Dissipative Descent: Rocking and Rolling down an incline

cs and
BC
Dissipative Descent: Rocking and
ommerfeld Rock’n’roll Experiments Contact Conclusions
Rolling down an incline
ore
Bill Young, Neil Balmforth,
John Bush & David Vener
hn
and

The snail ball from “Grand Illusions” - $50

“A small metallic gold ball just over 2cm in diameter ... the ball does roll, but does so
incredibly slowly. To an audience it seems baffling ... inside the ball, which is actually
hollow, there is a viscous liquid and a smaller ball which is heavy... it is the smaller heavier
ball which determines the pace and this is slow because of the viscous liquid.”

The Snail Cylinder from UBC - $10,000

le Oceanographic Institution, (Received for ?? and assistance in revised in building form ??)
2 the snail cylinder.
Department of Applied Mathema
The state of the snail cylinder is specified by
3 Scripps Institution of Oceanography, La Jolla
ppendix A. ɛ(t) Derivation , χ(t) of , theΩ equations a (t) , and of motion Ω b (t)
(Received ?? and in revised fo
A.1. Geometry
e geometry is shown in Figure 1. We use a Cartesian coordinate system attac
The flow in the gap is
lane which is inclined at an angle α to the –g
Ω b
ɛ(t)
horizontal;
, χ(t)
the unit vector î points
analyzed , Ωwith a (t) , the an
Ω a
α
lubrication approximation.
Rocking Rocking b& −rolling
a = δ& ≪rolling
a
φ (
B χ
g = −g sin α î − cos α ˆk
)
arrangement. arrangement. In total, In the total, apparatus the apparatus has . mass, has mass,
A
e slope and the unit normal ˆk is perpendicular to the plane. The unit vecto
o the page in Figure 1, and clockwise rotations are positive. The position ve
= Xî + Zˆk and the gravitational acceleration is −g, where
Rocking & rolling 3
. In total, the apparatus has mass,
φ = α + χ
M ≡ m
this (X, Z)-coordinate system, the two cylinders have centres a
M
+ m
≡ b
at m
+ m
a the + f .
mpositions,
b + m f .
î + Z aˆk and Xb = X
M O
b î +
≡
Z
m
The reduced bˆk, mass a + respectively.
m
of b + m
the inner f . The cylinder requirement is m that the outer cy
lls without slipping down
The
the
reduced
plane dictates
mass of
that
the inner cylinder ′ a ≡ m
is a −
m ′ m
a ≡ ′′ (2.1
a , wh
m a
mass of the inner displaced
displaced cylinder massis of
mass m fluid. ′
a ≡of mfluid.
a − m ′′
a , where m ′′
a ≡ πa 2 Lρ is th
ss of fluid.
The mathematical ˙ formulation of the model consists of equa
The C mathematical X b = bΩ b , formulation Z b = b .
positions and rotations of the cylinders, plus of the model Navier-Stokes consis
ematical formulation
ometry implies that the positions of the model
Appendixcenter A summarizes and of the α rotations consists
inner the cylinder full of of the equations
formulation iscylinders, of motion
given byof plus the problem. the forNav
the
d rotations of the
formulation
cylinders, Appendix in
plus
all Aits summarizes the
glory,
Navier-Stokes
but ( introduce the full equations
the formulation ) Stokes approxima
the of fluid the
summarizes the part full Xformulation of a = the formulation Xdynamics. b + ɛ, in
of
all However, ɛ ≡the itsɛ
glory,
problem. sin because χ î but + cos the We
introduce χarrangement ˆk do . not
the
attack
Stokes as a who thi
a
in all its glory, incline, but introduce some subtlety the Stokes is required approximation because oneto can simply impose the the fluid ap
k^
^

thank Keith Bradley of the Coastal Research Laboratory, Woods
Institution, for assistance in building the snail cylinder.
are d’Alembert forces arising because the position of the center of the outer cyl
which the coordinates ɛ and χ are
2.1. The equations of The
referred,
equations
does not lieof in an
motion
inertial frame. The
of the inner The cylinder configuration followsofrom
the apparatus is specified by four independent variab
Ω a (t) and Ω b (t). The dynamical considerations of Appendix A indicate th
–g
1
of the inner cylinder is determined 2 m aa 2 ˙Ωa by = theT a equations , of motion,
α
( )
where T
m a + m′′ 2
a is the fluid torque.
a
Assuming
(ɛ¨χ + 2 ˙ɛ ˙χ) = f χ + m ′ a
m g sin φ − m′ a b ˙Ω b cos χ ,
B χ
that the outer cylinder f
rolls without sliding down the inclined p
A
equation of motion of the outer cylinder can be similarly broken down. Imp
those equations contain
O
the
(
forces at
)
m a + m′′ 2 the contact point with the plane, which
a
eliminate from the problem algebraically (¨ɛ −given ɛ ˙χ 2 ) = that f ɛ −the m ′
m
ag rotation cos φ − rate, m ′ ab ˙Ω b Ωsin b , also χ ,
C
4 N. J. Balmforth, f J. W. M. Bush, D. Vener and W. R. Youn
the distance rolled. This leaves a single equation for Ω b (t), as described in App
α
where fAlternatively, χ and f ɛ represent the totalthe angular fluidmomentum forces in balance, the direction of the line
perpendicular
[
to that direction, respectively.
(
The)
second terms on the]
rig
d 1
(2.3) denote the Archimedean buoyancy force on the inner cylinder, and
dt
2 m aa 2 Ω a + (M + m b )b 2 Ω b + m a + m′′ 2
a
ɛ 2 ˙χ + m ′ a
m b d f dt (ɛ sin χ) are d’Alembert forces arising because the position of the center of the out
which the coordinates ɛ and χ are referred,
+m ′ does not lie in an inertial frame
ab ˙Ω b ɛ cos χ ≈ Mgb sin α + m ′ agɛ sin φ ,
Recall: of the inner cylinder follows from
rivation of the equations of motion
Ω b
Ω a
φ
A.1. Geometry
wn in Figure 1. We use a Cartesian coordinate system attached to
ined at an angle α to the horizontal; the unit vector î points down
nit normal ˆk is perpendicular to the plane. The unit vector ĵ is
ure 1, and clockwise rotations are positive. The position vector is
he gravitational acceleration is −g, where
(
g = −g sin α î − cos α ˆk
)
. (A 1)
k^
^
inate system, the two cylinders have centres at the positions, X a =
= X b î + Z bˆk, respectively. The requirement that the outer cylinder
g down the plane dictates that
X˙
b = bΩ b , Z b = b . (A 2)
at the center of the inner cylinder is given by
(
)
X a = X b + ɛ, ɛ ≡ ɛ sin χ î + cos χ ˆk . (A 3)
where φ = α+χ, which provides an equivalent equation. Though (2.2)-(2.4)
The only
final relation (2.5) contains our 1first 2 m approximation
aa 2 ˙Ωa = T
approximation
a ,
by neglecting the int
(so far)
momentum of the fluid in the frame of its centre of mass.
As the Stokes approximation describes the flow in the annular gap, the fl
etween the centers A and B is denoted by ɛ(t), and χ(t) is the angle
centers and the Z-axis. We denote the angle between the line of
ction of gravity by φ(t) ≡ α + χ(t).
s of the fluid is
∫
where T is the fluid torque.
The Assuming lubrication a state approximation that
of instantaneous
the outer the cylinder
force gap balance. expresses rolls
The
without the forces, hydrodynamic sliding
f ɛ and f
down χ , and
the
torque,
incli
T
equation forces computed and oftorques as
motionin functions
of terms theof of
outer the the four instantaneous geometrical arrangement (i.e.
cylinder independent can variables. be similarly broken down
those equations contain the forces at the contact point with the plane,
velocities of the cylinder surfaces (given by Ω a , Ω b , ˙ɛ and ˙χ). We do this v

inner centres cylinder of (i.e., the with on cylinders κ point ≡ Aare in not figure moving, 1). The and angle position θ is measured a polar c
ρνu ɛ/δ. zz = The a
nterclockwise at the inner direction cylinder with(i.e., θ = −1 result p θ , (Bp 6) z = is0,
obtained 1
on 0 running point Aalong in figure a u using
θ h(θ) + w the z = 0 integrals: .
∮
the∮
δ − ɛ cos θ
dθ
line 1). The of centers, angle through Lubrication theThese counterclockwise narrowest must
where in
bethe solved
δ ≡point gap:
subject
b −direction use of a the to
≪ the
a. gap. velocity
with We c. Then of then θ m.
boundary
= the frame 0use gap running
conditions
a width of ‘gap the along coordinate h(θ) fluid.
theis the
cylind
1 − κ cos θ = 2π
√ ,
dθ
app lin
In our new frame of reference, the fluid flow 1 − κ
2 (1 − κ cos θ)
is dictated by the motions 2 = (1
of
passing cylinders. through soThese that and ~ the narrowest point of the gap. Then the gap wid
b#
motions
b
(z, θ) can h(θ) = be (0, = divided 0) δ −is into ɛthe cos the θ point , rotations, O ˜Ω in a = Figure Ω a − ˙φ and 1. ˜Ω b =
∮
and a “squeeze
To a # ~ flow” which the cylinder movedθ
towards each other with a s
a leading r=a+z order in δ/a, h(θ) the= lubrication δ − ɛ cos θ , equation
≡ b − asqueezing ≪ a. We outthen fluid usefrom a ‘gap the narrowest coordinate’, part of0 the zgap. To h(θ), easedefined the constru by
(1 − κ cos θ) 3 = π(2 + κ2 )
(1 − κ 2 ) . 5/2
the fluid forces and torques, we split the lubrication problem into these two pa
d ! /dt
d ! /dt
(z, θ) where = (0, 0) δ ≡is bthe Returning − apoint P
≪ a. to OWe (Bin 5), then Figure anduse eliminating 1. a ‘gapq ρνu
coordinate’, 0 z h
zz = a −1 with (B 6), we obtain
construct the full solution via linear superposition. p θ , p z = 0,
adingso order that Bin (z, δ/a, θ) " = the (0, lubrication 0) is the point equations O in Figure take the1.
form
[ ( ) ]
! A O B.1. The p
To leading order in δ/a, the R
θ (θ) rotational = 12a2 ρν
flow, u
lubrication equations take the fo
~
b# ~
b
ρνu
a # zz = a −1 1
R (z, θ)
h 3 (˜Ω a + ˜Ω h 1 − κ
2
b )
2 − 2 + κ 2 δ
To leading order, the boundary p θ , conditions p z = 0, are
a
These must be solved subject to the velocity bound
In our new frame
ρνu zz = a −1 of reference, a u θ + w z = 0 .
u
the fluid 1
p θ , p z = 0,
d ! /dt
a u flow is
R (0, θ) = a˜Ω a , w R B.2. The
(0, θ) = 0 , and u R squeeze flow, u
(h, θ) = a˜Ω b , w S R (z, θ)
(h,
θ +
θ) =
w0 . z =
ust be solvedcylinders. subject The boundary toThese the condition velocity motions are boundary can beconditions divided into on the cyli rot
The solution is
r newThese framemust and of reference, be a u S solved “squeeze (0, θ) = the wsubject S (0, ( fluid flow” θ) = to flow in thewhich isvelocity dictated theboundary by cylinder motions conditio move
rs. The These rotational
Inmotions oursqueezing part:
newcan frame be u R (z, divided θ) = 1
out of the reference, into −
fluid z ) 0 , and
the a˜Ω
from the rotations, fluid the narrowest flow ˜Ω a = isΩdictated a −part ˙φ andof
by ˜Ω a + z u S (h, θ) = ˙ɛ sin θ , w S (h,
h h a˜Ω b −
pR θ
z(h − z) .
from which it follows that
2aρν b
“squeeze cylinders. flow”
the
in These fluid
which motions forces
the cylinder can andbe torques, divided move towards
we intosplit theeach rotations, the
other
lubricatio
with ˜Ω a =
ng The out squeeze
and
the
a
fluid part:
“squeeze
from the
construct flow”
narrowest u S (z, θ) = z
the in full which
part ˙ɛ sin θ
solution the
of −
cylinder
the pS θ
h gap.
via 2aρνlinear z(h − move z) To ease
superposition. towards and the pS cons (θ) =
eac
d forces
squeezing
and torques,
out the
we
fluid
split
from
the lubrication
the narrowest
problem
part of
into
the
these
gap.
two
B.3. Forces and torques To
The ct the pressure the
full
fluid
solution follows forces
via
and
linear
torques,
superposition.
Because from global the hydrodynamic mass conservation. we split forceB.1. Integration is the dominated lubrication Theround rotational by the the problem lubrication inner flowi
cylinder
construct
gives the forces ∮
the full
and torques.
B.1. solution The rotational via linear flow, superposition.
u R (z, θ)
To leading order, the boundary conditions are
F a = −aL
p(θ)(ˆɛ cos θ + ˆχ sin θ) dθ ≡ f ɛˆɛ + f

total angular momentum dt balance, f
2.1. The equations of motion m f dt
m a a 2 Ω a + (M + m(
b )b 2 Ω )
]
The configuration(
b + m a + m′′ 2
a
ɛ
of
M + m b )b 2 Ω b + m a + m′′ the2
apparatus )
2 ˙χ + m ′
a
ɛ ˙χ + m ′ a
m b d a
m is specified b d +m ′ ab ˙Ω b ɛ cos χ ≈ Mgb sin α + m
(ɛ sin χ)
by four independent
f dt (ɛ sin χ) +m ′ ab ˙Ω
variables, ɛ(t), χ(t),
Ω b ɛ cos χ ≈ Mgb sin α + m ′ a (t) and Ω b (t). The agɛ sin φ ,
m
dynamical considerations of Appendix A indicate that the position
a + m′′ 2 f where φ = α+χ, dt which provides an equivalent equation. Though
a
(¨ɛ final
The − ɛrelation ˙χ 2 ) full = f ɛ
(2.5)
Monty − m ′ contains
of the inner cylinder where is φ determined = m
ag cos φ − m our ′ first
ab ˙Ω b sinapproximation χ , (2.3) by neglect
+m ′ ab ˙Ω b f α+χ, ɛ cosmomentum χwhich ≈by Mgb theprovides sin equations ofα + m an fluid ′ agɛ of equivalent motion, sin the φ , frame equation. of(2.5)
its centre Though of mass. (2.2)-(2.4)
(
+m ′ final
ab ˙Ω relation )
φ = α+χ, where which f χ provides and f ɛ represent an equivalent (2.5) contains
b ɛ cos χ ≈ Mgb sin α + m ′ our first approximation by neglecting the int
momentum of the fluid agɛ
the
sin
frame
φ ,
of its centre
(2.5)
m a + m′′ 2 the equation. fluid As the forces Stokes
Though in approximation the(2.2)-(2.4) directionare of describes
a
(ɛ¨χ + 2 ˙ɛ ˙χ) = f χ + m ′ a
of mass.
which provides an equivalent m
As the equation. Stokes approximation Though (2.2)-(2.4) g sin describes are φ − exact, m′ a b ˙Ω
exact, the line this
the of flow centres the andannular
elation perpendicular (2.5) contains to our that firstdirection, approximation
arespectively. state of
by
instantaneous
neglecting The second the
force terms intrinsic
balance. on b cos angular theThe χ right , forces, of (2.2) f ɛ (2.2) and andf χ , and
ntum of(2.3) the fluid denote in the frame Archimedean of f
its centre buoyancy computed
of mass. force as functions the inner of the the cylinder, instantaneous flow this
and the the annular geometrical final terms gap, arrange the fl
he ) contains Stokes areapproximation d’Alembert our first approximation ( aforces state describes arising of instantaneous the ) because velocities
flow by neglecting the
of
force position annular
the the balance.
cylinder
gap, of intrinsic the the
surfaces
The center fluid angular forces, remains of (given thef ɛ outer by
and in
Ω
fcylinder, a , Ω
χ , and b , ˙ɛ and
torque, ˙χ). W
T
e fluid the frame ofcomputed its centre as of functions mass. of the instantaneous geometrical arrangement (i.e.
m
pproximation describes velocities the a + m′′ 2
of instantaneous which the coordinates force balance. ɛ and The a χforces, are theory, referred, f
as
(¨ɛ − ɛ ˙χ
flowof the thecylinder annular 2 ɛ and
outlined does f
) = f ɛ − χ , and notin m
surfaces gap, the ′ torque, lieAppendix T
m
ag cos φ − m
(given fluid remains ′ a inertial , can
ab
by Ω ˙Ω
B,
then
because frame. be The ourrotation
snail cylinder
b sin χ , (2.3)
ted as functions of the inner of the cylinder instantaneous follows f
from geometrical
gaps, although
arrangement
more general
(i.e.
results
ɛ and a , in
exist
Ωχ) b , and
(Finn & Cox 2002).
˙ɛ and ˙χ). We do this v
aneous ies of the force cylinder balance. 4 surfaces theory, TheN. (given forces, J. asBalmforth, by outlined f ɛ Ωand a , Ωf b χ , in J. , W. Appendix M. torque, Bush, B, TD. a , because can Vener then and our be
where f W. snail R. cylinders Young have rela
χ and f ɛ represent the fluid
1forces in the direction of the line of centres and
ctions of the instantaneous gaps, although geometrical more arrangement
2 m ˙ɛ and
aa 2 ˙χ). We
˙Ωa = T2.2. do
general results (i.e. a , Lubrication
this via lubrication
approximation and (2.4) a reduced
, as outlined Appendix B, because our snail cylinders have exist ɛ and relatively
perpendicular (Finn χ) and & narrow Cox 2002).
Alternatively, to that the direction, total angular respectively. momentum The second balance, terms on the right of (2.2) and
although ylinder (2.3) where surfaces more
denote Tgeneral a is [(given theresults fluid Archimedean by torque. Ωexist a , Ω b
(Finn
In lubrication
, buoyancy ˙ɛ and & Cox ˙χ). force ( We 2002).
approximation (δ ≡ (b − a) ≪ a), the fluid forces
T
do on this the inner ) via lubrication
a , take the form (see Appendix cylinder, B), and the]
final terms
ed in Appendix d B, because our snail cylinders 2.2. Lubrication have relatively approximation narrowand a reduced model
1
ore general results dt
2 m exist aa
In 2 Ω
lubrication (Finn a + (M
&
+
Cox
m
approximation b )b
2002).
2 Ω b + m a + m′′ 2
areAssuming d’Alembert that forces thearising outer cylinder because the rollsposition withoutof sliding
a the
ɛ
(δm≡ f (b − 2 center down
˙χ + m
of
a) ≪ ′ aa), b
the the d
dt the (ɛ
inclined outer sin fluid χ) cylinder, plane, the
2.2. Lubrication approximation and a reduced model
to
which equation the coordinates of motion of ɛ and the χouter are referred, cylinderdoes can not be similarly lie in an inertial brokenframe. down. The Importantly, forces rotation and torqu
rication
of those approximation
the inner equations (δ
cylinder Tcontain ≡ (b −
a , take follows the a) ≪
from forces a), the
form (see at the fluid
Appendix contact forces and pointtorques, f ɛ =
B), with the f − 12νam′′ a ˙κ
χ , plane, f ɛ and δ 2 which (1 −one κ 2 ) can
, 3/2
e2.2. the Lubrication form eliminate (see Appendix from approximation the B), problemand algebraically a reduced
+m
given ′ model
ab ˙Ω 1
b ɛ
that
cos χ
the
≈ Mgb
rotation
sin α
rate,
+ m
Ω
proximation (δ ≡ (b − a) ≪ a), the fluid forces andftorques, ɛ = − 12νam′′ f χ , fa
˙κ
′ b ,
agɛ sin
also
φ ,
dictates
(2.5)
the distance rolled. This leaves a single 2
where φ = α+χ, which provides m aaequation 2 ˙Ωa = T a for , Ω b (t), as described
an equivalent equation. ɛ and in Appendix
(see Appendix B),
δ 2 Though (2.2)-(2.4)
(1 − κ 2 ) , (2.4) A.
f ɛ = − 12νam′′ a ˙κ
f 3/2 are exact, this
where Tfinal a is the relation fluid torque. (2.5) contains δ 2
, χ = 12νam′′ a κ(Ω a + Ω b − 2 ˙χ)
(1 −our κ 2 )
3/2
δ 2 (2.6)
+ κ 2 ) √ 1 − κ 2
and
first approximation by neglecting the intrinsic angular
Assuming momentum that
f ɛ = − 12νam′′ the of the outer fluid cylinder
a ˙κ
therolls frame without of its centre sliding
δ 2
, f (2.6)
(1 − κ 2 )
3/2 χ = 12νam′′ ofdown mass.
a κ(Ωthe a + inclined Ω b − 2 ˙χ) plane, the
equation of Asmotion thef Stokes of the approximation outer cylinder describes can be thesimilarly flow in
δ 2 the broken annular down.
(2 + κ 2 ) √ χ = 12νam′′ a κ(Ω a + Ω b − 2 ˙χ)
gap, Importantly,
the
1 − κ 2 fluid remains in
δ
those equations a state ofcontain instantaneous the 2
forces
(2
force
+ κ
at 2 ) √ T (2.7)
balance. the
1 −
contact
κ 2 a = 12νam′′ a (1 − κ 2 )(Ω b − ˙χ) − (1 + 2κ 2 )(Ω a
The forces, point with
δ
f ɛ and thef χ plane, , and torque, which 3(2 + κone T 2 ) √ a , can 1 −
then
κ 2
be
eliminate computed from the and
f χ 12νam′′ asproblem a κ(Ω
functions algebraically
a + Ω
of where
b −
the
2 ˙χ)
instantaneous given thatgeometrical the rotationarrangement rate, Ω b , also (i.e. dictates ɛ and χ) and
the distance velocities rolled.
δ 2 of This the cylinder leaves
(2 + κ 2 ) √ asurfaces single equation (2.7)
1 −Tκ a 2 = 12νam′′ (given by afor (1Ω Ω κ 2 )(Ω b − − (1 + 2κ 2 a , b Ω(t), b , as ˙ɛ and described ˙χ). We )(Ω a − ˙χ)
T
δ
3(2 + κ 2 ) √ indo Appendix this via lubrication A.
a = 12νam′′ a (1 − κ 2 )(Ω b − ˙χ) − (1 + 2κ 2 )(Ω a − ˙χ)
,
theory, δ as outlined in3(2 Appendix + κ 2 ) √ , κ(t) (2.8) ≡ ɛ(t)
1B, − κbecause 2 our snail cylinders have 1 − κrelatively δ .
2 narrow
gaps, although
′′ 2where
more general The results equations exist of (Finn motion & Cox in (2.2), 2002). (2.3), (2.4) and (2.5), and t
tities defined 2 in (2.6) through (2.8), comprise a sixth-order dyn

Scripps Institution of Oceanography, La Jolla CA 92093-0213, USA
g
A non-accelerating solution
(Received ?? and in revised form ??)
ɛ(t) , χ(t) , Ω a (t) , and Ω b (t)
b − a = δ ≪ a
b − a = δ ≪ a
δ
a ≪ 1 and sin α ∼ δ δ
a
a ≪ 1 and sin α ∼ δ a
˙Ω a = ˙Ω a = ˙ɛ = ˙χ = 0
˙Ω a = ˙Ω a = ˙ɛ = ˙χ = 0
⇒
⇒
O
δ
a ≪ 1 and sin α ∼ δ a
A
B
˙Ω a = ˙Ω a = ˙ɛ = ˙χ = 0
⇒
Ω b & Ω a ∝ 1
sin α
(this is crazy)
Ω b & Ω a ∝ 1
sin α
C
α
We are relieved to
discover that this
solution is linearly
unstable
The c. of m. lies directly above the
point of contact and the line of
centers is horizontal.

Alternatively,
m a + m′′ 2
( In lubrication the total angular
a ) approximation momentum(δ balance, ≡ (b −
(ɛ¨χ + 2 ˙ɛ ˙χ) = f χ + m ′ a
m 2.1. The equations g sin of φ motion
− ] a) m′ a b ˙Ω
≪ a), the fluid forces and torques, f χ , f ɛ an
[ T a , take the form (see(
Appendix b cos χ , (2.2)
d
f
1
The dt configuration 2m (
aa 2 Ω a + (M of)
the + m apparatus b )b Ω b m
is specified a + m′′ 2B),
)
]
δ
M + m a
ɛ 2 ˙χ + m ′ by four independent m b d a ≪ 1 and sin f dt (ɛ α variables, sin ∼ δ χ) ⇒
The
b )b
main 2 Ω b + m
approximation:
a + m′′ 2
a
ɛ 2 ˙χ + m ′ a
m b d f dt (ɛ sin χ) +m ′ ab ˙Ω b ɛ cos χ ≈ Mgb sin α + m ′ agɛ sin φ ,
ɛ(t), χ(t),
f
Ω a (t) and Ω b (t). The dynamical considerations of Appendix A indicate that the position
of the inner m a cylinder + m′′ 2
ɛ = − 12νam′′ a ˙κ a
where φ = α+χ, which provides an equivalent
δ
a
is(¨ɛ determined − ɛ ˙χ 2 ) = fby ɛ −the m equations ′ of motion,
m
ag cos φ − m ′ ab ˙Ω 2 (1 − κ 2 )
3/2
equation. , Though (2.2)-(2.4)
b sin χ , (2.3)
+m ′ final
ab ˙Ω relation (2.5) contains
( f +m ′
)
ab ˙Ω b ɛ cos χ ≈ Mgb sin α m ′ b ɛ cos χ ≈ Mgb sin α + m ′ our first approximation by neglecting the int
agɛ sin φ , (2.5)
momentum of the fluid in agɛ
the
sin
m a + m′′ 2
where f χ
where and fφ ɛ
= represent α+χ, which the fluid aprovides (ɛ¨χ forces an
+ 2 ˙ɛ in equivalent
˙χ) = the
f
f direction χ = equation. 12νam′′ frame
φ ,
a κ(Ω of a its + Ω b − 2 ˙χ)
χ + m ′ a
m g sin of φ − the Though m′ a b line ˙Ω b cos of (2.2)-(2.4)
χ centres , and are exact, this
δ (2.2)
perpendicular final to relation that direction, (2.5) contains respectively. f our first The approximation 2 (2 + κ
second terms by on neglecting 2 ) √ centre
(2.5)
of mass.
which provides an equivalent
(2.
As the equation. Stokes approximation Though (2.2)-(2.4) describes are exact, the1flow −this
κ
the right of the 2 in the annular gap, the fl
(2.2) intrinsic and angular
) contains our first
(2.3) denote momentumand
approximation a state of instantaneous by neglecting forcethe balance. intrinsic Theangular
forces, f
the Archimedean of the fluid
(
buoyancy in the frame
)
force on of its thecentre inner cylinder, of mass.
ɛ and f χ , and torque, T
e fluid in the frame of
and the final terms
are d’Alembert As the forces Stokescomputed its centre
arising approximation because
m a + m′′ 2
as of functions mass.
the position of the center of the
a T
describes the flow the annular
(¨ɛ − ɛ ˙χ 2 ) = f ɛ − m ′
m
ag cos φ − m ′ ab ˙Ω
outer gap, cylinder, the fluid to remains in
a = 12νam′′ of
a (1 the − κinstantaneous 2 )(Ω b − ˙χ) − (1 + 2κ 2 )(Ω a − ˙χ)
which theacoordinates state of instantaneous ɛ and χ are referred, force balance. b sin χ , (2.3)
f
does not The δ lie forces, in an finertial ɛ and 3(2 fframe. + χ , κand 2 ) √ geometrical arrangement (i.e.
pproximation describes
, (2.
velocities the flowof inthe thecylinder annularsurfaces gap, the(given fluid remains by The torque, Ω1 a −, in
rotation κΩ 2 b T, a
˙ɛ , can andthen ˙χ). be We do this v
aneous of the inner force computed cylinder balance. as where follows functions theory, Thefrom
forces, of asthe outlined f ɛ instantaneous and f χ , and Appendix geometrical torque, B, T a arrangement , because can thenour be (i.e. snail ɛ and cylinders χ) and have rela
ctions ofwhere velocities the instantaneous f χ and of the f ɛ
gaps,
represent cylinder although geometrical surfaces the fluid forces in the direction of the line of centres and
perpendicular 4 to thatN. direction, J. Balmforth, 1
respectively. The second terms on the right of (2.2) and
2 m more (given arrangement general by Ω a , Ωresults b ,(i.e. ˙ɛ andexist ɛ ˙χ). We do this via lubrication
theory, as outlined in Appendix aa 2 ˙Ωa B, = J.
because TW. a , M.
our
Bush,
snail κ(t) D. ≡cylinders Vener ɛ(t)
and W. R. (2.4) Young
(2.3) denote the Archimedean buoyancy force on the inner cylinder, δ . (Finn χ) and & Cox 2002).
ylinder surfaces (given by Ω have relatively narrow (2.
a , Ω b , ˙ɛ and ˙χ). We do this via lubrication
and the final terms
ed where in
gaps,
TAppendix a are is theAlternatively, although more
d’Alembert fluidB, torque. because The the general total results angularexist momentum (Finn & balance, Cox 2002).
[
forces equations arising ourbecause snail of motion cylinders 2.2.
the position Lubrication (2.2), have
(
of (2.3), therelatively approximation
)
center (2.4) ofand thenarrow
outer (2.5), cylinder, andthe a reduced
]
tohydrodynamic model qua
ore Assuming general which that results the d
the coordinates tities outer exist Indefined cylinder ɛ and χrolls are referred, withoutdoes sliding not lie down in anthe inertial inclined frame. plane, The rotation the
1
equation ofthe motion inner of cylinder the outer follows cylinder from can be similarly broken down. Importantly,
dt
2 m aa 2 Ω a + (M + m b )b 2 Ω b + m a + m′′ 2
2.2. lubrication (Finn Lubrication
in &(2.6) Cox approximation through 2002).
(2.8), comprise (δ
aand ≡ a(b ɛ 2 reduced −a a) sixth-order ≪
˙χ + m ′ a
m b model d a), the dynamical fluid forces system. andHoweve
torqu
some
those equations
In lubrication
contain
approximation T further simplifications are afforded by virtue of
the forces at the
(δ ≡
contact
(b − a)
1
2 m aa 2 point
≪ a),
with
the f dt (ɛ δ sin ≪ χ) a. Moreover, unless the slop
a , take the form (see Appendix
the
fluid B),
plane,
forces
which
and torques,
one can
f χ , f ɛ and
2.2. Lubrication approximation is small, the cylinder and accelerates reduced model uncontrollably downhill. Hence, we focus on the disti
˙Ωa = T a , (2.4)
eliminate
T
from a , take
the
the
problem
form (see
algebraically
Appendix
given
B),
guished limit in which also that sinthe rotation rate, Ω b , also dictates
the distance where rolled. T a isThis the fluid leaves torque. a single equation
+m
for ′ α
ab
Ω ˙Ω ∼ δ/a. In this limit, the
b (t), b ɛ cos
as
χ
described
≈ Mgb sin
in
α
Appendix
+ m ′ equations are systematical
proximation (δ ≡ (b − a) ≪ a), the fluid forces and agɛ sin
A.
φ , (2.5)
simplified by first non-dimensionalizing
ftorques, Assuming wherethat φ = α+χ, the outer which cylinder provides f ɛ = − 12νam′′ a ˙κ ɛ
using
= − 12νam′′ f
the χ , fa
˙κ
time ɛ and
scale,
(see Appendix B),
rolls anwithout δequivalent 2 (1sliding − κ
equation. , δ 2 )
3/2 down the Though 2 (1 − κ
inclined (2.2)-(2.4) 2 ) , 3/2
plane, the are exact, (2.6) this
equation final of relation motion of (2.5) the contains outer cylinder our first canapproximation be similarly τ ≡ 12broken m′′ by a νaneglecting down. Importantly, the intrinsic angular
those equations fmomentum contain of the the fluid forces in the at the frame contact of itspoint centre with of mass. the plane, which one can
eliminateAs from the the Stokes problem approximation algebraically
f χ = 12νam′′ a κ(Ω a + Ω b − 2 ˙χ)
describes δ 2 given(2 that + the κ 2 the ) flow √ m ′ aδ 2 g . (2.1
ɛ = − 12νam′′ a ˙κ
δ
1rotation −in κthe 2 rate, annular Ω b , gap, also the dictates fluid remains
(2.7)
We next 2
, f
(1 − κ
introduce 2 )
3/2 χ = 12νam′′ a κ(Ω a + Ω b − 2 ˙χ)
(2.6)
δ
non-dimensional variables
2 (2 + κ 2 ) √ 1 − κ 2 in
the distance a staterolled. of instantaneous This leaves aforce singlebalance. equationThe for forces, Ω b (t), as f ɛ described and f χ , and in Appendix torque, TA.
a , can then be
and and
computed as functions of the instantaneous geometrical arrangement (i.e. ɛ and χ) and
velocities of Tthe a = cylinder 12νam′′ a (1 − κ 2 (ˆΩ
)(Ω b − ˙χ) a ,
− ˆΩ b ) ≡ τ(Ω
+ 2κ a , Ω
a b ) , (2.1
f χ = 12νam′′ a κ(Ω a + Ω b − 2 ˙χ)
˙χ)
surfaces (given by Ω a , Ω b , ˙ɛ and ˙χ). We do this via lubrication
theory, as outlined in
δ
Appendix B, 3(2 because + κ 2 ) √ , (2.8)
and
δ 2
a non-dimensional (2 + κ 2 ) √ (2.7)
1 −T time κ a 2 = 12νam′′ a (1 − κ 2 )(Ω b − ˙χ) − (1 + 2κ 2 )(Ω a − ˙χ)
ˆt ≡ t/τ. δ It is
our 1also −snail κ 2 convenient to work with φ = χ + α. The
cylinders 3(2 + κhave 2 ) √ ,
1relatively − κ 2 narrow
wheregaps, on
although
suppressing
more general
the hats
results
and to
exist
leading
(Finn
order
& Cox
in
2002).
δ/a,
12νam ′′
a (1 − where
κ2 )(Ω b − ˙χ) − (1 + 2κ 2 ˙κ )(Ω a −ɛ(t)
˙χ)

Solution of the reduced equations
Ω b & Ω a ∝ 1
sin α
If the slope is not too
αlarge, we find rocking
solutions in which the inner cylinder slowly
sediments towards the outer cylinder.
1 − κ(t) ∝ t −1
δ
a ≪ 1 and sin α ∼ δ a
˙Ω a = ˙Ω a = ˙ɛ = ˙χ = 0
Ω b & Ω a ∝ 1
sin α
⇒
If the slope is large, the system locks onto a
runaway rolling solution with concentric cylinders:
Ω b ∝ t −q , X b (t) ∝ t 1−q , q ≡ 3(1 + 4µ2 )
2(1 + 2µ) 2 , µ ≡ M + m b
˙ X b =
Mg sin α
M + m b + 1 2 m a
There is a range of slopes for which both rocking
and rolling solutions co-exist (depending Mgon sinICs).
α
t
α
1 − κ(t) ∝ t −1
Ω b ∝ t −q , X b (t) ∝ t 1−q , q ≡ 3(1 + 4µ2 )
2(1 + 2µ) 2 , µ
X˙
b =
M + m b + 1 2 m a
m a
t
The decisive slope parameter is:
s ≡ a δ
M
m a
′
sin α

T a , take the form (see Appendix B),
center of the
inner cylinder
Some ɛ(t) , details χ(t) of the rocking solutions:
a, ˙κ Ω a (t) , and Ω b (t)
f χ = 12νam′′ a κ(Ω a + Ω b − 2 ˙χ)
!(t)
"(t) and X b
(t)
# a
(t) and # b
(t)
δ 2 (2 + κ 2 ) √ (2.7)
δ
1 − κ 2
and
1
a ≪ 1 and b − sin a = α δ ∼ ≪ δ a
⇒
6
0.6
T a = 12νam′′ 0.8 a (1 − κ 2 )(Ω b − ˙χ) − (1 + 2κ 2 )(Ω a
a− ˙χ)
δ δ 3(2 + κ 2 ) √ , (2.8)
1 − κ 2 4
0.4
where
0.6
˙Ω a ≪ = 1 ˙Ω and a = ˙ɛ = sin ˙χ 2 = α ∼ δ ⇒
0 a
0.2
0.4 κ(t) ≡ ɛ(t)
δ . 0
(2.9)
The equations of motion0.2
in (2.2), (2.3), (2.4) and (2.5), and the hydrodynamic quantities
defined in (2.6) through (2.8), comprise a sixth-order !2dynamical system. However,
0
˙Ω a = ˙Ω a = ˙ɛ
0
!0.2
some further simplifications are afforded Ω by virtue of δ ≪ a. Moreover, unless the slope
0 b & Ω
10 a ∝ 1 = ˙χ = 0
20 0 10 20 0 10 20
is small, the cylinder accelerates uncontrollably
Time, t
downhill. sin Hence, α we focus
Time,
on
t
the distinguished
limit in which also sin α ∼ δ/a. InΩthis limit, the equations are systematically
b & Ω a ∝ 1
simplified by first non-dimensionalizing using the time
Time, t
scale,
outer
cylinder
Limiting
sedimentation
point
f ɛ = − 12νam′′
δ 2
, (2.6)
(1 − κ 2 )
3/2
τ ≡ 12 m′′ a νa
m ′ aδ 2 g . (2.10)
and a non-dimensional time ˆt ≡ t/τ. It is also convenient to work with φ = χ + α. Then,
on suppressing the hats and to leading order in δ/a,
˙κ
= − cos φ ,
(1 − κ 2 )
3/2
κ(Ω a + Ω b − 2 ˙φ) = − sin φ ,
(2 + κ 2 )(1 − κ 2 )
1/2
b − a = δ ≪ a
α
sin α
α
We next introduce non-dimensional variables
The gap closes: (ˆΩ a , 1 ˆΩ b ) −≡ τ(Ω κ(t) a , Ω b
∝) , t −1
(2.11)
1 − κ(t) ∝ t −1
Power-law deceleration:
X b (t) ∝ t q , q ≡ 3(1 + 4µ2 )
2(1 + 2µ) 2 , µ ≡ M + m b
Ω b ∝ t −q , X b (t) ∝ t 1−q , q ≡ 3(1 + 4µ2 ) m a
2(1 + 2µ) 2 , µ ≡ M + m b
m a

Experiments with the snail-cylinder
10 N. J. Balmforth, J. W. M. Bush, D. Vener and W. R. Young
40
90
80
4
3
35
5.6 °
70
2
30
Distance, X
60
50
40
1
0
0 2 4
30
20
(a) ν=2×10 −4 , Slope 2.4 ° 10
2.4 °
10
5
(b) ν=2×10 −4
0
0 20 40 60 80 100 120
0
0 5 10 15
Time, t
Figure 5. Distance travelled along the runway for a steel inner cylinder with various slopes.
Panel (a) shows a long run at 2.4 ◦ ; the inset shows a magnification of the path. Panel (b) shows
four different slopes, as marked. The dotted lines show the best linear fits calculated for all the
experiments. There is no indication of power-law deceleration.
The experimental results are (far) simpler
way (cm)
35
30
than 2×10the theoretical model.....
−4
3.5×10 −4
(gt 2 sinα)/3
25
20
15
5.0 °
3.3 °

˙
Mg sin α
X b =
M + m b + 1 2 m t
a
)(Ω Hypothesis
b − ˙φ) − (1 + 2κ 2 )(Ω a − ˙φ) { 1
3(2 Asperities + κ 2 ) √ prevent s ≡ a = 2 Υ ˙Ω a if κ <
1 − κ 2 1
M the closure 2
sin α of Υ the ˙Ω a + C
gap χ if κ =
and
e (C maintain an effective δ mminimum ′ ɛ , C χ ) denotes the (suitably a non-dimensionalized) separation: con
of motion (2.12)b remains
κ(t) ≡ ɛ(t)
unchanged
κ ∗ = ɛ because the fric
the fray when it becomes comparable ∗ to the viscous
ller than the lubricationδ
pressure force, δ f χ . Moreover,
ce (2.12)d is not affected by either of the contact forces.
tandard model of friction, the two components of the co
We include a contact force with a “friction angle”:
|C χ | |C ɛ | tan ψ,
ion angle. There are two options contained in (5.3): eith
yThe strong radial to lock equation the cylinder and the surface torque together, equation ormust
the su
eakly andbe slide modified over one once another contact with occurs. |C χ | = |C ɛ | tan

Intro The Fluid Cylinders Sommerfeld Rock’n’roll Experiments Contact Conclusio
Microscope images of the surfaces
Steel
Steel
Steel
Steel
Microscope images of
Microscope images of the
the
surfaces
surface
Steel
Perspex
Perspex
Perspex
Perspex
Perspex
Aluminium
Aluminium
Aluminium
Aluminium
Aluminium
The Thescale scaleis is 250 250 microns microns in in total tot
The scale
Suggests is 250 microns
the the roughness in total length
scale scale
The scale is 250 microns in total length is
Suggests the theorder order roughness of of tens tens scale of of microns. is microns. indeed of
Suggests the order the of tens roughness of microns. scale is indeed o
the order of tens of microns.

Steel, 50 cS
offers insight into the variations
0.01
ζ =0.4 *
of κ
Alum, 50 cS
∗ .
The Steel, fixed 60 cSpoint Figure to which 10. The the average solution speeds converges fitted to the
ζ =0.2 *
is given experiments. by Panel
Steel, Predictions against
200 cS of slope, the withcontact the different theory symbols corresponding to diffe
centipoise) 0.005 and inner cylinders. In panel (b) we scale the spee
Alum, 200 cS
and add errorbars basedκon ∗ sin theφ variations = −s , between several exp
Steel, 350 cS
theoretical Rocking & rolling
Alum, 350 cS 0
predictions assuming V/V ∗ = (ζ ∗ κ/2) ∗ (Ωsin a + α with Ω b ) ζ ∗ = 0
3 4 5 6
0 1 2 3 4 5 6
Slope (degrees) Steel, 500 cS
circles indicate measurementsintaken φ = for − 500 centiStoke (cS) oil in
Slope (degrees)
0.35
Alum, 500 cS
bubbles are entrained in the fluid and migrate (2 + κ 2 ∗ into )(1 − ,
the κ2 ∗ narrowe )1/2 Ω b
/(3ζ *
)
cylinders.
0.3
cos φ = −C ɛ ,
average speeds fitted to the experiments. Panel (a) shows the raw data, plotted
ith the different symbols 0.25corresponding (1 −to κdifferent 2 ∗ )Ω b −
viscosities
(1 + 2κ 2 ∗ (as )Ω alabelled in
inner cylinders. In panel (b) we
0.2
6. scaleComparison the3(2 speeds + κ 2 by the factor, V ∗ , in (6.2),
∗) √ = C χ .
of 1 −theory κ 2 ∗
and experiment
ictions assuming V/V ∗ = 0.15 (ζ ∗ /2) sin α with ζ ∗ = 0.1 and 0.2. The data6.1. shown Outer by cylinder speeds
easurements taken Thefor condition 500 centiStoke (5.5)(cS) oil which a large number of small
In dimensional
for locking
terms,
now
the
reduces
limiting
to s
cylinder
< 2 √ 1
speed
− s 2 /κ
with 2 ∗ tan
roug
φ
( √
0.05
Ω 1 1 − s
← Locking a = V ΩSliding b = 1
V→
∗
0
2κ
× 2 sin s(2 α + × κζ 2 ∗) √ ∗ Max 1 − κ
2 , 2 ∗. 1 −
∗
s
0 0.1 0.2 0.3 0.4 0.5
son of theory and experiment
s
Otherwise,
6.1. Outer cylinder Equation speeds (6.1) divides the speed into three factors: a veloci
igure 9. Scaled, steady rotation rates, Ω a /(3ζ ∗ ) and Ω b /(3ζ ∗ ), for ψ = 0.15 radians.
terms, the limiting cylinder speed with rough contact is expected to be
( √ )
V ∗ bMδg
Ω 4νm ′′ ,
a = s(1 − κ2 ∗) 3/2
2κ 2 − 3 √ 1 − κ 2 ∗ cos φ tan ψ, Ω b = s(2 + κ2 ∗) √
1 1 − s
V = V ∗ × sin α × ζ ∗ Max
2 , 1 − 2
a 1 −
∗
2κ
tan ψ . where
2 ∗
(6.1)
the mainsdependence on slope, sin α, and a final factor dep
When the minimum erties. gap On the is relatively smaller slopes, narrow, (6.1) andreduces ζ ∗ ≡ √ to1 V −= κ 2 ∗(V ∗ ζ ∗ /2)
edivides addition the speed of a rough into three contact factors: Average
therefore a velocity speeds
allows scale, of
the
the
cylinder
outer cylinder
arrangement
in the experiments
to approa
≪ 1, w
solutions in the compact form,
a
Scaled rotation rates
ars based on the variations between several experiments. The two lines show
0.1
Ω a
/(3ζ *
)
rained in the fluid and migrate into the narrowest part of the gap between the
dy rolling solution. In reality, the minimum gap size, κ , is unlikely to be unifor

particle size.
Wrap the inner-cylinder αwith sandpaper
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
50 grit
CAMI vs σ (mm)
(a) Scaled speed, V / V
80 grit
b *
1 − κ(t) ∝ t −1 1.5
120 grit
0.15
150 grit
1
220 grit
0.5
Smooth
0.1 0
0 0.2
Ω b ∝ t −q , X b (t) ∝ t 1−q , q ≡ 3(1 + 4µ2 )
X˙
b =
0
0 1 2 3 4 5
Slope (degrees)
(b) 2V
Rocking & rolling b
/ ζ *
V * 15
0
0 1 2 3 4 5
Figure 11. Cylinder speeds with sandpaper-coated inner cylinders. Panel (a) shows average
speeds of the outer
Grade
cylinder scaled by
220
V
150 120 80 50 Smooth
∗ against slope for the steel inner cylinder and 500
centiStoke silicone oil. Inferred The stars (mm) indicate 0.3the 0.4 speeds 0.5with 1the1.6 usual, smooth 0.05 cylinder. In panel
(b), a further scalingExpected of ζ ∗ /2 is(mm) used to0.07 collapse 0.09the 0.12 data. 0.2 The 0.36 values of ζ ∗ are calculated using
the roughness scales, Expected σ, listed = average in table particle 3. κ(t) The diameter inset ≡ ɛ(t) plots quoted
the by
κinferred the CAMI σ−values standard against the mean
∗ = ɛ ∗
particle size of the sandpaper (according to the American CAMI standard).
Table For 3. small Roughness slopes scales the for the various grades of sandpaper (as given by the “grit” value
microns. listed). The This “inferred” estimatevalue is consistent indicates with the number images used takento of collapse the surface the data of the in figure cylinders 11; the
“expected” contact value force isprevents
the number V quoted = bMgδ
with a microscope which revealed roughness by theof American × sin α
that order. CAMI × √ 1standard − κ and refers to average
4νm
′′
2 ∗
particle slipping size. and the system
a } {{ }
To explore further the dependence of } cylinder {{ } speed on surface ≡ζroughness, we conducted
more
rolls
experiments
with constant
in which
speed
∗
the steel cylinder ≡V ∗ was first covered with differing grades of
sandpaper (more specifically, we used 50, 80 120, 150 and 220 “grit”, American CAMI
0.08 50 grit CAMI vs σ (mm)
0.05
Mg sin α
M + m b + 1 2 m a
s ≡ a δ
δ
M
sin α
m
′ a
2(1 + 2µ) 2 , µ ≡ M + m
δ
t
← sinα
m a

Ω b ∝ t −q , X b (t) ∝ t 1−q , q ≡
Conclusions
3(1 + 4µ )
2(1 + 2µ) 2 , µ ≡ M + m
m a
Mg sin α
Sandpaper experiments X˙
show b = that surface roughness controls
the speed of descent
M +
in
mrocking b + 1 2 m t
aregime.
s ≡ a δ
M
sin α
m
′ a
The κ_✳ theory has some success in rationalizing
experimental results. But κ(t) κ_✳ ≡ ɛ(t) is not κ ∗ completely = ɛ ∗
convincing.
V = bMgδ
4νm a
′′
} {{ }
≡V ∗
Nonetheless, I am now officially declaring victory over
the snail cylinder and the snail ball.
δ
δ
× sin α × √ 1 − κ 2 ∗
} {{ }
≡ζ ∗
For example, the experimental
dependence on α is not this simple.
Other inclined-plane problems are diverting....

A Granular Snail Cylinder