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{SECT 0 {PARA 259 "" 0 "" {TEXT -1 13 "A Water Whirl" }}{PARA 256 ""
0 "" {TEXT -1 9 "Jim Herod" }}{PARA 257 "" 0 "" {TEXT -1 21 "School of
Mathematics" }}{PARA 258 "" 0 "" {TEXT -1 12 "Georgia Tech" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 22 "Describing the problem" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "restart; w
ith(plots): with(plottools):" }}}{PARA 0 "" 0 "" {TEXT -1 377 " A \+
cylinder which is partially filled with water spins about the major a
xis. As it spins, the water moves in reaction to two forces -- the gra
vitational force and the centrifugal force. Consider how the level of \+
the water will change. In the middle of the surface, the level will g
o down and at the outer edges the water will appear to climb up the si
des of the cylinder." }}{PARA 0 "" 0 "" {TEXT -1 436 " As the cylin
der begins to spin, the surface of the water changes shape in order to
accommodate the centrifugal force's tendency to sling the water to th
e sides. This change in the depth of the water will be the result of t
he surface of the water remaining perpendicular to the combination of \+
the gravitational and centrifugal forces. When there is no spin, the s
urface of the water is level -- perpendicular to gravitational force. \+
" }}{PARA 0 "" 0 "" {TEXT -1 517 " We wish to understand what the \+
shape of the spinning surface will be. The following animation simula
tes how a cross section of the water in the cylinder might look as the
cylinder begins to spin. The animation reminds us that if there is no
spin, the water lies flat, and that the faster the spin, the sharper \+
the shape of the surface appears as the water moves up the sides of th
e cylinder. In the animation, think of starting with the cylinder not \+
moving, and then beginning to spin at increasing velocities. " }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "animate(\{[-2,y,y=0..7],[2,
y,y=0..7],[x,1/64*w^2*x^2+3-1/32*w^2,x=-2..2]\},w=0..4*sqrt(6),color=B
LUE);" }}{PARA 13 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 96 " A three dimensional view, with \+
the sides of the cylinder removed. could be made as follows:" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "u:=(r,theta,w)->w^2*r^2/64+3-1/32*w
^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "animate3d([r,theta,
u(r,theta,w)],r=0..2,theta=0..2*Pi,w=0..5,\ncoords=cylindrical,color=B
LUE,axes=NORMAL,\norientation=[45,70]);" }}}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 " In what follows, we will argu
e that if the cylinder is spinning at " }{XPPEDIT 18 0 "omega" "6#%&om
egaG" }{TEXT -1 220 " revolutions per minute, then the height of the w
ater is a quadratic function of the distance r from the center of the \+
spin. With d the depth to the bottom at the middle and g the gravitati
onal constant, we will see that" }}{PARA 0 "" 0 "" {TEXT -1 42 " \+
" }{XPPEDIT 18 0 "h(omega,r,d)=ome
ga^2*r^2/(2*g)+d" "6#/-%\"hG6%%&omegaG%\"rG%\"dG,&*(F'\"\"#F(F,*&F,\"
\"\"%\"gGF.!\"\"F.F)F." }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0
"" {TEXT -1 33 "Finding the shape of the surface." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 228 " Choose a point on t
he surface of the water at a distance r from the center and begin the \+
rotation. The force acting on that point can be resolved into two comp
onents: The horizontal component of the force will be the vector" }}
{PARA 0 "" 0 "" {TEXT -1 46 " \+
" }{XPPEDIT 18 0 "[m*omega^2*r,0]" "6#7$*(%\"mG\"\"\"*$%&omegaG\"
\"#F&%\"rGF&\"\"!" }}{PARA 0 "" 0 "" {TEXT -1 21 "where m is the mass,
" }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 120 " is the rotation
al velocity, and r is the distance from the center. The downward force
is due to gravity and is given by" }}{PARA 0 "" 0 "" {TEXT -1 47 " \+
" }{XPPEDIT 18 0 "0,-m*g]
" "6#7$\"\"!,$*&%\"mG\"\"\"%\"gGF(!\"\"" }{TEXT -1 1 "." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 116 " In the followi
ng picture, the horizontal force, or centrifugal force, is green. The \+
gravitational force is red." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
202 "r:=4/3:\nG:=plot([x,x^2,x=-2..2],view=[-2..5,-2..5],scaling=const
rained):\nH:=arrow([r,r^2], [2*r,r^2], .01, .2, .1, color=green):\nJ:=
arrow([r,r^2], [r,r^2-1/2], .01, .2, .1, color=red):\ndisplay(\{G,H,J
\});" }}}{PARA 0 "" 0 "" {TEXT -1 88 " To find the total force at \+
that point on the surface, we add the two vectors to get" }}{PARA 0 "
" 0 "" {TEXT -1 44 " " }
{XPPEDIT 18 0 "[m*omega^2*r, -m*g]" "6#7$*(%\"mG\"\"\"*$%&omegaG\"\"#F
&%\"rGF&,$*&F%F&%\"gGF&!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "In the following picture, the s
um of the forces, or resolvent of the forces, is blue." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "K:=arrow([r,r^2], [2*r,r^2-1/2], .0
1, .2, .1, color=BLUE):\ndisplay(\{G,H,J,K\});" }}}{PARA 0 "" 0 ""
{TEXT -1 220 " Finally, the surface of the water must settle so th
at the pressure is in the direction of the sum of the forces. The resu
lt is that the tangent to the surface of the water is perpendicular to
the sum of the forces. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "
L:=plot([x,2*r*(x-r)+r^2,x=-2..2.5],view=[-2..5,-2..5],color=BLACK):\n
display(\{G,H,J,K,L\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 " \+
We have only to find the surface, h(r), whose tangent at each point i
s perpendicular to " }}{PARA 0 "" 0 "" {TEXT -1 46 " \+
" }{XPPEDIT 18 0 "[m*omega^2*r, -m*g]" "6#
7$*(%\"mG\"\"\"*$%&omegaG\"\"#F&%\"rGF&,$*&F%F&%\"gGF&!\"\"" }{TEXT
-1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
47 " A vector perpendicular to the above vector is " }{XPPEDIT 18 0 "[
1,omega^2*r/g]" "6#7$\"\"\"*(%&omegaG\"\"#%\"rGF$%\"gG!\"\"" }{TEXT
-1 79 " and a curve with this tangent vector at any distance r away fr
om the center is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 62 " \+
" }{XPPEDIT 18 0 "h(r)=omega^2*r^2/(2*g) + h[0]" "6#/-%\"hG6#%\"r
G,&*(%&omegaG\"\"#F'F+*&F+\"\"\"%\"gGF-!\"\"F-&F%6#\"\"!F-" }{TEXT -1
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "H
ere, h(r) is the height of the curve, and " }{XPPEDIT 18 0 "h[0]" "6#&
%\"hG6#\"\"!" }{TEXT -1 45 " is the height above the bottom of contain
er." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "W
e ask two questions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 15 "" 0
"" {TEXT -1 68 " What is the depth d of the low point of the water as \+
a function of " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 20 "? We
expect that as " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 41 " i
ncreases, then d decreases toward zero." }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 15 "" 0 "" {TEXT -1 23 " At what spin velocity " }{XPPEDIT
18 0 "omega" "6#%&omegaG" }{TEXT -1 48 " will will the depth at the lo
w point d be zero?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3
"" 0 "" {TEXT -1 98 "Relating the speed of the spin and the radius of \+
the cylinder with the minimum depth of the water." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 279 " If we keep the amou
nt of water in the cylinder unchanged as the container starts to spin,
the depth of the water drops in the middle and grows on the sides. We
find a correlation between these three relationships: spin velocity, \+
depth in the middle, and height on the sides." }}{PARA 0 "" 0 ""
{TEXT -1 185 " The quantity of water in this spinning parabolic so
lid can be found by finding the volume of the surface of revolution ob
tained by rotating the area under the graph of the function" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 " \+
" }{XPPEDIT 18 0 "
h(r,h[0],omega)" "6#-%\"hG6%%\"rG&F$6#\"\"!%&omegaG" }{TEXT -1 3 " = \+
" }{XPPEDIT 18 0 "omega^2*r^2/(2*g) + h[0]" "6#,&*(%&omegaG\"\"#%\"rGF
&*&F&\"\"\"%\"gGF)!\"\"F)&%\"hG6#\"\"!F)" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "about the vertical axis. This volu
me is computed by integration using the method of shells:" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 " \+
volume = " }{XPPEDIT 18 0 "2*Pi*int(r*h(r,h[0]
,omega),r=0..a)" "6#*(\"\"#\"\"\"%#PiGF%-%$intG6$*&%\"rGF%-%\"hG6%F+&F
-6#\"\"!%&omegaGF%/F+;F1%\"aGF%" }{TEXT -1 1 "," }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "Here is the volume of wat
er written as a function of the radius of the cylinder " }{TEXT 257 1
"a" }{TEXT -1 26 ", the depth at the center " }{TEXT 258 1 "d" }{TEXT
-1 4 " = " }{XPPEDIT 18 0 "h[0]" "6#&%\"hG6#\"\"!" }{TEXT -1 27 ", an
d the rate of the spin " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT
-1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "r:='r': g:=32;\nh
:=(r,d,w)->w^2*r^2/(2*g)+d;\nint(2*Pi*r*h(r,d,omega),r=0..a);" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 " We m
ake this into a function that depends on " }{XPPEDIT 18 0 "omega" "6#%
&omegaG" }{TEXT -1 11 ", a, and d." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 28 "vol:=unapply(%,(a,d,omega));" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 " Suppose the volume o
f the water is the constant V. There is a relationship between the rad
ius " }{TEXT 259 1 "a" }{TEXT -1 70 " of the cylinder, this constant V
, and the depth of the water at rest." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "b:=solve(V=vol(a,b,0),b)
;" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 68 " \+
Now, start a spin. The amount of the volume will stay the same. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 " W
e find the depth of the water at the low point as a function of the vo
lume of the water V, the radius of the cylinder " }{TEXT 260 1 "a" }
{TEXT -1 21 ", and the spin speed " }{XPPEDIT 18 0 "omega" "6#%&omegaG
" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 22 "solve(V=vol(a,d,w),d);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 29 "mindepth:=unapply(%,(w,a,V));" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot(mindepth(w,2,5),w=0..4);" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT
1 {PARA 3 "" 0 "" {TEXT -1 49 "Rate of spin to have the minimum depth \+
to be zero" }}{PARA 0 "" 0 "" {TEXT -1 146 "In this cylinder with radi
us a, and with initial depth of water b, what is the spin to get the \+
low point of the parabolic surface equal to zero? " }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 32 "sol:=solve(mindepth(w,a,V)=0,w);" }}}{PARA
0 "" 0 "" {TEXT -1 130 "Thus, the spin of the water has been found wh
en the height becomes zero. How high will the water be along the sides
at that time?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "h(a,0,sol[
1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}{PARA 0 "" 0 "" {TEXT 256 11 "Conclusion:" }{TEXT -1 137 " whe
n the cylinder spins fast enough that the bottom shows, the water will
have crept up the sides twice as high as it was to begin with." }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "E
xercise for the student" }}{PARA 0 "" 0 "" {TEXT -1 233 " Suppose \+
a deep cylinder has radius 5 inches and contains 7 cubic inches of wat
er. How deep is the water to begin with. How fast must the cylinder sp
in to have the depth in the middle zero? How high did the water go up \+
the sides?" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 0 }
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