Re: Fun problem
- From: "Ken S. Tucker" <dynamics@xxxxxxxxxxxx>
- Date: Sat, 29 Apr 2006 02:13:40 +0000 (UTC)
Jay R. Yablon wrote:
"Igor Khavkine" <igor.kh@xxxxxxxxx> wrote in message
news:1145745718.411724.269370@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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
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.
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
.
- References:
- Fun problem
- From: Igor Khavkine
- Re: Fun problem
- From: Jay R. Yablon
- Fun problem
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