Fun problem
Igor Khavkine
I found the solution a little surprising and intriguing. The problem
requires no more than high school classical mechanics and some
ingenuity. Here it is for your amusement.
Consider a point particle sliding on a flat table (ignore friction).
The table has a cylindrical hole of finite depth (vertical walls, flat
bottom). The particle can approach the hole with different velocities
and with different impact parameters (the particle's motion need not be
directed toward the center of the hole). As the particle falls into the
hole, it starts bouncing off the walls and the bottom (assume elastic
collisions). Sometimes it gets stuck in the hole forever, sometimes it
escapes (bounces out). Determine the relation between the depth of the
hole, its radius, the particle's initial velocity, and impact parameter
necessary for the particle to escape after it falls in.
Enjoy!
Igor
robert bristow-johnson
Igor Khavkine wrote:
>
> Consider a point particle sliding on a flat table (ignore friction).
> The table has a cylindrical hole of finite depth (vertical walls, flat
> bottom). The particle can approach the hole with different velocities
> and with different impact parameters (the particle's motion need not be
> directed toward the center of the hole). As the particle falls into the
> hole, it starts bouncing off the walls and the bottom (assume elastic
> collisions). Sometimes it gets stuck in the hole forever, sometimes it
> escapes (bounces out). Determine the relation between the depth of the
> hole, its radius, the particle's initial velocity, and impact parameter
> necessary for the particle to escape after it falls in.
imagine a tube of the same diameter and twice the length of the hole
depth. the bottom of the hole has an ideal reflection which can be
modeled by reflecting the hole cylinder to the other side of the bottom
surface. if the point particle makes an integer number of reflections
off of the sides of the tube (starting with zero axial velocity,
accelerating in the first half and decelerating by the same amount in
the 2nd half) and gets to the rim on the other side when it will be
back down to zero axial velocity, it will make it out. if, instead it
is not touching the rim on the other side of the tube, the particle
will find itself "up a creek without a paddle" and fall back into the
hole giving it another chance to get out.
that's about as far as i can take this at nearly 2 a.m.
r b-j
tttito
Here is my try.
Let v be the particle's speed, h the depth of the hole , r its radius.
Let's consider the projection of the particles's trajectory on the flat
table HOR. Let's take as impact parameter the angle alpha that the
particle's trajectory forms with the hole's radius on HOR. . Let C be
the projection on HOR of the chord that the particle trajectory cuts
on the hole. The length L of C is L= 2r*cos(alpha) . Note that every
subsequent chord traced by the particle's trajectory on HOR while it
bounces in the hole has the same length L.
Denote by K the length of the (possibly jagged) path described by the
particle's projection on HOR between its entry point on the hole and
the point where it first reaches (and bounces on) the bottom.
Subsequent bounces are at odd multiples of K. Crucially, the particle
will be at HOR's level at even multiples of K. Since the particle's
speed is constant along the possibly jagged path t describes on HOR, we
have K=v*srqt(2g*h)/g (that's v times the time it takes to fall from
height h). Both K and L have now been expressed using the depth of the
hole, its radius, the particle's initial velocity, and impact
parameter.
The idea now is that the particle can leave the hole only grazing the
hole's boundary (it can't go higher). In order for it to graze, L/K
must be a rational number 2n/m, n and m being integers. The particle
will leave the hole when distance D it has walked along its path on
HOR is D=2nK and D=mL.
IV
boson boss
This post has inspired me to think as it brought forth memories of the
good old russian school of problems. Perhaps the key spark for the mind
is the power of theoretical analysis to obtain plenitude of information
about a system from a very limited set of known variables.
The setting of this problem is a horizontal plane with a hole shaped
like a cylinder. While thinking about defining the "impact parameter"
it has occured to me that the simplest scenario is to put the particle
in question on the edge of the hole and study from there what would
happen if we launch or propell or simply shoot out the particle
anywhere in the direction if the hole.
In respect to the demand for the particle to leave the hole after it
was captured by the hole, I see 2 distinct extreme cases:
--->> 1 <<---
One extreme case is to shoot the particle directly towards the center
of the hole (center of upper circular opening of cylinder). In this
case the path of the particle can be considered as a series of bounces
against the wall of the cylinder back and forth on a plane normal to
the horizontal ground plane, and between the oposing "ends" of the
cylindric surface. In each bounce the particle would in theory (very
idealized case) pass through the imaginary central axis of the
cylinder. Once it would reach the bottom of cylinder there could exist
a case, in my mind, that the particle would bounce upwards through a
secondary upward series of bounces.
The way I imagine, very high initial velocity v0 would drive this
particle to a "near-paralel" motion - paralell to the plate on which it
was moving. The angle under which the particle would hit for the first
time the wall of the cylinder would be close to 0 degrees (remembering
that angle is formally measured from the axis normal to the plane).
That scenario guarantees that there would be many consequtive bounces.
On the other hand, low initial velocity would perhaps not bring any
bounce at all - it would just drop down at near free-fall. Particle
could fall according to the law:
h = r^2 * g / 2 * v0^2
where h is the height of cylinder, r is some "distance" 0<r<2R, where R
is radius of cylinder, and g is gravitational acceleration. In case
r>2R there would be at least one bounce against the wall of cylinder.
In each one the impulse is ideally preserved and not absorbed by the
walls; however, falling would bring rise of kinetic energy at the
expense of gravitational potential energy which was highest at the top:
E_p = m * h * g
where m is mass of particle.
As a result the initial kinetic energy
E_k = m * v0^2 / 2
would be higher at the bottom when the impact at the bottom occurs
E_k_bottom = m * v0^2 /2 + m * h * g
The path of the particle would not affect the increase due to the fact
that gravity is a conservative force (only height matters for potential
energy).
In my opinion, if the particle is bound to get out of the hole then it
must stay within certain parameters. If there is a high v0 then there
would be too many bounces and the impact at the bottom would have large
angle of impact leading to even more bounces firstly in the
predominantly upward direction. The problem in imagining this scenario
is that normally any ball would lose a lot of energy in each bounce.
Therefore the would be an optimal v0 which leads to optimal number of
bounces while still preserving enough kinetic energy to jump out of the
hole. In our case the particle ideally doesn't lose energy. Instead,
particle at free fall and E_p=mgh=mv^2/2 would bounce up again and
again to the same height. Therefore, pushing the particle in such
manner to hit the center of the bottom would bounce it out to the other
side of the hole-opening.
h = R^2 * g / 2 * v0^2
--->> 2 <<---
The second extreme case has provided more fun for me. It is an idea to
shoot the particle from the edge of the hole in near tangential manner,
but still so that it would enter the cylinder. In this case the
particle would spiral down to the bottom and bounce off only once at
impact from the bottom. Hopefully it would then spiral upward again and
leave the hole. (And, at our great great fortune this particle is not a
ball and has neither friction nor spin).
If the v0 is great, then there would be many rounds in the spiral until
it reaches the bottom. That would definitely not make it bounce out.
Very small v0 would also have negative effect.
Thus my first idea was that the interesting aspect of this problem is
to define the motion in the spiral as a mutiply of 2 vector components.
One component is along Y axis which is a vertical axis starting from
the center of the bottom. This component is given by the laws
v_y = sqrt(2gh) -free fall from up to down
h = v_y^2 / 2g -oposite of free fall from down to up
where v_y is the velocity component at the bottom right after impact.
The second component is the rotation of particle. Normal ball would
lose energy on friction, but not this one. This one stays in the circle
with the same kinetic energy
E_k = m * v0^2 / 2 = m * (v_x + v_y)^2 / 2 + mgh
Acctually the v_x and v_y are same at the bottom just before and after
bounce. Thus, if the above equation is true it represents the
condition: there should be enough of m*v0/2 to bounce the particle from
the bottom in any case if the angle of shooting is near tangential.
So, if the impact parameter is an angle counting from zero, meaning
that the particle is on the line which contains the center point of the
hole, to near 90 degrees meaning that particle is at near tangential
path towards the hole, v0 varies from something very small to v0 =
|sqrt(r^2 * g / 2h)|
~~~~~~~~~~~~~~~~~~~~~~~~~
Thank you for this fine problem and I do hope you will find plenty of
those olympic goodies. However, if I have to write this much every
time, I hope to live long and marry rich.
mahdiarnt
My idea is like that of IV (tttito). Just to illuminate what he/she has
said :
One should be able to appreciate that the motion of the point is of the
form depicted below, if one ignores the collisions ("*" shows
periodicity; the distance between *s is 2K). And when the collisions
are brought into account, the same picture survives with the
modification that the collisions will occure at periodic distances like
C1, C2, C3, etc (the distance between them being L):
____ ________ ________
\__ __/ \__ __/ \__
\_ _/ \_ _/ \_
\/ \/ \...
* C1 * C2 *C3
(hope the figure not be corrupted in the final view)
Now it's both necessary and sufficient that an * and a C_n eventually
overlap. This means that (L,2K) \neq 1, i.e., L and 2K must be
commensurable.
Roland Franzius
At least you should specify if you mean a force free 2-dimensional
motion on that 3 circle surface with prescribed trajectory behavior at
the boundaries or a motion in a plane in a gravitational field where
depth and walls of the hole have only energetic and no geometric
meaning. Or someting else.
--
Roland Franzius
Oh No
There are no energy losses, and no change in angular momentum, so if its
got enough energy to enter the hole it has enough to escape.
Regards
--
Charles Francis
substitute charles for NotI to email
John Bell
Firstly, the particle will not even fall into the hole unless we invoke
a gravitational field relative to the table.
If we do, then loss in gravitational potential energy on reaching the
bottom will then be reflected by a corresponding gain in upward kinetic
energy, if we assume perfectly elastic collisions. Therefore, I see no
reason why the particle should not eventually get out again.
Consequently I think you have defined too few parameters for the
problem to make sense (or too many).
John Bell
http://global.accelerators.co.uk
(Change John to Liberty to respond)
J. G. Waller
especulative/hypothetical
/idealized, so the solution must belong to the same kind. But, wait a
minute!,
what about the mass of that point particle?, what about the
gravitational
potential if any?.
- If we assume there exist a non null gravitational potential there,
then
1. if the point particle is endowed with a non zero mass value,
then it will fall passing through the plane table, as friction
is ignored,
and it will never reach the hole boundary at any non zero
initial velocity.
2. if the point particle is massless, then given any non zero
velocity,
it will fly over the hole boundaries without falling into it.
- If we assume there exist a null gravitational field, apply the above
2. solution point,
because the point particle can be regarded, in this case, as
massless.
Uncle Al
> > I recently ran into a fun problem from an old Russian physics olympiad.
> > I found the solution a little surprising and intriguing. The problem
> > requires no more than high school classical mechanics and some
> > ingenuity. Here it is for your amusement.
> >
> > Consider a point particle sliding on a flat table (ignore friction).
> > The table has a cylindrical hole of finite depth (vertical walls, flat
> > bottom). The particle can approach the hole with different velocities
> > and with different impact parameters (the particle's motion need not be
> > directed toward the center of the hole). As the particle falls into the
> > hole, it starts bouncing off the walls and the bottom (assume elastic
> > collisions). Sometimes it gets stuck in the hole forever, sometimes it
> > escapes (bounces out). Determine the relation between the depth of the
> > hole, its radius, the particle's initial velocity, and impact parameter
> > necessary for the particle to escape after it falls in.
You folks ought to get together, assemble a scholarly treatment with
dense equations and graphics, and submit to a major golfing magazine.
It would be amusing on so many levels.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz3.pdf
Richard Saam
Conditions for particle getting out of hole
Maybe something like "QM particle in the box"
2g/D = v^2 n^2 / r^2
g = gravity accleration
D = hole depth
v = initial particle horizontal velocity
n = integer
r = hole radius
Richard
Oh No
>I found the solution a little surprising and intriguing. The problem
>requires no more than high school classical mechanics and some
>ingenuity. Here it is for your amusement.
>
>Consider a point particle sliding on a flat table (ignore friction).
>The table has a cylindrical hole of finite depth (vertical walls, flat
>bottom). The particle can approach the hole with different velocities
>and with different impact parameters (the particle's motion need not be
>directed toward the center of the hole). As the particle falls into the
>hole, it starts bouncing off the walls and the bottom (assume elastic
>collisions). Sometimes it gets stuck in the hole forever,
It can only get stuck forever on a repeating path. There is no way it
can enter into such a path.
> sometimes it
>escapes (bounces out). Determine the relation between the depth of the
>hole, its radius, the particle's initial velocity, and impact parameter
>necessary for the particle to escape after it falls in.
>
>Enjoy!
>
>Igor
>
Regards
Cl.Massé
1145745718.4...@u72g2000cwu.googlegroups.com
> I recently ran into a fun problem from an old Russian physics olympiad.
> I found the solution a little surprising and intriguing. The problem
> requires no more than high school classical mechanics and some
> ingenuity. Here it is for your amusement.
>
> Consider a point particle sliding on a flat table (ignore friction).
> The table has a cylindrical hole of finite depth (vertical walls, flat
> bottom). The particle can approach the hole with different velocities
> and with different impact parameters (the particle's motion need not be
> directed toward the centre of the hole). As the particle falls into the
> hole, it starts bouncing off the walls and the bottom (assume elastic
> collisions). Sometimes it gets stuck in the hole forever, sometimes it
> escapes (bounces out). Determine the relation between the depth of the
> hole, its radius, the particle's initial velocity, and impact parameter
> necessary for the particle to escape after it falls in.
It seems not so difficult nor intriguing. We'll use symmetry everywhere.
The level of the table is the maximum height achievable by the particle,
therefore it escapes if and only if both at the table level and at the edge
of the hole. Consider the intersections of the cylinder and the initial
particle trajectory. The particle must bounce just at the vertical of the
middle of the two points. Between the bounce and the second point, the
trajectory is the same as between the first point and the bounce, but time
and space inverted/reflected. Now, given the impact parameter, it remains
to calculate the initial velocity so that the time to get at the middle
point from the first intersection for an horizontal motion is equal to the
time for falling the distance equal to the depth of the hole. The only
possible snag is that friction must be absolutely zero, and not merely
negligible.
Obviously, they aren't the only solutions. There is a denumerable
infinity of solutions because of multiple bouncing. For example, the speed
calculated above divided by two, three...
If the trajectory passes by the axis of symmetry of the hole, we have also
the solutions with each speed multiplied by two, three... for one, two...
bouncing of the wall, some of them having already been found.
There are still more solutions in the non symmetric case when the point
bounces off the wall, but I think it is enough for now.
--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.
John F
: I recently ran into a fun problem from an old Russian physics olympiad.
Let r = radius of hole,
h = height (depth) of hole,
v = initial velocity (in horizontal plane) of particle,
\theta = angle between v and radius vector at the moment
particle first crosses circumference of hole,
a = acceleration of gravity.
Starting out with no vertical component of velocity,
the particle will first reach the bottom of the hole at
time \sqrt{2h/a} _regardless_ of its collisions with the
sides of the hole. Then it will bounce up and reach the top
at T=2\sqrt{2h/a}. And, since the horizontal speed remains
v at all times, the total horizontal distance travelled
when it reaches the top is S=vT after the first bounce,
2S after the second, ..., nS after the nth (n an integer).
Now let's go back to the top of the hole, when the particle
first crosses its edge, making an angle \theta with a
radius vector. The particle begins falling, and collides
with the hole after travelling a horizontal distance
L=2r\cos{\theta}. It "reflects" making the same angle
\theta, and thus collides with the hole after travelling
horizontal distances L, 2L, ..., mL (m an integer).
To escape the hole, the particle must be exactly at
the hole's edge when it's at the top of one of its
bounces. That is, there must exist integers n and m
such that nS = mL. Or m = n(S/L). The necessary
and sufficient condition for this is that
S/L = (v\sqrt{2h/a})/(r\cos{\theta}) is rational.
--
John Forkosh ( mailto: j...@f.com where j=john and f=forkosh )
Jay R. Yablon
Actually, the angle does not matter. It will escpae if it hits bottom m/n
of the way across in whatever path it takes. Number of traversals will be
the same. The exit angle will depend only on the number of traversals, and
the entry angle.
Jay.
Boo
> can enter into such a path.
>
enough ?
--
Boo
John Bell
Perhaps this question would make more sense if you:
1) Specify a less than 100% efficiency in elastic collisions
2) Specify a finite height of the particle
3) Specify a gravitational field.
1) will admit the possibility that the particle can get trapped
2) will then re-admit the possibility that the particle can escape
regardless
3) will (a) ensure that the particle can fall into the hole in the
first place and
(b) introduce some relevance to the depth of the hole.
You would then also have a question with some physical significance.
Joe Warner
No, not with infinite time.
robert bristow-johnson
>
> imagine a tube of the same diameter and twice the length of the hole
> depth. the bottom of the hole has an ideal reflection which can be
> modeled by reflecting the hole cylinder to the other side of the bottom
> surface. if the point particle makes an integer number of reflections
> off of the sides of the tube (starting with zero axial velocity,
> accelerating in the first half and decelerating by the same amount in
> the 2nd half) and gets to the rim on the other side when it will be
> back down to zero axial velocity, it will make it out. if, instead it
> is not touching the rim on the other side of the tube, the particle
> will find itself "up a creek without a paddle" and fall back into the
> hole giving it another chance to get out.
since my last post submitted was not posted (and it's been at least a
day or two), here is my latest stab at it. it uses the concept above
of reflecting the cylinderical hole about its bottom surface (making it
a tube or twice the length) and separating the vertical and horizontal
motion of this particle. i am also assuming no kinetic energy goes to
rotation of this point particle and that all reflections are ideal
(angle of incidence = angle of reflection).
first, we determine the time it takes to go down the hole and bounce
back to the top: not too hard for bone-head enguneers like me:
T = 2 * sqrt(2*h/g)
where "h" is the height of the hole and "g" is the acceleration of
gravity.
now consider the one bounce case (one bounce off of the bottom of the
hole, perhaps multiple bounces off of the side) and view the motion of
the particle from directly above the center of the hole. now it's a 2
dimensional problem and it's whacking the side of this circle making
nice chords. now, if it gets out in 1 bounce (off of the bottom), it
will have to make an integer number of bounces or chords in exactly the
time it takes to go down and bounce back up (or T).
let "r" be the radius of the cylinderical hole, "s" be the distance
that the particle misses the cylinder axis when it first passes (s = 0
if you shoot the particle dead center at the hole, s = r if it hits the
outside rim, if s > r, you missed the hole), and "v" to be the
horizontal velocity. those are all of the parameters needed.
as a matter of geometry, the length of the chord, "c" is
c = 2 * sqrt(r^2 - s^2)
each reflection off of the side of the hole will result in another
chord trajectory of the very same length because the angle of incidence
= angle of reflection. in T seconds the horizontal distance travelled
is:
v*T = 2*v*sqrt(2*h/g)
now the number of chord lengths completed is:
v*T/c = (2*v/c)*sqrt(2*h/g) = (v/sqrt(r^2 - s^2))*sqrt(2*h/g)
= v*sqrt( (2*h) / (g*(r^2 - s^2) )
now, if that value is a positive integer, the particle makes it out
with one vertical bounce. that is:
v*sqrt( (2*h) / (g*(r^2 - s^2) ) = n = integer
now to make it out with "k" vertical bounces (k must be an integer),
then
v*(k*T) = k * v*sqrt( (2*h) / (g*(r^2 - s^2) ) = n = integer
or
v*sqrt( (2*h) / (g*(r^2 - s^2) ) = n/k = any positive rational
number.
now this becomes a real analysis problem. there are a fuck of a lot
more irrational numbers living in this universe than there are rational
numbers. (same is true about people, but that's another discussion.)
so does that mean we can never expect the quantity:
v*sqrt( (2*h) / (g*(r^2 - s^2) )
to be exactly rational? indeed sqrt( (2*h) / (g*(r^2 - s^2) ) is not
rational unless (2*h)/(g*(r^2 - s^2) is the square of something
rational, but since (2*h)/(g*(r^2 - s^2) is not dimensionless, i have
trouble thinking about that in terms of rational or irrational. if you
choose units of time and length and are lucky enough that (2*h)/(g*(r^2
- s^2) comes out to be the square of some rational number of units of
1/velocity, i'll just change the units so that my unit velocity will
sqrt(2) times your original unit.
so, even though i'm a bonehead engunneer, i think the particle *never*
gets out of the hole.
that's my story and i'm stickin' to it.
r b-j
boson boss
coming in?
Igor Khavkine
> Let v be the particle's speed, h the depth of the hole , r its radius.
> Let's consider the projection of the particles's trajectory on the flat
> table HOR. Let's take as impact parameter the angle alpha that the
> particle's trajectory forms with the hole's radius on HOR. . Let C be
> the projection on HOR of the chord that the particle trajectory cuts
> on the hole. The length L of C is L= 2r*cos(alpha) . Note that every
> subsequent chord traced by the particle's trajectory on HOR while it
> bounces in the hole has the same length L.
>
> Denote by K the length of the (possibly jagged) path described by the
> particle's projection on HOR between its entry point on the hole and
> the point where it first reaches (and bounces on) the bottom.
> Subsequent bounces are at odd multiples of K. Crucially, the particle
> will be at HOR's level at even multiples of K. Since the particle's
> speed is constant along the possibly jagged path t describes on HOR, we
> have K=v*srqt(2g*h)/g (that's v times the time it takes to fall from
> hole, its radius, the particle's initial velocity, and impact
>
> The idea now is that the particle can leave the hole only grazing the
> hole's boundary (it can't go higher). In order for it to graze, L/K
> must be a rational number 2n/m, n and m being integers. The particle
> will leave the hole when distance D it has walked along its path on
> HOR is D=2nK and D=mL.
Italo's solution is most like the one I found myself. However, I think
we both missed the limiting possibility pointed out by boson boss. That
is the case of the limit L -> 0, in other words, the particle is shot
into the hole right at its edge, so that it smoothly slides along the
cylindrical instead of bouncing off it. In this case, the particle
always escapes.
After everyone had some time to think about the problem, I can say more
about why I found it intriguing. There are two reasons and, as it is
not unusual for this kind of problem, they have to do with its
generalizations.
The first is that the answer depends crucially on the geometry of the
hole. The solution is simple to obtain because the of the hole's
symmetry. But the same question can be posed if it were square,
triangular, elliptical, or any irregular shape you care to consider.
What is interesting then is the inverse problem. Given just the
relationship between the the initial velocity and the impact parameter
(and, likely, the direction of approach) what can we say about the
geometry of the hole?
The second is that the answer has something to do with the rational
numbers, the ratio 2K/L must be rational. The rationals are conspicious
because they are dense in on the real line yet are a set of measure
zero. In other words, if we could perform the experiment described in
the problem by randonly selecting initial conditions for the particle's
motion, we would find zero probability of escape (as long as the impact
parameter is smaller than the hole radius). But there does exist this
singular or critical set of initial conditions that allow escape. This
criticality is closely tied to the idealizations present in the
problem. Analysis of idealized situations can often tell us much about
the limiting behavior of realistic ones, none of which actually exhibit
singularities. However, an important inverse problem appears here as
well. How do idealized singularities get resolved in more realistic
situations?
For instance, if we add friction or non-elastic collisions, then as
long as the particle falls into the hole it will never get out. It will
loose kinetic energy and thus well never be able to jump out. If make
the particle in to a ball or spherical shell of finite size, then my
suspicion is that it will also never escape, unless circumstances
conspire to convert all rotational kinetic energy into translational
vertical motion. If, on the other hand, the walls are made to slope
outward, there should be continuous ranges of initial velocity and
impact parameter for which escape is possible. What if the walls were
to slope inward instead? Or, one could keep all the idealizations, but
consider another resolver of singularities: quantum mechanics. The
table can be modeled as an impenetrable potential wall, while gravity
can be provided by a linear potential. What do solutions of the single
particle Schroedinger equation with such a potential tell us about the
escape probability? Is this information best encoded this potential's
stationary states, in the scattering cross section, or something else?
How does the quantum solution reduce to the classical one as hbar -> 0?
It seems that the most interesting of the above questions are quite
beyond high school mechanics.
Igor
boson boss
> 2g/D = v^2 n^2 / r^2
What kind of box?
boson boss
For example, what if the cylinder becomes a conus. For a conus with
small hole and big bottom it would be less likely to have a bounce-out.
For a big hole and small bottom there would be big chance for bounce.
And in case of perfect cylinder it would be "just right" - perfectly
fair.
If there is a balance or some type of dynamics between the potential
and kinetic energy, then the kinetic energy would be concentrated down
to the ground in case of ordinary conus. It would be more "fairly"
distributed in a cylinder. To be precise, from ground up there would be
a distribution of potential energy E_p=mgh, therefore linear.
Circle of the hole would rule the distribution of kinetic energy (we
fixate the size of bottom for consideration), therefore the impact
parameter would be defined as a line that goes across the circle of
hole 0<line<=2R. And v0 would rule the kinetic energy that goes inside
anyway. Mass would be lost somehow.
Now the question would be, what energy does it take for the particle to
stay, and what to go away.
In fact, projection of the light through the cylinder (normal to axis)
potraits just yet another function about which we don't have to know
everything.
(...)
Jay R. Yablon
news:1145745718.4...@u72g2000cwu.googlegroups.com...
Initial impression of how to approach this:
Start by looking at the case where the particle goes straight to the center
of the hole (direct impact parameter -- great putt!). If the hole has a
radius r, the particle will jump back out if its velocity is such that it
strikes the bottom of the hole at the exact center, because its upward path
after hitting the bottom will precisely reflect the downward path due to the
perfect elasticity assumption. It will bounce out at the opposite side of
its entry, and continue forward.
The next way for it to escape is to hit 1/3 of the way across. It will then
bounce up and reach escape height 2/3 of the way across, and hit bottom
again the whole way across. Then, it will precisely reverse trajectory and
come out the same way it came in, and continue back toward the golfer.
(Sounds like my putts.)
At 1/4, it will escape after a second bounce at 3/4, and move forward on
escape. At 1/5. escape but backwards. 1/6, escape but forward, etc. 1/n
where n is even is forward escape. 1/n where n is odd is backward escape.
Now, let's look at 2/3. Hits bottom at 2/3, 6/3, 10/3, 14/3 etc. Hits top
at 4/3, 8/3, 12/3. 12/3=4 is the answer. Goes back and forth twice (four
traversals), then escapes backwards.
Next, 3/4. Bottom at 3/4, 9/4, 15/4, 21/4. Top at 6/4, 12/4, Forward
escape after three traversals.
Next, 4/5. Top (that what counts) at 8/5, 16/5, 24/5, 30/5. Bingo. Six
traversals, backward escape.
Next, 3/5. Top at 6/5, 12/5, 18/5, 24/5, 30/5. Again, six traversals,
backward escape.
Next, 2/5. Top at 4/5, 8/5, 12/5, 16/5, 20/5. Backward escape, four
traversals. We did 1/5 before.
Next, 5/6. Top at 10/6, 20/6, 30/6=5 traversal, forward escape.
General rule: escapes if velocity, radius and depth (and the gravitational
acceleration) are such that it strikes the bottom m/n of the way across the
hole, where m and n are integers. If n is even it is forward escape. If n
is odd it is backward escape.
The impact parameter at first thought does not change anything: it still
needs top strike m/n of the way across the hole on whatever path it takes
toward the hole. This means that for a less than square impact, the
velocity will need to be reduced proportionally to the length of hole over
which it travels. BUT, on second thought: when it hits the hole wall, it
will not bounce straight back any more. It will go at an angle. I suspect
that you will then also need to consider paths which are triangles, squares,
pentagons, etc. That how I'd lay this out. The quantization m/n will carry
over to the angles of incidence, which will have similar quantum
constraints. That is step 2 of this problem. The interesting result here
will be the angles at which it bounces out.
One should also consider the extreme case of an exact tangential graze.
Then, as I think was pointed out earlier, it will spiral in, hit bottom,
then bounce back out. The exit angle will depend on depth, radius, and
velocity.
After all of this is laid out, then one reverse engineers the math to get
the formula involving the "depth of the hole, its radius, the particle's
initial velocity, and impact parameter," assuming, e.g. a sea-level
gravitational acceleration on earth at the equator, or some such thing.
Jay.
_____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
Henning Makholm
> It can only get stuck forever on a repeating path. There is no way it
> can enter into such a path.
It can also get stuck on a non-repeating path, if the ration between
the time one vertical bounce takes and the time it takes to move from
one wall to the other is irrational.
(We're clearly idealizing here - if we were not idealizing, the
particle would get stuck because the bounces cannot be completely
elastic and so drain the available energy).
--
Henning Makholm "Det er trolddom og terror
og jeg får en værre
ballade når jeg kommer hjem!"
John Bell
forever.
Cl.Massé
444d4159$0$20289$626a...@news.free.fr
> Obviously, they aren't the only solutions. There is a denumerable
> infinity of solutions because of multiple bouncing. For example, the
> speed calculated above divided by two, three...
>
> If the trajectory passes by the axis of symmetry of the hole, we have also
> the solutions with each speed multiplied by two, three... for one, two...
> bouncing of the wall, some of them having already been found.
Taking these two classes of solution together in the symmetrical case, we
see that each solution is labelled by a rational number, while remaining
denumerable. Consequently, almost all initial speeds cause the point to be
captured, but there is an infinity of solutions in any finite interval of
speed for which the point escape, the usual Q versus R stuff. I presume
that's what would make "the solution a little surprising and intriguing."
That's actually John Forkosh's solution, but less general and less elegant.
robert bristow-johnson
if and only if v^2*(2*h) / (g*(r^2 - s^2)) = the square of a
(dimensionless) rational number, the particle gets out of the hole.
otherwise no. at least in this mathematically perfect universe.
g = accelleration of gravity
h = hole depth
r = hole radius
s = minimum distance that the particle passes from the axis of the hole
v = horizontal speed of particle
r b-j
Ken S. Tucker
>From a wiki article about oscilloscopes,
http://en.wikipedia.org/wiki/Lissajous_curve
Odd and Even harmonics as viewed in the X-Y plane
with a stable Lissajous and the impact square, requires
a rational "a/b" in that article, (Y being up).
> The impact parameter at first thought does not change anything: it still
> needs top strike m/n of the way across the hole on whatever path it takes
> toward the hole. This means that for a less than square impact, the
> velocity will need to be reduced proportionally to the length of hole over
> which it travels. BUT, on second thought: when it hits the hole wall, it
> will not bounce straight back any more. It will go at an angle. I suspect
> that you will then also need to consider paths which are triangles, squares,
> pentagons, etc. That how I'd lay this out. The quantization m/n will carry
> over to the angles of incidence, which will have similar quantum
> constraints. That is step 2 of this problem. The interesting result here
> will be the angles at which it bounces out.
With a 2nd scope, a side view of the Lissajous can be generated
in the Z-Y plane that shows the side view motion, meaning we can
project the particle motion on two 2D planes.
I'm *guessing* that would be harmonic in the related Z-Y plane to
produce a particle escape as Jay and others found in the X-Y plane.
..
> Jay.
Regards
Ken
Richard Saam
"Consider a point particle sliding on a flat table (ignore friction).
The table has a cylindrical hole of finite depth (vertical walls, flat
bottom). The particle can approach the hole with different velocities
and with different impact parameters (the particle's motion need not be
directed toward the center of the hole). As the particle falls into the
hole, it starts bouncing off the walls and the bottom (assume elastic
collisions). Sometimes it gets stuck in the hole forever, sometimes it
escapes (bounces out). Determine the relation between the depth of the
hole, its radius, the particle's initial velocity, and impact parameter
necessary for the particle to escape after it falls in."
Basic Facts (To be appreciated by High Schoolers):
1. No matter how the point particle enters the "cylindrical hole"
with horizontal velocity v
because of elastic collision assumption
the particle will eventually come
to the top entry level "flat table surface" of the "cylindrical hole".
"conservation of energy and momentum"
In order to escape, the particle must come
to the "cylindrical hole" top at the hole rim.
"Cylindrical hole" depth is 'D'.
This defines a particular incremental constraint 'n'
Therefore
v = 2 r / (n t)
2. No matter how the point particle
enters the "cylindrical hole"
with horizontal velocity
and bounces on vertical walls
the time to reach the bottom 'D'
(and come to the top)
is the conventional "free fall time" t.
Therefore
D = 1/2 g t^2
then:
With Facts 1 and 2
2g/D = v^2 n^2 / r^2
g = gravity accleration
Ralph Hartley
Very good!
Now do the quantum version.
Assume the walls are totally impenetrable (zero probability of finding
the particle inside).
Ralph Hartley
Ralph Hartley
> It seems that the most interesting of the above questions are quite
> beyond high school mechanics.
It is indeed so.
> the answer depends crucially on the geometry of the
> hole. The solution is simple to obtain because the of the hole's
> symmetry. But the same question can be posed if it were square,
> triangular, elliptical, or any irregular shape you care to consider.
..
> The second is that the answer has something to do with the rational
> numbers, the ratio 2K/L must be rational. The rationals are conspicious
> because they are dense in on the real line yet are a set of measure
> zero. In other words, if we could perform the experiment described in
> the problem by randonly selecting initial conditions for the particle's
> motion, we would find zero probability of escape (as long as the impact
> parameter is smaller than the hole radius). But there does exist this
> singular or critical set of initial conditions that allow escape.
This is a *generic* property of this problem, and does not depend at all
on many of the details. It is a consequence of the Poincaré recurrence
theorem.
The hole can have any (possibly parameterized) shape, the particle can
have internal degrees of freedom (e.g. a rolling ball, an *elastic*
ball), or any of the changes you suggest, but *still* if you fix all the
parameters but one, then the remaining parameter will have a dense, but
measure zero, set of values for which the particle escapes.
> For instance, if we add friction or non-elastic collisions, then as
> long as the particle falls into the hole it will never get out. It will
> loose kinetic energy and thus well never be able to jump out. If make
> the particle in to a ball or spherical shell of finite size, then my
> suspicion is that it will also never escape, unless circumstances
> conspire to convert all rotational kinetic energy into translational
> vertical motion. If, on the other hand, the walls are made to slope
> outward, there should be continuous ranges of initial velocity and
> impact parameter for which escape is possible. What if the walls were
> to slope inward instead?
*None* of those things change the answer.
Even friction and inelastic collisions are OK, with two stipulations:
(1) Everything starts at absolute zero temperature. (otherwise the
particle always escapes).
(2) The number of degrees of freedom, including "thermal" ones, is
finite (if (1) is true but (2) is false then there may be only finitely
many escaping initial configurations, or none at all). This implies that
no heat escapes from the hole.
Note that almost all of the escaping paths are *very* long. Given only
that the particle does escape, and for any time T, there is a finite
number of paths shorter than T but infinitely many that are longer. In
fact, it is impossible to define the distribution of escape times (this
is OK since it is conditioned on an event with probability zero).
> Or, one could keep all the idealizations, but
> consider another resolver of singularities: quantum mechanics. The
> table can be modeled as an impenetrable potential wall, while gravity
> can be provided by a linear potential. What do solutions of the single
> particle Schroedinger equation with such a potential tell us about the
> escape probability?
That is an interesting question.
Ralph Hartley
Jay R. Yablon
"Igor Khavkine" <igo...@gmail.com> wrote in message
news:1145745718.4...@u72g2000cwu.googlegroups.com...
>I recently ran into a fun problem from an old Russian physics olympiad.
> I found the solution a little surprising and intriguing. The problem
> requires no more than high school classical mechanics and some
> ingenuity. Here it is for your amusement.
>
> Consider a point particle sliding on a flat table (ignore friction).
> The table has a cylindrical hole of finite depth (vertical walls, flat
> bottom). The particle can approach the hole with different velocities
> and with different impact parameters (the particle's motion need not be
> directed toward the center of the hole). As the particle falls into the
> hole, it starts bouncing off the walls and the bottom (assume elastic
> collisions). Sometimes it gets stuck in the hole forever, sometimes it
> escapes (bounces out). Determine the relation between the depth of the
> hole, its radius, the particle's initial velocity, and impact parameter
> necessary for the particle to escape after it falls in.
>
> Enjoy!
>
> Igor
The impact parameter at first thought does not change anything: it still
needs top strike m/n of the way across the hole on whatever path it
takes toward the hole. This means that for a less than square impact,
the velocity will need to be reduced proportionally to the length of
hole over which it travels. BUT, on second thought: when it hits the
hole wall, it will not bounce straight back any more. It will go at an
angle. I suspect that you will then also need to consider paths which
are triangles, squares, pentagons, etc. That how I'd lay this out. The
quantization m/n will carry over to the angles of incidence, which will
have similar quantum constraints. That is step 2 of this problem. The
interesting result here will be the angles at which it bounces out.
One should also consider the extreme case of an exact tangential graze.
Then, as I think was pointed out earlier, it will spiral in, hit bottom,
then bounce back out. The exit angle will depend on depth, radius, and
velocity.
After all of this is laid out, then one reverse engineers the math to
get the formula involving the "depth of the hole, its radius, the
boson boss
Here are some comments.
> Given just the relationship between
> the initial velocity and the impact
> parameter (and, likely, the direction
> of approach) what can we say about
> the geometry of the hole?
Lets see. Given the motion of the matter around a black hole, what can
we say about it's geometry? Well, its bright, hot, turning gas into
stars, it has magnetic field, it goes around...
> For instance, if we add friction or non-elastic collisions, then as
> long as the particle falls into the hole it will never get out. It will
> loose kinetic energy and thus well never be able to jump out. If make
> the particle in to a ball or spherical shell of finite size, then my
> suspicion is that it will also never escape, unless circumstances
> conspire to convert all rotational kinetic energy into translational
> vertical motion.
But then again! Bilyard doesn't exist without ALL these properties. As
a matter of fact ball spins all the time. Matter around a black hole
wouldn't be hot, bright and exciting. I estimate that from criticality
come sources of interesting things. It is away from the topic maybe,
but I think that's kind of intruige I like.
John Bell
> tttito wrote:
> > Let v be the particle's speed, h the depth of the hole , r its radius.
> > Let's consider the projection of the particles's trajectory on the flat
> > table HOR. Let's take as impact parameter the angle alpha that the
> > particle's trajectory forms with the hole's radius on HOR. . Let C be
> > the projection on HOR of the chord that the particle trajectory cuts
> > on the hole. The length L of C is L= 2r*cos(alpha) . Note that every
> > subsequent chord traced by the particle's trajectory on HOR while it
> > bounces in the hole has the same length L.
> >
> > Denote by K the length of the (possibly jagged) path described by the
> > particle's projection on HOR between its entry point on the hole and
> > the point where it first reaches (and bounces on) the bottom.
> > Subsequent bounces are at odd multiples of K. Crucially, the particle
> > will be at HOR's level at even multiples of K. Since the particle's
> > speed is constant along the possibly jagged path t describes on HOR, we
> > have K=v*srqt(2g*h)/g (that's v times the time it takes to fall from
> > hole, its radius, the particle's initial velocity, and impact
> >
> > The idea now is that the particle can leave the hole only grazing the
> > hole's boundary (it can't go higher). In order for it to graze, L/K
> > must be a rational number 2n/m, n and m being integers. The particle
> > will leave the hole when distance D it has walked along its path on
> > HOR is D=2nK and D=mL.
>
> Italo's solution is most like the one I found myself. However, I think
> we both missed the limiting possibility pointed out by boson boss. That
> is the case of the limit L -> 0, in other words, the particle is shot
> into the hole right at its edge, so that it smoothly slides along the
> cylindrical instead of bouncing off it. In this case, the particle
> always escapes.
This is just the obverse of the opposite limiting case possibility that
you have still missed. In the case you describe, there are effectively
infinite bounces from the sides, and the total transit time in the hole
is then, presumably, the same as if the particle makes one bounce at
the exact centre of the hole. Whether or not the particle hits the
sides, and how often, therefore appears to be irrelevant to the problem
as posed.
The other limiting case is an infinite number of bounces on the bottom,
which is why the particle always escapes eventually, if it has
sufficient horizontal velocity to enter the hole.
>
> After everyone had some time to think about the problem, I can say more
> about why I found it intriguing. There are two reasons and, as it is
> not unusual for this kind of problem, they have to do with its
> generalizations.
>
> The first is that the answer depends crucially on the geometry of the
> hole. The solution is simple to obtain because the of the hole's
> symmetry. But the same question can be posed if it were square,
> triangular, elliptical, or any irregular shape you care to consider.
> What is interesting then is the inverse problem. Given just the
> relationship between the the initial velocity and the impact parameter
> (and, likely, the direction of approach) what can we say about the
> geometry of the hole?
>
> The second is that the answer has something to do with the rational
> numbers, the ratio 2K/L must be rational. The rationals are conspicious
> because they are dense in on the real line yet are a set of measure
> zero. In other words, if we could perform the experiment described in
> the problem by randonly selecting initial conditions for the particle's
> motion, we would find zero probability of escape (as long as the impact
> parameter is smaller than the hole radius).
Untrue, given that you have not specified a finite time limit. Whilst
it is true that the solutions are (obviously) quantised to exact
integer numbers of bounces from the bottom, those integers constitute
an infinite set (from 1 to infinity). The particle, therefore, always
escapes eventually.
> But there does exist this
> singular or critical set of initial conditions that allow escape. This
> criticality is closely tied to the idealizations present in the
> problem. Analysis of idealized situations can often tell us much about
> the limiting behavior of realistic ones, none of which actually exhibit
> singularities. However, an important inverse problem appears here as
> well. How do idealized singularities get resolved in more realistic
> situations?
>
> For instance, if we add friction or non-elastic collisions, then as
> long as the particle falls into the hole it will never get out. It will
> loose kinetic energy and thus well never be able to jump out. If make
> the particle in to a ball or spherical shell of finite size, then my
> suspicion is that it will also never escape, unless circumstances
> conspire to convert all rotational kinetic energy into translational
> vertical motion.
Untrue again. Consider the simplest example: No initial particle spin,
and one bounce at the centre of the hole. If the energy loss on
crossing the hole is less than gr where r is the vertical distance from
the bottom of the body to its centre of gravity , it will roll out on
hitting the edge of the hole (even if the particle is not spherical).
Similarly for cases where it is shot over the hole with sufficient
momentum that it cannot fall a distane of r, before hitting the far
side of the hole.
Blackbird
> Igor Khavkine wrote:
[...]
>> The second is that the answer has something to do with the rational
>> numbers, the ratio 2K/L must be rational. The rationals are
>> conspicious because they are dense in on the real line yet are a set
>> of measure zero. In other words, if we could perform the experiment
>> described in the problem by randonly selecting initial conditions
>> for the particle's motion, we would find zero probability of escape
>> (as long as the impact parameter is smaller than the hole radius).
>
> Untrue, given that you have not specified a finite time limit. Whilst
> it is true that the solutions are (obviously) quantised to exact
> integer numbers of bounces from the bottom, those integers constitute
> an infinite set (from 1 to infinity). The particle, therefore, always
> escapes eventually.
You seem to believe that unless the particle is stuck on a repeating/cycling
path, it will eventually reach any position in the cylinder, including the
rim at the top. This is false. As numerous other posters have pointed out,
there are infinitely many paths that are not repeating, but still never
touches the rim.
Ralph Hartley
> This is a *generic* property of this problem, and does not depend at all
> on many of the details. It is a consequence of the Poincaré recurrence
> theorem.
>
> The hole can have any (possibly parameterized) shape, the particle can
> have internal degrees of freedom (e.g. a rolling ball, an *elastic*
> ball), or any of the changes you suggest, but *still* if you fix all the
> parameters but one, then the remaining parameter will have a dense, but
> measure zero, set of values for which the particle escapes.
That wasn't quite right.
It is true that the set of escaping solutions must be dense, but the in
the generic case the particle *always* escapes.
In the original problem there is no way for energy to transfer between
the vertical and and horizontal (or other) degrees of freedom, so the
particle can never rise above its original level. It must reach the edge
exactly at its highest point to escape.
If there is less symmetry, e.g. if the walls are sloped (in either
direction), then only the total energy is conserved, and there is more
than enough to escape. Sometimes the particle will rise *above* the
edge, while moving more slowly in the horizontal direction (or spinning
more slowly etc). Then there is a continuous range of positions in which
it escapes, which means that eventually it always will (with probability 1).
In general when it escapes it won't be sliding on the surface at its
original speed, it will be bouncing with less horizontal velocity.
Igor Khavkine wrote:
>> Or, one could keep all the idealizations, but
>> consider another resolver of singularities: quantum mechanics. The
>> table can be modeled as an impenetrable potential wall, while gravity
>> can be provided by a linear potential. What do solutions of the single
>> particle Schroedinger equation with such a potential tell us about the
>> escape probability?
I'm fairly sure that the quantum version always escapes, even with the
perfectly cylindrical hole.
Ralph Hartley
Cl.Massé
de news: 1146112139.9...@e56g2000cwe.googlegroups.com
> v*sqrt( (2*h) / (g*(r^2 - s^2) ) = n/k = any positive rational
> number.
This quantity is dimensionless, that is, independent of the physical unit
used. It can get any value by varying v or s, therefore may be rational or
not.
Cl.Massé
873bg16...@kreon.lan.henning.makholm.net
> It can also get stuck on a non-repeating path, if the ration between
> the time one vertical bounce takes and the time it takes to move from
> one wall to the other is irrational.
Strangely, if the path is non repeating, the point won't escape, but it will
escape for all repeating path. The explanation is that if the path is
repeating, it must go through its entry point or a symmetrical one at some
time. If not, any time it tries, each time differently, it'll fail because
there are infinitely more irrational numbers than rational ones.
Oh No
to quite such an extreme. However, it does occur to me that if one does
take things to that level of precision, one really ought to allow that
quantum effects should be taken into account and that the particle could
escape via quantum tunnelling if it gets close enough to a critical
escape point.
Regards
--
Charles Francis
substitute charles for NotI to email
Blackbird
Yes, this is correct. But now another question pops up: It it possible to
construct a cavity with a geometrical shape so that the measure of the set
of initial conditions that will make the particle escape is neither 0 nor 1,
but something in between?
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> I recently ran into a fun problem from an old Russian physics olympiad.
> I found the solution a little surprising and intriguing. The problem
> requires no more than high school classical mechanics and some
> ingenuity. Here it is for your amusement.
>
> Consider a point particle sliding on a flat table (ignore friction).
> The table has a cylindrical hole of finite depth (vertical walls, flat
> bottom). The particle can approach the hole with different velocities
> and with different impact parameters (the particle's motion need not be
> directed toward the center of the hole). As the particle falls into the
> hole, it starts bouncing off the walls and the bottom (assume elastic
> collisions). Sometimes it gets stuck in the hole forever, sometimes it
> escapes (bounces out). Determine the relation between the depth of the
> hole, its radius, the particle's initial velocity, and impact parameter
> necessary for the particle to escape after it falls in.
Since it's a point particle it has no rotational degrees of freedom. And
since the walls are perpendicular to the bottom (this is actually not spe-
cified, but I assume the bottom is not inclined, besides being flat), and
all are non-dissipative, the vertical motion is decoupled from the hori-
zontal one. So the particle will always come back to its entrance level
at z=0, the table-top surface. The time interval for this is tau_v =
2*sqrt(2*h/g), where h is the hole depth and g is the gravitational acce-
leration.
If it reaches the z=0 level somewhere _inside_ the hole, the particle will
just fall back into the hole to try again. But if reaches the vertical
wall at exactly the same instant, it will have made it out. With the walls
being elastic and the hole being a perfect circle, the particle will always
travel horizontally the same chord length (tho at different orientations)
with the same impact parameter b. If the hole radius is r, the chord
length is l = 2*sqrt(r^2 - b^2). Since the horizontal speed v is constant,
the bounce interval is tau_h = l/v. So the criterion for particle escape
is m*tau_v = n*tau_h, where m and n are integers - m vertical bounces, n
horizontal ones. This gives the relation m^2*(r^2 - b^2) = 2*h*v^2*n^2/g.
Cl.Massé
news: lxfJN7Ex$ZVE...@charlesfrancis.wanadoo.co.uk
> All too true. I just had some problem with idealising "point" particle
> to quite such an extreme. However, it does occur to me that if one does
> take things to that level of precision, one really ought to allow that
> quantum effects should be taken into account and that the particle could
> escape via quantum tunnelling if it gets close enough to a critical
> escape point.
In that case, every point will escape, since all non escaping trajectory
passes arbitrarily close to an escaping one. Even if the potential is
infinite, the possibility of diffraction remains.
Yvon Sauvageau
> I recently ran into a fun problem from an old Russian physics olympiad.
> I found the solution a little surprising and intriguing. The problem
> requires no more than high school classical mechanics and some
> ingenuity. Here it is for your amusement.
>
> Consider a point particle sliding on a flat table (ignore friction).
> The table has a cylindrical hole of finite depth (vertical walls, flat
> bottom). The particle can approach the hole with different velocities
> and with different impact parameters (the particle's motion need not be
> directed toward the center of the hole). As the particle falls into the
> hole, it starts bouncing off the walls and the bottom (assume elastic
> collisions). Sometimes it gets stuck in the hole forever, sometimes it
> escapes (bounces out). Determine the relation between the depth of the
> hole, its radius, the particle's initial velocity, and impact parameter
> necessary for the particle to escape after it falls in.
>
>
> Igor
There are two types of rebounds: rebound on the wall, and rebound on
the ground. I'll call the former a 'wall reflection', and the latter a
'rebound'.
Let the unit of distance be the length of the horizontal chord that the
particle travels through before hitting the wall. All subsequent wall
reflections will have this same length in the horizontal direction. So
all reflection lengths will be 1.
Because of its momentum, the particle will always make a full rebound
at the same horizontal distance k. So it comes back to the table's
height whenever it has travelled a path of length nk, where n is an
integer.
And it will come out of the hole as soon as nk is a multiple of the
reflection length (which is 1, in our distance unit). Therefore it will
come out at the smallest n such that nk is an integer. If k is a
rational number, it can be written as a/b where this fraction is in its
simplest form. And consequently, the smallest n is b. Therefore there
is a solution whenever k is rational, and in this case the particle is
guaranteed to exit eventually. And of course there is no solution
whenever k is irrational, and in this case the particle is trapped
forever in the hole. Now, whether k can physically be irrational is
probably a great philosophical question.
I suppose the olympiad contest required to find the conditions under
which k is irrational, but I'll stop here. I'm sure everyone can figure
out the rest.
Yvon Sauvageau
Ralph Hartley
> But now another question pops up: It it possible to
> construct a cavity with a geometrical shape so that the measure of the set
> of initial conditions that will make the particle escape is neither 0 nor 1,
> but something in between?
Taking the question exactly as worded it the answer is clearly "yes".
consider a hole with two disconnected parts, i.e. two holes. One of
which has vertical sides, and one with sloped walls. Some initial
conditions fall into the first hole, and (except for a set of measure
zero) never escape, and some fall into the second hole and always escape.
Ralph Hartley
Cl.Massé
1146638383....@j73g2000cwa.googlegroups.com
> Now, whether k can physically be irrational is
> probably a great philosophical question.
As long as there is no energy loss at all, k has a definite value that
can be rational or irrational. Now, when the speed is measured or the
particle is prepared, there is always an error. However small this
error is, it spans an infinity or rational values, and still more
irrational values. So if we give a value of the speed with the attached
error, all we can say is that the particle will "almost surely" be
trapped, that is, with probability 1.
Even if this error were zero, there are other errors in the size of the
hole, the planarity of the table, the inertiality of the referential and
so on, which amounts to the same.
Now, a question that isn't philosophical, is whether a zero energy loss
is possible, since it definitively isn't.
Yvon Sauvageau
I think this problem leads to the following question: how far can we
push idealization without falling into irrelevancy?
You're right to insist on the impossibility of a zero energy loss.
According to me, the spirit of this problem should really be
that---with conditions that LIMIT to perfection---the particle always
nearly exits the hole. This is definitely true for a rational k, and
the same can be said about an irrational k because there is always an
arbitrarily closer rational to it. To me, that's the true lesson I get
from this thought experiment. But I don't think anyone would have the
time to elaborate on this in the context of an olympiad, so they would
actually go for the "rational k exits the hole, and irrational k does
not" solution, and then move on to the next problem.
Yvon Sauvageau
Blackbird
Yes, I should have been clearer and of course required the hole to be
connected. Also, the "inside" of the hole should be topologically open with
every point an interior point (this should rule out two initially separate
holes connected by a set of mesure zero, etc.) Last, the hole should be
bounded, or else there will be trivial solutions where the particle
sometimes
gets trapped in an infinite tunnel.