#FyG\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 15 "Then the moment" }}
{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Mo:=crossprod(r,F);" "6#>%#MoG
-%*crossprodG6$%\"rG%\"FG" }}}{PARA 0 "" 0 "" {TEXT -1 64 "is the mome
nt on the mass particle acting force with respect to " }{TEXT 293 1 "O
" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 30 "Now we take Newtons s
econd law" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "Newton2:=F=m*rdd:
" "6#>%(Newton2G/%\"FG*&%\"mG\"\"\"%$rddGF)" }}}{PARA 0 "" 0 "" {TEXT 
-1 50 "with the second derivative of the position vector " }{TEXT 294 
1 "r" }{TEXT -1 22 " with respect to time " }{XPPEDIT 18 0 "rdd=diff(r
,t$2):" "6#/%$rddG-%%diffG6$%\"rG-%\"$G6$%\"tG\"\"#" }{TEXT -1 1 "." }
}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rdd:=vector(3,[rxdd,rydd,0]);
" "6#>%$rddG-%'vectorG6$\"\"$7%%%rxddG%%ryddG\"\"!" }}}{PARA 0 "" 0 "
" {TEXT -1 45 "Accordingly we write for the velocity vector " }
{XPPEDIT 18 0 "rd = diff(r,t):" "6#/%#rdG-%%diffG6$%\"rG%\"tG" }}
{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rd:=vector(3,[rxd,ryd,0]);" "6
#>%#rdG-%'vectorG6$\"\"$7%%$rxdG%$rydG\"\"!" }}}{PARA 0 "" 0 "" {TEXT 
-1 64 "Taking the cross product of  Newtons second law with the vector
 " }{TEXT 295 0 "" }{TEXT -1 0 "" }{TEXT 296 1 "r" }{TEXT -1 7 " yield
s" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "crossprod(r,m*rdd)=Mo;" "
6#/-%*crossprodG6$%\"rG*&%\"mG\"\"\"%$rddGF*%#MoG" }}}{PARA 0 "" 0 "" 
{TEXT -1 88 "Notice that the cross product of two vectors yields a vec
tor orthogonal to both of them." }}{PARA 0 "" 0 "" {TEXT -1 18 "For th
e left side " }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "crossprod(r,m*rdd);" 
"6#-%*crossprodG6$%\"rG*&%\"mG\"\"\"%$rddGF)" }}{PARA 0 "" 0 "" {TEXT 
-1 12 "we can write" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "diff(crossprod
(r,m*rd),t)=crossprod(rd,m*rd)+crossprod(r,m*rdd):" "6#/-%%diffG6$-%*c
rossprodG6$%\"rG*&%\"mG\"\"\"%#rdGF-%\"tG,&-F(6$F.*&F,F-F.F-F--F(6$F**
&F,F-%$rddGF-F-" }}{PARA 0 "" 0 "" {TEXT -1 9 "The term " }{XPPEDIT 
18 0 "crossprod(rd,m*rd)" "6#-%*crossprodG6$%#rdG*&%\"mG\"\"\"F&F)" }
{TEXT -1 4 " is " }{TEXT 297 1 "0" }{TEXT -1 61 ", because the cross p
roduct of two parallel vectors vanishes." }}{PARA 0 "" 0 "" {TEXT -1 
10 "We define " }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "Lo:=crossprod(r,m*r
d);" "6#>%#LoG-%*crossprodG6$%\"rG*&%\"mG\"\"\"%#rdGF+" }}{PARA 0 "" 
0 "" {TEXT -1 7 "as the " }{TEXT 307 16 "angular momentum" }{TEXT -1 
46 ". So the axiom of angular momentum is given by" }}{PARA 257 "" 0 "
" {XPPEDIT 18 0 "diff(Lo,t)=Mo" "6#/-%%diffG6$%#LoG%\"tG%#MoG" }}
{PARA 0 "" 0 "" {TEXT -1 19 "Next we consider a " }{TEXT 298 10 "syste
m of " }{TEXT 300 0 "" }{TEXT 301 1 "n" }{TEXT 299 15 " mass particles
" }{TEXT -1 45 ". The angular momentum of the total system is" }}
{PARA 257 "" 0 "" {XPPEDIT 18 0 "Lo := sum(crossprod(r[i],m[i]*rd[i]),
i = 1 .. n);" "6#>%#LoG-%$sumG6$-%*crossprodG6$&%\"rG6#%\"iG*&&%\"mG6#
F.\"\"\"&%#rdG6#F.F3/F.;F3%\"nG" }}{PARA 0 "" 0 "" {TEXT -1 43 "Differ
entiating with respect to time yields" }}{PARA 257 "" 0 "" {XPPEDIT 
18 0 "diff(Lo,t) = sum(crossprod(rd[i],m[i]*rd[i]),i = 1 .. n)+sum(cro
ssprod(r[i],m[i]*rdd[i]),i = 1 .. n);" "6#/-%%diffG6$%#LoG%\"tG,&-%$su
mG6$-%*crossprodG6$&%#rdG6#%\"iG*&&%\"mG6#F3\"\"\"&F16#F3F8/F3;F8%\"nG
F8-F+6$-F.6$&%\"rG6#F3*&&F66#F3F8&%$rddG6#F3F8/F3;F8F=F8" }}{PARA 0 "
" 0 "" {TEXT -1 25 "and because we know that " }{XPPEDIT 18 0 "crosspr
od(r[i],r[i])=0" "6#/-%*crossprodG6$&%\"rG6#%\"iG&F(6#F*\"\"!" }{TEXT 
-1 7 " we get" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "diff(Lo,t) = sum(cro
ssprod(r[i],m[i]*rdd[i]),i = 1 .. n)" "6#/-%%diffG6$%#LoG%\"tG-%$sumG6
$-%*crossprodG6$&%\"rG6#%\"iG*&&%\"mG6#F2\"\"\"&%$rddG6#F2F7/F2;F7%\"n
G" }}{PARA 0 "" 0 "" {TEXT -1 14 "with the mass " }{TEXT 303 0 "" }
{XPPEDIT 18 0 "m[i]" "6#&%\"mG6#%\"iG" }{TEXT -1 15 ", acceleration " 
}{XPPEDIT 18 0 "rdd[i]" "6#&%$rddG6#%\"iG" }{TEXT -1 14 " and position
 " }{XPPEDIT 18 0 "r[i]" "6#&%\"rG6#%\"iG" }{TEXT -1 22 " of the mass \+
particle " }{XPPEDIT 18 0 "i" "6#%\"iG" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 47 "The resultant force which acts on the particle " }
{XPPEDIT 18 0 "m[i]" "6#&%\"mG6#%\"iG" }{TEXT -1 11 " should be " }
{XPPEDIT 18 0 "F[i]" "6#&%\"FG6#%\"iG" }{TEXT -1 39 ", and all the inn
er passive forces are " }{XPPEDIT 18 0 "sum(F[j,i],j=1..n)" "6#-%$sumG
6$&%\"FG6$%\"jG%\"iG/F);\"\"\"%\"nG" }{TEXT -1 11 " This means" }}
{PARA 257 "" 0 "" {XPPEDIT 18 0 "m[i]*rdd[i]=F[i]+sum(F[j,i],j = 1 .. \+
n)" "6#/*&&%\"mG6#%\"iG\"\"\"&%$rddG6#F(F),&&%\"FG6#F(F)-%$sumG6$&F/6$
%\"jGF(/F6;F)%\"nGF)" }}{PARA 0 "" 0 "" {TEXT -1 11 "Thus we get" }}
{PARA 257 "" 0 "" {XPPEDIT 18 0 "diff(Lo,t) = sum(crossprod(r[i], F[i]
)+crossprod(r[i],sum(F[j,i],j = 1 .. n)),i = 1 .. n):" "6#/-%%diffG6$%
#LoG%\"tG-%$sumG6$,&-%*crossprodG6$&%\"rG6#%\"iG&%\"FG6#F3\"\"\"-F.6$&
F16#F3-F*6$&F56$%\"jGF3/F@;F7%\"nGF7/F3;F7FC" }}{PARA 0 "" 0 "" {TEXT 
-1 93 "The sum of moments of all inner forces are zero, because the in
ner forces cancel each other. " }}{PARA 0 "" 0 "" {TEXT -1 36 "We get \+
the axiom of angular momentum" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }
{XPPEDIT 18 0 "diff(Lo,t) = sum(crossprod(r[i], F[i]),i = 1 .. n):" "6
#/-%%diffG6$%#LoG%\"tG-%$sumG6$-%*crossprodG6$&%\"rG6#%\"iG&%\"FG6#F2/
F2;\"\"\"%\"nG" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "diff(Lo,t) = Mo[res
]:" "6#/-%%diffG6$%#LoG%\"tG&%#MoG6#%$resG" }}{PARA 0 "" 0 "" {TEXT 
-1 5 "Here " }{XPPEDIT 18 0 "Mo[res]" "6#&%#MoG6#%$resG" }{TEXT -1 
108 " is the resultant moment of all forces acting on the total system
 from outside with respect to the fixpoint " }{TEXT 304 1 "O" }{TEXT 
-1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 98 "So we have deduced the axiom \+
of angular momentum for systems of mass particles. The transition to \+
" }{TEXT 305 12 "rigid bodies" }{TEXT -1 103 " happens as follows. We \+
consider the rigid body as a system of mass particles with the positio
n vector " }{XPPEDIT 18 0 "r[i]" "6#&%\"rG6#%\"iG" }{TEXT -1 14 " and \+
velocity " }{XPPEDIT 18 0 "rd[i] = diff(r[i],t);" "6#/&%#rdG6#%\"iG-%%
diffG6$&%\"rG6#F'%\"tG" }{TEXT -1 114 ". From the definition above for
 the system of mass particles we get by the transition from the sum to
 the integral" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "Lo = int(crossprod(r
[i],rd[i]),m);" "6#/%#LoG-%$intG6$-%*crossprodG6$&%\"rG6#%\"iG&%#rdG6#
F.%\"mG" }}{PARA 0 "" 0 "" {TEXT -1 64 "So we get for the axiom of the
 angular momentum for rigid bodies" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 
"diff(Lo,t) = Mo[res]" "6#/-%%diffG6$%#LoG%\"tG&%#MoG6#%$resG" }}
{PARA 0 "" 0 "" {TEXT -1 153 "This formulation was first time done by \+
LEONHARD EULER. We need to continue this subject later. But first we n
eed to make some additional considerations." }}}}{PARA 3 "" 0 "" 
{TEXT -1 0 "" }}}{MARK "2 2 3 0" 88 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 
}{PAGENUMBERS 0 1 2 33 1 1 }