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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 -1 435 " As the cylin
der begins to spin, the surface of the water changes shape in order to
accommodate the centifugal force's tendency to sling the water to the
sides. This change in the depth of the water will be the result of th
e surface of the water remaining perpendicular to the combination of t
he gravitational and centrifugal forces. When there is no spin, the su
rface 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 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,t
heta,w)->w^2*r^2/64+3-1/32*w^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 124 "animate3d([r,theta,u(r,theta,w)],r=0..2,theta=0..2*Pi,w=0..1,
coords=cylindrical,color=BLUE,axes=NORMAL,orientation=[45,70]);" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 " In w
hat follows, we will argue that if the cylinder is spinning at " }
{XPPEDIT 18 0 "omega" "I&omegaG6\"" }{TEXT -1 220 " revolutions per mi
nute, then the height of the water is a quadratic function of the dist
ance r from the center of the spin. With d the depth to the bottom at \+
the middle and g the gravitational constant, we will see that" }}
{PARA 0 "" 0 "" {TEXT -1 42 " \+
" }{XPPEDIT 18 0 "h(omega,r,d)=omega^2*r^2/(2*g)+d" "/-%\"hG6%%&omega
G%\"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 s
urface." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
228 " Choose a point on the surface of the water at a distance r f
rom the center and begin the rotation. The force acting on that point \+
can be resolved into two components: The horizontal component of the f
orce will be the vector" }}{PARA 0 "" 0 "" {TEXT -1 46 " \+
" }{XPPEDIT 18 0 "[m*omega^2*r,0]" "7$
*(%\"mG\"\"\"*$%&omegaG\"\"#F%%\"rGF%\"\"!" }}{PARA 0 "" 0 "" {TEXT
-1 21 "where m is the mass, " }{XPPEDIT 18 0 "omega" "I&omegaG6\"" }
{TEXT -1 120 " is the rotational velocity, and r is the distance from \+
the center. The downward force is due to gravity and is given by" }}
{PARA 0 "" 0 "" {TEXT -1 47 " \+
" }{XPPEDIT 18 0 "0,-m*g]" "7$\"\"!,$*&%\"mG\"\"\"%\"gGF'!\"\""
}{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 116 " In the following picture, the horizontal force, or \+
centrifugal force, is green. The gravitational force is red." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 201 "r:=4/3: t:=1;\nG:=plot([x,x
^2,x=-2..2],view=[-2..5,-2..5],scaling=constrained):\nH:=arrow([r,r^2]
, [2*r,0], .01, .2, .1, color=green):\nJ:=arrow([r,r^2], [0,-1], .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]" "7
$*(%\"mG\"\"\"*$%&omegaG\"\"#F%%\"rGF%,$*&F$F%%\"gGF%!\"\"" }{TEXT -1
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "I
n the following picture, the sum of the forces, or resolvent of the fo
rces, is blue." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "K:=arrow([
r,r^2], [2*r,-1], .01, .2, .1, color=BLUE):\ndisplay(\{G,H,J,K\});" }}
}{PARA 0 "" 0 "" {TEXT -1 220 " Finally, the surface of the water \+
must settle so that the pressure is in the direction of the sum of the
forces. The result 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):\ndisplay(\{G,H,J,K,L\});" }}}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 " We have only to find the surf
ace, h(r), whose tangent at each point is perpendicular to " }}{PARA
0 "" 0 "" {TEXT -1 46 " \+
" }{XPPEDIT 18 0 "[m*omega^2*r, -m*g]" "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 t
o the above vector is " }{XPPEDIT 18 0 "[1,omega^2*r/g]" "7$\"\"\"*(%&
omegaG\"\"#%\"rGF#%\"gG!\"\"" }{TEXT -1 79 " and a curve with this tan
gent vector at any distance r away from 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]" "/-%\"hG6#%\"rG,&*(%&omegaG\"\"#F&F**&F*\"\"\"%\"gG
F,!\"\"F,&F$6#\"\"!F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 43 "Here, h(r) is the height of the curve
, and " }{XPPEDIT 18 0 "h[0]" "&%\"hG6#\"\"!" }{TEXT -1 45 " is the he
ight above the bottom of container." }}}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 21 "We ask two questions." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 15 "" 0 "" {TEXT -1 68 " What is the depth d o
f the low point of the water as a function of " }{XPPEDIT 18 0 "omega
" "I&omegaG6\"" }{TEXT -1 20 "? We expect that as " }{XPPEDIT 18 0 "om
ega" "I&omegaG6\"" }{TEXT -1 41 " increases, then d decreases toward z
ero." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 15 "" 0 "" {TEXT -1 23 "
At what spin velocity " }{XPPEDIT 18 0 "omega" "I&omegaG6\"" }{TEXT
-1 48 " will will the depth at the low 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 dep
th of the water." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 279 " If we keep the amount of water in the cylinder unch
anged 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 thes
e three relationships: spin velocity, depth in the middle, and height \+
on the sides." }}{PARA 0 "" 0 "" {TEXT -1 185 " The quantity of wa
ter in this spinning parabolic solid can be found by finding the volum
e of the surface of revolution obtained 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)" "-%\"hG6%%\"rG&F#6#
\"\"!%&omegaG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "omega^2*r^2/(2*g) + h
[0]" ",&*(%&omegaG\"\"#%\"rGF%*&F%\"\"\"%\"gGF(!\"\"F(&%\"hG6#\"\"!F(
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "about
the vertical axis. This volume is computed by integration using the m
ethod 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)" "*(\"\"#\"\"\"%#PiG
F$-%$intG6$*&%\"rGF$-%\"hG6%F*&F,6#\"\"!%&omegaGF$/F*;F0%\"aGF$" }
{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 112 "Here is the volume of water written as a function of the
radius of the cylinder a, the depth at the center d = " }{XPPEDIT
18 0 "h[0]" "&%\"hG6#\"\"!" }{TEXT -1 27 ", and the rate of the spin \+
" }{XPPEDIT 18 0 "omega" "I&omegaG6\"" }{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 make this into a function tha
t depends on " }{XPPEDIT 18 0 "omega" "I&omegaG6\"" }{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 -1 170 " Suppose the volume of the water is the constant V. \+
There is a relationship between the radius a of the cylinder, this con
stant 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 -1 148 " \+
We find the depth of the water at the low point as a function of \+
the volume of the water V, the radius of the cylinder a, and the spin \+
speed " }{XPPEDIT 18 0 "omega" "I&omegaG6\"" }{TEXT -1 1 "." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "so
lve(V=vol(a,d,w),d);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "min
depth:=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 radius a, and with initial depth o
f water b, what is the spin to get the low point of the parabolic sur
face 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, t
he spin of the water has been found when 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]);" }}}{PARA 0 "" 0 "" {TEXT
256 11 "Conclusion:" }{TEXT -1 137 " when the cylinder spins fast enou
gh 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 "Exercise for the student" }}
{PARA 0 "" 0 "" {TEXT -1 233 " Suppose a deep cylinder has radius \+
5 inches and contains 7 cubic inches of water. How deep is the water t
o begin with. How fast must the cylinder spin to have the depth in the
middle zero? How high did the water go up the sides?" }}}{PARA 0 ""
0 "" {TEXT -1 0 "" }}}{MARK "6" 0 }{VIEWOPTS 1 1 0 1 1 1803 }