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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 30 "Classical Mechanics with \+
Maple" }}{PARA 256 "" 0 "" {TEXT -1 47 "Section 3.2: Balance and Conse
rvation of Energy" }}{PARA 19 "" 0 "" {TEXT -1 41 "Dr. Harald Kammerer
\nmaple@jademountain.de" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 14 "Initi
alisation" }}{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(dynamic
s);with(figures_chapter_3);" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 38 "
3.2 Balance and Conservation of Energy" }}{PARA 0 "" 0 "" {TEXT -1 94 
"Above we distinguished active and passive forces. Additionally we sep
arate active forces into " }{TEXT 256 19 "conservative forces" }{TEXT 
-1 5 " and " }{TEXT 257 23 "non-conservative forces" }{TEXT -1 2 ". " 
}}{PARA 0 "" 0 "" {TEXT -1 119 "- A force is called conservative if it
 does no net work along a closed path. Examples are the gravitational \+
force (see " }{HYPERLNK 17 "gravitational_energy" 2 "gravitational_ene
rgy" "" }{TEXT -1 25 ") and spring forces (see " }{HYPERLNK 17 "spring
_energy" 2 "spring_energy" "" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" 
{TEXT -1 159 "- A force is called non-conservative if it does net work
 along a closed path. For example the sliding (friction) force and air
 resistance are non-conservative." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 16 "In other words,:" }}{PARA 0 "" 0 "" 
{TEXT -1 2 "- " }{TEXT 258 12 "conservative" }{TEXT -1 60 " systems ar
e systems in which the total energy is conserved," }}{PARA 0 "" 0 "" 
{TEXT -1 2 "- " }{TEXT 340 16 "non-conservative" }{TEXT -1 79 " system
s receive energy from the environment or lose energy to the environmen
t." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "Co
nsequently we can allocate a potential to conservative forces. For exa
mple such potentials are the potential of springs and the potential of
 the gravity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 61 "The result of all on a mass particle acting forces is the
 sum" }}{PARA 257 "" 0 "" {XPPEDIT 263 0 "F = F[conservative]+F[noncon
servative]+N;" "6#/%\"FG,(&F$6#%-conservativeG\"\"\"&F$6#%0nonconserva
tiveGF)%\"NGF)" }}{PARA 0 "" 0 "" {TEXT -1 23 "with the passive force \+
" }{TEXT 259 1 "N" }{TEXT -1 64 " which is caused by the boundary. New
ton's law of motion is then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
257 "" 0 "" {TEXT 262 1 "m" }{TEXT -1 1 " " }{XPPEDIT 261 0 "a = F[con
servative]+F[nonconservative]+N;" "6#/%\"aG,(&%\"FG6#%-conservativeG\"
\"\"&F'6#%0nonconservativeGF*%\"NGF*" }{TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 1 "(" }{TEXT 260 4 "Note" }{TEXT -1 63 ": the forces and th
e acceleration in the equations are vectors)" }}{PARA 0 "" 0 "" {TEXT 
-1 128 "Now we consider the motion of a mass particle along the trajec
tory of its motion between the two points (1) and (2) (see Fig. 6)" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "display(Fig_3_4(),scaling=co
nstrained,axes=none,title=\"Figure 6\");" }}}{PARA 0 "" 0 "" {TEXT -1 
71 "Multiplication of the law of motion with the incremental displacem
ent d" }{TEXT 264 1 "r" }{TEXT -1 36 " and integration yields the equa
tion" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "int(m*a,r=r1..r2)=int(F_c,r=r
1..r2)+int(F_nc,r=r1..r2)+int(N,r=r1..r2):" "6#/-%$intG6$*&%\"mG\"\"\"
%\"aGF)/%\"rG;%#r1G%#r2G,(-F%6$%$F_cG/F,;F.F/F)-F%6$%%F_ncG/F,;F.F/F)-
F%6$%\"NG/F,;F.F/F)" }}{PARA 0 "" 0 "" {TEXT -1 42 "with the result of
 the conservative force " }{TEXT 265 1 "F" }{TEXT 267 2 "_c" }{TEXT 
-1 47 " and the result of the non-conservative forces " }{TEXT 266 1 "
F" }{TEXT 268 3 "_nc" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 81 "On the right side is the sum of the wor
k of all acting forces. The passive force " }{TEXT 269 1 "N" }{TEXT 
-1 121 " is orthogonal to the trajectory and thus contributes no work.
 For the conservative forces there exist a total potential " }{TEXT 
271 1 "V" }{TEXT -1 1 "(" }{TEXT 270 1 "r" }{TEXT -1 50 "), which is d
efined as negative work. We can say: " }{TEXT 272 42 "The potential is
 the capability to do work" }{TEXT -1 105 ". The work that is done by \+
the conservative forces on the way from (1) to (2) is the potential di
fference" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "int(F_c,r = r1 .. r2)=V(r
1)-V(r2):" "6#/-%$intG6$%$F_cG/%\"rG;%#r1G%#r2G,&-%\"VG6#F+\"\"\"-F/6#
F,!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 10 "or with V(" }{TEXT 282 2 "r1" 
}{TEXT -1 2 ")=" }{TEXT 284 2 "V1" }{TEXT -1 7 " and V(" }{TEXT 283 2 
"r2" }{TEXT -1 2 ")=" }{TEXT 285 2 "V2" }}{PARA 257 "" 0 "" {XPPEDIT 
18 0 "int(F_c,r = r1 .. r2) = V1-V2:" "6#/-%$intG6$%$F_cG/%\"rG;%#r1G%
#r2G,&%#V1G\"\"\"%#V2G!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 58 "The calcul
ation of the work of the non-conservative forces" }}{PARA 257 "" 0 "" 
{XPPEDIT 18 0 "W[1..2]=int(F_nc,r = r1 .. r2):" "6#/&%\"WG6#;\"\"\"\"
\"#-%$intG6$%%F_ncG/%\"rG;%#r1G%#r2G" }}{PARA 0 "" 0 "" {TEXT -1 108 "
on the way form (1) to (2) requires that we know the force law. Furthe
r we need for the integration the form" }}{PARA 257 "" 0 "" {XPPEDIT 
18 0 "F_nc=F_nc(r):" "6#/%%F_ncG-F$6#%\"rG" }}{PARA 0 "" 0 "" {TEXT 
337 14 "So the motion " }{TEXT 273 1 "r" }{TEXT 338 1 "(" }{TEXT -1 1 
"t" }{TEXT 274 15 ") must be known" }{TEXT -1 113 ". In the simple cas
e of sliding the non-conservative force is constant and the integral c
an be easily calculated." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 52 "We transform the left side of the equation above  \+
to" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "int(m*a,r = r1 .. r2)=int(m*dv/
dt,r=r1..r2):" "6#/-%$intG6$*&%\"mG\"\"\"%\"aGF)/%\"rG;%#r1G%#r2G-F%6$
*(F(F)%#dvGF)%#dtG!\"\"/F,;F.F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 16 "and consequently" }}{PARA 257 "" 0 "" 
{XPPEDIT 18 0 "int(m*dv/dt,r = r1 .. r2) = int(m*dr/dt,v = v1 .. v2):
" "6#/-%$intG6$*(%\"mG\"\"\"%#dvGF)%#dtG!\"\"/%\"rG;%#r1G%#r2G-F%6$*(F
(F)%#drGF)F+F,/%\"vG;%#v1G%#v2G" }}{PARA 0 "" 0 "" {TEXT -1 5 "with " 
}{TEXT 275 1 "v" }{TEXT 286 1 "1" }{TEXT -1 1 "=" }{TEXT 277 1 "v" }
{TEXT -1 1 "(" }{TEXT 278 1 "r" }{TEXT -1 0 "" }{TEXT 279 1 "1" }
{TEXT -1 6 ") and " }{TEXT 276 1 "v" }{TEXT 287 1 "2" }{TEXT -1 1 "=" 
}{TEXT 280 1 "v" }{TEXT -1 1 "(" }{TEXT 281 1 "r" }{TEXT 288 1 "2" }
{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 9 "We know  " }{XPPEDIT 18 
0 "dr/dt = v;" "6#/*&%#drG\"\"\"%#dtG!\"\"%\"vG" }{TEXT -1 19 " and at
 last we get" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "int(m*a,r = r1 .. r2)
=m*v2**2/2-m*v1**2/2:" "6#/-%$intG6$*&%\"mG\"\"\"%\"aGF)/%\"rG;%#r1G%#
r2G,&*(F(F)*$%#v2G\"\"#F)F4!\"\"F)*(F(F)*$%#v1GF4F)F4F5F5" }}{PARA 0 "
" 0 "" {TEXT -1 14 "The expression" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 
"T = m*v^2/2:" "6#/%\"TG*(%\"mG\"\"\"*$%\"vG\"\"#F'F*!\"\"" }}{PARA 0 
"" 0 "" {TEXT -1 14 "is called the " }{TEXT 289 14 "kinetic energy" }
{TEXT -1 35 " (This relation is included in the " }{HYPERLNK 17 "dynam
ics" 2 "dynamics" "" }{TEXT -1 14 " package. See " }{HYPERLNK 17 "kine
tic_energy" 2 "kinetic_energy" "" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" 
{TEXT -1 38 "Insert all expressions and we get the " }{TEXT 290 26 "ba
lance of energy equation" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "T2+V2=T1+
V1+W[1 .. 2]:" "6#/,&%#T2G\"\"\"%#V2GF&,(%#T1GF&%#V1GF&&%\"WG6#;F&\"\"
#F&" }}{PARA 0 "" 0 "" {TEXT -1 16 "In words we say:" }}{PARA 258 "" 
0 "" {TEXT -1 160 "The sum of the kinetic and the potential energy at \+
point (2) is the sum of the kinetic and the potential energy at point \+
(1) increased or decreased by the work " }{XPPEDIT 18 0 "W[1 .. 2]" "6
#&%\"WG6#;\"\"\"\"\"#" }{TEXT -1 182 " of the nonconservative forces. \+
We get an increase of energy when the force points in the direction of
 the displacement and a decrease when the force points against the dis
placement." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 74 "In the special case that no non-conservative forces are acting \+
we get the " }{TEXT 316 22 "conservation of energy" }{TEXT -1 0 "" }}
{PARA 257 "" 0 "" {XPPEDIT 18 0 "T2+V2 = T1+V1:" "6#/,&%#T2G\"\"\"%#V2
GF&,&%#T1GF&%#V1GF&" }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 29 "Example: C
ar Braking Downhill" }}{PARA 0 "" 0 "" {TEXT -1 35 "Have a look at a c
ar with the mass " }{TEXT 305 1 "m" }{TEXT -1 53 " which is driving do
wnwards a hill as shown in Fig. 7" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 66 "display(Fig_3_5(),scaling=constrained,axes=none,title
=\"Figure 7\");" }}}{PARA 0 "" 0 "" {TEXT -1 42 "The car drives with t
he constant velocity " }{TEXT 291 2 "v0" }{TEXT -1 67 " in the directi
on of the street. The slope of the road is given by " }{TEXT 292 1 "a
" }{TEXT -1 36 ". When the car reaches the position " }{TEXT 293 3 "s=
0" }{TEXT -1 68 " the driver begins to stop the car. Then the constant
 sliding force " }{TEXT 294 1 "R" }{TEXT -1 97 " takes effect parallel
 to the trajectory of the motion, or in other words against the coordi
nate " }{TEXT 295 1 "s" }{TEXT -1 17 ". The coordinate " }{TEXT 296 1 
"z" }{TEXT -1 48 " points in the vertical direction and we choose " }
{TEXT 297 3 "z=0" }{TEXT -1 42 " at the center of gravity of the car w
hen " }{TEXT 298 3 "s=0" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 
17 "The question is: " }{TEXT 339 21 "What is the position " }{TEXT 
-1 3 "s2 " }{TEXT 299 25 "of the car when it stops?" }}{PARA 0 "" 0 "
" {TEXT -1 61 "We first consider the work that is done by the sliding \+
force " }{TEXT 300 1 "R" }{TEXT -1 12 ". The force " }{TEXT 301 1 "R" 
}{TEXT -1 22 " acts on the way from " }{TEXT 302 2 "s1" }{TEXT -1 4 " \+
to " }{TEXT 303 2 "s2" }{TEXT -1 14 ". We can write" }}{PARA 0 "" 0 "
" {TEXT -1 26 "for the work which is done" }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "W[1 .. 2] := int(-R,s = 0 .. s2);" "6#>&%\"WG6#;\"\"\"
\"\"#-%$intG6$,$%\"RG!\"\"/%\"sG;\"\"!%#s2G" }}}{PARA 0 "" 0 "" {TEXT 
-1 12 "Notice that " }{TEXT 304 1 "R" }{TEXT -1 72 " acts against the \+
direction of the motion, so we have the negative sign." }}{PARA 0 "" 
0 "" {TEXT -1 71 "Next we consider the potential of gravity. This pote
ntial is defined by" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "V:=m*g*
z:" "6#>%\"VG*(%\"mG\"\"\"%\"gGF'%\"zGF'" }}}{PARA 0 "" 0 "" {TEXT -1 
34 "(This relation is included in the " }{HYPERLNK 17 "dynamics" 2 "dy
namics" "" }{TEXT -1 14 " package. See " }{HYPERLNK 17 "gravitational_
energy" 2 "gravitational_energy" "" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" 
{TEXT -1 6 "Where " }{TEXT 306 1 "z" }{TEXT -1 62 " is measured from a
n arbitrary choosen horizontal. We had set " }{TEXT 307 3 "z=0" }
{TEXT -1 23 " at the beginning when " }{TEXT 308 4 "s=s1" }{TEXT -1 
11 ". So we get" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "z1:=0:" "6#
>%#z1G\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 75 "At the beginning of the b
raking process we get for the potential of gravity" }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 33 "V1:=gravitational_energy(m,g,z1);" }}}{PARA 
0 "" 0 "" {TEXT -1 96 "The altitude with respect to the initial height
 of the car when it reaches the unknown position " }{TEXT 309 4 "s=s2
" }{TEXT -1 3 " is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "z2:=-s2*
sin(alpha):" "6#>%#z2G,$*&%#s2G\"\"\"-%$sinG6#%&alphaGF(!\"\"" }}}
{PARA 0 "" 0 "" {TEXT -1 54 "With this relation we get for the potenti
al of gravity" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "V2:=gravita
tional_energy(m,g,z2);" }}}{PARA 0 "" 0 "" {TEXT -1 112 "At last we lo
ok at the kinetic energy. At the beginning of the braking process the \+
car drives with the velocity " }{TEXT 310 2 "v0" }{TEXT -1 34 " and we
 get for the kinetic energy" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "T1:=kinetic_energy(m,v0);" }}}{PARA 0 "" 0 "" {TEXT -1 21 "When th
e car reaches " }{TEXT 311 4 "s=s2" }{TEXT -1 29 " it stops and the ve
locity is" }{TEXT 312 0 "" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v
2:=0:" "6#>%#v2G\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 32 "So we get for t
he kinetic energy" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "T2:=kin
etic_energy(m,v2);" }}}{PARA 0 "" 0 "" {TEXT -1 17 "We write for the \+
" }{TEXT 313 17 "balance of energy" }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "Energy_Balance:=T2+V2=T1+V1+W[1..2];" "6#>%/Energy_Bala
nceG/,&%#T2G\"\"\"%#V2GF(,(%#T1GF(%#V1GF(&%\"WG6#;F(\"\"#F(" }}}{PARA 
0 "" 0 "" {TEXT -1 13 "Isolation of " }{TEXT 314 2 "s2" }{TEXT -1 45 "
 in this relation yields the braking distance" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 29 "s2:=solve(Energy_Balance,s2);" }}}{PARA 0 "" 0 "
" {TEXT -1 96 "At last we want to calculate the braking distance for t
he case of a horizontal road. Then we get" }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "alpha:=0:" "6#>%&alphaG\"\"!" }}}{PARA 0 "" 0 "" {TEXT 
-1 161 "Additionally we limit our consideration to the case that all w
heels are blocking. So we have the situation of sliding friction. That
 yields for the braking force" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 
1 "R:=m*g*mu:" "6#>%\"RG*(%\"mG\"\"\"%\"gGF'%#muGF'" }}}{PARA 0 "" 0 "
" {TEXT -1 33 "with the coefficient of friction " }{TEXT 315 1 "m" }
{TEXT -1 30 ". The braking distance is then" }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 9 "eval(s2);" }}}{PARA 0 "" 0 "" {TEXT -1 76 "We see th
at here the braking distance is independent of the mass of the car." }
}{PARA 0 "" 0 "" {TEXT -1 35 "The constant of gravity is known as" }}
{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "g:=9.81:" "6#>%\"gG-%&FloatG6$
\"$\")*!\"#" }}}{PARA 0 "" 0 "" {TEXT -1 75 "For example we get with t
he initial velocity of 75 km/h, which means in m/s" }}{EXCHG {PARA 0 "
> " 0 "" {XPPEDIT 19 1 "v0 := 75/3.6;" "6#>%#v0G*&\"#v\"\"\"-%&FloatG6
$\"#O!\"\"F," }}}{PARA 0 "" 0 "" {TEXT -1 31 "and the coefficient of f
riction" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "mu:=1:" "6#>%#muG\"
\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 27 "the braking distance (in m)" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(s2);" }}}{PARA 0 "" 0 
"" {TEXT -1 49 "When the car drives with the velocity of 150 km/h" }}
{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v0:=150/3.6;" "6#>%#v0G*&\"$]
\"\"\"\"-%&FloatG6$\"#O!\"\"F," }}}{PARA 0 "" 0 "" {TEXT -1 6 "we get
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(s2);" }}}{PARA 0 "
" 0 "" {TEXT -1 83 "This means that doubling the velocity yields a qua
drupling of the braking distance." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{SECT 0 {PARA 5 "" 0 "" {TEXT -1 57 "Example: The Mathematical Pendulu
m, Energy Considerations" }}{PARA 0 "" 0 "" {TEXT -1 67 "Now we consid
er the motion of the mathematical pendulum again (see " }{HYPERLNK 17 
"" 2 "" "Example1" }{HYPERLNK 17 "above" 2 "" "Example 1" }{TEXT -1 2 
")." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "unassign('t','a','m',
'L','T','g','v0','omega','v','phi0','phi_d_0','phi(t)'):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "display(Fig_3_6(),scaling=constrain
ed,axes=none,title=\"Figure 8\");" }}}{PARA 0 "" 0 "" {TEXT -1 99 "We \+
consider two different states. The first one is the initial state. The
 pendulum is displaced by " }{TEXT 317 1 "f" }{TEXT 326 1 "0" }{TEXT 
-1 34 " and has initial angular velocity " }{TEXT 325 7 "phi_d_0" }
{TEXT -1 62 ". Second we consider an arbitrary state with the displace
ment " }{TEXT 319 1 "f" }{TEXT -1 26 " and the angular velocity " }
{TEXT 318 1 "w" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 73 "Next we
 have to define a horizontal from which we measure the coordinate " }
{TEXT 320 1 "z" }{TEXT -1 43 " to describe the current height of the m
ass" }{TEXT 321 2 " m" }{TEXT -1 47 ". We choose the lowest point of t
he mass, i.e. " }{TEXT 323 1 "f" }{TEXT 322 2 "=0" }{TEXT -1 51 ". So \+
we get for the height of the mass (see Fig. 8)" }}{EXCHG {PARA 0 "> " 
0 "" {XPPEDIT 19 1 "z0:=L-L*cos(phi0):" "6#>%#z0G,&%\"LG\"\"\"*&F&F'-%
$cosG6#%%phi0GF'!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 56 "for the initial
 state and for the arbitrary second state" }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "z:=L-L*cos(phi):" "6#>%\"zG,&%\"LG\"\"\"*&F&F'-%$cosG6#
%$phiGF'!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 51 "Now we get the potentia
l of gravity for both states" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
33 "V0:=gravitational_energy(m,g,z0);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "V:=gravitational_energy(m,g,z);" }}}{PARA 0 "" 0 "" 
{TEXT -1 99 "The velocity of the mass particle points in the direction
 of the trajectory. We get for both states" }}{EXCHG {PARA 0 "> " 0 "
" {XPPEDIT 19 1 "v0 := L*phi_d_0:" "6#>%#v0G*&%\"LG\"\"\"%(phi_d_0GF'
" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "v := L*phi_d:" "6#>%\"vG*
&%\"LG\"\"\"%&phi_dGF'" }}}{PARA 0 "" 0 "" {TEXT -1 26 "and for the ki
netic energy" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "T0:=kinetic_
energy(m,v0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "T:=kinetic
_energy(m,v);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "In this example we have
 no non-conservative forces. So we write for the " }{TEXT 324 22 "cons
ervation of energy" }{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "Energy_conservation:=T+V=T0+V0;" }}}{PARA 0 "" 0 "" 
{TEXT -1 9 "The term " }{XPPEDIT 18 0 "m*g*L" "6#*(%\"mG\"\"\"%\"gGF%%
\"LGF%" }{TEXT -1 55 " is on both sides of the equation and can be eli
minated" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "EC:=expand(lhs(En
ergy_conservation)-m*g*L)=expand(rhs(Energy_conservation)-m*g*L);" }}}
{PARA 0 "" 0 "" {TEXT -1 84 "Have again a look at the equation of moti
on which we derived above, multiplied with " }{TEXT 328 1 "L" }{TEXT 
-1 5 " and " }{TEXT 327 1 "m" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
47 "EOM:=g*sin(phi(t))/L+diff(phi(t),`$`(t,2)) = 0;" }}}{PARA 0 "" 0 "
" {TEXT -1 20 "Multiplication with " }{XPPEDIT 18 0 "L**2" "6#*$%\"LG
\"\"#" }{TEXT -1 2 ", " }{TEXT 329 1 "m" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "diff(phi(t),t)" "6#-%%diffG6$-%$phiG6#%\"tGF)" }{TEXT 
-1 7 " yields" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "NEOM:=expan
d(L**2*m*EOM*diff(phi(t),t));" }}}{PARA 0 "" 0 "" {TEXT -1 45 "After i
ntegration with respect to time we get" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "E1:=int(lhs(NEOM),t);" }}}{PARA 0 "" 0 "" {TEXT -1 
12 "We see that " }{TEXT 331 2 "E1" }{TEXT -1 5 " and " }{TEXT 330 2 "
E0" }{TEXT -1 19 " are the same with " }{XPPEDIT 18 0 "phi_d=diff(phi(
t),t)" "6#/%&phi_dG-%%diffG6$-%$phiG6#%\"tGF+" }{TEXT -1 43 ". We say:
 \"The conservation of energy is a " }{TEXT 332 14 "first integral" }
{TEXT -1 93 " of the equation of motion\". The integration in the last
 step yields an integration constant " }{TEXT 333 5 "E1=E0" }{TEXT -1 
116 " which represents the total energy of the system at every time. I
t is given by the initial conditions of the motion:" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 12 "E0:=rhs(EC);" }}{PARA 0 "" 0 "" {TEXT -1 
170 "With the conservation of energy we can give an expression for the
 velocity of the mass particle for every position. For example we take
 the same concrete values as above." }}}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "g:=9.81:" "6#>%\"gG-%&FloatG6$\"$\")*!\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {XPPEDIT 19 1 "L:=10:" "6#>%\"LG\"#5" }}}{EXCHG 
{PARA 0 "> " 0 "" {XPPEDIT 19 1 "m:=10:" "6#>%\"mG\"#5" }}}{EXCHG 
{PARA 0 "> " 0 "" {XPPEDIT 19 1 "phi0 := .1:" "6#>%%phi0G-%&FloatG6$\"
\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "phi_d_0 := 0:" "
6#>%(phi_d_0G\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 34 "The conservation o
f energy is then" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Energy_c
onservation;" }}}{PARA 0 "" 0 "" {TEXT -1 57 "Solve this relation with
 respect to the angular velocity " }{XPPEDIT 18 0 "phi_d" "6#%&phi_dG
" }{TEXT -1 7 " yields" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "so
l:=solve(Energy_conservation,phi_d);" }}}{PARA 0 "" 0 "" {TEXT -1 63 "
We look at the graphical presentation of this solution (Fig. 9)" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "P1:=plot(sol[1],phi=-0.1..0.
1, color=green):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "P2:=plo
t(sol[2],phi=-0.1..0.1,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 55 "display(\{P1,P2\},title=\"Figure 9\", scaling=constra
ined);" }}}{PARA 0 "" 0 "" {TEXT -1 132 "In this diagram we get for ev
ery initial condition concentric curves with the common center (0,0). \+
This kind of diagram is called a " }{TEXT 334 14 "phase portrait" }
{TEXT -1 58 " of the undamped system. One discrete curve is called the
 " }{TEXT 335 5 "phase" }{TEXT -1 1 " " }{TEXT 336 6 "curve." }{TEXT 
-1 115 " Notice that the example represents a nonlinear equation of mo
tion, because we hadn't made here the simplification " }{XPPEDIT 18 0 
"sin(phi)=phi" "6#/-%$sinG6#%$phiGF'" }{TEXT -1 146 ". In the case of \+
linear motions, the phase curves are ellipses. Phase portraits are ver
y useful tools for comparing different types of vibrations." }}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "\027" }}}}{MARK "0 0 0" 7 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }