Pop Quiz for Science Buffs!
NoEinstein
only one correct, or more nearly correct answer. There is no time
limit for the quiz, so answer each question in the order written. The
correct answers are printed at the end, but please, no cheating! Each
answer has been authenticated by NoEinstein--who has no connection
whatsoever to, nor association with, the present science
establishment. Those who make perfect scores qualify as: Science
Gradually-rates (but no diplomas will be awarded)!
1. A 150 pound policeperson wishes to knock open a door by hitting
the door with his or her shoulder. Getting a running start from 10
feet away, the policeperson hits the door traveling 8 feet per
second. But the door doesn't bulge. A policeman suggests that the
150 pounder try again-this time taking a longer running start. The
policeperson moves away 100 feet, runs at the door and hits it
traveling 8 feet per second. In the second try the door was hit with
a force... A. Ten times greater. B. Exactly the same. C. I don't
know, I need to ask somebody (or IDKINTAS).
2. Another identical door, as in 1. above, needs to be knocked
open. The 150 pounder's shoulder is sore, so a 300 pound policeman
(or policewoman) is asked to do the honors. The 300 pounder steps
back ten feet and hits the second door traveling 8 feet per second.
The 300 pounder hits the second door... A. With exactly the same force
as the 150 pounder in question 1. hit. B. With twice the force,
because the 300 pounder is twice as heavy. C. IDKINTAS.
3. Over doughnuts and coffee, the 150 pounder and the 300 pounder
argue over which one is the better door opener. Soon, throughout that
police force, wagers are being made on each side, with a winner to be
determined in a shoulder against shoulder runoff at the department's
next picnic. There, the combatants face each other from 200 feet
apart. The Chief begins counting down: "10, 9, 8..." Just as the
count gets to zero, a policeman with a large bet on the 150 pounder
uses his small electric cattle prod to zap the butt of the 150
pounder, who is out of there in a flash!
The 150 pounder is traveling a whopping 16 feet per second when
meeting the shoulder of the 300 pounder who is traveling only 8 feet
per second. The 150 pounder hits with a force... A. Exactly the same
as he or she hit the door in question 1. B. Twice as hard as he or
she hit that door. C. IDKINTAS.
4. A spherical cannon ball weighing 150 pounds is dropped from an
adjustable height work platform. The height of the drop is set so
that the 150 pound cannon ball hits a thick, level bed of soft
modeling clay at exactly 8 feet per second. The ball leaves a dome
shaped indentation in the clay. For a second drop, the work platform
moves several feet to the side and increases the drop height such that
the same 150 pound cannon ball hits another part of the same clay bed
at 16 feet per second. Surprisingly, the second drop of the ball
leaves a dome shaped recess in the clay that is four times deeper than
the depth when the cannon ball was dropped at 8 feet per second. The
force of the second impact into the clay was... A. Four times greater,
because it left a dent in the clay four times as deep. B. Twice as
great, because the velocity was twice as great. C. IDKINTAS.
5. The same 150 pound cannon ball is accidentally placed on the soft
clay without dropping it. The cannon ball leaves a significant
indentation in the clay from its dead weight alone. From the above
one should conclude... A. The weight of the cannon ball at rest
doesn't have any effect on the depth that that same cannon ball, when
dropped, penetrates into the clay. B. The weight of the cannon ball,
just sitting there, has a significant effect on the depth the same
ball penetrates into clay at either velocity. C. IDKINTAS.
6. A young boy is pulling a new red wagon on a smooth, level, paved
driveway via a spring scale attached to the tongue of the wagon. The
wagon rolls when the spring scale reads one pound, but the force
reading on the scale drops to ½ pound as the boy pulls the wagon at a
uniform walking speed. Being a perfectionist, the boy tries to pull
the wagon fast enough so that the spring scale stays exactly on the
one pound indicator line. As he does that for a short distance, he
finds that he must run faster and faster each second, until he is
running his maximum speed. The one pound spring force caused the
wagon to speed up each second that he ran. In other words, the wagon
accelerated under a uniform force of one pound applied continuously
(though for only a limited distance). From the above one can
conclude... A. Uniformly accelerated bodies (such as a wagon) can have
widely varying forces causing the acceleration. B. Uniformly
accelerated bodies have only a single value of continuous force
maintaining that acceleration. C. IDKINTAS.
7. The young boy in question 6, above, doesn't need his old, but
otherwise identical wagon any more, so he decides to throw it off of a
high cliff near his home. His father, who is a science buff, too,
tells his son that the wagon will accelerate because of the force of
gravity and fall 32 feet more in any second than it fell in the
previous second. In other words, it will accrue 32 extra feet per
second of velocity over the velocity in the previous ½ second, and do
so successively. From the above one can conclude... A. The force of
gravity acting on the wagon increases the faster the wagon falls. B.
The force of gravity acting on the wagon is exactly equal throughout
the fall, regardless of the velocity of the wagon. C. IDKINTAS.
8. When our boy outgrows his wagon, he donates it to science. It
will be propelled into outer space beyond the pull of Earth's
gravity. Then, a tiny atomic rocket motor that exerts a continuous
one pound force will cause the wagon to accelerate away from the Sun.
The wagon goes faster and faster until it is approaching the speed of
light. But a scientist-of-old has said that the speed of light is the
fastest speed, anywhere. Yet, that little red wagon is accelerating
so smoothly... To stop it from breaking his speed limit, the scientist-
of-old concluded that that little one pound force rocket motor won't
cause any more acceleration if all of the applied energy from the
motor, somehow, converts into a more and more massive little red
wagon. Therefore, one can conclude... A. The little red wagon never
gets to the speed of light because the scientist-of-old said it can't
happen, and the ideas of scientists-of-old are absolute truths. B.
Who is the scientist-of-old? Has his concept ever been proven? If
not, then take what he said with a grain of salt. C. IDKINTAS.
9. A pendulum swings back and forth, back and forth. The weight
stores up gravitational potential as it rises on the upswing, then it
uses that potential to propel the weight back down. It has maximum
kinetic energy at the bottom of the swing. Physicists for nearly two
centuries have believed that kinetic energy increases in proportion to
the square of an object's velocity at any point in time. They accept
that the kinetic energy accrued as a pendulum descends exceeds the
potential energy gained as the weight ascends. Such is... A. OK if
physicists believe it is. B. Is wrong, because the uniform force of
gravity for near Earth objects causes a linear kinetic energy
increase, equal and opposite to the gravitational potential energy
being gained each time the weight ascends. C. IDKINTAS.
10. The previously mentioned scientist-of-old believed that not only
do objects become infinitely massive as they are forced closer and
closer to the speed of light, but he believed that TIME itself slows
down, and that the space near the object is compressed, or warped. A
very accurate atomic clock was carried on a long orbital space flight,
then, returned to Earth. When such clock was compared to a similar
clock back on Earth, the orbital clock had slowed down. This proves...
A. Time does stop, exactly as the scientist-of-old predicted!
Everything he said was true! B. The clock slowed, but there is no
proof that time stops at the speed of light. There is a probability
that any acceleration force (as on take off) or deceleration (as on re
entry) will cause mechanical or atomic devices, only, to slow, thus
accounting for the lost fractions of a second. C. IDKINTAS.
End of Test!
Answers: The correct answer to each question is B. If you got them
all right, pat yourself on the back for being an independent thinker!
Hope you enjoyed taking the quiz!
- Noeinstein -
Androcles
"NoEinstein" <noein...@bellsouth.net> wrote in message
news:1190839545.2...@w3g2000hsg.googlegroups.com...
INSTRUCTIONS: Each of the following multiple choice questions has
only one correct, or more nearly correct answer. There is no time
limit for the quiz, so answer each question in the order written. The
correct answers are printed at the end, but please, no cheating! Each
answer has been authenticated by NoEinstein--who has no connection
whatsoever to, nor association with, the present science
establishment. Those who make perfect scores qualify as: Science
Gradually-rates (but no diplomas will be awarded)!
1. A 150 pound policeperson wishes to knock open a door by hitting
the door with his or her shoulder. Getting a running start from 10
feet away, the policeperson hits the door traveling 8 feet per
second. But the door doesn't bulge. A policeman suggests that the
150 pounder try again-this time taking a longer running start. The
policeperson moves away 100 feet, runs at the door and hits it
traveling 8 feet per second. In the second try the door was hit with
a force... A. Ten times greater. B. Exactly the same. C. I don't
know, I need to ask somebody (or IDKINTAS).
B.
2. Another identical door, as in 1. above, needs to be knocked
open. The 150 pounder's shoulder is sore, so a 300 pound policeman
(or policewoman) is asked to do the honors. The 300 pounder steps
back ten feet and hits the second door traveling 8 feet per second.
The 300 pounder hits the second door... A. With exactly the same force
as the 150 pounder in question 1. hit. B. With twice the force,
because the 300 pounder is twice as heavy. C. IDKINTAS.
B.
3. Over doughnuts and coffee, the 150 pounder and the 300 pounder
argue over which one is the better door opener. Soon, throughout that
police force, wagers are being made on each side, with a winner to be
determined in a shoulder against shoulder runoff at the department's
next picnic. There, the combatants face each other from 200 feet
apart. The Chief begins counting down: "10, 9, 8..." Just as the
count gets to zero, a policeman with a large bet on the 150 pounder
uses his small electric cattle prod to zap the butt of the 150
pounder, who is out of there in a flash!
The 150 pounder is traveling a whopping 16 feet per second when
meeting the shoulder of the 300 pounder who is traveling only 8 feet
per second. The 150 pounder hits with a force... A. Exactly the same
as he or she hit the door in question 1. B. Twice as hard as he or
she hit that door. C. IDKINTAS.
D. Nine times as soft as your head.
E = 1/2mv^2
E1 = 1/2 * 150 * 8 * 8 = 4800
E2 = 1/2 * 150 * 16 * 16 = 19200
The 300 pounder is moving at 8 fps, so the 150 pounder hits at
24 fps as if he were standing still.
E3 = 1/2 * 150 * 24 * 24 = 43200
9 * 4800 = 43200
You seem to be NoNewton as well as NoEinstein.
Here's a more everyday problem.
How much harder is a pedestrian hit by a car moving at 40 mph
versus one moving at 28.28 mph?
Obey speed limits or a 450 pound pair of cops will come knocking
on YOUR door.
Uncle Al
>
> INSTRUCTIONS: Each of the following multiple choice questions has
> only one correct, or more nearly correct answer. There is no time
> limit for the quiz, so answer each question in the order written. The
> correct answers are printed at the end, but please, no cheating! Each
> answer has been authenticated by NoEinstein--who has no connection
> whatsoever to, nor association with, the present science
> establishment. Those who make perfect scores qualify as: Science
> Gradually-rates (but no diplomas will be awarded)!
>
> 1. A 150 pound policeperson wishes to knock open a door by hitting
> the door with his or her shoulder.
My, my - the bullshit smell is already clear.
> Getting a running start from 10
> feet away, the policeperson hits the door traveling 8 feet per
> second. But the door doesn't bulge. A policeman suggests that the
> 150 pounder try again-this time taking a longer running start. The
> policeperson moves away 100 feet, runs at the door and hits it
> traveling 8 feet per second. In the second try the door was hit with
> a force... A. Ten times greater. B. Exactly the same. C. I don't
> know, I need to ask somebody (or IDKINTAS).
Do you want a diverse answer, an Equal Opportunity answer, or the real
world answer?
> 2. Another identical door, as in 1. above, needs to be knocked
> open. The 150 pounder's shoulder is sore, so a 300 pound policeman
> (or policewoman)
You pig.
> is asked to do the honors. The 300 pounder steps
> back ten feet and hits the second door traveling 8 feet per second.
> The 300 pounder hits the second door... A. With exactly the same force
> as the 150 pounder in question 1. hit. B. With twice the force,
> because the 300 pounder is twice as heavy. C. IDKINTAS.
Fail the course to allow a diversity football jock to give the obvious
answer.
> 3. Over doughnuts and coffee, the 150 pounder and the 300 pounder
> argue over which one is the better door opener. Soon, throughout that
> police force, wagers are being made on each side, with a winner to be
> determined in a shoulder against shoulder runoff at the department's
> next picnic. There, the combatants face each other from 200 feet
> apart. The Chief begins counting down: "10, 9, 8..." Just as the
> count gets to zero, a policeman with a large bet on the 150 pounder
> uses his small electric cattle prod to zap the butt of the 150
> pounder, who is out of there in a flash!
Did it burn out his gaydar?
> The 150 pounder is traveling a whopping 16 feet per second when
> meeting the shoulder of the 300 pounder who is traveling only 8 feet
> per second. The 150 pounder hits with a force... A. Exactly the same
> as he or she hit the door in question 1. B. Twice as hard as he or
> she hit that door. C. IDKINTAS.
Do you know the mathematical physical definition of force? How about
that of impulse?
> 4. A spherical cannon ball
One typically utilizes a spherical isotropic homogeneous cow with zero
albedo.
> weighing 150 pounds is dropped from an
> adjustable height work platform. The height of the drop is set so
> that the 150 pound cannon ball hits a thick, level bed of soft
> modeling clay at exactly 8 feet per second. The ball leaves a dome
> shaped indentation in the clay. For a second drop, the work platform
> moves several feet to the side and increases the drop height such that
> the same 150 pound cannon ball hits another part of the same clay bed
> at 16 feet per second. Surprisingly, the second drop of the ball
> leaves a dome shaped recess in the clay that is four times deeper than
> the depth when the cannon ball was dropped at 8 feet per second. The
> force of the second impact into the clay was... A. Four times greater,
> because it left a dent in the clay four times as deep. B. Twice as
> great, because the velocity was twice as great. C. IDKINTAS.
You might want to consider the shape of the indentation in the first
case and contast it with the shape in the second case. Partial
penetration of a sphere into a layer of medium is not linear in
displaced mass vs. penetration depth.
> 5. The same 150 pound cannon ball is accidentally placed on the soft
> clay without dropping it.
How does one place a 150 pound ball "accidently?" 150 pounds implies
both awareness and volition. If a civil servant, at least volition.
> The cannon ball leaves a significant
> indentation in the clay from its dead weight alone.
YOU KILLED THE CANNON BALL?!!!
> From the above
> one should conclude... A. The weight of the cannon ball at rest
> doesn't have any effect on the depth that that same cannon ball, when
> dropped, penetrates into the clay. B. The weight of the cannon ball,
> just sitting there, has a significant effect on the depth the same
> ball penetrates into clay at either velocity. C. IDKINTAS.
[SNIP]
Save everybody a lot of needless waste. Become a mortgage broker.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
Eric Gisse
<noein...@bellsouth.net> wrote:
[...]
Why not ask something beyond the highschool physics level?
Here, I got one: Why is energy a conserved quantity in central force
potentials?
Uncle Al
Is there anything more dangerous than a question wholly stated in ten
words or fewer?
Igor
wrote:
> On Wed, 26 Sep 2007 13:45:45 -0700, NoEinstein
>
>
> [...]
>
> Why not ask something beyond the highschool physics level?
>
> Here, I got one: Why is energy a conserved quantity in central force
> potentials?
Are you sure you don't mean angular momentum? Energy can be conserved
in non-central potentials also.
sal
let's look at some possible answers.
On Thu, 27 Sep 2007 17:18:26 +0000, Igor wrote:
> On Sep 26, 10:15 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> wrote:
>> On Wed, 26 Sep 2007 13:45:45 -0700, NoEinstein
>>
>> <noeinst...@bellsouth.net> wrote:
>>
>> [...]
>>
>> Why not ask something beyond the highschool physics level?
>>
>> Here, I got one: Why is energy a conserved quantity in central force
>> potentials?
'cuz it's a potential, and if you can describe the force with a potential
function, then energy is conserved.
That's 19 words.
>
> Are you sure you don't mean angular momentum? Energy can be conserved
> in non-central potentials also.
'cuz theta's a cyclic coordinate.
That's 5 words. And 5 < 19.
See, your question was almost four times easier to answer than Eric's.
--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org
dlzc
On Sep 26, 1:45 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
...
> 1. A 150 pound policeperson wishes to knock
...
> In the second try the door was hit with
> a force... A. Ten times greater.
> B. Exactly the same. C. I don't
> know, I need to ask somebody (or IDKINTAS).
D. Collisions are not well described by "force", but by "impulse".
Policmen's shoulders behave differently upon second rammings, and
after having run farther.
> 2. Another identical door, as in 1. above,
...
> The 300 pounder hits the second door...
> A. With exactly the same force as the 150
> pounder in question 1. hit. B. With twice
> the force, because the 300 pounder is twice
> as heavy. C. IDKINTAS.
D. Collisions are not well described by "force", but by "impulse".
Secondly, the second policeman should be described as "more massive"
rather than "heavy". Thirdly, it is unlikely the 300 pounder has the
same skin-to-bone compression ratio that the 150 pounder does.
> 3. Over doughnuts and coffee, the 150
...
> per second. The 150 pounder hits with a
> force... A. Exactly the same as he or
> she hit the door in question 1. B.
> Twice as hard as he or
> she hit that door. C. IDKINTAS.
D. Collisions are not well described by "force", but by "impulse".
Policmen's shoulders behave differently upon being electrically
shocked.
> 4. A spherical cannon ball weighing 150
...
> force of the second impact into the clay
> was... A. Four times greater, because it
> left a dent in the clay four times as deep.
> B. Twice as great, because the velocity
> was twice as great. C. IDKINTAS.
D. Collisions are not well described by "force", but by "impulse".
Additionally, since the depth of the "dent" is four times deeper, and
the volume (and hence energy) displaced by the sphere is a cubic
function of depth, then the clay must be highly non-linear.
> 5. The same 150 pound cannon ball is
...
> one should conclude... A. The weight of
> the cannon ball at rest doesn't have any
> effect on the depth that that same cannon
> ball, when dropped, penetrates into the
> clay. B. The weight of the cannon ball,
> just sitting there, has a significant
> effect on the depth the same ball
> penetrates into clay at either velocity.
> C. IDKINTAS.
D. None of the above. The ball essentially "floats" after a finite
amount of time. Archimedes. It isn't "weight" per se, but density
difference and displaced volume.
> 6. A young boy is pulling a new red
...
> (though for only a limited distance).
> From the above one can conclude... A.
> Uniformly accelerated bodies (such as a
> wagon) can have widely varying forces
> causing the acceleration. B. Uniformly
> accelerated bodies have only a single
> value of continuous force maintaining that
> acceleration. C. IDKINTAS.
B. From the problem statement, and not your idiotic choices, it is
clear that the drag force on the red wagon is less than 1 lbf, hence
the non-zero acceleration.
> 7. The young boy in question 6, above,
...
>32 extra feet per second of velocity over
> the velocity in the previous ½ second
...
> so successively. From the above one can
> conclude... A. The force of gravity
> acting on the wagon increases the faster
> the wagon falls. B. The force of gravity
> acting on the wagon is exactly equal
> throughout the fall, regardless of the
> velocity of the wagon. C. IDKINTAS.
D. None of the above. The force of gravity near Earth's surface
increases the velocity of a falling object (neglecting air friction)
at a rate of 32 feet in each second. You describe a g value twice
that of Earth, the boy is dead due to lack of oxygen.
> 8. When our boy outgrows his wagon, he
...
> wagon. Therefore, one can conclude...
> A. The little red wagon never gets to
> the speed of light because the scientist-
> of-old said it can't happen, and the
> ideas of scientists-of-old are absolute
> truths. B. Who is the scientist-of-old?
> Has his concept ever been proven? If
> not, then take what he said with a grain
> of salt. C. IDKINTAS.
D. "Concepts" (also known as theories) are never proven in Science,
only disproven by experiment.
> 9. A pendulum swings back and forth,
...
> potential energy gained as the weight
> ascends. Such is... A. OK if
> physicists believe it is. B. Is wrong,
> because the uniform force of gravity for
> near Earth objects causes a linear
> kinetic energy increase, equal and
> opposite to the gravitational potential
> energy being gained each time the weight
> ascends. C. IDKINTAS.
D. None of the above. Gravitational acceleration does not require
force. Note that gravitationally accelerated charges do not radiate.
If gravitation were a force, they would.
> 10. The previously mentioned scientist-
...
> clock back on Earth, the orbital clock had
> slowed down. This proves...
> A. Time does stop, exactly as the
> scientist-of-old predicted! Everything he
> said was true! B. The clock slowed, but
> there is no proof that time stops at the
> speed of light. There is a probability
> that any acceleration force (as on take
> off) or deceleration (as on re entry) will
> cause mechanical or atomic devices, only,
> to slow, thus accounting for the lost
> fractions of a second. C. IDKINTAS.
D. None of the above. "A" is pretty close though, except for invoking
"truth".
> End of Test!
>
> Answers: The correct answer to each
> question is B.
Better check you answer key. Whoever made it hasn't the first clue
about even Newtonian mechanics.
David A. Smith
Eric Gisse
So can angular momentum.
:)
>
>
core duo
wrote:
while i doubt that any of you knows what
conserved angular momentum means
Eric Gisse
wrote:
Then enlighten us, dyslexic troll.
core duo
wrote:
> On Thu, 27 Sep 2007 14:12:19 -0700, core duo <e4yequ...@adexec.com>
you dont even know your tensors
how would you understand?
NoEinstein
Dear Class Cut-ups: Though some of your Pop Quiz answers were amusing
to read, they show a huge bias toward the status quo thinking in
Mechanics-namely, that KE = 1/2 mv^2 (sic). Actually, KE, under the
application of a uniform, continuous force, such as gravity or a
rocket engine, is: KE = a/g(m) + v/32.174(m). The correct answer to
"Question 1" is: B. The RELATIVE force of impact is not determined by
the DISTANCE that the policeperson ran, but solely by the CHANGE in
velocity. Since the velocity did not change, then the force of impact
didn't change, either.
Objects under the influence of Earth's gravity will fall distances
corresponding to Galileo's formula d = t^2. In this case, the "d"
unit is 32.174', and the time is in seconds. So, the plot of his
curve forms an inverted parabola. That's the same profile curve that
males create every time they... pee horizontally. Any water hose pipe
shooting horizontally makes the same profile arc.
The damning thing about Coriolis's KE equation is that it plots semi-
parabolic. The Law of the Conservation of Energy requires that
objects gaining energy must "get it" from somewhere. Where do falling
objects get their accruing KE? From the force of gravity, of course!
But because Coriolis's equation is semi-parabolic, he (and Einstein,
et al) require that gravity be capable of imparting more KE to a
faster falling object than to a slower falling one. NOTE: The longer
an object falls, the faster it falls, right? And such is true at
least up to the object's terminal velocity in air.
But there is no mechanism-no even warped space-time-that will allow
gravity to "know" what the velocity of an object is, let alone to be
able to keep pumping different amounts of energy into the potentially
huge numbers of objects that could be falling at the same time.
Gravity isn't a super computer with radar detectors aimed on every
falling pebble!
If KE did accrue semi-parabolically (but it doesn't), such energy
increase would plot similar to Galileo's "pee" profile; OR the
distance of fall in any second would be GREATER than the distance of
fall in the previous second. But, some of you will agree, that the
correct answer to "Question 1" is: B. Such answer teaches that force
of impact is INDEPENDENT of the distance traveled. The change in
velocity-that DOES correspond to KE-is represented by the SLOPE of
Galileo's parabola. And the rate of change of that slope (with
respect to time) is identical for each and every second-all the way to
infinity! If the rate of change of velocity is equal all the way to
infinity, then the rate of change of KE is equal all the way to
infinity, too-at least for a rocket ship in the vacuum of space.
The correct answer to "Question 1", B, invalidates Coriolis. Since
Coriolis's equation formed the basis for Einstein's ideas that: "There
isn't enough energy in all of the universe to cause even a speck of
matter to travel to velocity c.", then that B answer disproves
Einstein's theories of relativity!
Most of you "experts" on physics have one thing in common-
gullibility. If a thing was ever printed in a textbook, you accepted
it as the truth, without question. Your physics "expertise" is more
about your ability to memorize and repeat, than it is about your
ability to UNDERSTAND. If Coriolis's "law" of KE resulted from
dropping lead shots into soft beds of clay, and only from such
observations, that was fine with you, even though no other tests were
done-in violation of THE SCIENTIFIC METHOD.
If any of you "old dogs" would like to learn a "new trick", study the
rate-of-change of the slope of a parabola. Such corresponds to the
rate-of-change of velocity... AND the rate of change of KE! You will
see that my Pop Quiz is more about science, than it is about being
FUNNY. Critical thinking trumps rote "mastery" of science 24/7. -
NoEinstein -
bz
@hate.spam.net:
>> The 150 pounder is traveling a whopping 16 feet per second when
>> meeting the shoulder of the 300 pounder who is traveling only 8 feet
>> per second. The 150 pounder hits with a force... A. Exactly the same
>> as he or she hit the door in question 1. B. Twice as hard as he or
>> she hit that door. C. IDKINTAS.
None of the above because they meet at a relative velocity of 24 feet per
second and have unequal masses.
Writer of the test fails to meet their own criteria.
But may redeem his/her self if they
1) correctly compute the inertia of each at impact
2) correctly compute the velocity of each after impact (assume perfectly
elastic collision).
3) correctly compute the amount of tissue deformation of each (assume
perfectly NON elastic collision)
4) correctly state who spends more time in the hospital and why.
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+...@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
Tom Roberts
> On Sep 26, 10:15 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com>
> wrote:
>> Here, I got one: Why is energy a conserved quantity in central force
>> potentials?
>
> Are you sure you don't mean angular momentum?
[I am using the context of classical mechanics, as this
is implied by the problem statement.]
Hmmm. I can invent a central force for which energy is not conserved. I
make no claim this sort of central force is a good model of anything,
merely that one can invent such a force (just make it depend explicitly
on time).
On the other hand, I know of no way to invent a central force such that
angular momentum is not conserved (the problem statement implicitly
implies that no other interactions apply).
Tom Roberts
Bill Hobba
"core duo" <e4ye...@adexec.com> wrote in message
news:1190927539....@o80g2000hse.googlegroups.com...
Instead of being concerned about what others may or may not know be more
concerned that all you know is how to be a dyslexic troll.
Bill
Bill Hobba
"Eric Gisse" <jowr.pi...@gmail-nospam.com> wrote in message
news:se4mf3h4ib0k2se19...@4ax.com...
I think you mean one that has a time invariant Lagrangian. Noether's
theorem guarantees those have energy conserved.
Thanks
Bill
N:dlzc D:aol T:com (dlzc)
"NoEinstein" <noein...@bellsouth.net> wrote in message
news:1190933227....@d55g2000hsg.googlegroups.com...
On Sep 27, 3:28 pm, dlzc <dl...@cox.net> wrote:
...
>> > Answers: The correct answer to each
>> > question is B.
>
>> Better check you answer key. Whoever made
>> it hasn't the first clue about even Newtonian
>> mechanics.
> Dear Class Cut-ups: Though some of your Pop
> Quiz answers were amusing to read, they show
> a huge bias toward the status quo
You can't even get Newton right (9 out of 10 times), and we are
"biased towards the status quo"? I suppose having any education
at all is a mistake? Or is it only appropriate for *you* to be
both right (in your mind) and wrong (in fact) and blame others
for your mistakes?
David A. Smith
JM Albuquerque
"Tom Roberts" <tjrobe...@sbcglobal.net> escreveu na mensagem
news:xPXKi.2759$6p6....@newssvr25.news.prodigy.net...
> Igor wrote:
>> On Sep 26, 10:15 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com>
>> wrote:
>>> Here, I got one: Why is energy a conserved quantity in central force
>>> potentials?
>>
>> Are you sure you don't mean angular momentum?
>
> [I am using the context of classical mechanics, as this
> is implied by the problem statement.]
>
> Hmmm. I can invent a central force for which energy is not conserved. I
> make no claim this sort of central force is a good model of anything,
> merely that one can invent such a force (just make it depend explicitly on
> time).
Yup, that's why none of such "inventions" ever passed from
the paper to a commercial version.
Working commercial theories never depend explicitly on time.
It's amazing how simple rules like:
Energy in = energy out
Power in = power out
can be simply ignored.
The "time" with respect to which one take the derivative
is absolute time, taken on the stationary frame of reference.
So, a relativistic "time" can fool the energy conservation Law.
> On the other hand, I know of no way to invent a central force such that
> angular momentum is not conserved (the problem statement implicitly
> implies that no other interactions apply).
Translation:
Energy is the time derivative of angular momentum.
Or else, angular momentum the integral of energy over time.
Since time is out Tom Roberts cannot fool angular momentum.
With time in hand, Tom Roberts can fool the energy conservation Law.
Funny physics, ahaa...
>
>
> Tom Roberts
Dono
[snip crap]
> Energy is the time derivative of angular momentum.
> Or else, angular momentum the integral of energy over time.
No idiot, did you fail physics in high school? Don't answer that, you
failed.
JM Albuquerque
"Dono" <sa...@comcast.net> escreveu na mensagem
news:1190944917.9...@w3g2000hsg.googlegroups.com...
Look at the units - kg - m - s, and tell me I'm wrong.
Eric Gisse
<tjrobe...@sbcglobal.net> wrote:
>Igor wrote:
>> On Sep 26, 10:15 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com>
>> wrote:
>>> Here, I got one: Why is energy a conserved quantity in central force
>>> potentials?
>>
>> Are you sure you don't mean angular momentum?
>
> [I am using the context of classical mechanics, as this
> is implied by the problem statement.]
>
>Hmmm. I can invent a central force for which energy is not conserved. I
>make no claim this sort of central force is a good model of anything,
>merely that one can invent such a force (just make it depend explicitly
>on time).
Maybe I should have been explicit - central forces under the usual
definition that the force acts radially between the two particles and
that the potential is not an explicit function of time or velocity.
If you go with the Hamiltonian as your definition of total energy, my
condition that it be a central force is enough to ensure that energy
is a conserved quantity since dH/dt = @L/@t = 0.
Dono
> "Dono" <sa...@comcast.net> escreveu na mensagemnews:1190944917.9...@w3g2000hsg.googlegroups.com...
>
> > On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote:
> > [snip crap]
> >> Energy is the time derivative of angular momentum.
> >> Or else, angular momentum the integral of energy over time.
>
> > No idiot, did you fail physics in high school? Don't answer that, you
> > failed.
>
> Look at the units - kg - m - s, and tell me I'm wrong.
You are wrong and you are an idiot. happy?
Eric Gisse
wrote:
Of course the units are right but you are still wrong. TORQUE is the
time derivative of angular momentum. Same units, different concept.
JM Albuquerque
"Eric Gisse" <jowr.pi...@gmail-nospam.com> escreveu na mensagem
news:8qpof31r2rrbai9kd...@4ax.com...
It's my gyroscopic difficiency at work.
You got that right, it's torque, but it's funny how torque and energy
share the same units, and many more things to think about it.
Energy is force times displacement in a translational system.
What about rotation?
Well, in rotation we got:
Energy = Torque * angular displacement (rad).
You see, time passes, so does the radians, and... got energy.
JM Albuquerque
An idiot is someone with ideas. Then there are good and bad
ideas. I'm happy to bring you new good ideas. I'm good idiot.
Today, idiots rule the world Doc.
Dono
> "Dono" <sa...@comcast.net> escreveu na mensagemnews:1190946608.9...@50g2000hsm.googlegroups.com...
Not idiots like you, your ideas are shit.
JM Albuquerque
Look, Dono, this forum is about physics.
So, one is supposed to talk about physics.
I don't care, nor I want your opinion for nothing.
If I wanted your opinion I would had ask it for you.
Now, since you like opinions I also have an opinion
on you. You remaind me those litle dogs that come from
behind you, barckling and trying to bite the back of your foot.
But if you stop and look at him, he stops barckling and turns
around. Next, you go on your way and the cene repeats itself.
I like dogs very much, but the most funny about them is
that they don't recognize themselves in a mirror. The dog
really thinks the dog on the mirror is another dog.
So, get lost.
Tom Roberts
> "Tom Roberts" <tjrobe...@sbcglobal.net> escreveu na mensagem
> news:xPXKi.2759$6p6....@newssvr25.news.prodigy.net...
>> is implied by the problem statement.]
> is absolute time, taken on the stationary frame of reference.
Absolute time is implicit in classical mechanics, so it does not depend
on any "stationary frame of reference".
[This is not so in relativity, but THIS thread is in the
context of classical mechanics, as I said.]
> So, a relativistic "time" can fool the energy conservation Law.
No. But the conservation laws of relativity are DIFFERENT from those of
classical mechanics, even though they often share the same name (e.g.
"conservation of energy", ...).
> Energy is the time derivative of angular momentum.
False. Not even close.
> You got that right, it's torque, but it's funny how torque and energy
> share the same units,
It helps to STUDY the subject of the discussion before opening your
mouth (keyboard). While torque and energy share the same units, they are
completely different:
W = integral F.ds where W is work (=energy), F is the
3-vector force on an object, and ds
is the 3-vector displacement of it.
t = F x r where t is the torque, F is the 3-vector
force, r is the radius at which the force
is applied, and x is the cross product.
The difference between those two vector products (dot and cross) is
important, as is the integral and lack of integral, and the difference
in measuring ds vs r. And, of course, W is a scalar but t is a (pseudo)
3-vector.
Tom Roberts
Dono
> Energy is the time derivative of angular momentum.
No, imbecile. Energy is not the time derivative of any momentum, ok?
Dono
No, stooopid, the units for momentum still are (check any 9-th grade
physics book) kg*m*s^-1 , not the stooopidity you wrote. You are
showing the units for torque. Two people (Eric and Tom) have already
corrected you but you are toooo stoooopid to get it.
Either way, "energy is" NOT "the time derivative of angular momentum".
Now, go bark at the moon.
sal
Dimensions of angular momentum are mass * length^2 / time
Dimensions of kinetic energy are mass * length^2 / time^2
A factor of 1/time is hardly ignorable. So, by your own argument, you're
wrong.
In any case, if energy were dL/dt, then if energy were constant and
positive then L would be constantly increasing, which is nonsensical.
Worse, energy is a scalar, and dL/dt is a vector (commonly called
torque). Scalar and vector mismatch is about as fundamental as
mismatched dimensions -- maybe more so.
What ever gave you the idea that dL/dt = T?
Now if you want to turn it around, @T/@(theta-dot) is (one component of)
the angular momentum, at least if there aren't any velocity-dependent
potentials running around getting in the way. But that's rather
different from what you claimed.
sal
On Fri, 28 Sep 2007 04:03:01 +0000, sal wrote:
> On Fri, 28 Sep 2007 03:07:21 +0100, JM Albuquerque wrote:
>
>> "Dono" <sa...@comcast.net> escreveu na mensagem
>> news:1190944917.9...@w3g2000hsg.googlegroups.com...
>>> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip
>>> crap]
>>>> Energy is the time derivative of angular momentum. Or else, angular
>>>> momentum the integral of energy over time.
>>>
>>>
>>> No idiot, did you fail physics in high school? Don't answer that, you
>>> failed.
>>
>> Look at the units - kg - m - s, and tell me I'm wrong.
>
> Dimensions of angular momentum are mass * length^2 / time
>
> Dimensions of kinetic energy are mass * length^2 / time^2
You compared dL/dt, and I just cited L up there, not dL/dt -- the units
match for T and dL/dt, as you said. Whatever...
Darn I hate it when I do that.
Eric Gisse
Actually I didn't even notice they were the wrong units, all I did was
point out that the units of torque and energy were the same.
Bill Hobba
> On Fri, 28 Sep 2007 00:22:53 GMT, Tom Roberts
> <tjrobe...@sbcglobal.net> wrote:
>
>>Igor wrote:
>>> On Sep 26, 10:15 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com>
>>> wrote:
>>>> Here, I got one: Why is energy a conserved quantity in central force
>>>> potentials?
>>>
>>> Are you sure you don't mean angular momentum?
>>
>> [I am using the context of classical mechanics, as this
>> is implied by the problem statement.]
>>
>>Hmmm. I can invent a central force for which energy is not conserved. I
>>make no claim this sort of central force is a good model of anything,
>>merely that one can invent such a force (just make it depend explicitly
>>on time).
>
> Maybe I should have been explicit - central forces under the usual
> definition that the force acts radially between the two particles and
> that the potential is not an explicit function of time or velocity.
The last bit is the key point. If it is true Noethers theorem guarantees
energy is conserved - by the modern definition of energy as the conserved
Noether charge related to time symmetry.
Not meant for Eric - he already knows it - but for those interested in
following it up I highly recommend the following book. Reading it right
now - really really good.
http://www.colorado.edu/philosophy/vstenger/nothing.html
Thanks
Bill
JM Albuquerque
What's the beef around here?
Why have you changed angular momentum to momentum?
Let me correct myself and say it again:
Energy is the time derivative of angular momentum times a given
angular displacement.
Or else, angular momentum on a given angular displacement is
the integral of the energy that exist on that given angular
displacement over time.
If you guys don't get it, I will explain later and repply to
Tom Roberts as soon as I got the time.
NoEinstein
> Uncle Al <Uncle...@hate.spam.net> wrote in news:46FAEF9A.A4D91864
Dear bz: A person running into a stationary door impacts the door
with their own KE. But if two identical twins, running the same speed
collide, the "force" either would feel is identical to what they felt
running against the door. No "combined velocity" ever applies to
normal light objects of about the same size, because: The object with
the lest KE will always be "bounced away", and the force the latter
would experience is no greater than it's own KE-not the combined KE of
the two. This is the reason that a common housefly can sometimes
survive being swatted in mid air. The force on the fly is never much
more than its own weight. But that fly can still be batted across the
room. One such fly I hit impacted a soft curtain, fell to the floor,
and in a few seconds flew away as if nothing had happened. The
"physics" of things may not always be obvious. My advice: Don't speak
the first thing that comes into your head; REASON! - NoEinstein -
JM Albuquerque
Let me correct myself and say it again:
Energy is the time derivative of angular momentum times a given
angular displacement.
Or else, angular momentum on a given angular displacement is
the integral of the energy that exist on that given angular
displacement over time.
Energy = t . d(teta) = ( F x r ) . d(teta) = scalar
So, this all boils down to an angular displacement, that was
missing in my previous post, because was irrelevante for
the point I intended to make.
Therefore, my previous argument on time remains, even
if you made it disappear in your reply. Who cares anyway?
>
>
> Tom Roberts
JM Albuquerque
"sal" <pragm...@nospam.org> escreveu na mensagem
news:46fc7cf5$0$24907$ec3e...@news.usenetmonster.com...
> On Fri, 28 Sep 2007 03:07:21 +0100, JM Albuquerque wrote:
>
>> "Dono" <sa...@comcast.net> escreveu na mensagem
>> news:1190944917.9...@w3g2000hsg.googlegroups.com...
>>> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip
>>> crap]
>>>> Energy is the time derivative of angular momentum. Or else, angular
>>>> momentum the integral of energy over time.
>>>
>>>
>>> No idiot, did you fail physics in high school? Don't answer that, you
>>> failed.
>>
>> Look at the units - kg - m - s, and tell me I'm wrong.
>
> Dimensions of angular momentum are mass * length^2 / time
>
> Dimensions of kinetic energy are mass * length^2 / time^2
>
> A factor of 1/time is hardly ignorable. So, by your own argument, you're
> wrong.
How?
1/time is the derivative term to be "added".
> In any case, if energy were dL/dt, then if energy were constant and
> positive then L would be constantly increasing, which is nonsensical.
Look up how you do your derivatives.
> Worse, energy is a scalar, and dL/dt is a vector (commonly called
> torque). Scalar and vector mismatch is about as fundamental as
> mismatched dimensions -- maybe more so.
dL/dt is torque.
One needs to multiply torque by an angular displacement (the dot
product) to get energy as a scalar.
> What ever gave you the idea that dL/dt = T?
T = torque, and yes dL/dt = torque
Eric Gisse
wrote:
Really? Where does d\theta point?
>
>So, this all boils down to an angular displacement, that was
>missing in my previous post, because was irrelevante for
>the point I intended to make.
>Therefore, my previous argument on time remains, even
>if you made it disappear in your reply. Who cares anyway?
Scientists.
The units of torque and energy are the same but they are not the same
quantity no matter how hard you flap your arms or how shrilly you
scream.
>
>>
>>
>> Tom Roberts
>
JM Albuquerque
Well, theta is an angle so it's an angle, you know what an angle
is, don't you (expressed in radians)?
>>So, this all boils down to an angular displacement, that was
>>missing in my previous post, because was irrelevante for
>>the point I intended to make.
>>Therefore, my previous argument on time remains, even
>>if you made it disappear in your reply. Who cares anyway?
>
> Scientists.
>
> The units of torque and energy are the same but they are not the same
> quantity no matter how hard you flap your arms or how shrilly you
> scream.
Don't be a Policeman.
All I intended was to make a point about "time", but lower people
(far from being scientists) only care with silly details and even when
the detail is pointed out, several times, keep complaining to
avoid the really important issue.
If you didn't care about being my Policeman and just tried to
really understand the deep reasoning behind those statements
you would have learned something.
Angular momentum is more then momentum, since it implies
rotation. And energy can be defined on a rotating system as well
as in an inertial system, but that flew so high over your head...
...you know.
>
>>
>>>
>>>
>>> Tom Roberts
>>
Dono
[snip crap]
You couldn't even get the units right.
You couldn't tell the difference between the vector angular momentum
and the scalar energy.
You got everything else right :-)
Eric Gisse
wrote:
[...]
>>>
>>>Let me correct myself and say it again:
>>>Energy is the time derivative of angular momentum times a given
>>>angular displacement.
>>>Or else, angular momentum on a given angular displacement is
>>>the integral of the energy that exist on that given angular
>>>displacement over time.
>>>
>>>Energy = t . d(teta) = ( F x r ) . d(teta) = scalar
>>
>> Really? Where does d\theta point?
>
>Well, theta is an angle so it's an angle, you know what an angle
>is, don't you (expressed in radians)?
Not the question I asked, JM.
Where does d\theta point?
>
>
>>>So, this all boils down to an angular displacement, that was
>>>missing in my previous post, because was irrelevante for
>>>the point I intended to make.
>>>Therefore, my previous argument on time remains, even
>>>if you made it disappear in your reply. Who cares anyway?
>>
>> Scientists.
>>
>> The units of torque and energy are the same but they are not the same
>> quantity no matter how hard you flap your arms or how shrilly you
>> scream.
>
>Don't be a Policeman.
>All I intended was to make a point about "time", but lower people
>(far from being scientists) only care with silly details and even when
>the detail is pointed out, several times, keep complaining to
>avoid the really important issue.
Silly details like equating the quantities torque [a vector] and
energy [a scalar] because they have the same units?
>
>If you didn't care about being my Policeman and just tried to
>really understand the deep reasoning behind those statements
>you would have learned something.
I'm yet to see any deep reasoning at work here.
>Angular momentum is more then momentum, since it implies
>rotation. And energy can be defined on a rotating system as well
>as in an inertial system, but that flew so high over your head...
>...you know.
You don't need rotation to have angular momentum.
>
>
>
>>
>>>
>>>>
>>>>
>>>> Tom Roberts
>>>
>
JM Albuquerque
> On Sat, 29 Sep 2007 02:12:53 +0100, "JM Albuquerque" <jmD...@clix.pt>
> wrote:
> [...]
>
>>>>
>>>>Let me correct myself and say it again:
>>>>Energy is the time derivative of angular momentum times a given
>>>>angular displacement.
>>>>Or else, angular momentum on a given angular displacement is
>>>>the integral of the energy that exist on that given angular
>>>>displacement over time.
>>>>
>>>>Energy = t . d(teta) = ( F x r ) . d(teta) = scalar
>>>
>>> Really? Where does d\theta point?
>>
>>Well, theta is an angle so it's an angle, you know what an angle
>>is, don't you (expressed in radians)?
>
> Not the question I asked, JM.
>
> Where does d\theta point?
It's a scalar.
Ever seen a dot product of a vector times a scalar
to get an scalar?
Where does a scalar point?
A scalar points top-down and tilting a little to the left.
>
>>
>>
>>>>So, this all boils down to an angular displacement, that was
>>>>missing in my previous post, because was irrelevante for
>>>>the point I intended to make.
>>>>Therefore, my previous argument on time remains, even
>>>>if you made it disappear in your reply. Who cares anyway?
>>>
>>> Scientists.
>>>
>>> The units of torque and energy are the same but they are not the same
>>> quantity no matter how hard you flap your arms or how shrilly you
>>> scream.
>>
>>Don't be a Policeman.
>>All I intended was to make a point about "time", but lower people
>>(far from being scientists) only care with silly details and even when
>>the detail is pointed out, several times, keep complaining to
>>avoid the really important issue.
>
> Silly details like equating the quantities torque [a vector] and
> energy [a scalar] because they have the same units?
>
>>
>>If you didn't care about being my Policeman and just tried to
>>really understand the deep reasoning behind those statements
>>you would have learned something.
>
> I'm yet to see any deep reasoning at work here.
You won't because I don't care anymore.
>>Angular momentum is more then momentum, since it implies
>>rotation. And energy can be defined on a rotating system as well
>>as in an inertial system, but that flew so high over your head...
>>...you know.
>
> You don't need rotation to have angular momentum.
... and pink elephants fly too.
(only light pink, because dark pink fly very low).
JM Albuquerque
> On Sep 28, 6:12 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote:
> [snip crap]
>
> You couldn't even get the units right.
What units? Units don't exist. They are evil ideas of bad people.
> You couldn't tell the difference between the vector angular momentum
> and the scalar energy.
What angular momentum? I don't know any guy with that
name. Nor the so called "scalar". Energy is packed on
metal boxes and we have three sizes: small, regular and large.
> You got everything else right :-)
No I don't.
I got everything wrong, top to bottom.
Dono
> "Dono" <sa...@comcast.net> escreveu na mensagemnews:1191030352....@22g2000hsm.googlegroups.com...
>
> > On Sep 28, 6:12 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote:
> > [snip crap]
>
> > You couldn't even get the units right.
>
> What units? Units don't exist. They are evil ideas of bad people.
>
Classic crank vintage :-)
> > You couldn't tell the difference between the vector angular momentum
> > and the scalar energy.
>
> What angular momentum? I don't know any guy with that
> name. Nor the so called "scalar". Energy is packed on
> metal boxes and we have three sizes: small, regular and large.
>
Classic imbecile vintage :-)
> > You got everything else right :-)
>
> No I don't.
> I got everything wrong, top to bottom.
First correct words coming out of your mouth :-)
Eric Gisse
wrote:
>
>"Eric Gisse" <jowr.pi...@gmail-nospam.com> escreveu na mensagem
>news:njbrf39gkaf45m5u0...@4ax.com...
>> On Sat, 29 Sep 2007 02:12:53 +0100, "JM Albuquerque" <jmD...@clix.pt>
>> wrote:
>> [...]
>>
>>>>>
>>>>>Let me correct myself and say it again:
>>>>>Energy is the time derivative of angular momentum times a given
>>>>>angular displacement.
>>>>>Or else, angular momentum on a given angular displacement is
>>>>>the integral of the energy that exist on that given angular
>>>>>displacement over time.
>>>>>
>>>>>Energy = t . d(teta) = ( F x r ) . d(teta) = scalar
>>>>
>>>> Really? Where does d\theta point?
>>>
>>>Well, theta is an angle so it's an angle, you know what an angle
>>>is, don't you (expressed in radians)?
>>
>> Not the question I asked, JM.
>>
>> Where does d\theta point?
>
>It's a scalar.
>Ever seen a dot product of a vector times a scalar
>to get an scalar?
Nope.
You can't have a dot product with a vector and a scalar and get a
scalar.
>Where does a scalar point?
>A scalar points top-down and tilting a little to the left.
No, a scalar points nowhere because it has no directional information.
>
>>
>>>
>>>
>>>>>So, this all boils down to an angular displacement, that was
>>>>>missing in my previous post, because was irrelevante for
>>>>>the point I intended to make.
>>>>>Therefore, my previous argument on time remains, even
>>>>>if you made it disappear in your reply. Who cares anyway?
>>>>
>>>> Scientists.
>>>>
>>>> The units of torque and energy are the same but they are not the same
>>>> quantity no matter how hard you flap your arms or how shrilly you
>>>> scream.
>>>
>>>Don't be a Policeman.
>>>All I intended was to make a point about "time", but lower people
>>>(far from being scientists) only care with silly details and even when
>>>the detail is pointed out, several times, keep complaining to
>>>avoid the really important issue.
>>
>> Silly details like equating the quantities torque [a vector] and
>> energy [a scalar] because they have the same units?
>>
>>>
>>>If you didn't care about being my Policeman and just tried to
>>>really understand the deep reasoning behind those statements
>>>you would have learned something.
>>
>> I'm yet to see any deep reasoning at work here.
>
>You won't because I don't care anymore.
That makes the false presumption that you ever did. All you have done
so far is tell us how physics is "should" be done despite not really
knowing much about anything.
>
>
>>>Angular momentum is more then momentum, since it implies
>>>rotation. And energy can be defined on a rotating system as well
>>>as in an inertial system, but that flew so high over your head...
>>>...you know.
>>
>> You don't need rotation to have angular momentum.
>
>... and pink elephants fly too.
>(only light pink, because dark pink fly very low).
>
Show me where the rotation is in L = r x p.
Sam Wormley
Having a meltdown, I see....
Tom Roberts
>> > [I am using the context of classical mechanics.]
> Let me correct myself and say it again:
> Energy is the time derivative of angular momentum times a given
> angular displacement.
It seems that if you try often enough, sometimes you happen to hit upon
something that is essentially correct (or at least close enough so that
I can correct it without too much violence to your words). That seems to
have happened here (as you clearly do not yourself know how to fix up
what you said).
> Or else, angular momentum on a given angular displacement is
> the integral of the energy that exist on that given angular
> displacement over time.
This is just word salad, and I'm ignoring it.
> Energy = t . d(teta) = ( F x r ) . d(teta) = scalar
OK. Yes, the time derivative of angular momentum is indeed the applied
torque. The correct formulation is:
W = integral t . dq where W is the work achieved,
t is the applied torque, and dq is
the 3-vector whose direction is the
rotation axis [#] and whose magnitude
is the (infinitesimal) angle of
rotation.
But work is a type of energy, so this is essentially what you wrote,
fixed up. As dq is an angle, and is therefore unitless, we see that work
and torque must have the same units, even though they are not at all the
same thing.
[#] Use the same right-hand-rule used for cross products.
A few loose ends:
* dL/dt is indeed torque (L = angular momentum).
* here d(theta) [my dq above] is a 3-vector, not a scalar.
* d(theta) as a 3-vector points along the axis of rotation,
and the angle theta is its magnitude.
Tom Roberts
JM Albuquerque
Ok, the d(theta) is a vector, if you assume that the angle
theta is a generic angle which can point anywhere in space,
so that you need the vector notation to give him the right
direction.
I'm used to see the Euler angles (theta, phi and psi) so
that I have 3-orthogonal principal rotating axis and
3-orthogonal fixed axis. The origin is the same (for
the rotating and for the inertial coordinates).
The 3 Euler angles are defined relative from the
fixed to the rotating axis, so that my theta already
as a origin and a direction.
So, I'm used to see it as a scalar, since Euler angles
are a much better way of working. A generic angle
turns out to be a mess and a great confusion.
That's just a detail, but OK, you are right.
JM Albuquerque
> On Sat, 29 Sep 2007 03:03:19 +0100, "JM Albuquerque" <jmD...@clix.pt>
> wrote:
>
>>
>>"Eric Gisse" <jowr.pi...@gmail-nospam.com> escreveu na mensagem
>>news:njbrf39gkaf45m5u0...@4ax.com...
>>> On Sat, 29 Sep 2007 02:12:53 +0100, "JM Albuquerque" <jmD...@clix.pt>
>>> wrote:
>>> [...]
>>>
>>>>>>
>>>>>>Let me correct myself and say it again:
>>>>>>Energy is the time derivative of angular momentum times a given
>>>>>>angular displacement.
>>>>>>Or else, angular momentum on a given angular displacement is
>>>>>>the integral of the energy that exist on that given angular
>>>>>>displacement over time.
>>>>>>
>>>>>>Energy = t . d(teta) = ( F x r ) . d(teta) = scalar
>>>>>
>>>>> Really? Where does d\theta point?
>>>>
>>>>Well, theta is an angle so it's an angle, you know what an angle
>>>>is, don't you (expressed in radians)?
>>>
>>> Not the question I asked, JM.
>>>
>>> Where does d\theta point?
>>
>>It's a scalar.
>>Ever seen a dot product of a vector times a scalar
>>to get an scalar?
>
> Nope.
Euler angles.
> You can't have a dot product with a vector and a scalar and get a
> scalar.
Why not?
>>Where does a scalar point?
>>A scalar points top-down and tilting a little to the left.
>
> No, a scalar points nowhere because it has no directional information.
Euler angles have an origin an a direction already.
>>>>>>So, this all boils down to an angular displacement, that was
>>>>>>missing in my previous post, because was irrelevante for
>>>>>>the point I intended to make.
>>>>>>Therefore, my previous argument on time remains, even
>>>>>>if you made it disappear in your reply. Who cares anyway?
>>>>>
>>>>> Scientists.
>>>>>
>>>>> The units of torque and energy are the same but they are not the same
>>>>> quantity no matter how hard you flap your arms or how shrilly you
>>>>> scream.
>>>>
>>>>Don't be a Policeman.
>>>>All I intended was to make a point about "time", but lower people
>>>>(far from being scientists) only care with silly details and even when
>>>>the detail is pointed out, several times, keep complaining to
>>>>avoid the really important issue.
>>>
>>> Silly details like equating the quantities torque [a vector] and
>>> energy [a scalar] because they have the same units?
>>>
>>>>
>>>>If you didn't care about being my Policeman and just tried to
>>>>really understand the deep reasoning behind those statements
>>>>you would have learned something.
>>>
>>> I'm yet to see any deep reasoning at work here.
>>
>>You won't because I don't care anymore.
>
> That makes the false presumption that you ever did. All you have done
> so far is tell us how physics is "should" be done despite not really
> knowing much about anything.
You are mixing threads.
>>>>Angular momentum is more then momentum, since it implies
>>>>rotation. And energy can be defined on a rotating system as well
>>>>as in an inertial system, but that flew so high over your head...
>>>>...you know.
>>>
>>> You don't need rotation to have angular momentum.
>>
>>... and pink elephants fly too.
>>(only light pink, because dark pink fly very low).
>>
>
> Show me where the rotation is in L = r x p.
That's not L - angular momentum.
L = I omega ( kg m^2 s^-1 )
The rotation is defined by "omega" (and angular velocity).
Pmb
"JM Albuquerque" <jmD...@clix.pt> wrote in message
news:5m6sflF...@mid.individual.net...
>
> "Tom Roberts" <tjrobe...@sbcglobal.net> escreveu na mensagem
> news:3SjLi.9473$JD....@newssvr21.news.prodigy.net...
>> JM Albuquerque wrote:
>>>> > [I am using the context of classical mechanics.]
>>> Let me correct myself and say it again:
>>> Energy is the time derivative of angular momentum times a given
>>> angular displacement.
Do you have a source to a mechanics text that I might see this equality in?
IMHO that is, at best, an equality as applied to mechanics rather than an
actual definition.
I have no recollection of hearing such a definition (then again, as I say so
often, my memory sucks! lol!!). How is this "definition" supposed to be
applied to something like a static EM field where there is no angular
momentum? How does it apply to any system with no angular moment. E.g. the
rest energy of a particle in its rest frame?
Pete
Ken S. Tucker
> "Eric Gisse" <jowr.pi.nos...@gmail-nospam.com> escreveu na mensagemnews:ughrf3lr160assf41...@4ax.com...
>
>
>
> > On Sat, 29 Sep 2007 03:03:19 +0100, "JM Albuquerque" <jmDO...@clix.pt>
> > wrote:
>
> >>"Eric Gisse" <jowr.pi.nos...@gmail-nospam.com> escreveu na mensagem
> >>news:njbrf39gkaf45m5u0...@4ax.com...
> >>> On Sat, 29 Sep 2007 02:12:53 +0100, "JM Albuquerque" <jmDO...@clix.pt>
> >>> wrote:
> >>> [...]
>
> >>>>>>Let me correct myself and say it again:
> >>>>>>Energy is the time derivative of angular momentum times a given
> >>>>>>angular displacement.
> >>>>>>Or else, angular momentum on a given angular displacement is
> >>>>>>the integral of the energy that exist on that given angular
> >>>>>>displacement over time.
>
> >>>>>>Energy = t . d(teta) = ( F x r ) . d(teta) = scalar
>
> >>>>> Really? Where does d\theta point?
>
> >>>>Well, theta is an angle so it's an angle, you know what an angle
> >>>>is, don't you (expressed in radians)?
>
> >>> Not the question I asked, JM.
>
> >>> Where does d\theta point?
>
> >>It's a scalar.
> >>Ever seen a dot product of a vector times a scalar
> >>to get an scalar?
>
> > Nope.
>
> Euler angles.
>
> > You can't have a dot product with a vector and a scalar and get a
> > scalar.
A.B (scalar) = (scalar)
> Why not?
Hi JM,
Mathematicians have ways of doing things,
http://en.wikipedia.org/wiki/Dyadic_tensor
Don't be to hard in Gisse, she's a newbie:-).
Thought you might want to glance dyads kiddo.
Ken
JM Albuquerque
"Pmb" <som...@somewhere.net> escreveu na mensagem
news:xK2dnenl1edL3GPb...@comcast.com...
>
> "JM Albuquerque" <jmD...@clix.pt> wrote in message
> news:5m6sflF...@mid.individual.net...
>>
>> "Tom Roberts" <tjrobe...@sbcglobal.net> escreveu na mensagem
>> news:3SjLi.9473$JD....@newssvr21.news.prodigy.net...
>>> JM Albuquerque wrote:
>>>>> > [I am using the context of classical mechanics.]
>>>> Let me correct myself and say it again:
>>>> Energy is the time derivative of angular momentum times a given
>>>> angular displacement.
>
> Do you have a source to a mechanics text that I might see this equality
> in?
No, I don't have.
And you shouldn't relly on that too.
This is the old problem between rotating FoR and inertial FoR.
It's my way to see the problem in a rotating FoR.
Angular momentum implies a rotating FoR, even if some can think
about "r x p", but that "r" is a vector whose point of application
is at the center of the rotation (a rotating vector).
> IMHO that is, at best, an equality as applied to mechanics rather than an
> actual definition.
>
> I have no recollection of hearing such a definition (then again, as I say
> so often, my memory sucks! lol!!). How is this "definition" supposed to be
> applied to something like a static EM field where there is no angular
> momentum? How does it apply to any system with no angular moment. E.g. the
> rest energy of a particle in its rest frame?
I'm talking classical mechanics, only.
> Pete
>
Igor
wrote:
> On Sat, 29 Sep 2007 01:20:41 +0100, "JM Albuquerque" <jmDO...@clix.pt>
> wrote:
>
>
>
>
>
>
>
> >"Tom Roberts" <tjroberts...@sbcglobal.net> escreveu na mensagem
> >news:vp_Ki.55536$YL5....@newssvr29.news.prodigy.net...
> >> JM Albuquerque wrote:
Technically, perpendicular to the plane of rotation. It's an axial
vector. This is because r x dtheta is the displacement in the plane.
Not that JM actually understands that, but nice try on his part.
Igor
> "Eric Gisse" <jowr.pi.nos...@gmail-nospam.com> escreveu na mensagemnews:ughrf3lr160assf41...@4ax.com...
>
>
>
>
>
> > On Sat, 29 Sep 2007 03:03:19 +0100, "JM Albuquerque" <jmDO...@clix.pt>
> > wrote:
>
> >>"Eric Gisse" <jowr.pi.nos...@gmail-nospam.com> escreveu na mensagem
> >>news:njbrf39gkaf45m5u0...@4ax.com...
> >>> On Sat, 29 Sep 2007 02:12:53 +0100, "JM Albuquerque" <jmDO...@clix.pt>
Maybe it's time you learned the difference between defined quantities
and derived quantities.
PD
>
> It helps to STUDY the subject of the discussion before opening your
> mouth (keyboard). While torque and energy share the same units, they are
> completely different:
>
> W = integral F.ds where W is work (=energy), F is the
> 3-vector force on an object, and ds
> is the 3-vector displacement of it.
>
> t = F x r where t is the torque, F is the 3-vector
> force, r is the radius at which the force
> is applied, and x is the cross product.
>
r x R, actually.
Not that it's going to make any difference to JM.
Eric Gisse
wrote:
Really?
Show me where the dot product of a vector and a scalar gets you a
scalar in the usage of Euler angles.
>
>
>> You can't have a dot product with a vector and a scalar and get a
>> scalar.
>
>Why not?
The dot product maps two vectors to the real line. A scalar is not a
vector.
>
>>>Where does a scalar point?
>>>A scalar points top-down and tilting a little to the left.
>>
>> No, a scalar points nowhere because it has no directional information.
>
>Euler angles have an origin an a direction already.
Except they don't have any directional information. Being defined with
respect to something is not the same as having direction.
Nope. The principle definition for angular momentum is L = r x p.
That L = I.\omega is something that is derived from L = r x p.
Eric Gisse
Ken...I have no idea what you are good at but physics is not it. Go do
something else.
Timo A. Nieminen
> "JM Albuquerque" <jmD...@clix.pt> wrote:
>> "Eric Gisse" wrote:
>>> "JM Albuquerque" <jmD...@clix.pt> wrote:
>>>> "Eric Gisse" wrote:
>>>>>
>>>>> Where does d\theta point?
>>>>
>>>> It's a scalar.
>>>> Ever seen a dot product of a vector times a scalar
>>>> to get an scalar?
>>>
>>> Nope.
>>
>> Euler angles.
>
> Really?
>
> Show me where the dot product of a vector and a scalar gets you a
> scalar in the usage of Euler angles.
[cut]
>>>> Where does a scalar point?
>>>> A scalar points top-down and tilting a little to the left.
>>>
>>> No, a scalar points nowhere because it has no directional information.
>>
>> Euler angles have an origin an a direction already.
>
> Except they don't have any directional information. Being defined with
> respect to something is not the same as having direction.
What are Euler angles for? They're a (lousy) parameterisation of
rotations. In the magic 3D, a rotation can be described by a
(pseudeo)vector - the rotation axis - and Euler angles are just one way to
describe this rotation axis. Thus, a set of Euler angles certainly has
directional information, as they're equivalent to a vector.
An individual Euler angle is just a number (not even a scalar in the
geometric sense), and has no directional information.
Why Euler angles arose in this discussion is a mystery.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
Ken S. Tucker
wrote:
LOL, Gisse don't know dyads!
Pass HS, get a job, retire at 40.
Ken
Eric Gisse
<ti...@physics.uq.edu.au> wrote:
[...]
>Why Euler angles arose in this discussion is a mystery.
Agreed.
Igor
And you probably don't know your dyads from your tetrads, so be
careful.
Ken S. Tucker
LOL, that's what girls are for!
I haven't exploited them too much,
I mean dyads and tetrads.
I noticed Weinberg in his QFT book 3, is
using Tetrads (Veirbein) chp. 31 "super
gravity", looks ok to me so far.
I'm looking at other approaches too.
Regards
Ken S. Tucker
JM Albuquerque
"Timo A. Nieminen" <ti...@physics.uq.edu.au> escreveu na mensagem
news:Pine.WNT.4.64.07...@serene.st...
Assuming two sets of 3-orthogonal axis, both of which have
always the same origin for coordinates, being one set fixed,
and the other set rotating around the commum origin.
Actually its 3 sets of 3-orthogonal axis (all according to the
right hand-rule) 2 sets of (x,y,z) and one set of (teta, phi, psi)
that are rotating axis (for instance, the x axis also is an axis for
its own rotation, and so on for y and z, being x,y,z already
rotating - rotation over rotation and one fixed = 3 sets).
Euler angles (teta, phi, psi) are the VERY BEST OF ALL TIMES.
> An individual Euler angle is just a number (not even a scalar in the
> geometric sense), and has no directional information.
>
> Why Euler angles arose in this discussion is a mystery.
Because I've messed up with vectors.
The vectorial analysis is a very bad choice when things really
become hard.
For hard stuff one requires Euler angles.
Without Euler angles, nor gyroscopic momentum, nor
gyroscopic precession can be accounted for.
JM Albuquerque
"Igor" <thoo...@excite.com> escreveu na mensagem
news:1191086382....@50g2000hsm.googlegroups.com...
No problem to understand that.
Simply the vectors are colinear, so that cos(0) = 1
So, the "dot product" is there only to complicate what
is quite simple.
I was wrong with the "scalar" issue.
The vectors are colinear, that's all.
Therefore, make it a multiplication instead of the dot product,
and everything else is just smoke.
Igor
> "Timo A. Nieminen" <t...@physics.uq.edu.au> escreveu na mensagemnews:Pine.WNT.4.64.07...@serene.st...
>
>
>
>
>
> > On Sat, 29 Sep 2007, Eric Gisse wrote:
>
I can assure you that both angular momentum and precession, which is
just the vector addition of two or more angular momenta, can be
described quite well without using Euler angles.
Timo A. Nieminen
> "Timo A. Nieminen" wrote:
>>
>> What are Euler angles for? They're a (lousy) parameterisation of
>> rotations. In the magic 3D, a rotation can be described by a
>> (pseudeo)vector - the rotation axis - and Euler angles are just one way to
>> describe this rotation axis. Thus, a set of Euler angles certainly has
>> directional information, as they're equivalent to a vector.
>
> Assuming two sets of 3-orthogonal axis, both of which have
> always the same origin for coordinates, being one set fixed,
> and the other set rotating around the commum origin.
> Actually its 3 sets of 3-orthogonal axis (all according to the
> right hand-rule) 2 sets of (x,y,z) and one set of (teta, phi, psi)
> that are rotating axis (for instance, the x axis also is an axis for
> its own rotation, and so on for y and z, being x,y,z already
> rotating - rotation over rotation and one fixed = 3 sets).
> Euler angles (teta, phi, psi) are the VERY BEST OF ALL TIMES.
Yay! The Rotation Parameterisation Holy War returns!
Euler angles suck so much that some people resort to quaternions as a
substitute (to avoid gimbal lock in their equations of motion, if they
have a reason beyond "quaternions are cool!").
What's wrong with all of the various vector (i.e. axis-angle)
parameterisations?
Also, it's hard to beat the 3x3 rotation matrix as a parameterisation of
rotation. (Even more so if you go to higher dimensions.)
>> An individual Euler angle is just a number (not even a scalar in the
>> geometric sense), and has no directional information.
>>
>> Why Euler angles arose in this discussion is a mystery.
>
> Because I've messed up with vectors.
> The vectorial analysis is a very bad choice when things really
> become hard.
> For hard stuff one requires Euler angles.
>
> Without Euler angles, nor gyroscopic momentum, nor
> gyroscopic precession can be accounted for.
One can mess up with vectors, one can mess up with quaternions, one can
mess up Euler angles. One is least likely to mess up with the most
familiar.
But angular momentum and precession can be accounted for perfectly well in
a vector formulation.
(Hmm. Angular momentum and precession in 4D could be fun and educational
too. No more pretending that 3x3 antisymmetric tensors are vectors!)
JM Albuquerque
If so, please show me how you derive precession velocity for
a given torque N1 over the axis of spin of a spinning disk of
angular velocity w3 and inertia moment I3:
precession velocity =
=(I3*w3)/(2*I2*cos(teta)) * (1-sqrt(1 - (4*N1*I2)/(I3^2*w3^2)))
JM Albuquerque
"Timo A. Nieminen" <ti...@physics.uq.edu.au> escreveu na mensagem
news:Pine.WNT.4.64.07...@serene.st...
>
>> "Timo A. Nieminen" wrote:
>>>
>>> What are Euler angles for? They're a (lousy) parameterisation of
>>> rotations. In the magic 3D, a rotation can be described by a
>>> (pseudeo)vector - the rotation axis - and Euler angles are just one way
>>> to
>>> describe this rotation axis. Thus, a set of Euler angles certainly has
>>> directional information, as they're equivalent to a vector.
>>
>> Assuming two sets of 3-orthogonal axis, both of which have
>> always the same origin for coordinates, being one set fixed,
>> and the other set rotating around the commum origin.
>> Actually its 3 sets of 3-orthogonal axis (all according to the
>> right hand-rule) 2 sets of (x,y,z) and one set of (teta, phi, psi)
>> that are rotating axis (for instance, the x axis also is an axis for
>> its own rotation, and so on for y and z, being x,y,z already
>> rotating - rotation over rotation and one fixed = 3 sets).
>> Euler angles (teta, phi, psi) are the VERY BEST OF ALL TIMES.
>
> Yay! The Rotation Parameterisation Holy War returns!
>
> Euler angles suck so much that some people resort to quaternions as a
> substitute (to avoid gimbal lock in their equations of motion, if they
> have a reason beyond "quaternions are cool!").
Euler angles are a piece of cake.
I know nothing about quaternions (chinese to me).
> What's wrong with all of the various vector (i.e. axis-angle)
> parameterisations?
>
> Also, it's hard to beat the 3x3 rotation matrix as a parameterisation of
> rotation. (Even more so if you go to higher dimensions.)
>
>>> An individual Euler angle is just a number (not even a scalar in the
>>> geometric sense), and has no directional information.
>>>
>>> Why Euler angles arose in this discussion is a mystery.
>>
>> Because I've messed up with vectors.
>> The vectorial analysis is a very bad choice when things really
>> become hard.
>> For hard stuff one requires Euler angles.
>>
>> Without Euler angles, nor gyroscopic momentum, nor
>> gyroscopic precession can be accounted for.
>
> One can mess up with vectors, one can mess up with quaternions, one can
> mess up Euler angles. One is least likely to mess up with the most
> familiar.
>
> But angular momentum and precession can be accounted for perfectly well in
> a vector formulation.
Now I have to disagree.
The precession velocity for a given torque N1 applied over the axis
of spin of a spinning disk of angular velocity w3 and inertia moment I3,
is given by (more correct then the previous):
precession velocity =
=(I3*w3)/(2*I2*cos(teta)) * (1-sqrt(1 - (4*N1*I2*cos(teta))/(I3^2*w3^2)))
(case of w1 = 0, so that I1 is out of the problem)
With vectors you mean?
Never, ever.
JM Albuquerque
"JM Albuquerque" <jmD...@clix.pt> escreveu na mensagem
news:5madorF...@mid.individual.net...
> The precession velocity for a given torque N1 applied over the axis
> of spin of a spinning disk of angular velocity w3 and inertia moment I3,
> is given by (more correct then the previous):
> precession velocity =
> =(I3*w3)/(2*I2*cos(teta)) * (1-sqrt(1 - (4*N1*I2*cos(teta))/(I3^2*w3^2)))
> (case of w1 = 0, so that I1 is out of the problem)
If anyone wants to know how the above is derived I can send
the pdf file (about 850 kb).
It's from the book :
http://www.people.fas.harvard.edu/~djmorin/textbook.htm
It was free up to a few mounth ago.
Timo A. Nieminen
> "Timo A. Nieminen" wrote:
>> On Sun, 30 Sep 2007, JM Albuquerque wrote:
>>
>>> Euler angles (teta, phi, psi) are the VERY BEST OF ALL TIMES.
>>
>> Yay! The Rotation Parameterisation Holy War returns!
>>
>> Euler angles suck so much that some people resort to quaternions as a
>> substitute (to avoid gimbal lock in their equations of motion, if they
>> have a reason beyond "quaternions are cool!").
>
> Euler angles are a piece of cake.
> I know nothing about quaternions (chinese to me).
You know nothing about the alernatives to Euler angles but are willing to
claim that Euler angles are the VERY BEST OF ALL TIMES?
>> What's wrong with all of the various vector (i.e. axis-angle)
>> parameterisations?
>>
>> Also, it's hard to beat the 3x3 rotation matrix as a parameterisation of
>> rotation. (Even more so if you go to higher dimensions.)
[cut]
>>> Without Euler angles, nor gyroscopic momentum, nor
>>> gyroscopic precession can be accounted for.
>>
>> One can mess up with vectors, one can mess up with quaternions, one can
>> mess up Euler angles. One is least likely to mess up with the most
>> familiar.
>>
>> But angular momentum and precession can be accounted for perfectly well in
>> a vector formulation.
>
> Now I have to disagree.
>
> The precession velocity for a given torque N1 applied over the axis
> of spin of a spinning disk of angular velocity w3 and inertia moment I3,
> is given by (more correct then the previous):
> precession velocity =
> =(I3*w3)/(2*I2*cos(teta)) * (1-sqrt(1 - (4*N1*I2*cos(teta))/(I3^2*w3^2)))
> (case of w1 = 0, so that I1 is out of the problem)
>
> With vectors you mean?
> Never, ever.
L = angular momentum
w = angular velocity of disk
I = moment of inertia
W = angular velocity of precession
t = torque
L = Iw
t = dL/dt
t = I dw/dt
For the simple case of a disk spinning about its axis, and torque at right
angles to the spin (ie applied force and spin are co-planar), we have
|W| = |dw/dt|/|w|
|W| = |t|/(I3 w3).
JM Albuquerque
"Timo A. Nieminen" <ti...@physics.uq.edu.au> escreveu na mensagem
news:Pine.WNT.4.64.07...@serene.st...
That's the aproximate solution.
Your above solution has to produce exactly the same result
as "my" equation.
Now I'm going to explain what's missing in your vectorial
calculation:
1 - Assumes cos(teta) = 1 or else teta = 0
teta = 0 means maximum torque if the gyroscope is
precessing due to a gravity mass, whose torque will be:
t = m g r sin(teta)
(From the book)
2 - The error in the above analysis is that we omitted the
angular momentum arising from the x2 (defined in Section
8.7.1) component of the angular velocity due to the
precession of the top around the z-axis. This component
has magnitude (omega)sin(teta). The angular momentum
due to this angular velocity component has magnitude
L2 = I (omega) sin(teta)
(omega) = your value: |W| = |t|/(I3 w3).
Eric Gisse
[...]
>
>And you probably don't know your dyads from your tetrads, so be
>careful.
>
Replace "and" with "but". I do know dyads - just a fancy name for
tensor product.
Methinks "dyads" is a bit of terminology that is slightly dated.
Eric Gisse
<ti...@physics.uq.edu.au> wrote:
[...]
Now I want you to mention the Euler equations of motion, just so I can
see JM claim there are no solutions once again.
Eric Gisse
wrote:
[...]
>No problem to understand that.
>Simply the vectors are colinear, so that cos(0) = 1
>So, the "dot product" is there only to complicate what
>is quite simple.
>
>I was wrong with the "scalar" issue.
>The vectors are colinear, that's all.
>Therefore, make it a multiplication instead of the dot product,
>and everything else is just smoke.
>
Vector analysis is haaaarrrddd :(
JM Albuquerque
Bet my live on that.
JM Albuquerque
Translation (my translation):
The solution |W| = |t|/(I3 w3) assumes infinite gyroscopic
moment (it doesn't give under the torque).
If the gyroscopic moment I3*w3 is not enough for the
torque, or else, if the gyroscope is not free to precess
with zero friction and a torque around w2 exists, so that
all this starts to move, the above equation is just an
aproximation of the reality.
I've done this top-to-bottom.
JM Albuquerque
We can't know it all :)
Is just smoke, just smoke.
JM Albuquerque
Let me say it better.
Precession speed cannot be a very high velocity.
The solution |W| = |t|/(I3 w3) requires that torque "t"
doesn't produce a very high precession |W| velocity.
If the torque is high, so that precession |W| velocity
is a high value, obviously that friction around precession
causes a torque so that to make the gyroscope move
under the torque applied.
The limit situation where one blocks precession
simply causes the gyroscope to fall.
But even before that, at the instant t=0 when one
apply the torque "t", one needs to have energy to
account for precession kinetic energy.
Starting with the gyroscope standing still at zero
precession and no torque applied, next one apply
torque and the gyroscope starts to precess.
Where does the precession kinetic energy come
from?
The kinetic precession energy comes from the torque
itself TIMES a given ANGULAR DISPLACEMENT
along the torque.
Therefore, I repeat myself once again:
Energy is the time derivative of angular momentum times
a given angular displacement.
Or else, angular momentum on a given angular displacement
is the integral of the energy that exist on that given angular
displacement over time.
The displacement is along the force applied.
Timo Nieminen
You specified that the torque was N1, and the angular velocity was w3, by
which I assume you meant Cartesian vector components (N1,0,0) and
(0,0,w3) for the initial torque and angular velocity. Are you saying that
the magnitude of the initial precession angular velocity is _not_ equal to
|W| above, given the specified initial conditions?
Since you completely specified the initial torque, with no theta required,
theta should _not_ appear in the result, either.
Also note that if you're talking about a top or a gyroscope with
one end of the axle held in a fixed position, then you will either need to
start with the top/gyroscope precessing (not specified in the problem) or
you won't get regular precession (in which case you'd better be more
specific about what you mean by "precession speed").
Compare with a gyroscope held so that the CoM is fixed (and in which case,
the torques and angular velocity and momentum are taken about the CoM
rather than the point of contact with the table).
If you don't specify the problem precisely, don't complain if the "wrong"
problem is solved. If you mean a top, specify a top, not a disk.
The top/gyroscope is one of the best things to apply Euler angles to,
since, the rate of change of one is the spin, the rate of change of
another is the precession, and the rate of change of the last is the
nutation. But it can be done with vectors, too.
JM Albuquerque
"Timo Nieminen" <ti...@physics.uq.edu.au> escreveu na mensagem
news:Pine.LNX.4.50.0710010922240.23033-100000@localhost...
Yes.
> Are you saying that
> the magnitude of the initial precession angular velocity is _not_ equal to
> |W| above, given the specified initial conditions?
No. You are right if you assume the conditions you've pointed
below.
> Since you completely specified the initial torque, with no theta required,
> theta should _not_ appear in the result, either.
Right.
> Also note that if you're talking about a top or a gyroscope with
> one end of the axle held in a fixed position, then you will either need to
> start with the top/gyroscope precessing (not specified in the problem)
It could be a top (derived on the Harvard book I've mentioned), or it
could be a gyroscope with the center of mass fixed (there is fixed point
in space which is the center of mass of the system).
You assumed that top/gyroscope was already precessing.
> or you won't get regular precession (in which case you'd better be more
> specific about what you mean by "precession speed").
No. I can get perfect precession on gyroscope which was not
precessing previously to torque being applied.
Actually I can do it all the time with my gyroscope.
> Compare with a gyroscope held so that the CoM is fixed (and in which case,
> the torques and angular velocity and momentum are taken about the CoM
> rather than the point of contact with the table).
Right.
> If you don't specify the problem precisely, don't complain if the "wrong"
> problem is solved. If you mean a top, specify a top, not a disk.
Sorry about that.
> The top/gyroscope is one of the best things to apply Euler angles to,
> since, the rate of change of one is the spin, the rate of change of
> another is the precession, and the rate of change of the last is the
> nutation. But it can be done with vectors, too.
Yes, but there is "the other" equation.
So far I've presented the w2 equation.
Can you point out the w1 equation?
That is, the gyroscope is not precessing at t=0.
At t=0 the torque is applied and inertia around the precession
axis causes w1 to move along with N1.
Or else, torque applied over N2 causes precession w1 around
axis 1, can you point out the equation for w1, since it can
be done with vectors?
Timo Nieminen
> "Timo Nieminen" wrote:
> >> >
> >> > L = Iw
> >> > t = dL/dt
> >> > t = I dw/dt
[cut]
> > The top/gyroscope is one of the best things to apply Euler angles to,
> > since, the rate of change of one is the spin, the rate of change of
> > another is the precession, and the rate of change of the last is the
> > nutation. But it can be done with vectors, too.
>
> Yes, but there is "the other" equation.
> So far I've presented the w2 equation.
> Can you point out the w1 equation?
>
> That is, the gyroscope is not precessing at t=0.
> At t=0 the torque is applied and inertia around the precession
> axis causes w1 to move along with N1.
> Or else, torque applied over N2 causes precession w1 around
> axis 1, can you point out the equation for w1, since it can
> be done with vectors?
I used some ugly notation above, with t for torque and t for time at the
same time. Try T for torque, so
T = dL/dt
and
T = I dw/dt.
For the general case, with a fixed (ie non-rotating) coordinate system,
T = dL/dt gives you 3 coupled ODEs for w1, w2, w3. Solve them, numerically
if needed.
It can be convenient to use a coordinate system fixed to the object, so
that I is constant (and in this particular case, diagonal, with diagonal
elements I1, I1, I3. In the general case,
T = I dw/dt + wxL.
If w and L are parallel, as specified as initial conditions in the
question, then
T = I dw/dt.
How about an intermediate coordinate system, where the spin axis of the
top is along the z-axis, and the coordinate system precesses with the top,
but doesn't spin with it? In this case, w3 is the spin of the top, and
(w1,w2) describe the precession and nutation of the top. This leaves I the
same as in the axes-attached-to-body case, but we can choose axes such
that the applied torque is (T1,0,0).
To consider the non-initial motion, we have
T = I dw/dt + wxL.
The ODEs we get for the top are
I1 dw1/dt = w2 w3 ( I1 - I3 ) + T1
I1 dw2/dt = w3 w1 ( I3 - I1 )
I3 dw3/dt = 0.
So, the spin rate doesn't change. This reduces the equation we need to
work to solve to the first 2:
I1 dw1/dt = w2 w3 ( I1 - I3 ) + T1
I1 dw2/dt = w3 w1 ( I3 - I1 ).
Writing constant terms as A, B, C, we have
dw1/dt = A w2 + B
dw2/dt = -A w1.
For steady uniform precession, we need w2 = const, w1 = 0. w1 = 0 gives us
w2 = const, and we need w2 = -B/A. You can't get steady uniform precession
without the correct initial w2 (for a fast top, you can get close enough
so that you won't notice, but close to perfect in a real system with
friction is not the same as perfect in the ideal system).
For initial w2 = 0, just solve the equations.
For the precession we have
d^2 w2 / dt^2 + A^2 w2 + AB = 0,
for which a solution exists (ask a mathematician the details about the
existence theorem). Change of variable x2 = w2 + B/A gives you
d^2 x2 / dt^2 = -A^2 x2,
which is a simple harmonic oscillator, so the precession oscillates
harmonically about the value w2 = -B/A, the amplitude of the oscillation
depending on the initial "displacement" from this mean value.
For the nutation, you have
d^2 w1 / dt^2 = -A^2 w1,
so the nutation oscillates about 0 at the same frequency as the
oscillations of the precession.
So, it's pretty clear where the KE of the precession comes from - it
either comes from an initial push, or from the gravitational PE of the
top.
Eric Gisse
<ti...@physics.uq.edu.au> wrote:
[...]
>So, it's pretty clear where the KE of the precession comes from - it
>either comes from an initial push, or from the gravitational PE of the
>top.
ruh roh
You just touched upon JM's two major [well, among them...] peeves.
Rigid body motion and "where does the precessional energy come from?".
Have fun.
JM Albuquerque
> On Mon, 1 Oct 2007, JM Albuquerque wrote:
>
>> "Timo Nieminen" wrote:
>> >> >
>> >> > L = Iw
>> >> > t = dL/dt
>> >> > t = I dw/dt
> [cut]
>> > The top/gyroscope is one of the best things to apply Euler angles to,
>> > since, the rate of change of one is the spin, the rate of change of
>> > another is the precession, and the rate of change of the last is the
>> > nutation. But it can be done with vectors, too.
>>
>> Yes, but there is "the other" equation.
>> So far I've presented the w2 equation.
>> Can you point out the w1 equation?
>>
>> That is, the gyroscope is not precessing at t=0.
>> At t=0 the torque is applied and inertia around the precession
>> axis causes w1 to move along with N1.
>> Or else, torque applied over N2 causes precession w1 around
>> axis 1, can you point out the equation for w1, since it can
>> be done with vectors?
>
> I used some ugly notation above, with t for torque and t for time at the
> same time. Try T for torque, so
>
> T = dL/dt
>
> and
>
> T = I dw/dt.
Great reply Timo Nieminen, you helped me a lot.
I know you are excellent on this subject.
> For the general case, with a fixed (ie non-rotating) coordinate system,
> T = dL/dt gives you 3 coupled ODEs for w1, w2, w3. Solve them, numerically
> if needed.
Yes, from there I've derived Euler's equation of motion which are
the 3 coupled ODEs for w1, w2, w3 you mentioned.
They don't have analytic solution so one has to solve them
numerically, as I've said several times to Eric Gisse.
> It can be convenient to use a coordinate system fixed to the object, so
> that I is constant (and in this particular case, diagonal, with diagonal
> elements I1, I1, I3. In the general case,
>
> T = I dw/dt + wxL.
Yes, I've used a coordinate system fixed to the object too.
> If w and L are parallel, as specified as initial conditions in the
> question, then
>
> T = I dw/dt.
>
> How about an intermediate coordinate system, where the spin axis of the
> top is along the z-axis, and the coordinate system precesses with the top,
> but doesn't spin with it? In this case, w3 is the spin of the top, and
> (w1,w2) describe the precession and nutation of the top. This leaves I the
> same as in the axes-attached-to-body case, but we can choose axes such
> that the applied torque is (T1,0,0).
Well, here is the actual physics limit for the applied torque (T1,0,0).
I needed and I've solved the case (T1,T2,0).
That's the great part I've mentioned above.
You have showed that both precession axis are alike.
One cannot distinguish one from the other:
d^2 x2 / dt^2 = -A^2 x2
d^2 w1 / dt^2 = -A^2 w1
(x2 = w2 + B/A)
That was my conclusion too.
Nevertheless, you can't go for the case (T1,T2,0) from the
above, because the ODE are dependent.
Eric Gisse have already come to a similar situation a couple
of weeks ago.
> So, it's pretty clear where the KE of the precession comes from - it
> either comes from an initial push, or from the gravitational PE of the
> top.
Actually it comes from the potential energy of the top (your PE).
There is no initial push on this problem, but of course, one can
always give him a push by hand to help precession.
The all physics is a trade of between potential and kinetic energy.
That's so for light, electromagnetism, gyroscopes, and so on.
(It's the Lagrangian method at work).
The great thing about your excellent explanation is the fact that
both precession axis, w1, w2, are alike an one cannot distinguish
one from the other, so that one can use a vertical fixed axis - z
and solve the gyroscope relative to that fixed axis.
This problem can be solved by the Lagrangian method too.
One has to write T (kinetic energy) and V (potential energy)
and apply to formalism for each variable.
Both methods (the Newton's method T = I dw/dt) and the
Lagrange's method should give the same result.
Nevertheless they don't, because Newton's method becomes
indetermined due to a "sin(teta)" argument, and the Lagragian
simply don't give the same result as T = I dw/dt.
Newton's indetermination can be removed on the fixed
vertical axis.
The Lagragian method will give EXACTLY the same result
as Newton's method if one assumes the existence of a
potential energy for the vertical precession axis to undergo
precession, V = T2 phi cos(teta)
Based on that I've solved the gyroscope for (T1,T2,0).
And you can do it too, with an additional V = T2 phi cos(teta)
and taking into consideration that
Tz = T2 sin(teta)
and
wz = w2.
Meanwhile, once we knew w1, w2 equations:
w2 =
=(I3*w3)/(2*I2*cos(teta)) * (1-sqrt(1 - (4*N1*I2*cos(teta))/(I3^2*w3^2)))
w1=
= ...
With wz = w2 and Tz = Ta sin(teta) one concludes that:
T1 w1 = Tz wz
(It's a power conservation Law, which derives from the
energy conservation Law).
And the gyroscope becomes an harmonic oscillator for
the applied torque (T1,T2,0):
T1 = A sin(wt)
T2 = B cos(wt)
A/B = constant motion amplification value.
Now look at this:
T1 = Oceans Wave torque over a floater.
T1 = Mass of water over the floater * sin(wt)
Tz * wz = Electrical generator power
Patent: PCT/PT07/000007
I have a gyroscope working as the gearbox (A/B ratio)
within a mass-spring-damper system working at
resonance to capture Ocean Waves energy.
Mass inertia for the electrical generator, that must undergo
a sinusoidal motion (accelerations), has to come from
the spring springiness constant.
And this problem is blowing heads of, and it looks like
I'm in trouble for people to understand this mechanism.
sal
> "sal" <pragm...@nospam.org> escreveu na mensagem
> news:46fc7cf5$0$24907$ec3e...@news.usenetmonster.com...
>> On Fri, 28 Sep 2007 03:07:21 +0100, JM Albuquerque wrote:
>>
>>> "Dono" <sa...@comcast.net> escreveu na mensagem
>>> news:1190944917.9...@w3g2000hsg.googlegroups.com...
>>>> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip
>>>> crap]
>>>>> Energy is the time derivative of angular momentum. Or else, angular
>>>>> momentum the integral of energy over time.
>>>>
>>>>
>>>> No idiot, did you fail physics in high school? Don't answer that, you
>>>> failed.
>>>
>>> Look at the units - kg - m - s, and tell me I'm wrong.
>>
>> Dimensions of angular momentum are mass * length^2 / time
>>
>> Dimensions of kinetic energy are mass * length^2 / time^2
>>
>> A factor of 1/time is hardly ignorable. So, by your own argument,
>> you're wrong.
>
> How?
> 1/time is the derivative term to be "added".
Yeah I know that; I was looking at L not dL/dt -- I realized it shortly
after I hit "send" -- didn't you see the followup I posted to my own
post? (Are these posts coming through? Maybe I should be posting from
Google...)
As I said in the followup, "I hate it when I do that".
But that doesn't affect the rest of the post.
>> In any case, if energy were dL/dt, then if energy were constant and
>> positive then L would be constantly increasing, which is nonsensical.
>
> Look up how you do your derivatives.
Say what?
Integral over time of a positive constant is constantly increasing. Do
you disagree with that? Like, integral(k)dt = k*t.
You said: "angular momentum [is] the integral of energy over time".
So if energy is constant (e.g., in a system in which energy is conserved)
then angular momentum must be, by your claim, constantly increasing.
What's so hard to follow about that?
As I'm sure you realize, angular momentum is not, in general, constantly
increasing, so your assertion leads to the conclusion that either (a)
energy is identically zero or (b) energy is not dL/dt.
>> Worse, energy is a scalar, and dL/dt is a vector (commonly called
>> torque). Scalar and vector mismatch is about as fundamental as
>> mismatched dimensions -- maybe more so.
>
> dL/dt is torque.
You said:
"Energy is the time derivative of angular momentum."
Now you have said "dL/dt is torque".
Which do you intend? The latter is certainly true, the former certainly
is not, and they just as certainly can't both be true.
> One needs to multiply torque by an angular displacement (the dot
> product) to get energy as a scalar.
Yes, and angular displacement is dimensionless, as a result of which
torque and energy have the same dimensions. So call them homonyms if you
like; same dimensions but they mean different things.
>
>> What ever gave you the idea that dL/dt = T?
>
> T = torque,
That's not the intended reading. Sorry, I should have defined my terms:
T = kinetic energy
V = potential energy
Tau = torque
L = angular momentum
And if we write
{L} = the Lagrangian = T - V
theta-dot = rotation rate
@ = Cyrillic 'd' (partial derivative sign)
then we also have
L = @({L})/@(theta-dot)
and if V is independent of the rotational velocity then that's just
L = @T/@(theta-dot)
as I already mentioned, but again that's quite different from what you
said to start with.
> and yes dL/dt = torque
>
>
>> Now if you want to turn it around, @T/@(theta-dot) is (one component
>> of) the angular momentum, at least if there aren't any
>> velocity-dependent potentials running around getting in the way. But
>> that's rather different from what you claimed.
--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org
JM Albuquerque
I've seen, but I get confused.
>>> In any case, if energy were dL/dt, then if energy were constant and
>>> positive then L would be constantly increasing, which is nonsensical.
>>
>> Look up how you do your derivatives.
>
> Say what?
>
> Integral over time of a positive constant is constantly increasing. Do
> you disagree with that? Like, integral(k)dt = k*t.
>
> You said: "angular momentum [is] the integral of energy over time".
>
> So if energy is constant (e.g., in a system in which energy is conserved)
> then angular momentum must be, by your claim, constantly increasing.
Wait a minute.
The conserved quantity is angular momentum.
If you assume constant energy you are assuming a constant feed
of energy to the system, so that it was to do some work to
balance what goes in with what goes out.
> What's so hard to follow about that?
>
> As I'm sure you realize, angular momentum is not, in general, constantly
> increasing, so your assertion leads to the conclusion that either (a)
> energy is identically zero or (b) energy is not dL/dt.
Of course angular momentum is not constantly increasing.
Angular momentum is conserved.
If angular momentum L is not conserved, so that we have it
increasing or decreasing, then we know for sure there must
be an external torque T, so that:
T = dL/dt
>>> Worse, energy is a scalar, and dL/dt is a vector (commonly called
>>> torque). Scalar and vector mismatch is about as fundamental as
>>> mismatched dimensions -- maybe more so.
>>
>> dL/dt is torque.
>
> You said:
>
> "Energy is the time derivative of angular momentum."
>
> Now you have said "dL/dt is torque".
>
> Which do you intend? The latter is certainly true, the former certainly
> is not, and they just as certainly can't both be true.
Well, I've also corrected myself.
You need to read my other posts where correction was made.
>> One needs to multiply torque by an angular displacement (the dot
>> product) to get energy as a scalar.
>
> Yes, and angular displacement is dimensionless, as a result of which
> torque and energy have the same dimensions. So call them homonyms if you
> like; same dimensions but they mean different things.
Agree.
(snip)
sal
Say, rather, /a/ conserved quantity is angular momentum; there may be
more than one quantity which is conserved.
Suppose a ball on a rope is whirling around a post, rope attached to a
swivel on the post. Assume there's no air (we're on the moon) and
there's no friction (super-teflon bearings in the swivel).
Now the ball "orbits" the post at constant angular velocity, and angular
momentum and energy are _both_ constant (and conserved).
In this system, if the rope is of length r and is sticking out from the
post at a 90 degree angle (no gravity!), and if T=kinetic energy,
V=potential energy, L=magnitude of the angular momentum, and "omega" =
angular velocity, we'd say
V=0
T=1/2 m r^2 omega^2
L = m r^2 omega
All three are constant.
If I integrate the energy over time -- whether you mean kinetic energy,
kinetic plus potential, or kinetic minus potential, they're all the same
in this case -- we get
int(T)dt = 1/2 m r^2 omega^2 * (t - t0)
which doesn't mean much to me.
> If you assume constant
> energy you are assuming a constant feed of energy to the system,
No I am not! (Or, rather, I'm assuming a "constant feed" of ZERO energy
into the system.)
"Feed of energy" = power = derivative of energy with respect to time.
Power has dimensions energy/time. Energy has dimension of power*time.
For example, kilowatts are power, and kilowatt-hours are energy.
A 100 watt lightbulb draws 100 watts of _power_ and after 10 hours it's
converted 1 kilowatt-hour of electrical energy into (mostly) heat energy.
In general, public utilities charge for energy consumed, _not_ for power
used. You can fire up a 100 kW motor once a month for a few seconds and
it won't cost you as much as running a 100 watt lightbulb continuously,
day and night, all month long, even though the motor uses 1000 times as
much _power_ as the lightbulb.
Dimensions of force are
F = mass * distance / time^2
Dimensions of energy (using "T" for kinetic energy here) are
T = mass * distance^2 / time^2
Dimensions of power ("W" for "Work") are
W = mass * distance^2 / time^3
> so that
> it was to do some work to balance what goes in with what goes out.
If we assume energy is conserved, then "what goes out" is zero.
>> What's so hard to follow about that?
>>
>> As I'm sure you realize, angular momentum is not, in general,
>> constantly increasing, so your assertion leads to the conclusion that
>> either (a) energy is identically zero or (b) energy is not dL/dt.
>
> Of course angular momentum is not constantly increasing. Angular
> momentum is conserved.
>
> If angular momentum L is not conserved, so that we have it increasing or
> decreasing, then we know for sure there must be an external torque T, so
> that:
> T = dL/dt
>
>
>>>> Worse, energy is a scalar, and dL/dt is a vector (commonly called
>>>> torque). Scalar and vector mismatch is about as fundamental as
>>>> mismatched dimensions -- maybe more so.
>>>
>>> dL/dt is torque.
>>
>> You said:
>>
>> "Energy is the time derivative of angular momentum."
>>
>> Now you have said "dL/dt is torque".
>>
>> Which do you intend? The latter is certainly true, the former
>> certainly is not, and they just as certainly can't both be true.
>
> Well, I've also corrected myself.
> You need to read my other posts where correction was made.
OK, will do...
>
>
>>> One needs to multiply torque by an angular displacement (the dot
>>> product) to get energy as a scalar.
>>
>> Yes, and angular displacement is dimensionless, as a result of which
>> torque and energy have the same dimensions. So call them homonyms if
>> you like; same dimensions but they mean different things.
>
>
> Agree.
>
> (snip)
--
Timo A. Nieminen
> Timo Nieminen wrote:
>
>> For the general case, with a fixed (ie non-rotating) coordinate system,
>> T = dL/dt gives you 3 coupled ODEs for w1, w2, w3. Solve them, numerically
>> if needed.
>
> Yes, from there I've derived Euler's equation of motion which are
> the 3 coupled ODEs for w1, w2, w3 you mentioned.
> They don't have analytic solution so one has to solve them
> numerically, as I've said several times to Eric Gisse.
Analytical solutions are rare. Of course in the general case you need to
resort to numerical solution. It's even worse with PDEs. Nonetheless,
existence theorems are important, since knowing that a solution exists
makes it worthwhile attacking the problem numerically.
> That's the great part I've mentioned above.
> You have showed that both precession axis are alike.
> One cannot distinguish one from the other:
> d^2 x2 / dt^2 = -A^2 x2
> d^2 w1 / dt^2 = -A^2 w1
> (x2 = w2 + B/A)
>
> That was my conclusion too.
>
> Nevertheless, you can't go for the case (T1,T2,0) from the
> above, because the ODE are dependent.
> Eric Gisse have already come to a similar situation a couple
> of weeks ago.
They can be distinguished from each other - the precession oscillates
about a non-zero value, the nutation oscillates about zero. For your
(T1,T2,0) case, as long as the torque is time-independent, just make a
change of variables x1 = w1 + C, and you have an analytical solution
again. Then you need to seriously think about what constant (T1,T2,0)
means physically.
>> So, it's pretty clear where the KE of the precession comes from - it
>> either comes from an initial push, or from the gravitational PE of the
>> top.
>
> Actually it comes from the potential energy of the top (your PE).
> There is no initial push on this problem, but of course, one can
> always give him a push by hand to help precession.
>
> The all physics is a trade of between potential and kinetic energy.
> That's so for light, electromagnetism, gyroscopes, and so on.
> (It's the Lagrangian method at work).
>
> The great thing about your excellent explanation is the fact that
> both precession axis, w1, w2, are alike an one cannot distinguish
> one from the other, so that one can use a vertical fixed axis - z
> and solve the gyroscope relative to that fixed axis.
>
> This problem can be solved by the Lagrangian method too.
> One has to write T (kinetic energy) and V (potential energy)
> and apply to formalism for each variable.
Goldstein has a good coverage of the top (the mounted for rotation about
the CoM gyroscope is dismissed as too simple to bother with (not in
those words, but effectively)). Equations of motion in terms of both
vectors and Euler angles, and the Lagrangian method (iirc, this is what is
used to obtain the solution).
> Both methods (the Newton's method T = I dw/dt) and the
> Lagrange's method should give the same result.
> Nevertheless they don't, because Newton's method becomes
> indetermined due to a "sin(teta)" argument, and the Lagragian
> simply don't give the same result as T = I dw/dt.
If Newton becomes unstuck due to sin(theta), then that may well be an
Euler angle problem. Sin(theta) = 0 is when Euler angles traditionally
fail. Avoid theta. There's a force on the CoM, the CoM is at some r from
the pivot point. rxF might be zero and cause problems, but F is non-zero,
and r is non-zero.
[Will look at rest if time permits]
Igor
> "Igor" <thoov...@excite.com> escreveu na mensagemnews:1191086382....@50g2000hsm.googlegroups.com...
And you still don't get it. But then again, I'm not at all surprised.
Igor
> "Igor" <thoov...@excite.com> escreveu na mensagemnews:1191176963.0...@y42g2000hsy.googlegroups.com...
Look it up in Goldstein. That's what I usually do,
Igor
wrote:
> On Sun, 30 Sep 2007 16:44:24 -0000, Igor <thoov...@excite.com> wrote:
>
> [...]
>
>
>
> >And you probably don't know your dyads from your tetrads, so be
> >careful.
>
> Replace "and" with "but". I do know dyads - just a fancy name for
> tensor product.
>
> Methinks "dyads" is a bit of terminology that is slightly dated.
Bigtime. Plus the concept of dyads is a lot easier once you
understand tensors. At least it was for me. Prior to that, it was
just that section in Goldstein where the notation didn't make a lot of
sense.
Ken S. Tucker
I'll disagree. Dyads are fundamental to
understand antisymmetric metric tensors, for
example, ij =/= ji especially on a rotating
disk where it may not be possible to create
an orthogonal CS.
Regards
Ken S. Tucker
JM Albuquerque
"Timo A. Nieminen" <ti...@physics.uq.edu.au> escreveu na mensagem
news:Pine.WNT.4.64.07...@serene.st...
Very good point.
What constant (T1,T2,0) means physically?
First, constant means stationary in time (the angular momentum
conservation), required for each principal axis 1 and 2.
When both axis are taken into account, Physically it means that
both precession axis of the gyroscope are fully independent one
from each other.
That is, I can observe a constant angular precession w2 due
to torque T1 applied. And independently, I can have a torque
T2 applied that will produce a constant angular precession w1.
In practice, or else mechanically, this is not feasible because the
best mechanism we can have produces sinusoidal torque.
Why this is true.
This is true because at steady state a gyroscope converts
torque without motion (infinitely stiff and doesn't give under it)
into motion without torque (free-precession means no-torque T2).
Then, experimentally, when the gyroscope is precessing at constant
speed w2 due to a constant torque T1 applied, one can go to
the precessing motion and try to increase or decrease the speed
and feel how the gyroscope reacts and what's happening to
the angular speed w1.
It happens that w1 is constant too, and the gyroscope again
doesn't give under the force you try to apply by hand.
Hence, I'm quite confident that both precession axis 1,2 of the
gyroscope are fully independent one from each other.
Another excellent point.
When sin(theta) = 0 very strange things happens when I compute
the exact values of w1 and w2 into a graphical simulation.
Actually I don't remember if its dw1/dt and dw2/dt, but I can see
a function that goes from minus infinite to plus infinite is less then
a very small fraction of a second, then gets stable has predict.
Nevertheless the above is perfectly understandable (but hard).
What happens is that when sin(theta) = 0 the gyroscope
spin axis (3) is colinear with the precession axis w2.
So that, precession w2 is like trying to spin up the gyroscopic
mass w3, which is obviously impossible since no physical
connection exists (trough the bearings).
The picture is that of a gyroscope whose spin axis (3) also
rotates at constant speed w1 over time, so that :
1 - When sin(theta) = 0 the external sinusoidal torque is made
also to be zero (the problem disapears).
2 - Applied torque T1 is sinusoidal so that angular precession
velocity is also a sinusoidal motion that inverts direction every
180 degrees. Yes, one turn of w1 causes w2 to change direction
twice per turn.
3 - It's the harmonic oscillator working under an harmonic
sinusoidal external torque. It looks like an internal combustion
engine whose tie-rod lenght is infinite, so that when the piston
motion is a pure sinusoidal function, up and down, the
crankshaft speed is constant angular speed w1.
(The constant angular speed w1 can be seen mathematically
by means of derivatives too - the integral of arcsin or arc-cos,
if my memory doesn't fail, or something like that).
> There's a force on the CoM, the CoM is at some r from the pivot point. rxF
> might be zero and cause problems, but F is non-zero, and r is non-zero.
>
> [Will look at rest if time permits]
Actually I've already done both deviations (Newton's and Lagrange's
method), the way I've said (a new funny potential energy required,
plus the fixed vertical axis z where the output is located).
This is not from the book I've mentioned, it's mine.
(Microsoft Word 97 document).
So, I ask your kind permission to submit to you my deviations so
that you analyse then and give me some feed-back when time
permits.
(I will try to produce a pdf file with the scanner to avoid that
pages become messed-up with the symbols I used, but it
could take some time).
JM Albuquerque
"Igor" <thoo...@excite.com> escreveu na mensagem
news:1191256932.3...@d55g2000hsg.googlegroups.com...
I can get anything if you could explain what's wrong.
Energy = T . d(theta) = ( F x r ) . d(theta) = scalar
The external torque vector T is colinear with d(theta) vector, so
that the dot product is a scalar (cos(0)=1), a simple multiplication
where the sign counts (cos(180)=-1), since it could be an
acceleration or a decceleration..
Eric Gisse
Really, Ken? Dyad products are essential? Funny how I have only seen
dyads in classical mechanics textbooks.
Explain the difference between the dyad and tensor product.
Then show me an antisymmetric metric tensor. Then show me how you
obtain said tensor. While you are doing that, explain how come the dot
product suddenly became antisymmetric.
Eric Gisse
wrote:
[...]
>
>Great reply Timo Nieminen, you helped me a lot.
>I know you are excellent on this subject.
Don't be a smartass to someone who knows far more about this than you
do.
>
>
>> For the general case, with a fixed (ie non-rotating) coordinate system,
>> T = dL/dt gives you 3 coupled ODEs for w1, w2, w3. Solve them, numerically
>> if needed.
>
>Yes, from there I've derived Euler's equation of motion which are
>the 3 coupled ODEs for w1, w2, w3 you mentioned.
>They don't have analytic solution so one has to solve them
>numerically, as I've said several times to Eric Gisse.
Saying something you know to be wrong is known as "lying".
Why are you lying, JM?
http://groups.google.com/group/sci.physics/msg/a66bc9650330591d?dmode=source
[...]
>
>That's the great part I've mentioned above.
>You have showed that both precession axis are alike.
>One cannot distinguish one from the other:
> d^2 x2 / dt^2 = -A^2 x2
>d^2 w1 / dt^2 = -A^2 w1
>(x2 = w2 + B/A)
>
>That was my conclusion too.
>
>Nevertheless, you can't go for the case (T1,T2,0) from the
>above, because the ODE are dependent.
One of my more favorite saying is "Lying or stupid - which is it?"
You are either lying [you know this is not true - I proved it] or
stupid [you didn't understand the proof].
If you are lying, why...?
If you are stupid, why did you lie when you said you understood what I
did?
>Eric Gisse have already come to a similar situation a couple
>of weeks ago.
Why are you lying to Timo while invoking *my name* when you know I'm
watching the thread?
http://groups.google.com/group/sci.physics/msg/a66bc9650330591d?dmode=source
I went through all the math for the case of external torques. They
don't even have to be constant - you can trivially extend it to time
varying torques.
[...]
>Both methods (the Newton's method T = I dw/dt) and the
>Lagrange's method should give the same result.
Since they are completely and utterly equivalent, they should.
That you don't realize this tells me that you never had a
comprehensive education in classical mechanics.
I'll give you the cliffs notes version:
The Lagrangian [and by extension, Hamiltonian] formalism of classical
mechanics can be derived through F = ma and T = .5m|v|^2. Another
derivation, which also obtains you F = ma, is through variational
methods.
If you want to see a proof, look at Symon for the former and Goldstein
for the latter.
>Nevertheless they don't, because Newton's method becomes
>indetermined due to a "sin(teta)" argument, and the Lagragian
>simply don't give the same result as T = I dw/dt.
Really? Shit, that's news. Congrats - you have just proved that
classical mechanics is internally inconsistent.
On the other hand, it is far more likely that you simply screwed up.
Why don't you work the problem using both methods? I'm reasonably sure
either Timo or myself will spot the error.
>
>Newton's indetermination can be removed on the fixed
>vertical axis.
>
>The Lagragian method will give EXACTLY the same result
>as Newton's method if one assumes the existence of a
>potential energy for the vertical precession axis to undergo
>precession, V = T2 phi cos(teta)
Having to do that means you fucked up.
>
>Based on that I've solved the gyroscope for (T1,T2,0).
Or used my solution directly.
>And you can do it too, with an additional V = T2 phi cos(teta)
>and taking into consideration that
>Tz = T2 sin(teta)
THETA, it is spelled THETA.
[...]
>
>I have a gyroscope working as the gearbox (A/B ratio)
>within a mass-spring-damper system working at
>resonance to capture Ocean Waves energy.
Yea JM, nobody thought of capturing wave energy before you.
>Mass inertia for the electrical generator, that must undergo
>a sinusoidal motion (accelerations), has to come from
>the spring springiness constant.
>And this problem is blowing heads of, and it looks like
>I'm in trouble for people to understand this mechanism.
Be honest - do you actually believe this, or are you just trying to
fill your retirement years with a persecution complex?
[...]
Eric Gisse
wrote:
[...]
>
>(I will try to produce a pdf file with the scanner to avoid that
>pages become messed-up with the symbols I used, but it
>could take some time).
www.google.com "doc2pdf"
Eric Gisse
wrote:
[...]
>I can get anything if you could explain what's wrong.
Your response to my asking of "Where does d\theta point?":
"Well, theta is an angle so it's an angle, you know what an angle
is, don't you (expressed in radians)?"
Not once did you ever explain that it was a /vector/. In fact, you
made it sound exactly like a scalar. Angles are scalar quantities.
[...]
JM Albuquerque
> On Mon, 1 Oct 2007 15:31:13 +0100, "JM Albuquerque" <jmD...@clix.pt>
> wrote:
>
> [...]
>
>>
>>Great reply Timo Nieminen, you helped me a lot.
>>I know you are excellent on this subject.
>
> Don't be a smartass to someone who knows far more about this than you
> do.
>
>>
>>
>>> For the general case, with a fixed (ie non-rotating) coordinate system,
>>> T = dL/dt gives you 3 coupled ODEs for w1, w2, w3. Solve them,
>>> numerically
>>> if needed.
>>
>>Yes, from there I've derived Euler's equation of motion which are
>>the 3 coupled ODEs for w1, w2, w3 you mentioned.
>>They don't have analytic solution so one has to solve them
>>numerically, as I've said several times to Eric Gisse.
>
> Saying something you know to be wrong is known as "lying".
>
> Why are you lying, JM?
>
> http://groups.google.com/group/sci.physics/msg/a66bc9650330591d?dmode=source
Why didn't you show my reply?
You acuse me of lying when you showed nothing.
Your equations are dependent, moron.
How many times do I have to say to you hat you cannot go
out of there.
Actually you cannot even read what Timo Nieminen said
about it.
>
> [...]
>
>>
>>That's the great part I've mentioned above.
>>You have showed that both precession axis are alike.
>>One cannot distinguish one from the other:
>> d^2 x2 / dt^2 = -A^2 x2
>>d^2 w1 / dt^2 = -A^2 w1
>>(x2 = w2 + B/A)
>>
>>That was my conclusion too.
>>
>>Nevertheless, you can't go for the case (T1,T2,0) from the
>>above, because the ODE are dependent.
>
> One of my more favorite saying is "Lying or stupid - which is it?"
>
> You are either lying [you know this is not true - I proved it] or
> stupid [you didn't understand the proof].
>
> If you are lying, why...?
>
> If you are stupid, why did you lie when you said you understood what I
> did?
You are actually the biggest moron I ever seen.
Solve the problem or else fuck off.
Stupid imbecil kid which belives to rule the world.
>>Eric Gisse have already come to a similar situation a couple
>>of weeks ago.
>
> Why are you lying to Timo while invoking *my name* when you know I'm
> watching the thread?
>
> http://groups.google.com/group/sci.physics/msg/a66bc9650330591d?dmode=source
Again this shit.
Solve the problem.
Give me independent equations moron.
> I went through all the math for the case of external torques. They
> don't even have to be constant - you can trivially extend it to time
> varying torques.
You went trough shit.
So far you have showed to know shit about the subject.
Remember the kinetic energy that comes from the main
spin? What a moron.
And what about precession without torque?
Moron squared.
>
> [...]
>
>>Both methods (the Newton's method T = I dw/dt) and the
>>Lagrange's method should give the same result.
>
> Since they are completely and utterly equivalent, they should.
They should but they don't, moron.
And you will never know as far as it depends on me.
> That you don't realize this tells me that you never had a
> comprehensive education in classical mechanics.
Fuck you.
I wish to be who you think you are, moron.
> I'll give you the cliffs notes version:
>
> The Lagrangian [and by extension, Hamiltonian] formalism of classical
> mechanics can be derived through F = ma and T = .5m|v|^2. Another
> derivation, which also obtains you F = ma, is through variational
> methods.
>
> If you want to see a proof, look at Symon for the former and Goldstein
> for the latter.
You need a new potential energy to be added on the potential
energy term.
>>Nevertheless they don't, because Newton's method becomes
>>indetermined due to a "sin(teta)" argument, and the Lagragian
>>simply don't give the same result as T = I dw/dt.
>
> Really? Shit, that's news. Congrats - you have just proved that
> classical mechanics is internally inconsistent.
No, moron.
Just can't get the solution, that's all.
One needs to get means to solve the problem.
> On the other hand, it is far more likely that you simply screwed up.
The most likely to me is that you are stupid, how about that?
> Why don't you work the problem using both methods? I'm reasonably sure
> either Timo or myself will spot the error.
Can't you read moron.
I did it already.
>>Newton's indetermination can be removed on the fixed
>>vertical axis.
>>
>>The Lagragian method will give EXACTLY the same result
>>as Newton's method if one assumes the existence of a
>>potential energy for the vertical precession axis to undergo
>>precession, V = T2 phi cos(teta)
>
> Having to do that means you fucked up.
That's my problem, not yours, moron.
>>Based on that I've solved the gyroscope for (T1,T2,0).
>
> Or used my solution directly.
Your solution is horse dunk.
>>And you can do it too, with an additional V = T2 phi cos(teta)
>>and taking into consideration that
>>Tz = T2 sin(teta)
>
> THETA, it is spelled THETA.
So what?
Teta escreve-se teta, estupido idiota.
>
> [...]
>
>>
>>I have a gyroscope working as the gearbox (A/B ratio)
>>within a mass-spring-damper system working at
>>resonance to capture Ocean Waves energy.
>
> Yea JM, nobody thought of capturing wave energy before you.
Yea, you know lots of it too, I presume.
>>Mass inertia for the electrical generator, that must undergo
>>a sinusoidal motion (accelerations), has to come from
>>the spring springiness constant.
>>And this problem is blowing heads of, and it looks like
>>I'm in trouble for people to understand this mechanism.
>
> Be honest - do you actually believe this, or are you just trying to
> fill your retirement years with a persecution complex?
Fuck you moron, I'm not who you think I'm.
> [...]