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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 258 "" 0 
"" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT 256 12 "2D ROTATIONS" }}
{PARA 260 "" 0 "" {TEXT 259 11 "Tevian Dray" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 259 "" 0 "" {TEXT 257 33 "copyright 1998 & 2000 Tevian
 Dray" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT 258 
72 "In this worksheet, you will look at the world from a rotating turn
table." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" 
{TEXT -1 14 "Initialization" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
49 "restart:with(plots):with(plottools):with(linalg):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1264 "# 2Ddefs\n\n# Release 4:\n#Machin
e:=substring(interface(version),39..41):\n# Release 5.1:\n#Machine:=su
bstring(interface(version),41..43):\n# Fixed?\nMachine:=IBM:\n\nArrow:
=proc(base,dir,color)\n  local Dir:\n  global Machine:\n  if Machine=`
IBM` or Machine=`DEC` or Machine=`SUN` then\n    Dir:=base+dir\n  else
\n    Dir:=dir\n  fi;\n  arrow(base,Dir,0.02,0.05,0.1,color)\nend:\n\n
wb:=.02:\nwh:=.05:\nhh:=.1:\n\nd:=disk([0,0],1,color=wheat):\n\nQ:=[co
s(Pi/6),sin(Pi/6)]*2/5:\nP:=[cos(Pi/6)*2/5,4/5]:\nOO:=[0,0]:\n\nMyText
Plot:=proc(pt,label,alignment)\n  textplot([pt[1],pt[2],label],alignme
nt,font=[TIMES,BOLD,12])\nend:\n\nVvec:=\{Arrow(OO,P-OO,color=blue),\n
\011Arrow(OO,Q-OO,color=green),\n\011Arrow(Q,P-Q,color=green)\}:\nVlbl
:=\{MyTextPlot(Q/3,` R*r[hat] `,align=\{BELOW,RIGHT\}),\n\011\011MyTex
tPlot((P+Q)/2,` r[rel] `,align=RIGHT),\n\011\011MyTextPlot(P/2,` r[fix
] `,align=\{ABOVE,LEFT\}),\n\011\011MyTextPlot(OO,` O `,align=\{BELOW,
LEFT\}),\n\011\011MyTextPlot(P,`(x,y)`,align=ABOVE),\n\011\011MyTextPl
ot(Q,` Q `,align=\{BELOW,RIGHT\})\}:\n\nQx:=Q[1]:\nQy:=Q[2]:\nQcos:=0.
25*cos(Pi/6):\nQsin:=0.25*sin(Pi/6):\nRTvec:=\{Arrow(Q,[Qcos,Qsin],col
or=red),\n\011Arrow(Q,[-Qsin,Qcos],color=red)\}:\nRTlbl:=\n  \{textplo
t([Qx+Qcos,Qy+Qsin,` R `],align=\{BELOW,RIGHT\},font=[TIMES,ROMAN,12])
,\n   textplot([Qx-Qsin,Qy+Qcos,` Q `],align=\{BELOW,RIGHT\},font=[SYM
BOL,12])\}:\n" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 8 "Notation" }}
{PARA 0 "" 0 "" {TEXT -1 26 "Consider a fixed observer " }{XPPEDIT 18 
0 "Q" "6#%\"QG" }{TEXT -1 20 " located a distance " }{XPPEDIT 18 0 "R
" "6#%\"RG" }{TEXT -1 55 " from the origin, so that in polar coordinat
es we have " }{XPPEDIT 18 0 "Q=[R,theta]" "6#/%\"QG7$%\"RG%&thetaG" }
{TEXT -1 42 ".  The natural orthonormal polar basis at " }{XPPEDIT 18 
0 "Q" "6#%\"QG" }{TEXT -1 4 " is:" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 70 "r[hat]:=[cos(Theta),sin(Theta)];\ntheta[hat]:=[-sin(T
heta),cos(Theta)];" }}}{PARA 262 "" 1 "" {TEXT -1 48 "With respect to \+
this basis, the position vector " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT 
-1 19 " can be written as:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "r[fix]:=R*'r[hat]'+r[rel];" }}}{PARA 0 "" 0 "" {TEXT -1 33 "as sho
wn in the following figure:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
18 "plotsetup(inline):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "d
isplay(Vvec,Vlbl,RTvec,d,axes=NONE);" }}}{PARA 0 "" 0 "" {TEXT -1 66 "
In order to consider a rotating observer, all we have to do is set" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Theta:=Omega*t;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "Velocity and Acceleration" }}
{PARA 0 "" 0 "" {TEXT -1 50 "Now consider a moving object with positio
n vector " }{XPPEDIT 18 0 "r(t)" "6#-%\"rG6#%\"tG" }{TEXT -1 72 ".  In
 the nonrotating frame, the velocity and acceleration are given by " }
{XPPEDIT 18 0 "v[fix](t)=diff(r[fix](t),t)" "6#/-&%\"vG6#%$fixG6#%\"tG
-%%diffG6$-&%\"rG6#F(6#F*F*" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "a[fix
](t)=diff(v[fix](t),t)" "6#/-&%\"aG6#%$fixG6#%\"tG-%%diffG6$-&%\"vG6#F
(6#F*F*" }{TEXT -1 139 ", respectively.  In the rotating frame, things
 are not quite so simple, since the rotating observer is not aware of \+
the rotation.  Thus, if" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "r
[rel]:=X(t)*'r[hat]'+Y(t)*'theta[hat]';" }}}{PARA 0 "" 0 "" {TEXT -1 
92 "then the rotating observer computes the velocity and acceleration \+
simply by differentiating " }{XPPEDIT 18 0 "f(t)" "6#-%\"fG6#%\"tG" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "g(t)" "6#-%\"gG6#%\"tG" }{TEXT -1 
37 ".  The rotating observer thus obtains" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 59 "v[rel]:=\n  diff(X(t),t)*'r[hat]'+diff(Y(t),t)*'theta
[hat]';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "a[rel]:=\n  diff
(X(t),t,t)*'r[hat]'+diff(Y(t),t,t)*'theta[hat]';" }}}{PARA 0 "" 0 "" 
{TEXT -1 68 "These differ from the true velocity and acceleration for \+
2 reasons: " }{XPPEDIT 18 0 "r[rel]" "6#&%\"rG6#%$relG" }{TEXT -1 20 "
 is not the same as " }{XPPEDIT 18 0 "r[fix]" "6#&%\"rG6#%$fixG" }
{TEXT -1 47 ", and the time dependence of the basis vectors " }
{XPPEDIT 18 0 "R" "6#%\"RG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "Theta
" "6#%&ThetaG" }{TEXT -1 27 " has been ignored.  We have" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "v[Fix]:=expand(diff(r[fix],t)):" }}
}{PARA 0 "" 0 "" {TEXT -1 54 "which we expand with respect to the orth
onormal basis " }{XPPEDIT 18 0 "\{R,Theta\}" "6#<$%\"RG%&ThetaG" }
{TEXT -1 1 ":" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "v[fix]:=\n
  simplify(dotprod(v[Fix],r[hat],orthogonal))*'r[hat]'+\n  simplify(do
tprod(v[Fix],theta[hat],orthogonal))*'theta[hat]';" }}}{PARA 0 "" 0 "
" {TEXT -1 44 "Similarly, computing the acceleration yields" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 152 "a[Fix]:=diff(v[Fix],t):\na[fix]:=
\n  simplify(dotprod(a[Fix],r[hat],orthogonal))*'r[hat]'+\n  simplify(
dotprod(a[Fix],theta[hat],orthogonal))*'theta[hat]';" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Linear Motion" }}{PARA 0 "" 0 "" 
{TEXT -1 93 "Consider now the case of linear motion, for which the tru
e acceleration must vanish, so that " }{XPPEDIT 18 0 "a[fix]=0" "6#/&%
\"aG6#%$fixG\"\"!" }{TEXT -1 60 ".  The previous equation yields a dif
ferential equation for " }{XPPEDIT 18 0 "X(t)" "6#-%\"XG6#%\"tG" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "Y(t)" "6#-%\"YG6#%\"tG" }{TEXT -1 
17 " which determine " }{XPPEDIT 18 0 "r[rel" "6#&%\"rG6#%$relG" }
{TEXT -1 57 ".  There are 2 ways to solve this differential equtation.
" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 11 "Using Maple" }}{PARA 0 "" 0 "
" {TEXT -1 103 "Maple has a command \"dsolve\" which can solve many di
fferential equations.  For further information, try" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 7 "?dsolve" }}}{PARA 0 "" 0 "" {TEXT -1 23 "Her
e our equations are " }{XPPEDIT 18 0 "a[fix]" "6#&%\"aG6#%$fixG" }
{TEXT -1 39 " and our \"variables\" are the functions " }{XPPEDIT 18 
0 "X(t)" "6#-%\"XG6#%\"tG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "Y(t)" "
6#-%\"YG6#%\"tG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 118 "eq1:=simplify(dotprod(a[Fix],r[hat])):\neq2:=simplify(dotprod
(a[Fix],theta[hat])):\nsols:=dsolve(\{eq1,eq2\},\{X(t),Y(t)\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 42 "Converting the Equatio
n of a Straight Line" }}{PARA 0 "" 0 "" {TEXT -1 186 "A more intuitive
 way to solve this differential equation is to realize that the soluti
ons must be straight lines, but expressed in the rotating frame.  The \+
equation of a straight line is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 26 "r[fix]:='r[0]'+t*'v[fix]';" }}}{PARA 0 "" 0 "" {TEXT -1 5 "where
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "r[0]:=[x[0],y[0]];\nv[fi
x]:=[v[x],v[y]];" }}}{PARA 0 "" 0 "" {TEXT -1 30 "In the rotating fram
e, we have" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "r[REL]:='r[0]'
+t*'v[fix]'-R*'r[hat]';" }}}{PARA 0 "" 0 "" {TEXT -1 17 "or in other w
ords" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "r[Rel]:=\n  simplif
y(dotprod(r[REL],r[hat],orthogonal))*'r[hat]'+\n  simplify(dotprod(r[R
EL],theta[hat],orthogonal))*'theta[hat]';" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 
4 "" 0 "" {TEXT -1 10 "Comparison" }}{PARA 0 "" 0 "" {TEXT -1 59 "The \+
constants in these 2 approaches are related as follows:" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "d1:=expand(subs(t=0,subs(sols,r[rel
])));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "d2:=expand(subs(t=
0,r[Rel]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "const1:=solv
e(\n  \{d1[1]=d2[1],d1[2]=d2[2]\},\{_C1,_C2\});" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 65 "d3:=expand(subs(const1,\n  subs(t=Pi/2/Omega,s
ubs(sols,r[rel]))));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "d4:
=expand(subs(t=Pi/2/Omega,r[Rel]));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "const2:=solve(\n  \{d3[1]=d4[1],d3[2]=d4[2]\},\{_C3,_
C4\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}}{MARK "5" 
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