Hourglass of constant weight

0.06 MB·English

by  Volker Becker

Hourglass of constant weight
Volker Becker and Thorsten Po¨schel
Charit´e, Augustenburger Platz 1, 13353 Berlin, Germany
(Dated: February 6, 2008)
Incontrasttoastillcommonbelieve,asteadilyflowinghourglasschangesitsweightinthecourse
1
oftime . Wewillshowthat,nevertheless,itispossibletoconstructhourglasses thatdonotchange
7 theirweight.
0
0
2 In the state of steady flow, the center of mass of the the containedsand. For the definition of y1, y2, a, and u
sand in an hourglass is permanently moving downward. see Fig. 1. The velocity of the center of mass reads
n
Theabsolutevelocityofthecenterofmassdecreaseslead-
a
J ingtoanupwardaccelerationofthecenterofmass. This My˙ =ρ(u˙ +y˙ )(u+y )A (y )−ρu˙uA (0)
cm 1 1 1 1 1
1 simpleargumentwasworkedoutbyShenandScott1 and u+y1
3 led to the conclusion that a suspended at a balance run- dA1
+ρy˙ (a+y )A (y )+m˜ u˙ +ρu˙ dyy
t] npiunrgewhoeiugrhgtlaosfsthaeppheoaurrsgltaossbaendofthlaercgoenrtwaienigedhtsatnhdan,ththaet 2 2 2 2 1 Zu du (cid:12)(cid:12)(u−y1)
of is, a flowing hourglassappears to be heavier than a non- (cid:12)(cid:12) (2)
flowing one by a certain quantity ∆m. While the time
s Using dA /du| = −dA /dy| the last term
. dependence of ∆m(t) depends on the specific construc- 1 (y−u) 1 (y−u)
at tionofthehourglass,itwasshown1 that∆m(t)isalways reads m1(t)u˙ + ρu˙uA1(0) − ρu˙(u+y1)A1(y1), where
m a positive function that decreases with time. m1(t) is the mass of the sand in the lower compartment.
With the definition of the mass flow
- The finding by Shen and Scott naturally provokes the
d
questionwhetherwecaninventanhourglasswhosetotal
n j =ρA (y )y˙ =−ρA (y )y˙ (3)
weightinthesteadyflowingstateisindependentoftime. 1 1 1 2 2 2
o
Itturnsoutthatsuchaconstructionispossible. Consider
c we obtain
[ the hourglass shown in Fig. 1.

The vertical component of the center of mass of the My˙ =[m (t)+m˜ ]u˙ +j(u−a+y −y ) (4)
1 cm 1 1 1 2
hourglass, y , is described by
v cm
and, thus,
6
7 u+y1 a+y2
7 j2 1 1
My = dyρyA (y−u)+ dyρyA (y−u) My¨ =[m (t)+m˜ ]u¨+2ju˙+ + .
1 cm 1 2 cm 1 1 ρ A (y ) A (y )
0 Zu Za (cid:20) 1 1 2 2 ((cid:21)5)
7 +m˜ y(1)+m˜ y(2), (1) Forthecaseofconstantcross-section,assketchedinFig.
0 1 cm 2 cm
1, A (y ) = A (y ) = A and with m (t) = jt, the
/ 1 1 2 2 1
at whereA1(y)andA2(y)arethecross-sectionsofthelower change of weight of the hourglass as measured by the
m and upper container, m˜1 and m˜2 are the masses of the balance is
(1) (2)
(empty) containers,y andy are the verticalcompo-
- cm cm 2j2
d nent of their center of masses, ρ is the mass density of My¨ =[jt+m˜ ]u¨+2ju˙ + . (6)
cm 1
n the sand, and M is the total mass of the hourglass and ρA
o
This equation provides a condition for the time-
c
v: (cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1) dependentpositionu(t)ofthelowercompartmenttokeep
balance the weight of the hourglass constant, My¨ =0. A par-
i cm
X 0 ticular solution of the latter equation is
r
a j
y2 u˙ =− , (7)
ρA
a
j
that is,the lowercontainermovesdownwardatconstant
(k/2,η/2) (k/2,η/2)
velocity.
y Assume the lower container is suspended via damped
1
u linear springs as shown in Fig. 1. Let us see whether
this system fulfills the condition Eq. (7) as a stable so-
FIG. 1: Hourglass in steady-state motion. The lower con- lution. Newton’s equation of motion for the position of
tainer is suspended at the upper one by damped springs.
the container u(t) reads
Choosing appropriate parameters of thespring constants the
weight of thehourglass is independent of time. [m (t)+m˜ ]u¨+ηu˙ +ku=−m (t)g−F −m˜ g, (8)
1 1 1 j 1
2
where k/2 and η/2 are the elastic and dissipative con- in Eq. (10) decays as t−1/4. Therefore, for stability of
stants of the springs and F is the force that originates thesolution,the prefactorofthis termmustnotincrease
j
fromtheimpactofthefreefallingparticlesatthesurface steeperthant1/4,i.e.,η/j >1/2. Withthiscondition,we
accordingtothe constantflowratej. Usingthe previous obtain from Eq. (10) for the asymptotic velocity u˙∞ =
expressionsform (t)andabbreviatingthelasttwoterms jg/k. Comparing with Eq. (7) we obtain k =ρAg.
1
by the constant K we obtain
Inconclusion,wefindthatalthoughasimplehourglass
inthestateofsteadyflowchangesitsweightinthecourse
(jt+m˜ )u¨+ηu˙ +ku=−jtg+K (9)
1 of time1, it is possible to construct hourglasses whose
with the solution weightremainsconstant. Tothisend,assketchedinFig.
1, the lower container should be suspended by damped
linear elastic springs of total force
j−η jkt+km˜1
u(t)=(jt+m˜1) 2j "C1J1−ηj 2s j2 !
F (u,u˙)=−ku−ηu˙ (12)
jkt+km˜ jtg+K ηjg
1
+C2Y1−ηj 2s j2 !#− k + k2 , (10)
with elastic and dissipative constants
where J (x) is the Bessel function of order n and Y (x)
n n
is corresponding Weber function and C and C are in-
1 2 k =ρAg
tegration constants. For large t these functions behave (13)
as2 η >j/2.
2 nπ π
J (x)≃ cos x− −
n
πx 2 4
r (11)
h (cid:16) (cid:17)i
2 nπ π
Y (x)≃ sin x− − ,
n
πx 2 4
r
h (cid:16) (cid:17)i
that is,the absolute value ofthe termin squarebrackets
1
K.Y.ShenandB.L.Scott,“Thehourglassproblem,”Am. Mathematical Functions,” Dover(New York,1972).
J. Phys. 53 (8), 787-788 (1985).
2 M. Abramowitz and I. A. Stegun (eds.), “Handbook of