It's all a fallacy, right? What if we added in the possibility that you are playing a biased wheel and win often and big. One night, someone was tracking you instead of the wheel, then follows you home and robs you. Can we really put any number of variables in to determine probability from an equation that is completely random? So, when I hear the million spin argument, it completely misses the waitress who spilled a drink on you just as you were attempting to lay down big on that biased wheel. Did the drink spill just on you or across the table? Did they close the table for cleanup? Change dealers? It also misses the RNG machine that just shit the bed as you had 25 down on your 1/5/25 prog. Add in any other random event you like to the mix. There are millions of possibilities.
People believe what they will. A few years back I thought I had the HG. It lasted for over 3 years and I was able to pay 2 of my kids student loans off (over$200K combined). Still have another one to go. I was riding high. It crashed and crashed big. But, the loans were paid off. I stopped playing for a few years. Did I win? I don't know yet, I'm not dead. If I never play again, which is highly unlikely, I would say yes. So, the loss was big, but the win was bigger over the long term. It's all risk versus reward with some serious amount of greed in there.
I've been on this board for quite a few years and it's more of a bitching session now than a friendly forum where folks openly share their ideas for what they believe is or will be successful. My hope is that everyone wins in the long term, regardless of what way they play.
I don't post much, but will end with this. You only need to prove anything to yourself. It doesn't matter what others think or do and it only matters when you have real skin in the game. My experience has been that simulations are great, but that's all they are, simulations. Nothing will prepare you for the real event, as there are events that will occur that you've not encountered before and variables you'll need to take into considerations that can't possibly be addressed through simulation.
Wally, well said my friend. Sounds like you had a great run! There are many variables that simulations cannot account for. One I think that has me helped many times is intuition.
I'm not saying running simulations are a bad thing when testing. It's good to know what to expect. Just saying that only looking at the very end result of 1 million spins is pointless.
Wally gator, finally some real talk.
The permutation from hell can hit you any time. Doesn't matter how much time or how many spins you delay between sessions - your luck will run out. Therefore, simulating 1 mill spins is perfect to find out if your system is a winner. To be a winner you need edge that is at least the % of the house advantage; in the long run you would then be guaranteed to be in profit. If you don't have that edge then hiding in-between sessions is not going to help. Either that or you need some way to "recover" and turn a losing session into a winning session before the next permutation from hell comes. However, so far all attempts at avoiding variance or trying to "dodge" it somehow, has failed in my attempts by the hundreds.
Tinsoldiers, I think you are asking about a system to test? Not so much a strategy. When I talk about strategy, I'm referring to my own personal rules which is almost impossible to simulate because they may change depending on how the game is going.
My strategy is to bring no more than 20-25% of my bankroll to casino. This way I can never lose my entire bankroll in a single session if variance takes an ugly turn. My strategy is sometimes start with a parachute system, playing many numbers with few chips to grind out small gains. Normally, I'll always switch to a repeater system playing few numbers.
My goal may be 500 units, and leave. But at times I hit this early and raise my goal to 1000 units using a up as you win positive progression. If it goes back down to 500, I'll stop. If it reaches 1000, I may even change my goal to 1500, with a stopping point at +1000.
Some days wins don't come so quickly. I'll play for up to 6 to 8 hours alot. Usually when I get to that 6 hour mark, that's when I'm looking for a good point to exit the game. Main reason for this is really because I'm tired of playing at that point.
My strategy will never involve steep negative progressions. I will never go from betting 1 unit a number to 50 units. If I do a negative progression, it will only be after a hit and only if there was a long gap between hits. Normally, I will never go more than 1 unit to 5 units max.
Positive progressions are a different story. If variance is on my side, I will keep adding to my wagers as long as the wins keep coming.
But looks like you are looking for a system to test. I like playing recent repeaters, no more than 4 numbers. Check out the post for Repeaters and Gaps.
You are now deviating from what you started this whole thing or at least it sounded like. Intuition and Biased wheel conditions cannot be simulated and no one is saying simulate that to death. People only say simulation when someone claims there is a mechanical method and it will never lose. I am sure anyone will agree this is completely different from intuition. There is no strategy here. Strategy can be defined, it can be repeatable executed and it can be simulated. That’s different from when I play what my mind says, that’s not strategy - as you rightly put it is intuition and no one is saying here simulate it.
General, you can't sidestep probability right? Someone can claim that 20 repeaters in a row can never happen. But you can claim that if you simulate a statistically revelent number spins, at some point it will. So, with randomness and enough test anything is possible. So, that means even your chi square method of spotting bias wheels at some point will fail in the long run. Random can spit out numbers way outside standard deviation on a wheel that isn't biased at all. You can't get around it. Your bankroll will eventually all be lost to the house edge :wink:
I don't play on a biased wheel unless I'm absolutely sure that it's biased. However, contrary to popular believe, the wheel is found before the numbers are written.
As the topic here is simulation.
General can you show us some differential equations of a sphere rolling with friction on a cone ?
Considering tilt would be nice too.
Quote from: Herby on Jul 15, 05:45 AM 2018
As the topic here is simulation.
General can you show us some differential equations of a sphere rolling with friction on a cone ?
Considering tilt would be nice too.
Here you go.
Sphere rolling on the surface of a cone
I Campos,†J L Fern´andez-Chapou,‡ A L Salas-Brito,*‡ C A
Vargas,‡§
†Facultad de Ciencias, Universidad Nacional Aut´onoma de M´exico, Apartado Postal
21-939, Mexico City 04000 D F, M´exico
‡ Departamento de Ciencias B´asicas, Universidad Aut´onoma Metropolitana, Unidad
Azcapotzalco, Apartado Postal 21-267 Coyoacan, Mexico City 04000 D F, M´exico
* Nonlinear Dynamical Systems Group, Department of Mathematics and Statistics,
San Diego State University, 5500 Campanile Drive, San Diego, CA 92182-7720, USA
Abstract. We analyse the motion of a sphere that rolls without slipping on a conical
surface having its axis in the direction of the constant gravitational field of the Earth.
This nonholonomic system admits a solution in terms of quadratures. We exhibit that
the only circular of the system orbit is stable and furthermore show that all its solutions
can be found using an analogy with central force problems. We also discuss the case
of motion with no gravitational field, that is, of motion on a freely falling cone.
Submitted to: Eur. J. Phys.
‡ On sabbatical leave from Laboratorio de Sistemas Din´amicos, UAM-Azcapotzalco, Mexico City D F,
Mexico, email: asb@correo.azc.uam.mx
§ Corresponding author. E-mail: cvargas@correo.azc.uam.mx
Sphere rolling on the surface of a cone 2
1. Introduction
Rigid body motion has always been an interesting and very usable subject of classical
mechanics [1, 2]. Many subtle points of dynamics and of mathematical techniques for
studying the behaviour of physical systems can be learned by studying the motion of
rigid bodies. On this matter see, for example, [3, 4, 5, 6, 7, 8, 9, 10]. Part of the interest
of today comes from the insight that can be gained on the behaviour of spinning asteroids
or artificial satellites and, furthermore, rigid body motion can be chaotic [11, 12, 13].
Besides, many everyday phenomena can be understood in terms of rigid body motion
at least in a first approximation. For example, balls rolling on inclines, motion in
toboggans, bowls in a bowling alley, the motion of snow boarders, the dynamics of
bicycles or wheels, the behaviour of billiard balls, and so on. Though the importance
of rigid bodies is clear, some of the problems involving rolling particles on a surface are
often modelled in beginning courses as point particles sliding on surfaces [6, 14]. This
modelling is an appropriate pedagogical device in introductory courses but we want to
show here that the problems can be addressed using a rigid body approach in more
advanced courses.
In this work we analyse the motion of a sphere that rolls without slipping on the
inside of a right circular cone under the influence of a uniform gravitational field acting
verically downwards, in the direction of the symmetry axis of the cone. The motion
of a spherical body rolling without slipping on surfaces of revolution has been recently
studied with the purpose of illuminating control processes [10]. Here our aim is to
study the motion of a sphere on the inner surface of a conical surfaceas an exactly
solvable example of rigid body motion. We obtain the general solution of this problem
expressing it in quadratures. We analyse certain qualitative features of the motion, like
the existence of a stable circular orbit, establishing an analogy with particle motion in a
central force field in two dimensions. For the case of a conical surface in free fall we find
that the general solution of the problem can be casted in terms of expressions similar
to those defining sets of ellipses and hyperbolas. We also calculate the apsidal (apogee)
angle of the center of mass (CM) motion to find the orbit’s symmetry axes and argue
that the sphere’s CM trajectory densely fills a strip on the conical surface.
2. The equations of motion
To describe the position of the sphere’s centre of mass (CM), we choose a Cartesian
coordinate system (x, y, z) such that the origin of coordinates is at the position of the
sphere’s centre of mass (CM) when the sphere simply rests on the cone (that is, it
corresponds to the vertex of the imaginary cone â€"shown dashed in figure 1â€" on which
the CM moves); the coordinates are then (see Figure 1)
x = r sin α cos Ï',
y = r sin α sin Ï',
z = r cos α,
(1)
Sphere rolling on the surface of a cone 3
Figure 1. The cone on which the sphere of radius a rolls is shown in a continuous
black line. The dashed cone is the imaginary cone on which the sphere’s centre of mass
(CM) moves. The imaginary cone is the same as the actual one but displaced upwards
a distance a cscα. The origin of coordinates is chosen as the vertex of the imaginary
cone so the vertex of the actual cone is at z0 = âˆ'a cscα. The figure also shows the
CM generalised coordinates r and Ï'. The radius of the parallel circle containing the
CM is symbolised by Ï.
where α is half the angle of aperture of the cone â€"therefore, 0 ≤ α ≤ Ï€/2, α = Ï€/2
corresponds to motion on a plane, whereas α = 0 to motion on a cylindrical surfaceâ€",
r is the distance from the origin to the CM, and Ï' is the polar angle of the sphere’s CM.
All these relations may be seen from figure 1. The tangential components of the CM
velocity, in the, respectively, meridional and parallel directions to the conical surface
(using a sort of geographical terminology), are
u = ˙r,
v = r sin αÏ'.Ë™
(2)
Therefore, Lagrangian of the system can be written as
L =
1
2
M(u
2 + v
2
) + 1
2
I[φ˙2 + 2φ˙ψ˙ cos θ + ˙θ
2 + φ˙2
] âˆ' Mgz (3)
where M is the mass, I = 5Ma2/2, is the moment of inertia of the sphere, a its radius;
g is the acceleration of gravity, and φ, θ and ψ are Euler’s angles [1]. Lagrangian (3)
with u, v, and z defined appropriately, is fairly general; the description it gives is valid
for an arbitrary surface of revolution with a sphere rolling on its surface.
The motion of the sphere is further constrained according to u âˆ' aω · eˆ1 = 0,
and v + aω · eˆ2 = 0, where ω is the angular velocity vector of the sphere, and
eˆ1 = (sin α cos Ï', sin α sin Ï', cos α), eˆ2 = (âˆ' sin Ï', cos Ï', 0) are unit vectors in the,
respectively, direction of the meridional line and the parallel circle on the cone. The
Sphere rolling on the surface of a cone 4
vector ω can be written in terms of the Euler angles, an expression that can be found in
many textbooks [1, 2, 3]. If we employ equations (2) for substituting u and v, we obtain
rË™
a
âˆ' ˙θ sin(φ âˆ' Ï') + ψ˙ sin θ cos(φ âˆ' Ï') = 0,
Ï'Ë™ sin α
r
a
+ ˙θ sin α cos(φ âˆ' Ï') âˆ' φ˙ sin α +
ψ˙
[sin α sin θ sin(φ âˆ' Ï') + cos α cos θ] = 0; (4)
these two relations just mean that the sphere rolls with no slipping on the surface of
the cone.
The equations of motion of the sphere can be obtained, from L and constrictions
(4), using Lagrange’s undetermined coefficient method[2], to get
uË™ + v
v
Ï
sin α âˆ'
2
7
aκω
= âˆ'
5
7
g cos α,
vË™ âˆ' u
v
Ï
sin α
= 0,
aω˙ + uvκ = 0, (5)
where Ï is the radius of curvature of the parallel circle at point of contact: Ï = r sin α,
κ = cos α/Ï, and ω ≡ ω3 is the component of ω normal to the cone.
The first two equations above are essentially Newton’s second law for the centre of
mass of the sphere, the third one is its rotating counterpart for a rotating rigid sphere.
With the help of the constrictions, the Euler’s angles have been eliminated from the
description.
3. Conserved Quantities in the Motion of the Sphere
It should be clear that the total energy, E, of the sphere is a constant of motion since we
are neglecting any dissipative effects. The energy of the sphere can be obtained directly
from equations (5),
E =
7
10
M(u
2 + v
2
) + 7
10
Iω2 +
7
5
Mgz. (6)
We note in passing that this expression (6) is valid, again with u and v properly defined,
for any surface of revolution. The point z = 0 has been chosen as the point of zero
potential energy [1, 10]. It can be worth noting that the factor 7/10 in the CM’s kinetic
energy, instead of the usual 1/2, comes from effects associated with the rolling-but-noslipping
condition.
For arbitrary surfaces of revolution, the energy is the only constant of motion. But
in our conical case it is simple to show, using the equations of motion (5), that the
z-component of the angular momentum of the center of mass (CM) is also a constant
of the motion,
Lz = MÏv = MÏ2Ï'.Ë™ (7)
Sphere rolling on the surface of a cone 5
Moreover, for spheres rolling on flat surfaces â€"with symmetry axis as z-axisâ€" there
exist one more constant of motion given by [10]
L+ = l âˆ'
2ab2
5(z âˆ' c)
Lz (8)
with b and c constants coming from the parametrization, in cylindrical coordinates, of
such surfaces as
z = bÏ + c, (9)
where l = Iω is the angular momentum of the sphere respect to the normal to the
surface of revolution at the point of contact and ω the rotational angular speed around
this same axis. In the conical surface b is the slope of the generatrix and c is the zcoordinate
of its vertex. The easiest way of showing that (8) is a constant is by taking
its t-derivative and using (5) to show that it vanishes. In is also worth pinpointing
that z ≥ 0 ≥ z0 ≡ âˆ'a csc α since the sphere is geometrically constrainedâ€"i. e. we are
assuming that it is not possible for the sphere to cut through the surface of the cone; z0
being the z-coordinate of the vertex of the cone on which the sphere is rolling.
Using equation (2) and the expression for Ï in the integrals of motion (6), (7) and (8),
we obtain
1
2
M( ˙r
2 + r
2Ï'Ë™2
sin2 α) + 1
7
Ma2ω
2 +
5
7
Mgr cos α =
5
7
E, (10)
Mr2Ï'Ë™ sin2 α = Lz, (11)
aω âˆ'
Lz cos α
Mr sin2 α
=
2 L+
5 Ma
≡ h, (12)
where h is a new constant. Notice that equation (11) shows that Ï' never changes sign,
i. e. the change in Ï' is monotonic.
4. The equivalent radial problem
Combining the previous equations, we can get a new expression for r which can be
expressed as
E =
1
2
MrË™
2 +
A
r
2
+
B
r
+ Cr, (13)
with E, A, B and C constants defined by
E =
5
7
E âˆ'
1
7
Mh2
,
A =
L
2
z
(1 + 2
7
cot2 α)
2M sin2 α
,
B =
2
7
Lzh
sin α
cot α,
C =
5
7
Mg cos α,
(14)
Sphere rolling on the surface of a cone 6
Figure 2. The effective potential Veff(r) is plotted against r in arbitrary units. A
particular value of the constant ‘energy’ E = âˆ'1 is also shown. The points marked
r1 and r2 are turning points, that is, points in which ˙r = 0. As the plot exhibits the
effective potential has only one minimum and thus just two turning points for any
E > Veff(rc). The plot also exhibits that Veff(r) attains its minimum value at rc and
that this is a stable minimum. The plot was made using A = 1, B = âˆ'3.5, C = 2 in
equation (15).
in the case the sphere is rotating around the cone axis [Lz 6= 0 and 0 < α < π/2] both
A and C are strictly positive constants.
Expression (13) can be regarded as the energy equation for an equivalent, purely
‘radial’ problem, with an effective potential energy term given by
Veff(r) = A
r
2
+
B
r
+ Cr, (15)
which we show in figure 2. In this figure we have also traced a constant E(=
âˆ'1, in arbitrary units) value as the horizontal line crossing Veff in two points, r1 and
r2â€"the turning points where Ë™r vanishes. This figure shows the typical behaviour of Veff.
A glance to figure 2 thus shows that the motion is restricted between the two turning
points, meaning that the sphere’s CM moves on a strip bounded by two parallel circles
on the conical surface.
The graph of Veff, figure 2, also shows a minimum value at rc. That is, an stable
circular orbit is possible at this distance rc. The actual radius of the circular orbit, as
measured from the cone’s symmetry axis, is Ïc = rc sin α.
Moreover, equation (13) can be directly integrated to yield the general solution of
the problem as
t = ±
Z r
r1
dr′
q
2
M
[E âˆ' (
A
r
′2 +
B
r
′ + Cr′)]
, (16)
Sphere rolling on the surface of a cone 7
where we have chosen the reference point as one of the turning points of the motion.
Taking into account that Ë™r = (dr/dt)Ï'Ë™
, we obtain, from equations (11) and (13), the
orbit equation as
Ï' = ±
1
sin2 α
√
2M
Z r
r1
Lzdr′
r
′2
q
E âˆ' (
A
r
′2 +
B
r
′ + Cr′
)
, (17)
where r1 corresponds to the first turning point and the double sign corresponds to the
clockwise and counterclockwise rotations, respectively. Equations (16) and (17) are the
general form of the solution of the problem in terms of quadratures. The integrals (17)
and (16) can be expressed in terms of elliptic integrals of the first and of the second
kind, but such expressions do not offer much insight into the properties of the motion.
However, a simple analysis of figure 2 says that the motion is always bounded between
r1 and r2, and that, for given E, h, and Lz, there can be only one circular orbit which
besides is stable. To determine the radius, rc, of this orbit we must take Ë™u = 0 â€"which
means that the acceleration, and thus the force, in the r-direction vanishesâ€" in (5) to
get
7Ï'Ë™2
c
rc sin2 α âˆ'
h
5g âˆ' 2aωcÏ'Ë™
c
i
cos α = 0, (18)
the suffix c meaning magnitudes evaluated on the parallel circle r = rc. Since the first
term in equation (18) is always positive and cos α > 0 because 0 < α < π/2, the circular
orbit is only possible when the initial conditions are such that
Ï'Ë™
cωc <
5g
2a
. (19)
Equation (18) is quite useful since it allows evaluating any one of the three quantities
rc, Ï'Ë™
c, or ωc as a function of initial conditions of the remaining two for starting the
motion on the circular orbit. If we know the initial values ω0 and Ï'Ë™
0, we could ascertain
whether equation (18) has a solution or not, just by checking if those initial values satisfy
equation (19). Clearly equation (19) holds if ω0 and Ï'Ë™
0 have contrary signs independent
of their magnitudes. Now equation (18) has one solution for ωc with initial conditions
r0 and Ï'Ë™
0 while it has two solutions with initial conditions r0 and ω0 one of opposite
sign to ω0 and another of the same sign, but satisfying equation (19).
Equation (18) in terms of the constants of motion Lz and h is
5 r
3
c M2
g sin2 α cos αâˆ'2 rc MLzh sin2 α cos αâˆ'7L
2
z
1 âˆ'
5
7
cos2 α
= 0.(20)
This is a cubic equation for rc that has always a real positive root; but let us pinpoint
that, due to condition (19), this equation is only valid when 2hLz < 5Mr2
c
g sin2 α. The
existence of only one real and positive root, follows from the fact that the last term in
(20) is less than zero, whereas the coefficient of the first term is positive [15]. However,
it can be much simpler to just take a look at figure 2, as it exhibits that there is one
and only one circular orbit on the surface.
Sphere rolling on the surface of a cone 8
There is a further point worth mentioning, the existence of a relationship between
the radius of the circular orbit and the r-values of the turning points, ri
, i = 1, 2,
namely
r
3
c =
2r
2
1
r
2
2
r1 + r2
+
B
C
rc âˆ'
r1r2
r1 + r2
. (21)
This is the sort of property called universal in [6]; in fact, (21) is very easily shown to
reduce to equation (13) of reference [6] in the point particle limit (that is, when B = 0).
5. Analogy with an Non Homogeneous Harmonic Oscillator and the Orbit
on a Freelly Falling Cone.
Let us substitute equations (2) in the first of equations (5), to get
r¨ âˆ' rÏ•Ë™ =
2
7
Ï•Ë™
sin α
(aω) cosα âˆ'
5
7
g cos α, (22)
where Ï• ≡ Ï' sin α. Multiplying by M and using equations(7) and (8) in the right hand
side of equation (22), we get
M(¨r âˆ' rÏ•Ë™
2
) = f(r), (23)
where
f(r) ≡
B′
r
2
+
A′
r
3
âˆ'
5
7
Mg cos α, (24)
and
B
′ ≡
2
7
Lh cot α
A
′ ≡
2
7
L
2
cot2 α
M
L ≡ Mr2ϕ˙ =
Lz
sin α
;
(25)
therefore the equations of motion of the sphere are found, as we did in section 4,
analogous to the equations of motion of a particle in a central force field. The variables
r(t) and Ï•(t) can be considered as the polar coordinates of a particle moving on a plane
under the action of the “central†force f(r) given in equation (23). If, additionally we
assume the cone is in free fall, that is, that g = 0, then the orbit of the center of the
sphere, r = r(Ï•), can be easily obtained by rewriting equation (22) in terms of Ï•, using
the relations
d
dt
= ˙ϕ
d
dϕ =
L
Mr2
d
dϕ (26)
and
d
2
dt2
=
L
2
Mr2
d
dϕ
1
r
2
d
dϕ
. (27)
Sphere rolling on the surface of a cone 9
Now, introducing Binet’s variable W ≡ 1/r, the equation of motion (23) becomes, with
g = 0 and using (26) and (27),
d
2W
dϕ2
+
1 +
2
7
cot2 α
W = âˆ'
2
7
Mh cot α
L
(28)
or, in terms of the angle in the polar plane, Ï' = Ï•/ sin α, and the vertical component of
the angular momentum, Lz,
d
2W
dÏ'2
+
1 âˆ'
5
7
cos2 α
W = âˆ'
2
7
Mh cos α sin2 α
Lz
. (29)
This is the well-known inhomogeneous differential equation for harmonic motion. If we
define â,,¦2 ≡ (1 âˆ' (5/7) cos2 α), the general solution can be written
W ≡
1
r
= A cos [â,,¦(Ï' âˆ' Ï'0)] âˆ'
2Mh sin2 α cos α
7Lzâ,,¦2
. (30)
The constants of integration A and Ï'0 can be determined by the initial conditions.
To clasify the solutions of the problem in free fall, we recognize two cases,
A) when h and Lz have different signs, i.e. hLz < 0; or
B) when h and Lz have the same sign, i.e., hLz > 0.
In any case, the general solution can be written in the form
W ≡
1
r
=
cos α
p
e cos [â,,¦(Ï' âˆ' Ï'0) ± 1] (31)
with
p ≡
7|Lz|â,,¦
2
2M|h|sin2 α
(32)
or in the form
r(Ï') = rmin
e ± 1
e cos â,,¦(Ï' âˆ' Ï'0) ± 1
(33)
with
rmin =
p
cos α(e ± 1), (34)
where e and Ï'0 are constants of integration. The sign in equation (34) is plus if hLz < 0
and minus if hLz > 0. Without loss of generality we can assume that e ≥ 0, since its
sign can be changed by substituting Ï' â†' Ï'+Ï€, that is, just by changing the orientation
of the coordinates.
The possible motions are best analysed case by case, as follows.
A) In the case that hLz > 0, we have the minus sign in equation (34) and therefore
e > 1. In this case there are two values of Ï' for which e cos â,,¦(Ï' âˆ' Ï'0) = 1. When Ï'
approaches any of these two values r â†' ∞; therefore they define two asymptotes
and the orbit corresponds to a branch of an hyperbola.
Sphere rolling on the surface of a cone 10
B) In the case that hLz < 0, the plus sign is in order in equation (34). We find now
four possibilities, e = 0, 0 < e < 1, e = 1 and e > 1. The motion in these cases is
as we describe in what follows.
B1. If e = 0, the orbit is the circle
r(Ï') = rc = p csc α
1 âˆ'
5
7
cos2 α
; (35)
this relation coincides with (20) if g = 0.
B2. If 0 < e < 1, the orbit is confined to the strip defined by the two parallel
circles on the cone, having radii
r = rmin (36)
and
r = rmax = rmin
1 + e
1 âˆ' e
. (37)
The CM of the sphere touches the circle with radius r = rmax when
Ï' âˆ' Ï'0 =
Ï€
q
1 âˆ'
5
7
cos2 α
, (38)
which means that the apsidal angle ψ (the angle swept by the CM’s radius
vector in going from rmin to rmax) is
ψ =
Ï€
q
1 âˆ'
5
7
cos2 α
> π, (39)
thus the apsides advance in each period of rotation. As the orbit is symmetric
with respect to the polar axis (r(âˆ'Ï') = r(Ï')), the angular displacement of the
apsides in each rotation of the center of the sphere around the z-axis is
∆ψ = 2Ï€[(1 âˆ'
5
7
cos2 α)
âˆ'1/2 âˆ' 1] (40)
and the angular velocity of precession of the apsides is
∆ψ
T
=
2Ï€
T
[(1âˆ'
5
7
cos2 α)
âˆ'1/2âˆ'1] = hÏ'Ë™i[(1âˆ'
5
7
cos2 α)
âˆ'1/2âˆ'1](41)
where T is the period of rotation of the CM around the z axis, and
hÏ'Ë™i ≡ 1
T
Z T
0
Ï'dt Ë™ ≡
2Ï€
T
(42)
is the mean value of Ï'Ë™
.
Since (1 âˆ' 5 cos2 α/7)1/2
can be shown to be an irrational number for 0 < α <
Ï€/2, the orbit never repeats itself thus filling completely the strip that contains
the orbit as t â†' ∞. We may say that the orbit is ergodic on the strip, that is,
that eventually the orbit would densely fill all of the strip’s surface.
Sphere rolling on the surface of a cone 11
B3. If e = 1, we can easily verify that r â†' ∞ as
Ï' â†' ±
Ï€
q
1 âˆ'
5
7
cos2 α
+ Ï'0. (43)
Therefore, the orbit is a branch of an hyperbola with asymptotes defined by
Ï' = ±
Ï€
q
1 âˆ'
5
7
cos2 α
+ Ï'0. (44)
B4. If e > 1, then there exist two values of Ï' for which
e cosr
1 âˆ'
5
7
cos2 α(Ï' âˆ' Ï'0) = âˆ'1, (45)
and therefore r â†' ∞ when Ï' approaches any of these values, which simply
define the asymptotes of the corresponding branch of the hyperbola.
The only thing that remains to be done is to obtain the value of e in terms of the
constants of motion. For this we have only to susbstitute the values of rmin and of p
into equation (13) (with C = 0) to get
e =
s
1 +
49(1 âˆ'
5
7
cos2 α)E
2Mh2 cos2 α
. (46)
Note that e depends only on E and h, and, furthermore, that 0 < e < 1, if E < 0; that
e = 1, if E = 0; and that e > 1, if E > 0. But, if hLz > 0 then B > 0 and therefore
E > 0, from which we have e > 1 always, consequently this case is equivalent to motion
in a repulsive force field.
6. Conclusions
We have studied the motion of a sphere rolling on the inner surface of a right circular cone
with opening angle α under the assumption that it does not slip. The motion was first
considered under the action of gravity and we managed to obtain an implicit solution in
quadratures. We established the existence of bounded motions and of circular orbits in
the system analysing an effective potential in which the equivalent single particle system
moves. With the help of this effective potential we were able to reduce the problem to
quadratures obtaining implicitly r(t) and thus θ(t).
However, to actually integrate the equations of motion for obtaining the centre’s
of mass orbit in terms of elementary functions, we required changing to a freely falling
frame. In such a frame we have been able to obtain solutions for the CM orbit analogous
to ones in a related central problem. We have been able to show, with the help of this
relation, that the CM can move on hyperbolic or on preceding elliptic, but never on
parabolic orbits.
The analogy between motions in our system with central-field motion and specially
with an inhomogeneous oscillator in the freely falling case, should be pointed out as
interesting and quite amusing. This aspect of the behaviour has a certain pedagogical
significance as can serve to emphasise that the same equations have always the same
solutions; it does not really matter that we are describing different physical systems.
Sphere rolling on the surface of a cone 12
Acknowledgments
ALSB has been partially supported by a PAPIIT-UNAM grant (108302), he expresses
his thaks to the Computational Science Research Center of SDSU and particularly
to Jose E Castillo and Ricardo Carretero-Gonz´alez for computational support. We
acknowledge with thanks the help of H. N. N´u˜nez-Y´epez. The cheerful collaboration of
A Simon and the late M Nick of Rolando California, and of H Kranken, P M Lobitta,
D Yoli, V Binny of the Monte Bello gang, is also gratefully acknowledged. This paper
is dedicated to the memory of our beloved friends P M Botitas, F C Sadi, M K Dochi,
and C Suri.
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If your balls are charged they will lose energie in the bend and everything falls down due to gravity :lol:
If my balls are charged, I'd be banging.