Thread: the concept of angular momentum

hallo, I'm trying to understand the concept of L= angular momentum related to momentum and the concept of the lever. If someone has studied this problem, please tell me the solution, otherwise, : imagine

a sinker (a ball): mass =1 attached to a hanging line : length = 10 (meters or dimensionless) with a hook at the other end, flying frictionless at speed = 5 (m/s) along x axis =>=>=>=>=>

it has momentum = 5 (M*L/T), right? now imagine the hook gets caught on a pivot and starts rotating frictionless at speed 5 in a vertical plane (xz),
1) its (angular) momentum increases to 50? (m*L²/2) (r * p), how is that possible?
2) a vector τ , its torque, αppears on the y axis. They say it is the time derivative of angular momentum (M*L²/T²) the variation of angular momentum, but does L vary with time?
3) they say torque is the measure of the turning force, why is it represented as a vector? a vector is not only the measure of speed?
4) they say: apply the right hand rule, what would happen if we point that vector in the opposite direction?

thanks

all right, folks, let's forget about the conceptual problem. Can you tell me if the practical part, the math is ok?

I couldn't picture what you were trying to describe. The hook was hanging down below the path of the sinker, or was it traveling in the same path? What is the velocity?

Originally Posted by logic

a sinker (a ball): mass =1 attached to a hanging line : length = 10 (meters [or dimensionless]) with a hook at the other end, flying frictionless at speed = 5 (m/s) along x axis =>=>=>=>=>

hi Harold, thanks for your attention, the speed is 5m/s, I thought it was clear, probably it was not . We have a ball flying horizontally, there is a string hanging down from it, and the hook gets caught.

Originally Posted by logic

Originally Posted by logic

a sinker (a ball): mass =1 attached to a hanging line : length = 10 (meters [or dimensionless]) with a hook at the other end, flying frictionless at speed = 5 (m/s) along x axis =>=>=>=>=>

hi Harold, thanks for your attention, the speed is 5m/s, I thought it was clear, probably it was not . We have a ball flying horizontally, there is a string hanging down from it, and the hook gets caught.
If it is simpler,
let's start from a ball [mass=1]rotating on a line 10 m long, tangential velocity is 5m/s , L (r * p) = 50 [M*L/T) , right?

The angular momentum L is 50. To answer the original questions.
(1) the sinker did not suddenly gain angular momentum when the hook caught on the branch. It already had angular momentum with respect to the axis of rotation because of its velocity and distance of its path from the branch.
(2)L does not vary in time in the problem you have described. There is no torque.
(3)Torque is a vector because it can be applied in different directions.
(4)To point the angular momentum vector in the opposite direction you would have to apply a torque in the opposite direction and accelerate it so it spins the other way about its axis.

Originally Posted by Harold14370

The angular momentum L is 50. To answer the original questions.
(1) the sinker did not suddenly gain angular momentum when the hook caught on the branch. It already had angular momentum with respect to the axis of rotation because of its velocity and distance of its path from the branch.
(2)L does not vary in time in the problem you have described. There is no torque.
(3)Torque is a vector because it can be applied in different directions.
(4)To point the angular momentum vector in the opposite direction you would have to apply a torque in the opposite direction and accelerate it so it spins the other way about its axis.

Thanks Harold, for your reply:
1) so: the rotating sinker has L = 50 and (momentum) p= 5,
a bird flying over my head has (absolute) momentum and (relative to my head) angular momentum? but, If I got it right
the bird's angular momentum decreases when he's approching me and increases when he overtakes me, so L is changing all the time and there should be a vector torque, is that right?

2) you are right, there should be not torque, that's what I thought and that's why I came here: but in chemistry and atomic physics they say that in H1 there is a vector pointing (right, that is opposite to magnetic vector) from the nucleus, and speed is not changing and radius is not changing, L is not changing, don't they?

thanks again

In your problem, angular momentum is 50 and momentum is 5.

A bird has angular momentum. It does not increase or decrease as he passes over you head, unless he has changed course or velocity. The radius used in the angular momentum calculation would be the distance from you to the bird's flight path at his closest approach, not from you to the bird.

Not only torque is a vector, so is angular momentum. The Bohr model of an atom with an electron orbiting around the nucleus is not a very useful model, and it has been replaced by quantum mechanics.

Originally Posted by Harold14370

1) In your problem, angular momentum is 50 and momentum is 5.
2) The radius used in the angular momentum calculation would be the distance from you to the bird's flight path at his closest approach, not from you to the bird.
3) The Bohr model of an atom with an electron orbiting around the nucleus is not a very useful model, and it has been replaced by quantum mechanics.

Thanks, Harold,
1) an 2) are settled,and 3) but for a puzzling detail: if you check the animation at http://en.wikipedia.org/wiki/angular_momentum, you'll see that vector τ (in blue) disappears with acceleration, and vector L stays there even when v (and p) is constant.
3) can you give me a link to see how QM deals with L and m ?
That's the math.

Now comes the most difficult part, the original conceptual question. I hope you can spare some more time.
A rotating body as both L and p, if we cut the line the sinker retains its p = 5 and its L = 50 (re: pivot) to infinity; if it collides with equal sinker transfers these properties to it : can we conclude that the concept of L has nothing to do with p, except that its value depends on the value of p
5) now, if you say in atoms it has been replaced, is this vector L unnecessary?, what is for , in the first place?
6) why should it point outward from pivot, one verse or other?
7) why in that direction? what makes you determine that direction. If it can be replaced, it might have been (and then replaced) in the opposite direction
a bird flying over my head makes a vector L spring horizontally from my head, why should this be in a particular direction, this vector is not velocity, is not force, is not momentum, what does the vector angular momentum point, mean ?

Edit: bogus smilie

Originally Posted by logic

1) an 2) are settled,and 3) but for a puzzling detail: if you check the animation at http://en.wikipedia.org/wiki/angular_momentum, you'll see that vector τ (in blue) disappears with acceleration, and vector L stays there even when v (and p) is constant.

The animation shows an oscillating system, possibly like a balance wheel and spring on a watch. When the system is at its maximum rotation the velocity and angular momentum are zero, and the spring is stretched in one direction. Since the spring is deflected the maximum from its rest position, force and torque are maximum. In the mid travel of the system, the spring is relaxed and the velocity and angular momentum are maximum. At the other end of travel, the spring is compressed in the other direction so the torque is maximum in the opposite direction. Torque/force and angular momentum/velocity are out of phase.

3) can you give me a link to see how QM deals with L and m ?
That's the math.

That same Wikipedia article has a section on quantum mechanics.

Now comes the most difficult part, the original conceptual question. I hope you can spare some more time.
A rotating body as both L and p, if we cut the line the sinker retains its p = 5 and its L = 50 (reivot) to infinity; if it collides with equal sinker transfers these properties to it : can we conclude that the concept of L has nothing to do with p, except that its value depends on the value of p
5) now, if you say in atoms it has been replaced, is this vector L unnecessary?, what is for , in the first place?
6) why should it point outward from pivot, one verse or other?
7) why in that direction? what makes you determine that direction. If it can be replaced, it might have been (and then replaced) in the opposite direction
a bird flying over my head makes a vector L spring horizontally from my head, why should this be in a particular direction, this vector is not velocity, is not force, is not momentum, what does the vector angular momentum point, mean ?

That is what I do not understand

The direction of the vector of angular momentum depends upon the pivot point you choose to calculate the angular momentum about. It makes sense to align the vector with the axis of rotation. The direction it points along the axis of rotation is just a mathematical convention. We could have used the left hand rule just as well.

"Angular-momentum" and "-acceleration" has a vector because rotation has directions and need acceleration to change directions just like a normal motion do. Vector in rotation is not something intuitive at first but understanding it will explain many stuff that's happening is our hi-tech (wheel laden) world. For example you can see one way of demonstrating this thing in this video: MIT Physics Demo -- Bicycle Wheel Gyroscope - YouTube , and another one here you could see how angular vector is made to work: Conservation of Angular Momentum - YouTube ;in the first video you can see a wheel defying gravity, and in the second video you see how a girl can change his own rotation by changing a rotation of another rotation.

I suspect you haven't seen this stuff before. When you did, you can try to use them to make sense of the notes and the math that the lecturer gave you: and maybe you could even answer the problem that you posted here (honestly I can't understand it). The way I first learn angular vector is from a book which used "Gyroscope precession" as a example, (precession is literally out of this world, you have to see it).

Originally Posted by Harold14370

It makes sense to align the vector with the axis of rotation. The direction it points along the axis of rotation is justa mathematicalconvention. We could have used the left hand rule just as well.

thanks, Harold, so, if I got it right: the direction of the vector is arbitrary, the existence of a vector L (and τ) is a matter of logic.
I will not ask you to expand on theoretical issues, but I'll question the concrete validity of that, in a while, about torque.
I'll skip for now angular momentum, because I need some technical terms, to express my thought. You are an engineer, Harold tell, me, when you pitch a tent,

1) what do tou call two opposite vectors that stabilize it ? <= | => tensional, tensorial, stabilizing vectors?
b) what do you call a vector trying to rotate a flywheel in 3rd dimension, the same as you rotate the wheel of a bycicle when you lean left/right or make a turn left/ right?
2) the arrow is used to represent velocity, momentum, angular momentum, force is there a convention to tell one from another?.
3) can we conclude that the concept of L has been devised only to explain the stabilizing property of a flywheel, gyroscope etc? and has no relation to linear momentum?

now we come to torque, the moment of force, math description of the efficiency of lever, of steelyard balance.
we have a yard (on x axis) with a pivot at (origin) o , 2 (moment) a[rms] -a , a , (-10, +10) and a force F (9.8 N) at -10.
........................................ vectors pointing up (can't find arrow up)
_ -10_[1Kg]_________o_|_ | _ | _ | _ | _ | _ | _
vector ↓ pointing down

---

Now, on the +x axis you have a bundle of vectors ( 4 do you call it a tensor?) pointing up, exerting a force (5 do you call it charge, pressure, tension?) different in value in every point, describing the curve of a hyperbola with equation y = [1*10] 10/x
6) what is the sense of aligning a vector torque with the axis of rotation?, why a cross product? the right formula should be (moment of force) τ = r * F / a
that is the sense of my original question, what has this concept to do with the concept of momentum and angular momentum?.

Edit: corrected drawing

Originally Posted by logic

Originally Posted by Harold14370

It makes sense to align the vector with the axis of rotation. The direction it points along the axis of rotation is justa mathematicalconvention. We could have used the left hand rule just as well.

thanks, Harold, so, if I got it right: the direction of the vector is arbitrary, the existence of a vector L (and τ) is a matter of logic.
I will not ask you to expand on theoretical issues, but I'll question the concrete validity of that, in a while, about torque.
I'll skip for now angular momentum, because I need some technical terms, to express my thought. You are an engineer, Harold tell, me, when you pitch a tent,

1) what do tou call two opposite vectors that stabilize it ? <= | => tensional, tensorial, stabilizing vectors?

I would call them torque vectors. The torque is the result of the tension in the ropes.

b) what do you call a vector trying to rotate a flywheel in 3rd dimension, the same as you rotate the wheel of a bycicle when you lean left/right or make a turn left/ right?

Torque.

2) the arrow is used to represent velocity, momentum, angular momentum, force is there a convention to tell one from another?.

They are labeled with a v, tau, p, L, F, etc as shown in the Wikipedia article you were looking at before, sometimes with an arrow above the letter symbol to indicate vector.

3) can we conclude that the concept of L has been devised only to explain the stabilizing property of a flywheel, gyroscope etc? and has no relation to linear momentum?

No, we cannot conclude that. It is related to the linear momentum by the radius. It doesn't only explain the stabilizing property of a flywheel. It would be used to calculate the acceleration of a rotating mass upon the application of a torque, for example.

now we come to torque, the moment of force, math description of the efficiency of lever, of steelyard balance.
we have a yard (on x axis) with a pivot at (origin) o , 2 (moment) a[rms] -a , a , (-10, +10) and a force F (9.8 N) at -10.

_ -10_[1Kg]_________o__________|+10^[1Kg]
_ F |

---

Now, on the +x axis you have a bundle of vectors ( 4 do you call it a tensor?) pointing up, exerting a force (5 do you call it charge, pressure, tension?) different in value in every point, describing the curve of a hyperbola with equation y = 10/x [a]
6) what is the sense of aligning a vector torque with the axis of rotation?, why a cross product? the right formula should be (moment of force) τ = r * F / a
that is the sense of my original question, what has this concept to do with the concept of momentum and angular momentum?.

You have stated that in a very confusing manner. I think you are just showing a lever with two 10 kg masses each at a distance of 10 in opposite directions from the fulcrum. Where are the 4 vectors? Do you mean two forces and two torques? There are no charges, pressures or tensions involved in this example.

In this case the concept of momentum and angular momentum don't really apply, or rather they would be zero. This problem can be easily solved by just using scalar values for the torque. If you had a torque applied in a different plane, the concept of cross product may prove useful to determine the direction of the resulting torque.

Originally Posted by Harold14370

I would call them torque vectors. The torque is the result of the tension in the ropes.

3) can we conclude that the concept of L has been devised only to explain the stabilizing property of a flywheel, gyroscope etc? and has no relation to linear momentum?

No, we cannot conclude that. It is related to the linear momentum by the radius. It doesn't only explain the stabilizing property of a flywheel. It would be used to calculate the acceleration of a rotating mass upon the application of a torque, for example.

( 4 do you call it a tensor?) pointing up, exerting a force (5 do you call it charge, pressure, tension?) different in value in every point, describing the curve of a hyperbola with equation y = 10/x [a]
6) what is the sense of aligning a vector torque with the axis of rotation?,?.

you are just showing a lever with two 10 kg masses each at a distance of 10 in opposite directions from the fulcrum. Where are the 4 vectors? Do you mean two forces and two torques? There are no charges, pressures or tensions involved in this example.

In this case the concept of momentum and angular momentum don't really apply, or rather they would be zero. This problem can be easily solved by just using scalar values for the torque. If you had a torque applied in a different plane, the concept of cross product may prove useful to determine the direction of the resulting torque.

1) so if I got it right, people playing tug-of-war are exerting a torque?
2) rotating a flywheel/lever is a torque but the concepts are different:
the lever is like amagnifying: efficiency/resistance varies with distance a from fulcrum
the resistance of a flywheel to a rotation has no definition, because the principle of the Gyroscope has not been described ,has it got a name? that resistance has unique behaviour . [If you know Walter Lewin's lectures at MIT he says that gyroscope "is chasing angular momentum", but I do not want to drag you into theorethical discussion]

3) I am sorry about thedrawing, but that's all my keyboard allows me [I can't make arrows up and down, ASCII code does not work here],
4 is the number of the question
the weigth at -10 should have an arrow down, from o to +10 [or infinity] you should imagine a bundle of vectors pointing up, a hyperbola of vectors produced by weight [like the stress on a dam/arch] a tensor
Thanks

Originally Posted by logic

1) so if I got it right, people playing tug-of-war are exerting a torque? if so , this was the main obstacle in understanding the issue, but I can't digest it, yet
2) rotating a flywheel, is a torque, like the lever, but I thought we should differentiate the concepts because they are different:
the lever is just a magnifying lens that concentrates energy, and the resistance it encounters varies with the distance from fulcrum
the resistance of a flywheel to a rotation has no definition, because the principle of the Gyroscope has not been described , has it got a name? that resistance has a strange, unique behaviour that has not been yet understood. [If you know Walter Lewin's lectures at MIT he says that gyroscope "is chasing angular momentum" but that is risible. But I do not want to drag you into theorethical discussion]

3) I am sorry about the lousy drawing, but that's all my keyboard allows me [I can't make arrows up and down, ASCII code does not work here], I hoped you would
understand reading the text. 4 is just the number of question
the weigth is only at -10 that should be an arrow down, from o to +10 or infinity you should imagine a bundle of vectors pointing up , a hyperbola of vectors produced by weight , I supposed like the charge on a dam, an arch, a tensor
I hope now you are ableunderstand and comment the drawing
Thanks

People playing tug of war are normally pulling in a straight line opposite to each other. Usually we wouldn't call that a torque, as there is no radius involved. Although, I suppose we could arbitrarily define some point about which to calculate a torque (sort of like calculating the angular momentum of the bird flying overhead). Now, if they were pulling on opposite ends of a rope looped around a wheel, then we could say that they are exerting opposing torque, that torque being equal to the tension in the rope multiplied by the radius of the wheel. And if there were two ropes, one looped around a larger diameter wheel, the tug-of-war team whose rope was on the larger wheel would have a mechanical advantage, so that they would exert greater torque for a given force, and would win the tug-of-war.

When you say the principle of the gyroscope has not been described, I think you are referring to the weird kind of counter-intuitive thing that happens when we try to turn a gyroscope in a direction that would change its axis of rotation. If you are spinning up or slowing down the gyroscope I hope you can understand that this is done by applying a torque and that the angular momentum gained or lost is analogous to the way that straight-line momentum is gained or lost by the application of a force.

As a matter of fact it is not only analogous to it, the angular momentum is just simply the accumulated result of a bunch of individual particles of mass, each obeying Newtons ordinary, straight line, laws of motion.

This might help you understand the gyroscope effect.
HowStuffWorks "How Gyroscopes Work"

Originally Posted by Harold14370

I think you are referring to the weird kind of counter-intuitive thing...
the angular momentum is ...individual particles of mass, each obeying Newtonsordinary, *straight line, laws of motion.

[I do not think what happens is counter-intuitive,[nothing is], when you understand what is really happening. People thought "earth is still" only because had never been on a jet plane. (I won't tell you I fully understand the issue because I do not wish you to take your hat off)
What happens in a flywheel is surely that * , [but that* is called inertia not angular momentum], but it is not only that*, because * would cause only resistance and not wobbling, and wouldn' explain why a gyroscope °doesn't fall.
what's L the explanation for that°?
Is there a better explanation than Lewin's for precession? answer only if you are interested]
----------------

I am a bit confused by the fact that pickets exert a torque and tug-of-war does not. Now consider this:
a saucepot is hanging from to 2 strings tied to each handle. we pull strings apart until pot is level and stays up, are we exerting a torque when we pull strings?
1) when pot is level strings are exerting a torque? if not, what do you call suchforces? if yes, where is L and τ ?
whould you agree that force can be only a pull, a linear push and a tangential push (torque)?
the lever concept:
I corrected my drawing, I hope it is clear now, if it is not, just imagine 10 Kg on a steelyard and describe what happens on other arm a:
a bundle of vectors is a tensor, is that called stress, charge in engineering?
is a "hyperbolic tensor" the torque in a lever?

Originally Posted by logic

[I do not think what happens is counter-intuitive,[nothing is], when you understand what is really happening. People thought "earth is still" only because had never been on a jet plane. (I won't tell you I fully understand the issue because I do not wish you to take your hat off)
What happens in a flywheel is surely that * , [but that* is called inertia not angular momentum],

What is the definition of inertia in physics? Do you think it is different than momentum?

but it is not only that*, because * would cause only resistance and not wobbling, and wouldn' explain why a gyroscope °doesn't fall.
what's L the explanation for that°?
Is there a better explanation than Lewin's for precession? answer only if you are interested]

I can't explain it better than the "How Stuff Works" article did.

----------------

I am a bit confused by the fact that pickets exert a torque and tug-of-war does not.

What's pickets?

Now consider this:
a saucepot is hanging from to 2 strings tied to each handle. we pull strings apart until pot is level and stays up, are we exerting a torque when we pull strings?
1) when pot is level strings are exerting a torque? if not, what do you call suchforces? if yes, where is L and τ ?
whould you agree that force can be only a pull, a linear push and a tangential push (torque)?

The pot will never get level, there will always be some sag in the line.
Let's take an example. The pot weighs 1 lb and hangs half way between two pegs, which are 10 feet apart, such that the lines are at a 45 degree angle to horizontal. The weight of the pot is supported by the vertical component of the tension in each line. Since it is equally supported by both lines, each line supports a weight of 1/2 lb. This is also the downward force on each peg. The string tension can be calculated as the hypotenuse of a triangle of forces consisting of the vertical component of the tension (which we said was 1/2 lb), the horizontal component and the string tension T. Therefore sin 45 degrees = 0.5 lb/T = 0.707 and T=0.707 lb.
The magnitude of the force on the left peg is 0.707 and this force points down and to the right. The force on the right peg is .707 down and to the left.
We know that the floppy rope cannot exert a torque, but we could imagine that it is a frozen rope and calculate the torque about the left hand peg. We would find that the pot is exerting a downward force of 1 lb at a horizontal distance of 5 feet, or 5 lb-ft of torque clockwise, whereas the right peg is exerting an upward force of 1/2 lb at a 10 feet, or 5 lb-ft of torque counterclockwise. Net torque on the left peg = 0.

the lever concept:
I corrected my drawing, I hope it is clear now, if it is not, just imagine 10 Kg on a steelyard and describe what happens on other arm a:
a bundle of vectors is a tensor, is that called stress, charge in engineering?
is a "hyperbolic tensor" the torque in a lever?

I still don't understand your drawing and don't know what is hyperbolic about it or why you want to call it a tensor.

Originally Posted by Harold14370

Originally Posted by logic

is called inertia not angular momentum],

1) What is the definition of inertia in physics? Do you think it is different than momentum?
2) I still don't understand your drawing and don't know what is hyperbolic about it or why you want to call it a tensor.

1) I didn't know inertia is angular momentum, but I do not know what angular momentum is.
They say inertia is a property of mass, never heard is a vector.
momentum is a vector, the force that velocity exerts on another body

2) each vector on the right side of o is the upward force that weight at -10 is exerting on each point of right lever arm. It's a common balance scale, at o force isit is infinite, at +10 is 1 kg, at 100 is 0.1. The vectors describe a hyperbola y=10/x, I can't explain it any better, sorry
thank you, Harold, for your competence and patience .

Originally Posted by logic

1) I didn't know inertia is angular momentum, but I do not know what angular momentum is.
They say inertia is a property of mass, never heard is a vector.
momentum is a vector, the force that velocity exerts on another body

I didn't say inertia was momentum. It actually doesn't have a mathematical definition. Why do you say you do not know what angular momentum is? You looked it up on the internet and found the math expression for it. That's what it is. Momentum is not a force. It's momentum - mass times velocity.

2) each vector on the right side of o is the upward force that weight at -10 is exerting on each point of right lever arm. It's a common balance scale, at o force isit is infinite, at +10 is 1 kg, at 100 is 0.1. The vectors describe a hyperbola y=10/x, I can't explain it any better, sorry

Okay, I think you want to figure out the torque due to a distributed mass which is distributed according to y=10/x. There is nothing exotic about what you are describing. In fact, it is more common for masses to be distributed than concentrated at a single point. All you would do is integrate the product of the function multiplied by its distance from the fulcrum, from zero to 10.

Hi Harold
1) I thougth "inertia is not different from momentum" meant momentum is inertia
then if momentum is not a force, not inertia, not velocity what is it? its math expression says it is mass times vector-velocity
2) an entity cannot be just its math expression in principle, a fortiori it in dimensional analysis where different entities(quantities) have same dimensions.Then however you define it you must wxplain why you can cross-product it
3) my idea is not exotic, is just reality, a hard fact known for centuries: if you hang a (steel) yard from its middle and put a weight W on one arm at -10 cm from pivot, you generate a force [on the other arm at +10cm] which is exactly W. If you want to balance the yard at 5 cm you must put 2W, what is exotic about it? if you consider all [virtual] vectors-force together, you get abundle of vectors, [is it a tensor]? it has the shape of a hyperbola with equation [F]y= [F=W]*r/ x[=a]
4) you said that the direction of L is a convention, IF direction is arbitrary, than L itself must be fictitious, because if vector had math necessity it would have a mandatory direction. too. Now, in conclusion, when you exert a torque on a lever what do you get on the other arm, if it is not a hyperbola , what?

Logic,
I think a lot of the questions you are asking could be cleared up by taking a class in physics, or getting a textbook and studying it. You will never really understand the concepts of mass, momentum, torque, etc., until you have worked through some exercises which use these concepts. After you have solved a few problems of masses colliding elastically and inelastically, projectiles being launched, and things like that, and used the concepts of conservation of momentum and conservation of energy, then you will suddenly realize, aha, that's what momentum is. Until then, you will only have a vague idea of it.

Originally Posted by Harold14370

Not only 0) torque is a vector, so is angular momentum..

Originally Posted by Harold14370

0)Torque/force and angular 1b) momentum/velocity are out of phase.
The direction of the vector of angular momentum ... points along the axis of rotation is just a mathematical convention. We could have used the left hand rule just as well.

Originally Posted by Harold14370

What is the definition of inertia in physics? Do you think it is 1c)different than momentum?

Originally Posted by Harold14370

I didn't say inertia was momentum. 1a)Momentum is not a force. It' s momentum- mass times velocity

Originally Posted by Harold14370

...inertia....doesn't have a mathematical 4) definition

Originally Posted by Harold14370

I would call them torque vectors. 2) The torque is the result of the tension in the rope

Originally Posted by Harold14370

The weight of the pot is supported by the vertical component of 2) the tension in each line.
I still don't understand your drawing and don't know what is hyperbolic about it or why you want to call it a 3)tensor.

Originally Posted by Harold14370

....could be cleared up by *taking a class in physics, or getting a textbook .

[Hi Harold, * If you look at your answers, probably you will see why I am confused. ]

1) a) "[angular]momentum is [angular]momentum" is not a definition, is a tautology: not-science, not even meaningful language: like ["a bluge is a bluge , its math is = age times beauty"]. if you check "scientific method" at wiki, you'll see you(science/physics) have mandatory duty to clearly define your terms.
b) "momentum is m * v", is not a math definition, it is meaningless, too, if you do not say what is the result of math: what does "momentum/velocity" mean (1b) "Momentum= velocity", or momentum=inertia"? [1c] is it a force, or it is a [vector (0) not a force(0,4)].?
Velocity and force are the only defined concepts in physics, you have no other choice: a) momentum is a force, exerted by what? b) it is velocity
2) is torque "force times radius" or the "result of tension"(2)? tension is a force, since it supports the pot (2)
now in real world,
3) in a lever: you have a force W pushing/pulling down the yard. If there is no centre/pivot, the yard falls down. If you steady the fulcrum, the yard itself cannot fall down any more, and
force W cannot point down any more, but must follow a circle, must point as a radian. That is a torque , a vector-radian-force and its math is F * r[-a], on the other arm a, is force*radius[ -a ]/[radius] a
force-torque on a points up, in opposite direction, its math description/ value depends on the position on [length of radius] a. The result is a hyperbola of, not mass distribution, but forces-torques.
can you refute this?

A definition can only explain one thing in terms of some other thing you already understand. If I say that torque is force multiplied by a radius, and if I show how it is used in a calculation, that's about all the definition one can give. And if engineers can successfully use these concepts to build large structures that do not collapse, to make machinery that transports tons of freight, and so forth, then I don't think the science is seriously flawed. If you do not like it, I can't help you.

Originally Posted by Harold14370

1) If I say that torque is force multiplied by a radius, and 2)if I show how it is used in a calculation, that's about all the definition one can give. ..and if engineers can successfully use these concepts ...., then I don't think the science is seriously flawed...

1) that's right , Harold, if you say F*r! but if you mean F x r you must explain it.
You said "[angular] momentum is a vector "and then "momentum is not different from inertia" and then again corrected yourself as you realize (point(4) in previous post) that" inertia is not defined", what a flaw/mess!.
Momentum is inertia if you call it by name and not by surname: force of [inertia/poltergeist*]. On one hand you deny momentum is force, and on the other you affirm it, that is your problem. Momentum is force, of course, because all vectors is force. You are confused because you were using a wrong surname.
[angular]momentum/torque are/is a force[vector], put its position along axis is wrong/false, angular momentum is just an alias, it is bogus (fictitious, like centrifugal) force When you understand this you can understand/explain the behaviour of gyroscope and will change "precess" to proceed"
(there is a vector-force acting on fulcrum but it is linear andis pointing down, [aligned with] gravity)

2) of course Harold, you are an engineer [and a very competent one, my hat off to you], a practical man, that's why I hesitated to drag you in a theoretical discussions, and I am grateful you accepted.
BUT,that is reductive and dangerous, because if you do not fully understand what you are doing, you can blow the world up.
Laws of motions work in praxis, sure, but they would work as well if you substitute "force-of-inertia" with a "*poltergeist" or "fluid-of-universal-spirit". You can use a "bluge" all right, if you manage to see how it works in certain situations, but you might discover it is lethal in certain conditions
thanks again, Harold

Now if you are willing to explain another point:

our sinker/ball [ m=1,v=5, KE 12,5, p=5 ] is flying ,what energy is necessary to make its direction deflect by 1°?
If I hit it with a tennis-racket how much energy is necessary to increase v =6?

Originally Posted by logic

Momentum is force, because all vectors ar force. You are confused [and I disagreed] because you are using a wrong surname.
So, in conclusion, momentum and torque are force[vectors] but vector torque at fulcrum is bogus

If you say momentum and torque are force, then you are inventing your own terminology. If you want to communicate with physicists or engineers, you will have to use the same language. There are certain technical terms that are always used the same way, no exceptions. It is really necessary for proper communication.

Originally Posted by Harold14370

Originally Posted by logic

Momentum is force, because all vectors ar force. You are confused [and I disagreed] because you are using a wrong surname.
So, in conclusion, momentum and torque are force[vectors] but vector torque at fulcrum is bogus

If you say momentum and torque are force, then you are inventing your own terminology. If you want to communicate with physicists or engineers, you will have to 1)use the same language. There are certain technical terms that are always used the same way, no exceptions. It is really necessary for proper communication.

No Harold, the logic of the issue is that:
you say you do not know what is momentum, I am just filling a gap saying it is force.
You may refute my definition with arguments [you say are non-existant], but you cannot question my right to do that and (1) I cannot talk same language because on the other side there is no-language
moreover, you haven't yet justified the cross-product.

Would you like to answer the new point?, It is my answer to your "frame of reference" reply in the other thread
Thanks again

Originally Posted by logic

Originally Posted by Harold14370

Originally Posted by logic

Momentum is force, because all vectors ar force. You are confused [and I disagreed] because you are using a wrong surname.
So, in conclusion, momentum and torque are force[vectors] but vector torque at fulcrum is bogus

If you say momentum and torque are force, then you are inventing your own terminology. If you want to communicate with physicists or engineers, you will have to 1)use the same language. There are certain technical terms that are always used the same way, no exceptions. It is really necessary for proper communication.

No Harold, the logic of the issue is that:
you say you do not know what is momentum, I am just filling a gap saying it is force.
You may refute my definition with arguments [you say are non-existant], but you cannot question my right to do that and (1) I cannot talk same language because on the other side there is no-language
moreover, you haven't yet justified the cross-product.

If you think a new word is needed for it, then call it something else. Force is already taken for something that has a different meaning.
Maybe you could call it "poltergeist"

Would you like to answer the new point?, It is my answer to your "frame of reference" reply in the other thread
Thanks again

our sinker/ball [ m=1,v=5, KE 12,5, p=5 ] is flying ,what energy is necessary to make its direction deflect by 1°?
If I hit it with a tennis-racket how much energy is necessary to increase v =6?

It could be deflected by 1 degree and either gain or lose energy depending on the final speed, so there is no single answer. If it deflected off another object to change course 1 degree, it could have given up energy, not gained it.
If the mass is 1 and velocity is 6, the KE is 1/2*1*6^2=18 so the energy added is 5.5.

Originally Posted by Harold14370;286451[QUOTE

our sinker/ball [ m=1,v=5, KE 12,5, p=5 ] is flying ,what energy is necessary to make its direction deflect by 1°?
If I hit it with a tennis-racket how much energy is necessary to increase v =6?

1) It could be deflected by 1 degree and either gain or lose energy depending on the final speed, so there is no single answer.
2)If the mass is 1 and velocity is 6, the KE is 1/2*1*6^2=18 so the energy added is 5.5.[/QUOTE]

1) only deflected, no racket, no hit, no collision, same v, same KE just its trajectory is bent. [To get the right picture imagine an asteroid and a rocket thrusting tangentially, or a planet in an orbit, where gravity does zero work.] everything stays equal, only new direction makes an angle 1° with old direction
2)so I add 5.5 work,KE, Now imagine same ball and me on a pickup and a me on the side of the road hitting it to v=6 how much work do I do and how much KE ball gets. And now imagine ball and me on pickup, me pushing ball to v=6. How does frame of reference work?

I can see you are not convinced that kinetic energy is relative to the frame of reference. But since it is a function of velocity, which is relative, it must be relative as well.
If the trajectory is bent, there must have been a force, and if the force is at right angles to the velocity, it will impart a component of velocity at right angles to its original trajectory. The velocity in the direction of the original trajectory does not change, as there is no component of force in that direction. Therefore the final velocity is found by the pythagorean theorem, and the KE calculated accordingly.
I had to think a little bit about the planet in orbit, because it does not gain any energy (relative to earth) after an acceleration at right angles to its trajectory. But this KE calculation only works relative to an inertial frame of reference. After some acceleration at an instant in time, then some time later there is a velocity, and KE, relative to the inertial reference frame, which continues along the tangent line after that instant.

Yes, you will have to do more work to achieve the same acceleration if you are working from the ground than if you are on the pickup truck. When you are on the pickup truck, the truck is doing some of the work for you. It is applying a force to counteract the reaction force and keep from slowing the vehicle down due to the recoil. Later on, I may do some calculations to show you how it all works out.

Edit to add:

Let's assume the pickup truck with its load weighs 1000 kg and is going at 5 meters per second. Without using the engine to accelerate the vehicle we throw the 1 kg mass 1 meter per second forward, giving it a final velocity of 1 meter per second relative to the trucks original frame, or 6 relative to the road.

From the pickup truck's original frame of reference:
Neither the truck nor the ball had any KE initially. After the throw of the ball, the ball has KE of 1/2*1*1^2=0.5 Joule and momentum of mv=1*1=1. By conservation of momentum, the truck has momentum of -1 and so the velocity is -1/999. Its kinetic energy is 1/2*m*v^2= 0.0005 joules, which is negligible so net gain is 0.5J.

From the road frame of reference, we had the pickup and its load weighing 1000 kg and going 5 meters per second. KE=1/2 m v^2 = 1/2 *1000*5*5=12500. Of this, the ball itself has KE of 1/2 *1*5*5=12.5.
After the ball is thrown we see that the ball has KE of 1/2*1*6*6 = 18. The pickup truck has a velocity of 5-1/999 meters per second, mass of 999 kg and energy of 0.5*999*(5-1/999)^2 = 12482.5
Therefore from the road frame of reference the total KE is 12482.5+18=12500.5, and increase of 0.5 Joules. This is the same change in energy for the total system that was calculated in the truck frame.

Originally Posted by Harold14370

. The velocity in the direction of the original trajectory does not change, as there is no component of force in that direction. Therefore the final velocity is found by the pythagorean theorem, and the KE calculated accordingly.
1)I had to think a little bit about the planet in orbit, because it does not gain any energy (relative to earth) after an acceleration at right angles to its trajectory. But this KE calculation only works relative to an inertial frame of reference. After some acceleration at an instant in time, then some time later there is a velocity, and KE, relative to the inertial reference frame, which continues along the tangent line after that instant.
.

1)That is the point I am making, and probably you are missing. There is no applying to frame of reference, because KE is an absolute value, and velocity is its sqrt, so the sqrt of an absolute value is an absolute value. you cannot apply vector-sum, you do not do it for planets, electrons, why poor sinkers?
I hope sometime you'll begin to admit "you have a point there, logic"
sinker, asteroid, planet makes no difference if you say normal force does zero work, it must apply for every body.

2) the pickup example was not clear, you have one ball 1)flying over your head at v=5, one ball 2) at rest on a pickup running at same v=5, one ball 3) at rest on the ground, frame of reference is the ground, you hit them with a racket or push them as you like, from the ground or from pickup.Explain why you should make less effort in any of these instances

3)angular momentum cannot be a poltergeist, but only force because it has what you call the math definition of a force, Harold, kg*m²/s, M*L²/T

4) when do you use cross product, in physics, what does it mean concretely in applied physics, in engineering?
why can you use it for L and torque?

Originally Posted by logic

Originally Posted by Harold14370

. The velocity in the direction of the original trajectory does not change, as there is no component of force in that direction. Therefore the final velocity is found by the pythagorean theorem, and the KE calculated accordingly.
1)I had to think a little bit about the planet in orbit, because it does not gain any energy (relative to earth) after an acceleration at right angles to its trajectory. But this KE calculation only works relative to an inertial frame of reference. After some acceleration at an instant in time, then some time later there is a velocity, and KE, relative to the inertial reference frame, which continues along the tangent line after that instant.
.

1)That is the point I am making, and probably you are missing. There is no applying to frame of reference, because KE is an absolute value, and velocity is its sqrt, so the sqrt of an absolute value is an absolute value. you cannot apply vector-sum, you do not do it for planets, electrons, why poor sinkers?
I hope sometime you'll begin to admit "you have a point there, logic"
sinker, asteroid, planet makes no difference if you say normal force does zero work, it must apply for every body.

2) the pickup example was not clear, you have one ball 1)flying over your head at v=5, one ball 2) at rest on a pickup running at same v=5, one ball 3) at rest on the ground, frame of reference is the ground, you hit them with a racket or push them as you like, from the ground or from pickup.Explain why you should make less effort in any of these instances

3)angular momentum cannot be a poltergeist, but only force because it has what you call the math definition of a force, Harold, kg*m²/s, M*L²/T

4) when do you use cross product, in physics, what does it mean concretely in applied physics, in engineering?
why can you use it for L and torque?

1. I just did apply frame of reference to KE and showed that conservation of energy worked the same in both frames of reference.
2. Work it out yourself, just like I did. In order to accelerate the ball from the ground you will have to do some work on it. You will apply a force over a distance. To you it will appear to be a longer distance than it will to a person on the truck, so you will calculate more work done in your reference frame.
3. Wrong. Force is kilogram meters per second squared and angular momentum is kilogram meters squared per second. SI derived unit - Wikipedia, the free encyclopedia.
4. You can use it because it works and gives you the right answer.

Originally Posted by Harold14370

1) I can see you are not convinced that kinetic energy is relative to the frame of reference. But since it is a function of velocity, which is relative, it must be relative as well.
2)If the trajectory is bent, there must have been a force, and if the force is at right angles to the velocity, it will impart a component of velocity at right angles to its original trajectory. The velocity in the direction of the original trajectory does not change, as there is no component of force in that direction. Therefore the final velocity is found by the pythagorean theorem, and the KE calculated accordingly.

3) the planet in orbit, because it does not gain any energy (relative to earth) after an acceleration at right angles to its trajectory. But this KE calculation only works relative to an inertial frame of reference. After some acceleration at an instant in time, then some time later there is a velocity, and KE, relative to the inertial reference frame, which continues along the tangent line after that instant.
.

1) KE is absolute and so is velocity. If you consider the fiying ball 1) and ball on the pickup 2), they have same KE and velocity=5. You may say 2) is at rest, when relate it to vehicle, because pickup has same velocity = 5 and 5-5=0 , but you can relate it to a faster vehicle and you get negative velocity and KE which is absurd. the two balls have same v and KE. so to accelerate them to 6 and 6 [or 6 and 1], you require same KE 5.5

2) if velocity increases in the sinker-ball it must increase in an asteroid, a planet or electron. Consider frame of reference the solar system.

I asked you to calc what force deflects the sinker-ball, or any body you choose, by 1°, can you do that?

3)sorry, this period is incomprehensible, probable because of haste.

4)[L has units of action, the force of photon, Planck's unit,] 4) math definition of force applies only to gravity, does not apply to Coulomb, magnetic, or inertia force. Do you know other visible forces? I'll make a thread on Coulomb to discuss that. L is a force, a tensional force and has different units, also inertia/poltergeist has different units from gravity
5) dimensional analysis is not reliable. you say velocity has units L/T and then say velocity is dimensionless : α [in orbit H¹], these DA units are both false, but you must admit that at least one must be false, that this is contradictory?

6) answer to cross product is not an answer. I ask you again: if we abolish vector[not-force] L and force-torque on fulcrum/axis, what happens? If you do not specify any consequence they are bogus. You must admit that a vector force means force exerted and consequent effect, usualy velocity or deformation etc. It you cannot describe any missing effect, your vector-force is bogus, like centrifugal.

[] Edit

Originally Posted by logic

1) KE is absolute and so is velocity. If you consider the fliing ball 1) and ball on the pickup 2), they have same KE and velocity=5. You may say it is at rest, when relate it to vehicle, because pickup has same velocity = 5 and 5-5=0 , but you can relate it to a faster vehicle and you get negative velocity and KE which is absurd. the two balls have same v and KE.so to accelerate them to 6 and 1, you require same KE 5.5

Velocity is direction and speed. There is nothing absurd about the concept of negative velocity as it is applied to the direction component of the velocity and whether or not the direction is negative or positive is a matter of arbitrary choice as to which direction is considered positive, and this is usually done as a matter of convenience. The KE would never be negative because it is a function of the square of velocity and squares of negative values are positive. In addition, if you insist that velocity is absolute, why are you ignoring the rotation of the Earth, its orbital motion, etc? Your "absolute" velocity is nothing but relative velocity with the Earth chosen as the relative reference.

2) if velocity increases in the sinker-ball it must increase in an asteroid, a planet or electron. Consider frame of reference the solar system.

The only case in which you would not see an change in speed would be a perfectly circular orbit, which is unique in that the force is always perpendicular to the direction of the velocity. However, perfectly circular orbits do not exist in nature and all orbits are eccentric to some degree. The Earth does increase velocity as it goes from aphelion to perihelion as do all planets and asteroids. The only difference between the sinker ball and these objects is that the eccentricity of the sinker ball's trajectory is much greater, so the increase in velocity is more noticeable over the relatively small segment of its trajectory. But to be frank, I think is best for you to avoid the complexities of orbital mechanics until you haver a much greater grasp of the basics.

4)math definition of force applies only to gravity. does not apply to Coulomb, magnetic, or inertia force.Do you know other forces?I'll make a thread on Coulomb to discuss that. L is a force, a tensional force and has different units, also inertia/poltergeist has different units from gravity

This is just a false statement. All forces reduce dimensionally to the definition given by Harold14370.


At this point I'm going to put my moderator hat on:

What exactly is your intent in this thread? Is it to learn what the accepted science of this subject is, or is it to argue against it. If it is the former, then your attitude has not been very conducive to this end. If it the latter, let me know so that I can move the thread to the appropriate sub-forum.

respectfully....maybe this could be a mis interpretation.if you look at through the action on a flat plane.ie a pool table with the balls representing 2 dimensional aspects,and compare to 3d then many different results are seen...but the concept of 2d action is fundamental in relation to 3d...in 3d the angle of incedence can be off the gravitaional centre, without precise formula to explain then problems in calculations can occur...its a matter of learning all the terminolgy used in 3d space to fully appreciate the simplicity of this action......and thats before you apply general relativty