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Natural Science Forum / Physics / General Physics / October 2007Pop Quiz for Science Buffs!
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). 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 - 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. > INSTRUCTIONS: Each of the following multiple choice questions has > only one correct, or more nearly correct answer. There is no time [quoted text clipped - 7 lines] > 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 [quoted text clipped - 4 lines] > 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 [quoted text clipped - 5 lines] > 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 [quoted text clipped - 9 lines] > 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.  SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 Uncle Al <UncleAl0@hate.spam.net> wrote in news:46FAEF9A.A4D91864 @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. Signaturebz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap > Uncle Al <Uncle...@hate.spam.net> wrote in news:46FAEF9A.A4D91864 > @hate.spam.net: [quoted text clipped - 25 lines] > > bz+...@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap 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 - [...] Why not ask something beyond the highschool physics level? Here, I got one: Why is energy a conserved quantity in central force potentials? > [...] > > Why not ask something beyond the highschool physics level? > > Here, I got one: Why is energy a conserved quantity in central force > potentials? Is there anything more dangerous than a question wholly stated in ten words or fewer? SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 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 > [quoted text clipped - 6 lines] > 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. I dunno, Igor -- I think Eric's question was harder than yours. Here, let's look at some possible answers. >> On Wed, 26 Sep 2007 13:45:45 -0700, NoEinstein >> [quoted text clipped - 6 lines] >> 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. SignatureNospam becomes physicsinsights to fix the email I can be also contacted through http://www.physicsinsights.org >On Sep 26, 10:15 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> >wrote: [quoted text clipped - 11 lines] >Are you sure you don't mean angular momentum? Energy can be conserved >in non-central potentials also. So can angular momentum. :) On Sep 27, 9:45 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> wrote: > >On Sep 26, 10:15 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> > >wrote: [quoted text clipped - 15 lines] > > :) while i doubt that any of you knows what conserved angular momentum means >On Sep 27, 9:45 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> >wrote: [quoted text clipped - 20 lines] >while i doubt that any of you knows what >conserved angular momentum means Then enlighten us, dyslexic troll. On Sep 27, 11:32 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> wrote: > >On Sep 27, 9:45 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> > >wrote: [quoted text clipped - 22 lines] > > Then enlighten us, dyslexic troll. you dont even know your tensors how would you understand? > On Sep 27, 9:45 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> > wrote: [quoted text clipped - 20 lines] > while i doubt that any of you knows what > conserved angular momentum means 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 > 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 >> On Sep 26, 10:15 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> >> wrote: [quoted text clipped - 10 lines] > 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 [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. > [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. > > [snip crap] > >> Energy is the time derivative of angular momentum. [quoted text clipped - 4 lines] > > Look at the units - kg - m - s, and tell me I'm wrong. You are wrong and you are an idiot. happy? >> > [snip crap] >> >> Energy is the time derivative of angular momentum. [quoted text clipped - 6 lines] > > You are wrong and you are an idiot. happy? 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. > >> > [snip crap] > >> >> Energy is the time derivative of angular momentum. [quoted text clipped - 11 lines] > > Today, idiots rule the world Doc. Not idiots like you, your ideas are sh.t. >> >> > [snip crap] >> >> >> Energy is the time derivative of angular momentum. [quoted text clipped - 14 lines] > > Not idiots like you, your ideas are sh.t. 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. >> [snip crap] >>> Energy is the time derivative of angular momentum. [quoted text clipped - 4 lines] > >Look at the units - kg - m - s, and tell me I'm wrong. Of course the units are right but you are still wrong. TORQUE is the time derivative of angular momentum. Same units, different concept. >>> [snip crap] >>>> Energy is the time derivative of angular momentum. [quoted text clipped - 7 lines] > Of course the units are right but you are still wrong. TORQUE is the > time derivative of angular momentum. Same units, different concept. 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. [snip crap] > Look at the units - kg - m - s, and tell me I'm wrong. 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. >[snip crap] >> [quoted text clipped - 6 lines] >Either way, "energy is" NOT "the time derivative of angular momentum". >Now, go bark at the moon. 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. > [snip crap] >> [quoted text clipped - 6 lines] > Either way, "energy is" NOT "the time derivative of angular momentum". > Now, go bark at the moon. 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. >> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip >> crap] [quoted text clipped - 5 lines] > > 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. 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. SignatureNospam becomes physicsinsights to fix the email I can be also contacted through http://www.physicsinsights.org Oops. >>> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip >>> crap] [quoted text clipped - 9 lines] > > 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. > In any case, if energy were dL/dt, then if energy were constant and > positive then L would be constantly increasing, which is nonsensical. [quoted text clipped - 9 lines] > potentials running around getting in the way. But that's rather > different from what you claimed. SignatureNospam becomes physicsinsights to fix the email I can be also contacted through http://www.physicsinsights.org >>> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip >>> crap] [quoted text clipped - 12 lines] > 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 > 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. >>>> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip >>>> crap] [quoted text clipped - 15 lines] > 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. SignatureNospam becomes physicsinsights to fix the email I can be also contacted through http://www.physicsinsights.org >>>>> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip >>>>> crap] [quoted text clipped - 24 lines] > > But that doesn't affect the rest of the post. 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. [quoted text clipped - 10 lines] > 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 [quoted text clipped - 10 lines] > 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) >>>>>> On Sep 27, 6:45 pm, "JM Albuquerque" <jmDO...@clix.pt> wrote: [snip >>>>>> crap] [quoted text clipped - 45 lines] > Wait a minute. > The conserved quantity is angular momentum. 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? >> [quoted text clipped - 27 lines] > 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. [quoted text clipped - 6 lines] > > (snip) SignatureNospam becomes physicsinsights to fix the email I can be also contacted through http://www.physicsinsights.org >> [I am using the context of classical mechanics, as this >> is implied by the problem statement.] > The "time" with respect to which one take the derivative > 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 >>> [I am using the context of classical mechanics, as this >>> is implied by the problem statement.] [quoted text clipped - 36 lines] > measuring ds vs r. And, of course, W is a scalar but t is a (pseudo) > 3-vector. 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 >>>> [I am using the context of classical mechanics, as this >>>> is implied by the problem statement.] [quoted text clipped - 45 lines] > >Energy = t . d(teta) = ( F x r ) . d(teta) = scalar 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 >>>>> [I am using the context of classical mechanics, as this >>>>> is implied by the problem statement.] [quoted text clipped - 50 lines] > > 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)? >>So, this all boils down to an angular displacement, that was >>missing in my previous post, because was irrelevante for [quoted text clipped - 7 lines] > 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 [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 :-) > [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. > > [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 :-) >> [snip crap] >> [quoted text clipped - 13 lines] > No I don't. > I got everything wrong, top to bottom. Having a meltdown, I see.... [...] >>>Let me correct myself and say it again: >>>Energy is the time derivative of angular momentum times a given [quoted text clipped - 9 lines] >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 [quoted text clipped - 13 lines] >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 > [...] > [quoted text clipped - 15 lines] > > 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 [quoted text clipped - 22 lines] > > 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). >> [...] >> [quoted text clipped - 19 lines] >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 [quoted text clipped - 24 lines] > >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 [quoted text clipped - 5 lines] >... 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. >>> [...] >>> [quoted text clipped - 21 lines] > > 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 [quoted text clipped - 28 lines] > 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 [quoted text clipped - 7 lines] > > 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). > >>> [...] > [quoted text clipped - 26 lines] > > 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 >> >>> [...] >> [quoted text clipped - 38 lines] >Thought you might want to glance dyads kiddo. >Ken Ken...I have no idea what you are good at but physics is not it. Go do something else. On Sep 29, 4:33 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> wrote: > On Sat, 29 Sep 2007 05:26:29 -0700, "Ken S. Tucker" > [quoted text clipped - 45 lines] > Ken...I have no idea what you are good at but physics is not it. Go do > something else. LOL, Gisse don't know dyads! Pass HS, get a job, retire at 40. Ken > On Sep 29, 4:33 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> > wrote: [quoted text clipped - 52 lines] > Pass HS, get a job, retire at 40. > Ken- And you probably don't know your dyads from your tetrads, so be careful. > > On Sep 29, 4:33 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> > > wrote: [quoted text clipped - 55 lines] > And you probably don't know your dyads from your tetrads, so be > careful. 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 [...] >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. On Sep 30, 5:04 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> wrote: > [...] > [quoted text clipped - 5 lines] > > 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. > On Sep 30, 5:04 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> > wrote: [quoted text clipped - 13 lines] > just that section in Goldstein where the notation didn't make a lot of > sense. 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 >> On Sep 30, 5:04 pm, Eric Gisse <jowr.pi.nos...@gmail-nospam.com> >> wrote: [quoted text clipped - 21 lines] >Regards >Ken S. Tucker 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. > >>> [...] > [quoted text clipped - 88 lines] > > The rotation is defined by "omega" (and angular velocity. Maybe it's time you learned the difference between defined quantities and derived quantities. >>>> [...] >>>> [quoted text clipped - 23 lines] > >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. >> 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. >>>>>>>So, this all boils down to an angular displacement, that was >>>>>>>missing in my previous post, because was irrelevante for [quoted text clipped - 48 lines] > >The rotation is defined by "omega" (and angular velocity). 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. >>>>> Where does d\theta point? >>>> [quoted text clipped - 10 lines] > 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. [quoted text clipped - 5 lines] > 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. SignatureTimo 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 >>>>>> Where does d\theta point? >>>>> [quoted text clipped - 26 lines] > 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. > 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. > >>>>>> Where does d\theta point? > [quoted text clipped - 49 lines] > Without Euler angles, nor gyroscopic momentum, nor > gyroscopic precession can be accounted for. 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. >> >>>>>> Where does d\theta point? >> [quoted text clipped - 55 lines] > just the vector addition of two or more angular momenta, can be > described quite well without using Euler angles. 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))) > >> >>>>>> Where does d\theta point? > [quoted text clipped - 61 lines] > precession velocity = > =(I3*w3)/(2*I2*cos(teta)) * (1-sqrt(1 - (4*N1*I2)/(I3^2*w3^2)))- Look it up in Goldstein. That's what I usually do, >> What are Euler angles for? They're a (lousy) parameterisation of >> rotations. In the magic 3D, a rotation can be described by a [quoted text clipped - 11 lines] > 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. [quoted text clipped - 8 lines] > 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!) SignatureTimo 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 >>> What are Euler angles for? They're a (lousy) parameterisation of >>> rotations. In the magic 3D, a rotation can be described by a [quoted text clipped - 18 lines] > 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? [quoted text clipped - 21 lines] > 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. > (Hmm. Angular momentum and precession in 4D could be fun and educational > too. No more pretending that 3x3 antisymmetric tensors are vectors!) > 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. >>> Euler angles (teta, phi, psi) are the VERY BEST OF ALL TIMES. >> [quoted text clipped - 6 lines] > 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. [quoted text clipped - 17 lines] > 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). SignatureTimo 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 >>>> Euler angles (teta, phi, psi) are the VERY BEST OF ALL TIMES. >>> [quoted text clipped - 54 lines] > |W| = |dw/dt|/|w| > |W| = |t|/(I3 w3). 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). >>>>> Euler angles (teta, phi, psi) are the VERY BEST OF ALL TIMES. >>>> [quoted text clipped - 80 lines] > L2 = I (omega) sin(teta) > (omega) = your value: |W| = |t|/(I3 w3). 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. >>>>>> Euler angles (teta, phi, psi) are the VERY BEST OF ALL TIMES. >>>>> [quoted text clipped - 93 lines] > all this starts to move, the above equation is just an > aproximation of the reality. 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. > >>>> Without Euler angles, nor gyroscopic momentum, nor > >>>> gyroscopic precession can be accounted for. [quoted text clipped - 57 lines] > L2 = I (omega) sin(teta) > (omega) = your value: |W| = |t|/(I3 w3). 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. SignatureTimo 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 >> >>>> Without Euler angles, nor gyroscopic momentum, nor >> >>>> gyroscopic precession can be accounted for. [quoted text clipped - 66 lines] > which I assume you meant Cartesian vector components (N1,0,0) and > (0,0,w3) for the initial torque and angular velocity. 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. Ye |
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