so you start with
W = F x D
F = ma
D = 1\2vt
a= v\t
it follows that
w = m(v\t) x 1\2vt
which simplifies as
w = 1\2mv^2
Well, here comes the question:
isn't the total force in a system f=mv? I thought f=ma was just the amount of force that was continually added to a system that was accelerating constantly, so why do they substitute "ma" and not "mv"?
Thanks in advance guys
QUOTE (NeutrinoQ+Dec 10 2006, 04:14 PM)
so you start with
W = F x D
F = ma
D = 1\2vt
a= v\t
it follows that
w = m(v\t) x 1\2vt
which simplifies as
w = 1\2mv^2
Well, here comes the question:
isn't the total force in a system f=mv? I thought f=ma was just the amount of force that was continually added to a system that was accelerating constantly, so why do they substitute "ma" and not "mv"?
Thanks in advance guys
mass*velocity is known as momentum
mass*acceleration is known as force
total, or net force in a system is still f=ma
W = F x D
F = ma
D = 1\2vt
a= v\t
it follows that
w = m(v\t) x 1\2vt
which simplifies as
w = 1\2mv^2
Well, here comes the question:
isn't the total force in a system f=mv? I thought f=ma was just the amount of force that was continually added to a system that was accelerating constantly, so why do they substitute "ma" and not "mv"?
Thanks in advance guys
mass*velocity is known as momentum
mass*acceleration is known as force
total, or net force in a system is still f=ma
I'm pretty sure net force is not given by f=ma, but is given by f=(ma)t= mv, although I could be wrong
Example to illustrate
i have a 10 kg car accelerating from 0 to 15 in three seconds at constant acceleration
f=(10)(5)= 50N; which means (i think) for every second, the car has 50N more force
So if you put a wall at then end of the cars acceleration, the wall is going to feel 150N of force, not 50N, right?
Example to illustrate
i have a 10 kg car accelerating from 0 to 15 in three seconds at constant acceleration
f=(10)(5)= 50N; which means (i think) for every second, the car has 50N more force
So if you put a wall at then end of the cars acceleration, the wall is going to feel 150N of force, not 50N, right?
You are getting "force" (how much push is applied) mixed up with kinetic energy.
If the acceleration doesn't change (you keep picking up speed at the same rate) that means the force ("push") is constant: in your example F = m a = (10)(5) = 50 N. The unit "newton" has the idea of acceleration built into it.
What the object does get more of as it is subject to force is kinetic energy (the 1/2 mv^2). Those units are "joules".
The kinetic energy an object has (1/2 mv^2) is the same as the "work" (F x D) that was performed on it (sort of makes sense, you work on something, so it has more energy).
So we can figure the joules both ways in this case:
Using 1/2 m v^2 we have 1/2 (10)(15^2) = 1/2 (10)(225) = 1/2 (2250) = 1125 joules
Using F x D we need D so since in 3 seconds the object traveled 1/2 a t^2 meters, then
D = 1/2 (5)(3^2) = 1/2 (5)(9) = 1/2 (45) = 22.5 meters
so
F x D = (10 kg) (5 m/s) (22.5 meters) = 1125 joules
"Power" is the measure of how fast the 1125 joules was expended. In this case it was over a 3 seconds period, so power = 1125 joules / 3 seconds = 375 watts
If the acceleration doesn't change (you keep picking up speed at the same rate) that means the force ("push") is constant: in your example F = m a = (10)(5) = 50 N. The unit "newton" has the idea of acceleration built into it.
What the object does get more of as it is subject to force is kinetic energy (the 1/2 mv^2). Those units are "joules".
The kinetic energy an object has (1/2 mv^2) is the same as the "work" (F x D) that was performed on it (sort of makes sense, you work on something, so it has more energy).
So we can figure the joules both ways in this case:
Using 1/2 m v^2 we have 1/2 (10)(15^2) = 1/2 (10)(225) = 1/2 (2250) = 1125 joules
Using F x D we need D so since in 3 seconds the object traveled 1/2 a t^2 meters, then
D = 1/2 (5)(3^2) = 1/2 (5)(9) = 1/2 (45) = 22.5 meters
so
F x D = (10 kg) (5 m/s) (22.5 meters) = 1125 joules
"Power" is the measure of how fast the 1125 joules was expended. In this case it was over a 3 seconds period, so power = 1125 joules / 3 seconds = 375 watts
to noclevername:
I see what your saying about the energy, and I agree. But what I'm really trying to ask is "total" force that an object has due to its motion, almost like an impact force. I should be more clear:
a 10kg car that goes from 0 to 15 in 3 seconds has a force of 50N
a 10 kg car that goes from 0 to 10 in 2 seconds has a force of 50N
But if they both hit a wall, and the force is decelerated over one second, we have the first car "hitting" with a force of 150N and the second hitting with a force of 100N, correct?
If my above calculation is true, then f=ma isn't the Total force an object can impart onto another object, you need to factor in the amount of time the object has been accelerating and add up all of the pushes to get the total force. which brings me back to my original question, why do they use f=ma to derive the formula, when that isnt the total force?
I see what your saying about the energy, and I agree. But what I'm really trying to ask is "total" force that an object has due to its motion, almost like an impact force. I should be more clear:
a 10kg car that goes from 0 to 15 in 3 seconds has a force of 50N
a 10 kg car that goes from 0 to 10 in 2 seconds has a force of 50N
But if they both hit a wall, and the force is decelerated over one second, we have the first car "hitting" with a force of 150N and the second hitting with a force of 100N, correct?
If my above calculation is true, then f=ma isn't the Total force an object can impart onto another object, you need to factor in the amount of time the object has been accelerating and add up all of the pushes to get the total force. which brings me back to my original question, why do they use f=ma to derive the formula, when that isnt the total force?
Replace your words "has a force of ...." with "has momentum of" (and forgetting the newtons, because those are the units of force, not momentum) and your intuitive feel will be validated.
An object wouldn't "have a force of 50 N"; instead it would have been "subject to a force of 50 N. Applying a force for a short duration would result in a small final velocity (v = a t) and a modest momentum (m v). Likewise, apply the force for a longer period and you will get a larger final velocity and correspondingly larger momentum.
An alternative definition of "force" includes the idea that momentum is gained when a force is applied for a duration of time: F = d mv / dt, which is to say, "force is the rate at which momentum is gained or lost".
When an object strikes another it will potentially lose momentum. The force of the collision is still F = m a with the decelleration (in this case) dependent on how long the collision took. The fact that momentum changes sometimes allows you to determine what the force must have been by simply looking at the total momentum of all the objects after a collision (because of the "change in momentum" definition of force).
An object wouldn't "have a force of 50 N"; instead it would have been "subject to a force of 50 N. Applying a force for a short duration would result in a small final velocity (v = a t) and a modest momentum (m v). Likewise, apply the force for a longer period and you will get a larger final velocity and correspondingly larger momentum.
An alternative definition of "force" includes the idea that momentum is gained when a force is applied for a duration of time: F = d mv / dt, which is to say, "force is the rate at which momentum is gained or lost".
When an object strikes another it will potentially lose momentum. The force of the collision is still F = m a with the decelleration (in this case) dependent on how long the collision took. The fact that momentum changes sometimes allows you to determine what the force must have been by simply looking at the total momentum of all the objects after a collision (because of the "change in momentum" definition of force).
QUOTE (NeutrinoQ+Dec 10 2006, 05:37 PM)
I'm pretty sure net force is not given by f=ma, but is given by f=(ma)t= mv, although I could be wrong
Example to illustrate
i have a 10 kg car accelerating from 0 to 15 in three seconds at constant acceleration
f=(10)(5)= 50N; which means (i think) for every second, the car has 50N more force
So if you put a wall at then end of the cars acceleration, the wall is going to feel 150N of force, not 50N, right?
If acceleration is constant, then the force generated is constant. It doesn't gain force for every unit of time. F=ma is the definition of force, and net force. Net force is merely the sum of all the forces acting on the body. Thus if one has a 4 newton force acting one way, and a 2 Newton force acting the opposite, the net force is 2 newtons. Either way its still F=ma. Momentum=mv. Just look at the units, they aren't the same.
Example to illustrate
i have a 10 kg car accelerating from 0 to 15 in three seconds at constant acceleration
f=(10)(5)= 50N; which means (i think) for every second, the car has 50N more force
So if you put a wall at then end of the cars acceleration, the wall is going to feel 150N of force, not 50N, right?
If acceleration is constant, then the force generated is constant. It doesn't gain force for every unit of time. F=ma is the definition of force, and net force. Net force is merely the sum of all the forces acting on the body. Thus if one has a 4 newton force acting one way, and a 2 Newton force acting the opposite, the net force is 2 newtons. Either way its still F=ma. Momentum=mv. Just look at the units, they aren't the same.
to oomchu:
did you even bother reading the two replies above yours?
to noclevername:
Thanks for the explanation; the word force brings up many preconceived notions, but when looking at force as the rate of change of momentum, everything makes sense.
did you even bother reading the two replies above yours?
to noclevername:
Thanks for the explanation; the word force brings up many preconceived notions, but when looking at force as the rate of change of momentum, everything makes sense.
Believe me, NQ, the pleasure is mine since getting a clear explanation down on paper forces one to clearly get it straight in my own mind, too. (I haven't had to look at these things for years --- make that decades 
)
)
Let's add a historical dimension to the discussion. What is the historical development of the concepts of energy and momentum?
Shortly after Newton, the development of physics took two branches. In England, a group of physicists following Newton became isolated from the continental European physicists (primarily in France at that time). The reasons for the isolation were political: England and France had been at war on and off for most of the past century, and to the French, nothing English was acceptable, and vice versa (exception made for Newton, or course).
At the time, physics was in an early state, and the primary tools were the concept of force and the laws of kinematics (based on the infinitesimal calculus) and the universal gravitational formula of Newton. In other words, physics had Newton's three laws and gravity, and little else. A burning question of the day concerned the everyday concept of "effort" and how to quantify it physically so that it could be worked with.
There were two schools of thought about effort. One group (primarily English) thought that the best way to quantify effort was as the product of force and time. This makes intuitive sense, because if someone were to you that he was tired because he pushed a heavy wheelbarrow for an hour, you would be enclined to agree that he was justified in being tired, because he had put out a lot of effort.
Another group (primarily French) though that it was better to quantify effort as theh product of force and distance. They also had intuitive good sense on their side, because if someone were to you that he was tired because he pushed a heavy wheelbarrow for a kilometer, you would again be enclined to agree that he was justified in being tired, because he had put out a lot of effort.
These two schools of thought evolved independently because of the relative lack of communication (and positive hostility) between England and France. When relations finally warmed up again and scientific contact became more frequent, they found themselves in conflict, each side claiming that their own way was the "right" way and that the other side had it wrong. It was eventually realized that both methods of converting the concept of effort into a technically useful form were correct but independent. In other words, that there was more than one technically precise notion hiding in the vague everyday concept of effort. This happens commonly in science -- precise definition points out distinctions that aret thought of in everyday conversation.
Now, what did each of these schools of thought develop by way of theory? In both cases, one fundamental question they wanted to solve was that of finding a formula for the total amout of effort that had been invested in moving an object, WITHOUT a detailed accounting of individual contributions. In other words, if several people had moved the wheelbarrow for different distances and times, how could you, just by looking at the final effect, deduce the total efford, without knowing who did how much?
The answer comes from the fundamental theorem of calculus: the total effort must be the integral of the infinitesimal contributions that happen moment by moment, and so therefore, the rate of effort must be the derivative of the total. Alternatively (for those whose calculus is rusty or nonexistent), one can think of the central insight in terms of a bank account. Suppose you have a savings account to which you make deposits but never withdrawals. Then for each deposit D the balance B goes up, and the amount it goes up is just the size of the deposit. In other words, D = change in B. On the other hand, the balance is also the sum total of all the deposits, so B = sum of D.
Therefore, to try to find a formula for the total by adding up the efforts, they could try to find it by writing down a formula whose change was equal to one single effort contribution. This is not as tricky as it sounds, because the laws of kinematics can be used. Specifically, it was known that change of velocity = acceleration * time interval, so change(v) = a*t, and also it was known that the change of v^2 = 2*a*change(x), where x is distance.
What does this give to the English physicists? Their concept of effort was named impulse, and is usually denoted by J, so J = F*t. Since F=ma, it must be true that J = mat, and since at = change(v), it follows that J = m*change(v). Since the mass of the pushed object is constant (forget about rockets for now), m*chang(v) = m*(v_final - v_initial) = m*v_final - m*v_initial = change(mv). Since J = change(mv) it follows that mv must be the sum of the J's. Since it is the formula that they were trying to derive, it deserved a name, and it was called "momentum" and designated by P, so they proved in short that P = mv.
What did the French physicists get? Their concept of effort was called work and is usually denoted by W, so W = F*d, where d is the distance the object was pushed, and where we further must stipulate that the force is parallel to d. This is because pushing sideways on a moving object does not speed it up or slow it down, so it is wasted effort and should not be counted in the total. The effective part of the force is only the part alighned with the motion. Since F=ma, it must be true that W = mad = ma*change(x), and since 2a*change(x) = change(v^2), it follows that W = m/2*change(v^2) = m/2*(v_final^2 - v_initial^2) = m/2*v_final^2 - m/2*v_initial^2 = change(m/2*v^2). Since W = change(1/2*mv^2), it follows that 1/2*mv^2 must be the sum of the W's. Since it is the formula that they wer trying to derive, it also deserved a name, and it was called "energy" and was designated by E, so they proved that E = 1/2*mv^2.
This is the origin of both the momentum and energy formulas. They weren't just made up and then studied, but they were the outcome of a process of investigation. Of course nowadays we know that energy and momentum are components of one single thing, the relativistic energy momentum 4-vector, so the old dispute over who's really right is resolved once and for all.
Hope this was interesting or amusing!
--Stuart Anderson
Shortly after Newton, the development of physics took two branches. In England, a group of physicists following Newton became isolated from the continental European physicists (primarily in France at that time). The reasons for the isolation were political: England and France had been at war on and off for most of the past century, and to the French, nothing English was acceptable, and vice versa (exception made for Newton, or course).
At the time, physics was in an early state, and the primary tools were the concept of force and the laws of kinematics (based on the infinitesimal calculus) and the universal gravitational formula of Newton. In other words, physics had Newton's three laws and gravity, and little else. A burning question of the day concerned the everyday concept of "effort" and how to quantify it physically so that it could be worked with.
There were two schools of thought about effort. One group (primarily English) thought that the best way to quantify effort was as the product of force and time. This makes intuitive sense, because if someone were to you that he was tired because he pushed a heavy wheelbarrow for an hour, you would be enclined to agree that he was justified in being tired, because he had put out a lot of effort.
Another group (primarily French) though that it was better to quantify effort as theh product of force and distance. They also had intuitive good sense on their side, because if someone were to you that he was tired because he pushed a heavy wheelbarrow for a kilometer, you would again be enclined to agree that he was justified in being tired, because he had put out a lot of effort.
These two schools of thought evolved independently because of the relative lack of communication (and positive hostility) between England and France. When relations finally warmed up again and scientific contact became more frequent, they found themselves in conflict, each side claiming that their own way was the "right" way and that the other side had it wrong. It was eventually realized that both methods of converting the concept of effort into a technically useful form were correct but independent. In other words, that there was more than one technically precise notion hiding in the vague everyday concept of effort. This happens commonly in science -- precise definition points out distinctions that aret thought of in everyday conversation.
Now, what did each of these schools of thought develop by way of theory? In both cases, one fundamental question they wanted to solve was that of finding a formula for the total amout of effort that had been invested in moving an object, WITHOUT a detailed accounting of individual contributions. In other words, if several people had moved the wheelbarrow for different distances and times, how could you, just by looking at the final effect, deduce the total efford, without knowing who did how much?
The answer comes from the fundamental theorem of calculus: the total effort must be the integral of the infinitesimal contributions that happen moment by moment, and so therefore, the rate of effort must be the derivative of the total. Alternatively (for those whose calculus is rusty or nonexistent), one can think of the central insight in terms of a bank account. Suppose you have a savings account to which you make deposits but never withdrawals. Then for each deposit D the balance B goes up, and the amount it goes up is just the size of the deposit. In other words, D = change in B. On the other hand, the balance is also the sum total of all the deposits, so B = sum of D.
Therefore, to try to find a formula for the total by adding up the efforts, they could try to find it by writing down a formula whose change was equal to one single effort contribution. This is not as tricky as it sounds, because the laws of kinematics can be used. Specifically, it was known that change of velocity = acceleration * time interval, so change(v) = a*t, and also it was known that the change of v^2 = 2*a*change(x), where x is distance.
What does this give to the English physicists? Their concept of effort was named impulse, and is usually denoted by J, so J = F*t. Since F=ma, it must be true that J = mat, and since at = change(v), it follows that J = m*change(v). Since the mass of the pushed object is constant (forget about rockets for now), m*chang(v) = m*(v_final - v_initial) = m*v_final - m*v_initial = change(mv). Since J = change(mv) it follows that mv must be the sum of the J's. Since it is the formula that they were trying to derive, it deserved a name, and it was called "momentum" and designated by P, so they proved in short that P = mv.
What did the French physicists get? Their concept of effort was called work and is usually denoted by W, so W = F*d, where d is the distance the object was pushed, and where we further must stipulate that the force is parallel to d. This is because pushing sideways on a moving object does not speed it up or slow it down, so it is wasted effort and should not be counted in the total. The effective part of the force is only the part alighned with the motion. Since F=ma, it must be true that W = mad = ma*change(x), and since 2a*change(x) = change(v^2), it follows that W = m/2*change(v^2) = m/2*(v_final^2 - v_initial^2) = m/2*v_final^2 - m/2*v_initial^2 = change(m/2*v^2). Since W = change(1/2*mv^2), it follows that 1/2*mv^2 must be the sum of the W's. Since it is the formula that they wer trying to derive, it also deserved a name, and it was called "energy" and was designated by E, so they proved that E = 1/2*mv^2.
This is the origin of both the momentum and energy formulas. They weren't just made up and then studied, but they were the outcome of a process of investigation. Of course nowadays we know that energy and momentum are components of one single thing, the relativistic energy momentum 4-vector, so the old dispute over who's really right is resolved once and for all.
Hope this was interesting or amusing!
--Stuart Anderson
Thank you very much stuart! That was really great!
Sorry to throw a wrench into your final conclusions but it has not been “resolved once and for all”. Very simple experiments prove that the Law of Conservation of Energy is false, or they negate Newtonian Physics. One or the other formulas has to be false.
A cylinder that is spun on its central axis can be stopped by tethered masses first held on the cylinders surface (at 180°) and then the masses are allowed to swing out. As the tethered masses swing out the cylinder will stop. When the tethered masses have all the motion they can conserve only one of the two quantities of motion, kinetic energy (1/2mv²), or Newtonian momentum (mv). The reverse of this process is the ballistics pendulum; it conserves Newtonian momentum.
Such experiments can be assembled for less than $25, and require only slightly sophisticated video capturing and playback equipment.
A cylinder that is spun on its central axis can be stopped by tethered masses first held on the cylinders surface (at 180°) and then the masses are allowed to swing out. As the tethered masses swing out the cylinder will stop. When the tethered masses have all the motion they can conserve only one of the two quantities of motion, kinetic energy (1/2mv²), or Newtonian momentum (mv). The reverse of this process is the ballistics pendulum; it conserves Newtonian momentum.
Such experiments can be assembled for less than $25, and require only slightly sophisticated video capturing and playback equipment.
^ Guess who doesn't know anything about angular momentum or rotational velocity, doesn't understand there's other forms of energy and wouldn't know Newtonian mechanics even if it gave him kicked him in the nuts...
Still haven't asked your professor friend about Lagrangian mechanics
Happy in your ignorance.
Still haven't asked your professor friend about Lagrangian mechanics
Happy in your ignorance.
NeutrinoQ: Don't mind Limon; he's pretty smart but a little misguided. I think he's into perpetual motion, you know. Now Mr_Homm and Alpha, on the other hand, you can trust them.
Sorry your thread has attracted some flies.
And to you Mr_H: I hope you're writing a book.
Sorry your thread has attracted some flies.
And to you Mr_H: I hope you're writing a book.
Oh, and thanks for the feedback, NQ. 
Limon..... I don't even know what to say to you... ::shakes head::
I guess maybe its just best to say nothing... crying babies just want attention.
I wonder what would happen if we put limon and zephir along with the other nutcases in the same room together.
Now thats what I call a singularity.
I guess maybe its just best to say nothing... crying babies just want attention.
I wonder what would happen if we put limon and zephir along with the other nutcases in the same room together.
Now thats what I call a singularity.
QUOTE (NeutrinoQ+Dec 11 2006, 07:53 PM)
to oomchu:
did you even bother reading the two replies above yours?
yes, and?
did you even bother reading the two replies above yours?
yes, and?
NeutrinoQ; You speak of increments of force. Newton’s increments are a function of time.
If you drop a one kilogram object one meter you get (d=1/2v²/a), √2*1*9.81= 4.43 units of momentum. Time equals .4515 sec., this is 9.81 newtons working for .4515 sec. = 4.43
Place a 9 kilogram block on a frictionless plane, connect a string, position the sting over a frictionless pulley, connect a one kilogram mass to the other end, let the mass drop one meter and you will get 14.007 units of momentum. Time equals 1.4278 sec., this is 9.81 newtons working for 1.4278 sec. = 14.007
If you can give 4.43 units of momentum to the lowered one kilogram object the object will return to the top and start the system over again, which gives you 9.587 units extra. Some call it perpetual motion, I call it Newtonian Physics.
What law of physics prevents 14.007 units of momentum from being used to produce 4.43. I know of a law of physics that says this is absolutely possible, in fact it is required that it is doable. It is the Law of Conservation of Momentum and all of Newtonian Physics (concerning motion) hinges upon it being achievable. Anyone who denies that it is possible is taking the continental view (according to Homm) of physics, that kinetic energy (or any assortment of forms of energy) will take precedence over momentum conservation.
The cylinder and spheres gives all the motion of four units of mass to one unit of mass. Momentum is conserved or energy is conserved; one or the other is false.
QUOTE (Limon+Dec 12 2006, 11:16 PM)
The cylinder and spheres gives all the motion of four units of mass to one unit of mass. Momentum is conserved or energy is conserved; one or the other is false.
Maybe one day you'll learn how to work out the dynamics and energy of rotating systems.
Maybe one day you'll learn how to work out the dynamics and energy of rotating systems.
Too bad "momentum" isn't work or energy, Limon.
Let's take your 10 kg worth of blocks. After the 1 m drop they are all moving at 1.4 m/s. Therefore the total kinetic energy is 1/2 m v^2 or (1/2)(10)(1.4^2) = 9.8 joules. That number is, surprise, surprise, exactly the same as the potential energy of a 1 kg block raised 1 m ( P.E. = m g h - (1)(9.8)(1) =9.8 joules).
Now the 9 kg block after it gets accelerated to 1.4 m/s will have 8.8 joules of kinetic energy. Lifting the 1 kg block again to its starting position will require W = F d of work = m a d = (1)(9.8)(1) = 9.8 joules.
So it's going to take 9.8 joules of work to add each 8.8 joules to the 9 kg block. Not exactly perpetual motion, is it?
Momentum, of course, is a property more associated with "constant velocity" not work, since work requires force (which wil change momentum).
But I know you've always been trying to squeeze work out of momentum because force is a change in momentum and of course force over distance is work. So your quest for free momentum goes on. To bad you need work to acquire momentum in the first place (like the 9.8 joules you need to "give" the 4.43 units of momentum to the block to get it back up a meter), so you've got to look at the energy balance, after all.
Let's take your 10 kg worth of blocks. After the 1 m drop they are all moving at 1.4 m/s. Therefore the total kinetic energy is 1/2 m v^2 or (1/2)(10)(1.4^2) = 9.8 joules. That number is, surprise, surprise, exactly the same as the potential energy of a 1 kg block raised 1 m ( P.E. = m g h - (1)(9.8)(1) =9.8 joules).
Now the 9 kg block after it gets accelerated to 1.4 m/s will have 8.8 joules of kinetic energy. Lifting the 1 kg block again to its starting position will require W = F d of work = m a d = (1)(9.8)(1) = 9.8 joules.
So it's going to take 9.8 joules of work to add each 8.8 joules to the 9 kg block. Not exactly perpetual motion, is it?
Momentum, of course, is a property more associated with "constant velocity" not work, since work requires force (which wil change momentum).
But I know you've always been trying to squeeze work out of momentum because force is a change in momentum and of course force over distance is work. So your quest for free momentum goes on. To bad you need work to acquire momentum in the first place (like the 9.8 joules you need to "give" the 4.43 units of momentum to the block to get it back up a meter), so you've got to look at the energy balance, after all.
So you are saying that kinetic energy conservation overrides momentum conservation. I am aware that the kinetic energy of the two systems (14.007 and 4.43) is equal, but since when is kinetic energy conserved, there is no law of conservation of kinetic energy and now you are acting as if there is. You take the French view, fine. Now prove it.
The spheres in the cylinder and spheres experiment can have one and only one final velocity; surely you would like to prove your theory. We should be able to tell the difference between 2 m/sec and 4 m/sec.
The spheres in the cylinder and spheres experiment can have one and only one final velocity; surely you would like to prove your theory. We should be able to tell the difference between 2 m/sec and 4 m/sec.
QUOTE (Limon+Dec 14 2006, 03:35 AM)
So you are saying that kinetic energy conservation overrides momentum conservation. I am aware that the kinetic energy of the two systems (14.007 and 4.43) is equal, but since when is kinetic energy conserved, there is no law of conservation of kinetic energy and now you are acting as if there is. You take the French view, fine. Now prove it.
Nobody ever said there was a law of conservation of kinetic energy, Limon --- but there is a law of conservation of total energy. All I showed was that the P.E. from the raised block was entirely converted to K.E. after it dropped so that the total energy of 9.8 J was conserved. (And I showed it with your numbers, so don't complain).
And no one said anything about one law "overriding" another. I don't see any problem with the momentum of these systems. Forces were applied and momentums created (d mv = F dt). Conservation of momentum isn't even an issue here --- unless you somehow feel the momentums should be equal in some way because the 1 kg block was subject to the acceleration "g" in both runs. Too bad it wasn't because you forgot about the vertical force from the tension on the cord that reduced the acceleration via mg-T.
I suppose you'll be bringing up your silly "line of force" with the 10 1 kg masses pretty soon ...
... but let's not pollute this poor guy's thread with that nonsense, shall we?
Unless is this some misguided attempt to re-open the 300 year-old controversy noted by Mr_H as I see you computed "momentum" as "impulse" with your "force x time". You have some problem with Work = F x D, then? Can't handle that linkage between momentum and energy: maybe because you don't like to include potential energy like it's some sort of second-class energy?
Nobody ever said there was a law of conservation of kinetic energy, Limon --- but there is a law of conservation of total energy. All I showed was that the P.E. from the raised block was entirely converted to K.E. after it dropped so that the total energy of 9.8 J was conserved. (And I showed it with your numbers, so don't complain).
And no one said anything about one law "overriding" another. I don't see any problem with the momentum of these systems. Forces were applied and momentums created (d mv = F dt). Conservation of momentum isn't even an issue here --- unless you somehow feel the momentums should be equal in some way because the 1 kg block was subject to the acceleration "g" in both runs. Too bad it wasn't because you forgot about the vertical force from the tension on the cord that reduced the acceleration via mg-T.
I suppose you'll be bringing up your silly "line of force" with the 10 1 kg masses pretty soon ...
... but let's not pollute this poor guy's thread with that nonsense, shall we?
Unless is this some misguided attempt to re-open the 300 year-old controversy noted by Mr_H as I see you computed "momentum" as "impulse" with your "force x time". You have some problem with Work = F x D, then? Can't handle that linkage between momentum and energy: maybe because you don't like to include potential energy like it's some sort of second-class energy?
If the 1kg block has a momentum of 1.4Ns Down, an instantaneous impulse of 4.43Ns Up will result in a final momentum of 3.03Ns Up, which is not enough to return the 1kg block to its' starting position.
Or did I miss something?
Or did I miss something?
QUOTE (Benny+Dec 14 2006, 09:48 AM)
If the 1kg block has a momentum of 1.4Ns Down, an instantaneous impulse of 4.43Ns Up will result in a final momentum of 3.03Ns Up, which is not enough to return the 1kg block to its' starting position.
Or did I miss something?
No, you didn't.
Limon is envisioning a "perpetual motion machine" whereby the 1 kg block gets the 9 kg blok moving, then the 1 kg block probably comes up against a table or something to stop its motion. Then he wants to "give" the 1kg block a shot of 4.43 momentum to get it up to a level where he can give the 9 kg block another boost.
He figures he is getting the 10 or so units of 9 kg momentum for "free" or at least with the "expenditure" of only 4.43 units of 1 kg block momentum per cycle.
Or did I miss something?
No, you didn't.
Limon is envisioning a "perpetual motion machine" whereby the 1 kg block gets the 9 kg blok moving, then the 1 kg block probably comes up against a table or something to stop its motion. Then he wants to "give" the 1kg block a shot of 4.43 momentum to get it up to a level where he can give the 9 kg block another boost.
He figures he is getting the 10 or so units of 9 kg momentum for "free" or at least with the "expenditure" of only 4.43 units of 1 kg block momentum per cycle.
That is the idea; the 9 kg block (on the frictionless plane) moving 1.4 m/sec has a momentum of 12.6 units. It needs to transfer only 3.03 units to the dropped 1 kg (moving 1.4 m/sec) in order to return the one kilogram object back to the top. The motion of the one kilogram going down can be conserved by catching it in a pendulum arraignment. It will sweep through the pendulum motion and soon be moving up with the same velocity that is was moving down and at the same height, then add the 3.03 units. The Law of Conservation of Momentum seems to predict that this addition is achievable.
Let’s place the same amount of momentum in a different object. Lets place ten spheres spinning in the same circle equally spaced (36°) from each other. Let’s give them a velocity of 1.4 m/sec around the arc of the circle. If just one of the spheres can flare out on the end of a tether and absorb only 3.03 units of momentum from the other nine, that lone sphere would have enough momentum to cause all of the starting motion of all ten rotating spheres. That would leave the other nine with 9.58 units of momentum.
I have done experiments like this using two spheres and not just a portion of the motion is given to the tethered spheres but all the motion is given to the tethered spheres.
To eliminate the need for a center bearing I used two spheres mounted at 180°. Each tethered sphere would absorb half the momentum as they flare out from the cylinder. In this situation that would be 7.00 units of momentum each, only one of the two spheres needed only 4.43 units of momentum to start the system spinning at the original rate of 1.4 m/sec. 7.00 is greater than 4.43 and you have two spheres at 7.00 not just one.
One sphere moving 4.43 m/sec will rise 1 m and it can restart the system.
Two spheres moving 7.00 m/s will rise 2.50 m (each) and will have 5 times as much energy as is necessary to restart the system.
The only thing that would prevent this is that the Law of Conservation of Momentum does not apply in this situation. And if it does not apply wouldn’t that mean that Newtonian Physics is false? After all; it is a universal law not a pick and choose law. Momentum conservation applies to many similar situations; such as the ballistic pendulum, air table experiments, etc.
If the cylinder and spheres experiment conserves kinetic energy instead of momentum the final velocity of the two spheres would be only 3.13 m/sec (when the cylinder is stopped) instead of 7.00 m/sec.
And the reverse process of a ballistic pendulum with a 2 kilogram mass moving at 3.13 m/sec striking an 8 kilogram mass would not give you 14.007 units of momentum, but only 6.26.
Here is a curious thought. If the cylinder had a center bearing and could give all the motion to one kg it would rise 10 meters. I have not done this experiment; and I think that air resistance would be a factor.
Let’s place the same amount of momentum in a different object. Lets place ten spheres spinning in the same circle equally spaced (36°) from each other. Let’s give them a velocity of 1.4 m/sec around the arc of the circle. If just one of the spheres can flare out on the end of a tether and absorb only 3.03 units of momentum from the other nine, that lone sphere would have enough momentum to cause all of the starting motion of all ten rotating spheres. That would leave the other nine with 9.58 units of momentum.
I have done experiments like this using two spheres and not just a portion of the motion is given to the tethered spheres but all the motion is given to the tethered spheres.
To eliminate the need for a center bearing I used two spheres mounted at 180°. Each tethered sphere would absorb half the momentum as they flare out from the cylinder. In this situation that would be 7.00 units of momentum each, only one of the two spheres needed only 4.43 units of momentum to start the system spinning at the original rate of 1.4 m/sec. 7.00 is greater than 4.43 and you have two spheres at 7.00 not just one.
One sphere moving 4.43 m/sec will rise 1 m and it can restart the system.
Two spheres moving 7.00 m/s will rise 2.50 m (each) and will have 5 times as much energy as is necessary to restart the system.
The only thing that would prevent this is that the Law of Conservation of Momentum does not apply in this situation. And if it does not apply wouldn’t that mean that Newtonian Physics is false? After all; it is a universal law not a pick and choose law. Momentum conservation applies to many similar situations; such as the ballistic pendulum, air table experiments, etc.
If the cylinder and spheres experiment conserves kinetic energy instead of momentum the final velocity of the two spheres would be only 3.13 m/sec (when the cylinder is stopped) instead of 7.00 m/sec.
And the reverse process of a ballistic pendulum with a 2 kilogram mass moving at 3.13 m/sec striking an 8 kilogram mass would not give you 14.007 units of momentum, but only 6.26.
Here is a curious thought. If the cylinder had a center bearing and could give all the motion to one kg it would rise 10 meters. I have not done this experiment; and I think that air resistance would be a factor.
Still haven't worked out the difference between angular and linear systems yet then Limon
Suppose you have some model rocket engines, that have a thrust of 10 newtons and a firing duration of 1 sec. If you mount it and fire it on a one kilogram frictionless sled, you will have 10 units of velocity and 10 units of momentum, and (1/2mv²) 50 units of kinetic energy.
If you mount the same model rocket engine and fire it on a 5 kilogram frictionless sled, you will have 2 units of velocity, 10 units of momentum, and (1/2mv²) 10 units of kinetic energy.
If you mount the same model rocket engine and fire it on a 10 kilogram frictionless sled, you will have 1 units of velocity, 10 units of momentum, and (1/2mv²) 5 units of kinetic energy.
So the certain mass of chemical energy in the rocket is equal to 10, and 10, and 10 units of momentum. This is a one to one relationship between energy and momentum.
So the certain mass of chemical energy in the rocket is equal to 50, and 10, and 5 units of kinetic energy. There is not a consistent relationship between energy and kinetic energy.
What is the true energy of the systems.
If you mount the same model rocket engine and fire it on a 5 kilogram frictionless sled, you will have 2 units of velocity, 10 units of momentum, and (1/2mv²) 10 units of kinetic energy.
If you mount the same model rocket engine and fire it on a 10 kilogram frictionless sled, you will have 1 units of velocity, 10 units of momentum, and (1/2mv²) 5 units of kinetic energy.
So the certain mass of chemical energy in the rocket is equal to 10, and 10, and 10 units of momentum. This is a one to one relationship between energy and momentum.
So the certain mass of chemical energy in the rocket is equal to 50, and 10, and 5 units of kinetic energy. There is not a consistent relationship between energy and kinetic energy.
What is the true energy of the systems.
QUOTE (Limon+Dec 17 2006, 03:02 AM)
Suppose you have some model rocket engines, that have a thrust of 10 newtons and a firing duration of 1 sec. If you mount it and fire it on a one kilogram frictionless sled, you will have 10 units of velocity and 10 units of momentum, and (1/2mv²) 50 units of kinetic energy.
If you mount the same model rocket engine and fire it on a 5 kilogram frictionless sled, you will have 2 units of velocity, 10 units of momentum, and (1/2mv²) 10 units of kinetic energy.
If you mount the same model rocket engine and fire it on a 10 kilogram frictionless sled, you will have 1 units of velocity, 10 units of momentum, and (1/2mv²) 5 units of kinetic energy.
So the certain mass of chemical energy in the rocket is equal to 10, and 10, and 10 units of momentum. This is a one to one relationship between energy and momentum.
So the certain mass of chemical energy in the rocket is equal to 50, and 10, and 5 units of kinetic energy. There is not a consistent relationship between energy and kinetic energy.
What is the true energy of the systems.
Why this is so:
F = m(d²x)/(dt²)
F = 10 N
Assuming v starts at zero, v after t = 1 s is v=at and x = 1/2 at²
If m = 1 kg, a = 10 N / 1 kg = 10 m/s² so v = 10 m/s so KE = 50 J but x = 5 m
If m = 5 kg, a = 10 N / 5 kg = 2 m/s² so v = 2 m/s so KE = 10 J but x = 1 m
If m = 10 kg, a = 10 N / 10 kg = 1 m/s² so v = 1 m/s so KE = 5 J but x = 0.5 m
Force is the change in momentum per unit time : F = d(mv)/dt
But it the change in Kinetic energy is called Work, and it's formula is: F dot x
If m = 1 kg, x = 5 m so KE = 0 + Work = 10 N * x = 50 J
If m = 5 kg, x = 1 m so KE = 0 + Work = 10 N * x = 10 J
If m = 10 kg, x = 0.5 m so KE = 0 + Work = 10 N * x = 5 J
We can even solve for the time it would take each mass to reach a KE of 50 J.
x = 50 J / 10 N = 5 meters in all cases.
For m = 1 kg, we have already done this.
For m = 5 kg, a = 2 m/s², so t = sqrt(2x/a) = sqrt(5) seconds, so v = 2sqrt(5) m/s so KE = 1/2 (5kg) (2sqrt(5) m/s)² = 50 J
For m = 10 kg, a = 1 m/s², so t = sqrt(2x/a) = sqrt(10) seconds, so v = sqrt(10) m/s so KE = 1/2 (10kg) (sqrt(10) m/s)² = 50 J
It's obvious that units of F * t = ( kg m / s² ) * (s) = kg m / s which are units of momentum, but F dot x = (kg m / s²) * (m) = kg m² / s² = J which are units of energy. It's also obvious that not all forces are created equal. If you mount TWO of these rockets, but on opposite sides, and fire them at the same time, the sled doesn't move at all. If you fire them one after the other the sled moves, but then stops. The concept of the dot product solves this situation. The generalized dot product is the tensor product which is used at university level to do mechanics and electromagnetism and particle physics and General Relativity...
If you mount the same model rocket engine and fire it on a 5 kilogram frictionless sled, you will have 2 units of velocity, 10 units of momentum, and (1/2mv²) 10 units of kinetic energy.
If you mount the same model rocket engine and fire it on a 10 kilogram frictionless sled, you will have 1 units of velocity, 10 units of momentum, and (1/2mv²) 5 units of kinetic energy.
So the certain mass of chemical energy in the rocket is equal to 10, and 10, and 10 units of momentum. This is a one to one relationship between energy and momentum.
So the certain mass of chemical energy in the rocket is equal to 50, and 10, and 5 units of kinetic energy. There is not a consistent relationship between energy and kinetic energy.
What is the true energy of the systems.
Why this is so:
F = m(d²x)/(dt²)
F = 10 N
Assuming v starts at zero, v after t = 1 s is v=at and x = 1/2 at²
If m = 1 kg, a = 10 N / 1 kg = 10 m/s² so v = 10 m/s so KE = 50 J but x = 5 m
If m = 5 kg, a = 10 N / 5 kg = 2 m/s² so v = 2 m/s so KE = 10 J but x = 1 m
If m = 10 kg, a = 10 N / 10 kg = 1 m/s² so v = 1 m/s so KE = 5 J but x = 0.5 m
Force is the change in momentum per unit time : F = d(mv)/dt
But it the change in Kinetic energy is called Work, and it's formula is: F dot x
If m = 1 kg, x = 5 m so KE = 0 + Work = 10 N * x = 50 J
If m = 5 kg, x = 1 m so KE = 0 + Work = 10 N * x = 10 J
If m = 10 kg, x = 0.5 m so KE = 0 + Work = 10 N * x = 5 J
We can even solve for the time it would take each mass to reach a KE of 50 J.
x = 50 J / 10 N = 5 meters in all cases.
For m = 1 kg, we have already done this.
For m = 5 kg, a = 2 m/s², so t = sqrt(2x/a) = sqrt(5) seconds, so v = 2sqrt(5) m/s so KE = 1/2 (5kg) (2sqrt(5) m/s)² = 50 J
For m = 10 kg, a = 1 m/s², so t = sqrt(2x/a) = sqrt(10) seconds, so v = sqrt(10) m/s so KE = 1/2 (10kg) (sqrt(10) m/s)² = 50 J
It's obvious that units of F * t = ( kg m / s² ) * (s) = kg m / s which are units of momentum, but F dot x = (kg m / s²) * (m) = kg m² / s² = J which are units of energy. It's also obvious that not all forces are created equal. If you mount TWO of these rockets, but on opposite sides, and fire them at the same time, the sled doesn't move at all. If you fire them one after the other the sled moves, but then stops. The concept of the dot product solves this situation. The generalized dot product is the tensor product which is used at university level to do mechanics and electromagnetism and particle physics and General Relativity...
Thanks for taking the first whack at Limon, rpenner; let me smash him up a bit more!
By the way, I'm not going to dredge up all the math ...
Anyway, Limon's problem has always been between conservation of momentum and conservation of energy. In this problem, he is posing the "dilemma" that the chemistry in the rocket has inherent "energy" that should be totally transformed into kinetic energy of motion, regardless the mass of the object being moved. Thus, he wonders why momentum is conserved, but seemingly "total energy" isn't conserved.
First of all, and here is where I'm not about the bring in the math, your problem formuation is fundementally flawed: you have postulated a "magic" rocket engine that works without reaction mass. In a real engine, there's propellant mass shot out the back. Not only does this have momentum, it also had mass that had to be moved via F = m a in the first place.
This, when the engine was first fired, the sled had mass = (sled + propellant) and at the end of the shot had just mass = (sled + no propellant). So you can't just come up with your final velocity as if the force was external to the system as in the case of someone just giving the sled a push.
This also brings to light the fact that the chemical energy was used not only to push the sled forward but also push the propellant mass back. I suspect if you bothered to get down and dirty calculating where all the K.E. was going, you would find it all adds up. It'll help in those calculations to remember that momentum is conserved by the sled and the exhaust, although the decreasing sled mass will thrill you. There is this "rocket equation" which should prove useful.
rpenner has already aptly described why you need proportionally more energy to go faster the faster you go, but I suspect the fact that more "work" was required per increment of velocity was not at the heart of your problem. There is obviously a major difference between an external force that is able to "maintain the necessary force" by using whatever energy it needs (the pushing hand) versus a strapped-on engine that has limited energy resources. You can't run the math as if they were equals. Or pretend you have a "magic" engine. Even in the "pushing hand" case, you could figure it out by giving the hand a "limited energy budget". If you did that, those final velocities wouldn't be so high, would they? You know, 9.8 newtons with a limited budget of 100 joules or so. Try that on for size.
Here's the thing, Limon; I knew what was wrong right away, but I haven't done anything like this in decades, so I had to look up a couple of things. Now why should I do any more work for you? It's all there in textbooks and online. I don't feel like I have to beat myself up regainng long lost skills just to prove you wrong, as you can more easily do that yourself!
By the way, I'm not going to dredge up all the math ...
Anyway, Limon's problem has always been between conservation of momentum and conservation of energy. In this problem, he is posing the "dilemma" that the chemistry in the rocket has inherent "energy" that should be totally transformed into kinetic energy of motion, regardless the mass of the object being moved. Thus, he wonders why momentum is conserved, but seemingly "total energy" isn't conserved.
First of all, and here is where I'm not about the bring in the math, your problem formuation is fundementally flawed: you have postulated a "magic" rocket engine that works without reaction mass. In a real engine, there's propellant mass shot out the back. Not only does this have momentum, it also had mass that had to be moved via F = m a in the first place.
This, when the engine was first fired, the sled had mass = (sled + propellant) and at the end of the shot had just mass = (sled + no propellant). So you can't just come up with your final velocity as if the force was external to the system as in the case of someone just giving the sled a push.
This also brings to light the fact that the chemical energy was used not only to push the sled forward but also push the propellant mass back. I suspect if you bothered to get down and dirty calculating where all the K.E. was going, you would find it all adds up. It'll help in those calculations to remember that momentum is conserved by the sled and the exhaust, although the decreasing sled mass will thrill you. There is this "rocket equation" which should prove useful.
rpenner has already aptly described why you need proportionally more energy to go faster the faster you go, but I suspect the fact that more "work" was required per increment of velocity was not at the heart of your problem. There is obviously a major difference between an external force that is able to "maintain the necessary force" by using whatever energy it needs (the pushing hand) versus a strapped-on engine that has limited energy resources. You can't run the math as if they were equals. Or pretend you have a "magic" engine. Even in the "pushing hand" case, you could figure it out by giving the hand a "limited energy budget". If you did that, those final velocities wouldn't be so high, would they? You know, 9.8 newtons with a limited budget of 100 joules or so. Try that on for size.
Here's the thing, Limon; I knew what was wrong right away, but I haven't done anything like this in decades, so I had to look up a couple of things. Now why should I do any more work for you? It's all there in textbooks and online. I don't feel like I have to beat myself up regainng long lost skills just to prove you wrong, as you can more easily do that yourself!
Addendum
Well, I did break down and do some math. The equation for the propellant mass m_p needed to be exhausted to power payload m is:
m_p = m*(e^(v/v_e) - 1)
where v is the velocity we want and v_e is the engine exhaust velocity.
For the 1 kg payload to achieve 10 m/s using an engine with v_e of 2500 m/s we need to expel 0.004 kg of propellant exhausted. Turning that around, 0.004 kg mass at 2500 m/s has momentum (surprise, surprise) of -10! And the thrust is 10 N.
For the 5 kg block the propellant exhausted is also 0.004 kg.
The K.E. for a 2,500 m/s stream of 0.004 kg is 12,500 joules.
So, OK, the sled isn't going to be significantly fattened up with the addition of the rocket fuel. Obviously for bigger "sleds" (like the space shuttle) fuel will be more significant.
What we can see, however, is that the engine has a "choice" of what to work on: its own exhaust gas or the sled. Most of the work appears to go to the gas. Because the gas is so light, it takes a lot of v^2 and thus a lot of work, to generate enough backwards momentum to get the same forwards momentum out of the sled.
Just doing gross calculations, if the 0.004 kg was exhausted instantaneously then its backward velocity would be -2490 m/s (as the sled moved forward at 10 m/s) and it would have a K.E. of 12,400 J. So we add in the 50 J for the moving sled and I've got pretty close to the 12,500 right there.
You can take it from here, Limon.
Well, I did break down and do some math. The equation for the propellant mass m_p needed to be exhausted to power payload m is:
m_p = m*(e^(v/v_e) - 1)
where v is the velocity we want and v_e is the engine exhaust velocity.
For the 1 kg payload to achieve 10 m/s using an engine with v_e of 2500 m/s we need to expel 0.004 kg of propellant exhausted. Turning that around, 0.004 kg mass at 2500 m/s has momentum (surprise, surprise) of -10! And the thrust is 10 N.
For the 5 kg block the propellant exhausted is also 0.004 kg.
The K.E. for a 2,500 m/s stream of 0.004 kg is 12,500 joules.
So, OK, the sled isn't going to be significantly fattened up with the addition of the rocket fuel. Obviously for bigger "sleds" (like the space shuttle) fuel will be more significant.
What we can see, however, is that the engine has a "choice" of what to work on: its own exhaust gas or the sled. Most of the work appears to go to the gas. Because the gas is so light, it takes a lot of v^2 and thus a lot of work, to generate enough backwards momentum to get the same forwards momentum out of the sled.
Just doing gross calculations, if the 0.004 kg was exhausted instantaneously then its backward velocity would be -2490 m/s (as the sled moved forward at 10 m/s) and it would have a K.E. of 12,400 J. So we add in the 50 J for the moving sled and I've got pretty close to the 12,500 right there.
You can take it from here, Limon.
NoCleverName quote: What we can see, however, is that the engine has a "choice" of what to work on: its own exhaust gas or the sled. Most of the work appears to go to the gas. Because the gas is so light, it takes a lot of v^2 and thus a lot of work, to generate enough backwards momentum to get the same forwards momentum out of the sled.
The exhaust velocity in all situations would be equal. Whatever the physics of the rocket is, it will be equal in all situations.
The engine has no ability to make choices. We can choose to see the truth or we can ignore it. The Law of Conservation of Energy is false.
The exhaust velocity in all situations would be equal. Whatever the physics of the rocket is, it will be equal in all situations.
The engine has no ability to make choices. We can choose to see the truth or we can ignore it. The Law of Conservation of Energy is false.
QUOTE (Limon+Dec 18 2006, 12:01 PM)
The exhaust velocity in all situations would be equal. Whatever the physics of the rocket is, it will be equal in all situations.
The engine has no ability to make choices. We can choose to see the truth or we can ignore it. The Law of Conservation of Energy is false.
The "choice", in case you haven't figured it out yet Limon, is actually the different velocities directed by the conservation of momentum (which you so dearly love) setup by the equal and opposite reaction to the pressure inside the engine chamber being released at the nozzle. The energy is directed to "support" the velocities.
Just like your silly swinging 1 kg mass "going back up to the level it was" because it has the same speed going up as down --- sure does, but it isn't going to get very far moving up at 1.4 m/s! And the 9 kg block "giving" momentum to raise the lower block: I'm sure you've already calculated that even if the block gave up all its momentum in an effort to raise the block it would come up short.
I suppose it never occured to you that if energy conservation is false, then momentum conservation is false, too, since the former is a mathematical consequence of the latter.
In every one of your experiments you find "energy is not conserved". But that's because in every one of your experiments you manage to let momentum slip through your fingers.
But enough of your pithy comments and poorly thought out experiments: let's see some real algebraic analysis of your "theories".
The engine has no ability to make choices. We can choose to see the truth or we can ignore it. The Law of Conservation of Energy is false.
The "choice", in case you haven't figured it out yet Limon, is actually the different velocities directed by the conservation of momentum (which you so dearly love) setup by the equal and opposite reaction to the pressure inside the engine chamber being released at the nozzle. The energy is directed to "support" the velocities.
Just like your silly swinging 1 kg mass "going back up to the level it was" because it has the same speed going up as down --- sure does, but it isn't going to get very far moving up at 1.4 m/s! And the 9 kg block "giving" momentum to raise the lower block: I'm sure you've already calculated that even if the block gave up all its momentum in an effort to raise the block it would come up short.
I suppose it never occured to you that if energy conservation is false, then momentum conservation is false, too, since the former is a mathematical consequence of the latter.
In every one of your experiments you find "energy is not conserved". But that's because in every one of your experiments you manage to let momentum slip through your fingers.
But enough of your pithy comments and poorly thought out experiments: let's see some real algebraic analysis of your "theories".
I clearly stated that the 9 kg block has more than enough momentum to bring the 1 kilogram mass back up to the starting position. If the 9 kg block gave the 1 kg only half its momentum it would rise 3 meters, and it only dropped one.
Ballistic pendulums were used for centuries before the introduction of strobe light photography and electronic timing equipment.
One ballistic pendulum experiment was to suspend a rifle from two strings. The height of the rise was used to determine the momentum of the rifle, from that they calculated the momentum of the bullet. It is never mentioned that the mass of the rifle (sled) changes the momentum of the bullet (exhaust gasses). Doubling the mass of the rifle would not change the momentum of the bullet. The velocity of the rifle (sled) would drop to half to maintain the same momentum as the bullet (exhaust gasses). The kinetic energy formula was not used in these ballistic pendulum calculations.
So; a certain quantity of gun powder produced a certain quantity of momentum not kinetic energy. As the gun powder explodes in the chamber a force F is exerted against the bullet and an equal and opposite force F is exerted against the gun, the time is of course also equal. The relationship F = ma predicts that, when the bullet is expelled from the muzzle, both bullet and gun will have the same momentum. The centuries of ballistic pendulum experiments; strobe light photography; and electronic timing equipment seem to confirm this theory.
The gun powder and the rocket propellant are certain quantities of energy that conserve momentum not energy.
What I am saying is: if one takes precedence over the other it is always momentum that takes precedence over kinetic energy, not the other way around. And what should take precedence in the cylinder and spheres experiment is momentum not kinetic energy.
Energy conservation has application in some instances but it is not a universal law.
there wre 3 teams... one of 5 and two of 4... same location as the D-day story... my team split up into two groups... one of the groups went to flank the more expierenced team and my group went for the other... we pulled back, climed down the hill and flanked the team of 5, i took the first shot of the game, hit one in the back, my partner shot at another but his gun jammed, he took cover to try and fix his gun... so i hurridly took out as many as i could, hit two more...the other two took cover behind some trees... and i was ducking behind a log... i shot at one i saw starting to pop around the corner,hit him the the elbow, and the other guy being a newb shot a few times and reached over and picked up dead partners angel... and as he popped out to try and unload to unload both guns on me at the same time, i popped him between the eyes as he came around the tree... so as soon as i took out the whole team my partner got his gun unjammed...
so we started to charge the remaining team... i moved slowly forward wondering wher the were hiding and where the rest of my team was... i quickly found the other team in a forgotten bunker we built... so as i was stranded between and abondoned fridge and stove... i fired a few blind shots... and my partner fired some supressive fire to provide cover while i ran outta range... as i stood up i saw a guy stand up and take aim... i heard a shot fire... but it wasnt his, it was from one of my missing team mates, he had covertly taken the high ground above thier bunker, and fire fast burst took out two of them... as the other two ran my final team mate popped up from across the bank and fired fast and finished off the remainder of that team....
great day, great scirmish
so we started to charge the remaining team... i moved slowly forward wondering wher the were hiding and where the rest of my team was... i quickly found the other team in a forgotten bunker we built... so as i was stranded between and abondoned fridge and stove... i fired a few blind shots... and my partner fired some supressive fire to provide cover while i ran outta range... as i stood up i saw a guy stand up and take aim... i heard a shot fire... but it wasnt his, it was from one of my missing team mates, he had covertly taken the high ground above thier bunker, and fire fast burst took out two of them... as the other two ran my final team mate popped up from across the bank and fired fast and finished off the remainder of that team....
great day, great scirmish
QUOTE (Limon+Dec 18 2006, 10:51 PM)
I clearly stated that the 9 kg block has more than enough momentum to bring the 1 kilogram mass back up to the starting position. If the 9 kg block gave the 1 kg only half its momentum it would rise 3 meters, and it only dropped one.
Oh, really? You have the 9 kg block moving at 1.4 m/s and it's going to start pulling on the 1 kg mass (which I assume has stopped dead because of some table or another, otherwise you are even deader meat), so now the whole 10 kg is going to be slowed by the "weight" of the 1 kg mass (9.8N):
F = m a
9.8 N = 10 a
a = 0.98 m/s
since v - 1.4 m/s, it will take t = v/a seconds to stop the motion or 1.43 sec.
And with s = 1/2 a t^2 the whole system will stop after:
s = (0.5)(.98)(1.43^2) = 1 meter
Momentum is conserved and the whole thing stops dead, ready to go downhill the other way. It's a pendulum, isn't it?
Even if you "gave" the stopped 1 kg block 3.03 of momentum, that would only give it a velocity of 3 m/s would would stop it after 0.3 seconds of rising to the grand height of only 0.46 meters. You really need to "give" it 4.43 units of momentum.
Now here is where you come in saying, but because m_1v_1 + m_2V_2 must be a constant, I've got 12.6-4.43 = 8.17 units of momentum left over in the 9 kg block! True, if you could transfer the momentum via an elastic collision ... but it's not an elastic collision situation. There's this extra nagging force of gravity that you have to fight against to get the block up to the needed 4.3 m/s upward velocity. So it's really a 9 kg block moving at 1.4 m/s being slowed by a 9.8 N force.
And I didn't even have to bring in kinetic energy.
You're obviously wrong about that, too. Just too a few bullets in a fire and light 'em off so see if it's momentum of energy that's constant. In the gun, the recoil velocites are so low that the difference in bullet velocities isn't much but it's there.
You can also tell that a gun isn't a simple recoil problem because the recoil is produced by two separate events: the back pressure while the bullet is still in the barrel and then the exhaust gas momentum after the bullet leaves the barrel. For example, it's obvious when you're shooting blanks versus a bullet. And you said momentum was in the charge?
Oh, really? You have the 9 kg block moving at 1.4 m/s and it's going to start pulling on the 1 kg mass (which I assume has stopped dead because of some table or another, otherwise you are even deader meat), so now the whole 10 kg is going to be slowed by the "weight" of the 1 kg mass (9.8N):
F = m a
9.8 N = 10 a
a = 0.98 m/s
since v - 1.4 m/s, it will take t = v/a seconds to stop the motion or 1.43 sec.
And with s = 1/2 a t^2 the whole system will stop after:
s = (0.5)(.98)(1.43^2) = 1 meter
Momentum is conserved and the whole thing stops dead, ready to go downhill the other way. It's a pendulum, isn't it?
Even if you "gave" the stopped 1 kg block 3.03 of momentum, that would only give it a velocity of 3 m/s would would stop it after 0.3 seconds of rising to the grand height of only 0.46 meters. You really need to "give" it 4.43 units of momentum.
Now here is where you come in saying, but because m_1v_1 + m_2V_2 must be a constant, I've got 12.6-4.43 = 8.17 units of momentum left over in the 9 kg block! True, if you could transfer the momentum via an elastic collision ... but it's not an elastic collision situation. There's this extra nagging force of gravity that you have to fight against to get the block up to the needed 4.3 m/s upward velocity. So it's really a 9 kg block moving at 1.4 m/s being slowed by a 9.8 N force.
And I didn't even have to bring in kinetic energy.
QUOTE
So; a certain quantity of gun powder produced a certain quantity of momentum not kinetic energy.
You're obviously wrong about that, too. Just too a few bullets in a fire and light 'em off so see if it's momentum of energy that's constant. In the gun, the recoil velocites are so low that the difference in bullet velocities isn't much but it's there.
You can also tell that a gun isn't a simple recoil problem because the recoil is produced by two separate events: the back pressure while the bullet is still in the barrel and then the exhaust gas momentum after the bullet leaves the barrel. For example, it's obvious when you're shooting blanks versus a bullet. And you said momentum was in the charge?
By the way, like I said before, I'm no longer interested in your numbers or your anecdotes: let's see some equations that back you up. I'm not in the business of correcting your schoolboy mistakes. If you've got something "important" to say, say it in algebra.
NoCleverName Quote: If you've got something "important" to say, say it in algebra.
Limon's answer; F = ma The force F in the tether as the spheres feed out is equal in both directions, therefore the momentum change in the spheres will equal the momentum change in the cylinder. Yes I know you have worked it so you don’t even think a spinning cylinder has momentum, but I suggest you read Principia again.
I do not think that 1/2mv² is a derivative of mv, I think history proves that they were develop independently in an attempt to describe the motion of objects. This is covered under the topic Leibniz vs. Newton.
Leibniz noticed that an object moving twice as fast rises four times as high. This he knew meant that if momentum is conserved energy could be made, and there was common understanding in that time (as this) that no such production of energy could occur. So Leibniz came up with a formula (mv²) to explain away this production, which gained common acceptance without significant experimental proof. Imaginary heat in ballistic pendulums, vector cancellation used in rotating wheels so that momentum is allegedly zero, formulas for the motion of satellites applied to pucks on an air table, ignoring the fact that there is no one to one relationship between chemical energy and kinetic energy, refusal to do simple experiments that will prove Newton correct like the cylinder and spheres experiment, these are all used so that science can ignore the obvious conclusion of Newtonian Physics, which is that Leibniz was wrong and that the Law of Conservation of Energy is false.
With the 9 kilograms on the frictionless plane moving 1.4 m/sec and the one kilogram dropping at 1.4 m/sec, catch the one kilogram in a 3.0 meter pendulum and allow it to drop one half meter more to the bottom of the pendulum. While it is dropping it will not be under gravitational declaration, add only half of the momentum that the 9 kilograms has and it will rise 3 meters (1.5 meters above where it started). In this way the force of gravity will not eat up the momentum being applied to the one kilogram mass, in fact it will add some more motion to it.
Limon's answer; F = ma The force F in the tether as the spheres feed out is equal in both directions, therefore the momentum change in the spheres will equal the momentum change in the cylinder. Yes I know you have worked it so you don’t even think a spinning cylinder has momentum, but I suggest you read Principia again.
I do not think that 1/2mv² is a derivative of mv, I think history proves that they were develop independently in an attempt to describe the motion of objects. This is covered under the topic Leibniz vs. Newton.
Leibniz noticed that an object moving twice as fast rises four times as high. This he knew meant that if momentum is conserved energy could be made, and there was common understanding in that time (as this) that no such production of energy could occur. So Leibniz came up with a formula (mv²) to explain away this production, which gained common acceptance without significant experimental proof. Imaginary heat in ballistic pendulums, vector cancellation used in rotating wheels so that momentum is allegedly zero, formulas for the motion of satellites applied to pucks on an air table, ignoring the fact that there is no one to one relationship between chemical energy and kinetic energy, refusal to do simple experiments that will prove Newton correct like the cylinder and spheres experiment, these are all used so that science can ignore the obvious conclusion of Newtonian Physics, which is that Leibniz was wrong and that the Law of Conservation of Energy is false.
With the 9 kilograms on the frictionless plane moving 1.4 m/sec and the one kilogram dropping at 1.4 m/sec, catch the one kilogram in a 3.0 meter pendulum and allow it to drop one half meter more to the bottom of the pendulum. While it is dropping it will not be under gravitational declaration, add only half of the momentum that the 9 kilograms has and it will rise 3 meters (1.5 meters above where it started). In this way the force of gravity will not eat up the momentum being applied to the one kilogram mass, in fact it will add some more motion to it.
QUOTE (Limon+Dec 19 2006, 10:27 PM)
I do not think that 1/2mv² is a derivative of mv
Strickly speaking, no , because it is the integral of momentum with respect to velocity during application of a force F=ma. The math is shown in any physics textbook.
Like haven't you ever noticed that d K.E dv = mv ?
Hasn't it ever occurred to you that as a force continues on an object, the incrementally increasing momentum adds up to something? Integrate the force over a distance. ...
Integral of F ds = Integral of mv dv = 1/2 mv^2
So kinetic energy exists, at least as the definition of how much force over distance was used against a body. That in itself is useful enough.
Clearly, k.e. is conserved in totally elastic collisions because there is no mathematical way out . Your problem seems with other experiments when the transformation of k.e. to other forms of energy (like gravitational potential energy) occurs. The deal is that total energy is conserved, not each type of energy individually.
But you sure do rely on your pet pendulums where energy is cycling between 1/2 mv^2 (K.E.) and mgh (P.E), don't you. Pretty much depend on total energy being conserved there, right?
While granted there is some small benefit to me by doing your study for you, you could break down and hit the books a bit yourself, don't you think?
Strickly speaking, no
, because it is the integral of momentum with respect to velocity during application of a force F=ma. The math is shown in any physics textbook.Like haven't you ever noticed that d K.E dv = mv ?
Hasn't it ever occurred to you that as a force continues on an object, the incrementally increasing momentum adds up to something? Integrate the force over a distance. ...
Integral of F ds = Integral of mv dv = 1/2 mv^2
So kinetic energy exists, at least as the definition of how much force over distance was used against a body. That in itself is useful enough.
Clearly, k.e. is conserved in totally elastic collisions because there is no mathematical way out . Your problem seems with other experiments when the transformation of k.e. to other forms of energy (like gravitational potential energy) occurs. The deal is that total energy is conserved, not each type of energy individually.
But you sure do rely on your pet pendulums where energy is cycling between 1/2 mv^2 (K.E.) and mgh (P.E), don't you. Pretty much depend on total energy being conserved there, right?
While granted there is some small benefit to me by doing your study for you, you could break down and hit the books a bit yourself, don't you think?
Let's begin with the 1kg and 9kg blocks at their start position and ready to be released. The total momentum in the system at this time is 0. When we release the blocks they accelerate and gain a total of 14 units of momentum. Due to the law of conservation of momentum the blocks should not have gained any momentum. Therefore, the law of Conservation of Momentum is false! Now that we know that the "law" is false you cannot say that the 9kg block is able to give the 1kg block 4.43 (or 3.03 or whatever) units of momentum without loss.
^ You are forgetting that when gravity is involved, you can't consider the blocks an isolated system. The force the Earth exerts on the blocks (ie gravity) is equal and opposite (Newton's 3rd Law) to the force the blocks exert on the Earth. You'll find when you consider the entire system, the Earth and the blocks, momentum is zero both before and afterwards.
Did you not stop for a moment to think that in the last 350 years noone would have thought of that experiment?! Or did you think that all the great minds of the last 3 centuries were somehow blind to such an obvious experiment? One which has been taught to school children for more than 50 years?! Did you not stop to think for just a tiny fraction of a second that perhaps, just perhaps, it was an error in your thinkings?
Cranks are too ready to jump in with "Ah ha, you're wrong because of...." without even realising their 'reason' is something a 14 year old school kid would know to be wrong
Did you not stop for a moment to think that in the last 350 years noone would have thought of that experiment?! Or did you think that all the great minds of the last 3 centuries were somehow blind to such an obvious experiment? One which has been taught to school children for more than 50 years?! Did you not stop to think for just a tiny fraction of a second that perhaps, just perhaps, it was an error in your thinkings?
Cranks are too ready to jump in with "Ah ha, you're wrong because of...." without even realising their 'reason' is something a 14 year old school kid would know to be wrong
I am going to assume that I went over this to fast for an unfamiliar scientist to understand.
This is a common experiment, NoCleverName even said so; (One which has been taught to school children for more than 50 years?!) and then he shows that he does not understand (NoCleverName quote: momentum is zero both before and afterward). Poor Newton; if he only knew what NoCleverName knows he wouldn't have wasted his time on the Three Laws of Motion.
NQ; You speak of increments of force. Newton’s increments were a function of time.
If you drop a one kilogram object one meter you get (d=1/2v²/a), √(2*1*9.81)= 4.43 units of momentum. Time equals .4515 sec., this is 9.81 newtons working for .4515 sec. = 4.43
Place a 9 kilogram block on a frictionless plane, connect a string, position the sting over a frictionless pulley, connect a one kilogram mass to the other end, let the mass drop one meter and you will get 14.007 units of momentum. Time equals 1.4278 sec., this is 9.81 newtons working for 1.4278 sec. = 14.007
The nine kilograms on the plane and the one kilogram that is dropping over the frictionless pulley is being accelerated by the 9.81 newtons of force that is contained in the dropping one kilogram. This is F = ma 9.81 N = 10 kg * .981m/sec/sec d = 1/2v²/a or √(1m * 2*.981) = 1.4007 m/sec. This is both the 9 kilogram and the one kilogram moving 1.4 m/sec for a momentum of 14, The nine kilograms is not under gravitational acceleration and the one kilogram is detached from the string and caught in a pendulum system.
With the 9 kilograms on the frictionless plane moving 1.4 m/sec and the one kilogram dropping at 1.4 m/sec, catch the one kilogram in a 3.0 meter pendulum and allow it to drop one half meter more to the bottom of the pendulum. While it is dropping it will not be under gravitational declaration, add only half of the momentum that the 9 kilograms has and it will rise 3 meters (1.5 meters above where it started). In this way the force of gravity will not eat up the momentum being applied to the one kilogram mass, in fact it will add some more motion to it.
If you can give 4.43 units of momentum to the lowered one kilogram object the object will return to the top and start the system over again, which gives you 9.587 units extra. Some call it perpetual motion, I call it Newtonian Physics.
What law of physics prevents 14.007 units of momentum from being used to produce 4.43. I know of a law of physics that says this is absolutely possible, in fact it is required that it is doable. It is the Law of Conservation of Momentum and all of Newtonian Physics (concerning motion) hinges upon it being achievable. Anyone who denies that it is possible is taking the continental view (according to Homm) of physics, that kinetic energy (or any assortment of forms of energy) will take precedence over momentum conservation.
The cylinder and spheres gives all the motion of four units of mass to one unit of mass. Momentum is conserved or energy is conserved; one or the other is false.
NoCleverName quote: Did you not stop for a moment to think that in the last 350 years none would have thought of that experiment?! Or did you think that all the great minds of the last 3 centuries were somehow blind to such an obvious experiment? One which has been taught to school children for more than 50 years?! Did you not stop to think for just a tiny fraction of a second that perhaps, just perhaps, it was an error in your thinking?
Limon's answer: If we all had these negative attitudes nothing new would ever be invented. The Wrights, Edison, Marconi, Gates, these men are within the last century, don’t tell me new things are not happening.
Granted; the sum of gravitational potential energy and kinetic energy is conserved, but I think it is honest to say that that is the definition of kinetic energy (how high will an object rise that has a certain velocity). The height of an object can be looked at as the energy of the object. Objects of velocity (v) (V) can rise to a certain height according to the relationship: 1/2mv², 1/2mV². But the distance formula (d = 1/2 at², or d = 1/2 v²/a) tells us the same thing (without reference to mass). Acceleration in the kinetic energy formula is 1 instead of 9.81 m/sec as in the distance formula. Now has anyone got any idea why multiplying the distance formula by mass should give us a conserved quantity, that overrules Newton’s Three Laws of Motion?
Limon, i have no idea what you mean by the pendulum with the 9 and 1 kg masses. But you dont seem to understand momentum, because it is mass*velocity, is a vector!
Because the velocities of the 9 and 1 kg masses are at right angles, so are their momentums, thus you have to do a vector addition.
Total momentum of the system, P = squareroot( p1^2 + p2^2 ) = 12.68
It also breaks conversation of energy, if the system ends up with more energy than it started with.
Because the velocities of the 9 and 1 kg masses are at right angles, so are their momentums, thus you have to do a vector addition.
Total momentum of the system, P = squareroot( p1^2 + p2^2 ) = 12.68
It also breaks conversation of energy, if the system ends up with more energy than it started with.
QUOTE (Limon+Dec 21 2006, 04:10 AM)
(NoCleverName quote: momentum is zero both before and afterward). Poor Newton; if he only knew what NoCleverName knows he wouldn't have wasted his time on the Three Laws of Motion.
I said it, not NCN You can call the momentum zero if nothing is moving relative to one another. You can call the momentum any value you want, you end up with the same result at the end.
Just because you forget to include the Earth's effect in your calculations isn't my fault.
But the distance formula (d = 1/2 at², or d = 1/2 v²/a) tells us the same thing (without reference to mass). Acceleration in the kinetic energy formula is 1 instead of 9.81 m/sec as in the distance formula. Now has anyone got any idea why multiplying the distance formula by mass should give us a conserved quantity, that overrules Newton’s Three Laws of Motion?
Oh, really?
d = (a t^2) / 2
d = (v t) / 2
F d = (F v t) / 2
F d = (m a v t) / 2
F d = (m v^2) / 2
and by definition work "F d" = k.e. So it would seem the distance formula and F = m a get along quite nicely, thankyou. And remember you admitted total energy was conserved: what we have here is kinetic energy alone, which no one is saying is individually conserved. Therefore, even by your own words in the same paragraph ("Granted; the sum of gravitational potential energy and kinetic energy is conserved") the statement "why multiplying the distance formula by mass should give us a conserved quantity" is meaningless.
You can also get from step 1 to step 2 by substituting a = F/m and going from there.
In fact, I'm heading off for a couple of days of energy conservation using exactly that same formula of K.e = p.e on the ski slopes right now, so see you all later.
(“Granted; the sum of gravitational potential energy and kinetic energy is conserved") is in reference to a pendulum. If the pendulum bob strikes an object at the bottom of the swing much of your kinetic energy disappears, and therefore your gravitational potential energy disappears. In other words your theory goes up in smoke, or shall I say heat that does not exist.
No I have never said that total energy is conserved, I don’t makeup things like heat being produced in ballistic pendulums. I was only referring to potential and kinetic energy in pendulums.
I have stated many times that the cylinder and spheres can not conserve both kinetic energy (which is all there is of your forms) and momentum. One or the other has to be false.
Sorry; for mixing a quote from, AlphaNumeric with NoCleverName I shall try to be more careful.
I said it, not NCN
You can call the momentum zero if nothing is moving relative to one another. You can call the momentum any value you want, you end up with the same result at the end. Just because you forget to include the Earth's effect in your calculations isn't my fault.
QUOTE (Limon+Dec 21 2006, 04:10 AM)
Limon's answer: If we all had these negative attitudes nothing new would ever be invented. The Wrights, Edison, Marconi, Gates, these men are within the last century, don’t tell me new things are not happening.
Once again, I said that quote you attributed to NCN. It's not about saying to people "Your new idea is crap", it's about the fact Benny thought that millions of people, including people who took us to the Moon and back, missed that 'obvious' violation of the conservation of momentum Rather than think "This is so obvious, there must be something wrong with my logic" he thought "Ah ha, I've seen something noone else has, despite all school childrens in the Western world knowing about this!" If someone is being ignorant, I'll call them so. Benny was. The Wrights, Edison, Marconi and Gates didn't do that.
All your 'experiments' just show you've a terrible grasp of Newtonian physics. I see you still haven't gone to your professor friend to have him explain Lagrangian mechanics to you You're a crank Limon, don't think you've got us on the ropes. You're like Nick, another poster here, you think holes in your understanding are the theories fault. No, it couldn't be you, you're perfect. So it must be someone else's fault. Delusions of perfection it would seem
Rather than think "This is so obvious, there must be something wrong with my logic" he thought "Ah ha, I've seen something noone else has, despite all school childrens in the Western world knowing about this!" If someone is being ignorant, I'll call them so. Benny was. The Wrights, Edison, Marconi and Gates didn't do that.All your 'experiments' just show you've a terrible grasp of Newtonian physics. I see you still haven't gone to your professor friend to have him explain Lagrangian mechanics to you
You're a crank Limon, don't think you've got us on the ropes. You're like Nick, another poster here, you think holes in your understanding are the theories fault. No, it couldn't be you, you're perfect. So it must be someone else's fault. Delusions of perfection it would seem
QUOTE (Limon+Dec 21 2006, 03:10 AM)
But the distance formula (d = 1/2 at², or d = 1/2 v²/a) tells us the same thing (without reference to mass). Acceleration in the kinetic energy formula is 1 instead of 9.81 m/sec as in the distance formula. Now has anyone got any idea why multiplying the distance formula by mass should give us a conserved quantity, that overrules Newton’s Three Laws of Motion?
Oh, really?
d = (a t^2) / 2
d = (v t) / 2
F d = (F v t) / 2
F d = (m a v t) / 2
F d = (m v^2) / 2
and by definition work "F d" = k.e. So it would seem the distance formula and F = m a get along quite nicely, thankyou. And remember you admitted total energy was conserved: what we have here is kinetic energy alone, which no one is saying is individually conserved. Therefore, even by your own words in the same paragraph ("Granted; the sum of gravitational potential energy and kinetic energy is conserved") the statement "why multiplying the distance formula by mass should give us a conserved quantity" is meaningless.
You can also get from step 1 to step 2 by substituting a = F/m and going from there.
In fact, I'm heading off for a couple of days of energy conservation using exactly that same formula of K.e = p.e on the ski slopes right now, so see you all later.
(“Granted; the sum of gravitational potential energy and kinetic energy is conserved") is in reference to a pendulum. If the pendulum bob strikes an object at the bottom of the swing much of your kinetic energy disappears, and therefore your gravitational potential energy disappears. In other words your theory goes up in smoke, or shall I say heat that does not exist.
No I have never said that total energy is conserved, I don’t makeup things like heat being produced in ballistic pendulums. I was only referring to potential and kinetic energy in pendulums.
I have stated many times that the cylinder and spheres can not conserve both kinetic energy (which is all there is of your forms) and momentum. One or the other has to be false.
Sorry; for mixing a quote from, AlphaNumeric with NoCleverName I shall try to be more careful.
QUOTE (NoCleverName+Dec 21 2006, 11:01 AM)
d = (a t^2) / 2
....
F d = (m v^2) / 2
Which works by the coincidence that the starting velocity was zero and the acceleration was constant. In the real world v and d and F are vectors, so while v^2 = v^2, F dot d does not always equal Fd.
....
F d = (m v^2) / 2
Which works by the coincidence that the starting velocity was zero and the acceleration was constant. In the real world v and d and F are vectors, so while v^2 = v^2, F dot d does not always equal Fd.
Limon, at 0s your 1kg block-9kg block system has a total momentum of 0Ns and at 1.4s P(system) = (12.68Ns, -6.34°). Since (0Ns, 0°) does not equal (12.68Ns, -6.34°) you have an open system and momentum is free to enter (obviously, since it just did) and is also free to leave. If you want momentum conservation to apply, you must fix this problem first.
Enhance your calm AlphaNumeric. I'm not actually THAT stupid.
Enhance your calm AlphaNumeric. I'm not actually THAT stupid.
First; about vector addition. Are you people aware that one of the devices that is used to prove that F=ma is a device called an Atwood’s machine. An Atwood is simply a string draped over a pulley with equal masses on each side such as 5 kilogram on both sides, and then an extra one kilogram is placed on one of the sides. The acceleration is used to prove F =ma. Well 5 kilograms is going up and 5 kilograms is going down, if they used vector cancellation F would not equal ma. So your reference to the 1 kilogram going in a different direction than the 9 kilograms on the plane is a complete misunderstanding about vectors. They are added together. You have 14.007 units of momentum.
The question; is the numeric value from the distance formula a consequence of motion or a quantity of motion?
A 1 kg mass moving 4 m/sec will rise .8155 m, if this mass collides with a 3 kg mass at rest the 4 kg combined mass will move away at 1 m/sec, this 4 kilogram mass will rise .05097 m (there are 4 so this is the same as 1 kg at .20388 m) . The numeric value presented by the distance formula is reduced to 25%. The value presented by the kinetic energy formula is also reduced to 25%. The momentum remains at 100%.
It appears to me that the distance formula ½ v²/a and the kinetic energy formula ½ mv²/1 are consequences of motion not quantities of motion.
Now why would you predict that there is a 75% loss of motion to heat, when there is no change in momentum? Are you that eager to protect the distance formula?
The question; is the numeric value from the distance formula a consequence of motion or a quantity of motion?
A 1 kg mass moving 4 m/sec will rise .8155 m, if this mass collides with a 3 kg mass at rest the 4 kg combined mass will move away at 1 m/sec, this 4 kilogram mass will rise .05097 m (there are 4 so this is the same as 1 kg at .20388 m) . The numeric value presented by the distance formula is reduced to 25%. The value presented by the kinetic energy formula is also reduced to 25%. The momentum remains at 100%.
It appears to me that the distance formula ½ v²/a and the kinetic energy formula ½ mv²/1 are consequences of motion not quantities of motion.
Now why would you predict that there is a 75% loss of motion to heat, when there is no change in momentum? Are you that eager to protect the distance formula?
Limon, all I'm asking is that you demonstrate momentum conservation during the 1m drop of the 1kg block. That's it. For me the vector addition and Atwood machines and everything else can wait until I get this cleared up.
rpenner: I was hoping to "get away" with the non-vector interpretation given that even the so-called "distance formula" really doesn't enforce a "straight line" path. In any event, I think I was able to show the two formula are at least "compatible".
limon: I, too, am not happy with the idea of vector-adding the momenta in this problem as this almost seems to fall within the category of the so-called "constrained motion" problem, so I'll give you your 14 units of momentum.
That being said, think you are still in the position of "just tossing off an opinion" about k.e., rather than providing anything that resembles rigorous analysis. So far, all you have is some "experiments" whose results you interpret as defying accepted theory. At this point, everyone else has to do the heavy-lifting of producing the experimental description and calculations using accepted theory to which you merely dismiss without showing any analytical reasons.
I guess what I don't understand is what motivates you: Is it the experimental results themselves or something you have against accepted theory from "first principles".
Now, let us limit ourselves to mechanics in the "ordinary" range of experience from as small as gases to maybe as large as a galaxy because there could very well be breakdowns outside that domain.
You are going to have to come up with, I'm sorry to say, some "new" formulas to use and analyze your problem (and maybe some of ours) with them.
limon: I, too, am not happy with the idea of vector-adding the momenta in this problem as this almost seems to fall within the category of the so-called "constrained motion" problem, so I'll give you your 14 units of momentum.
That being said, think you are still in the position of "just tossing off an opinion" about k.e., rather than providing anything that resembles rigorous analysis. So far, all you have is some "experiments" whose results you interpret as defying accepted theory. At this point, everyone else has to do the heavy-lifting of producing the experimental description and calculations using accepted theory to which you merely dismiss without showing any analytical reasons.
I guess what I don't understand is what motivates you: Is it the experimental results themselves or something you have against accepted theory from "first principles".
Now, let us limit ourselves to mechanics in the "ordinary" range of experience from as small as gases to maybe as large as a galaxy because there could very well be breakdowns outside that domain.
You are going to have to come up with, I'm sorry to say, some "new" formulas to use and analyze your problem (and maybe some of ours) with them.
I would like to give you a few experimental examples in an attempt to correct the concepts some of you have concerning vector cancellation.
This is a real experiment, I have done most of it, the portion I haven’t done can be found in the literature.
An Atwood’s machine is a pulley wheel with masses suspended on both sides of the pulley. Let’s mount a pulley wheel near the ceiling and let’s drape an inelastic string over the pulley and hang one 5 kg mass on each end of the string. What happens?
Answer: Nothing happens the system is balanced with 49.05 N of force (5 kg each side) on both sides of the pulley, the forces cancel each other.
Now let’s add a 1 kg mass to the right side just below the 5 kg mass. What happens?
Answer: We now have 58.86 newtons of force on one side and 49.05 newtons on the other. So we have 9.81 newtons accelerating 11 kg. F = ma 9.81 = 11 * a, a = .8918 m/sec/sec. After the 1 extra kg side falls one meter the whole 11 kg has a velocity of: d = ½ v²/a , √(2*1*.8918) = 1.3355 m/sec. This is 6 kg going down and 5 kg going up.
After a 1 m drop, let’s pick up a 1 kg mass off of the floor by attaching a string to the left (ascending) side 5 kg mass. What happens?
Answer: We now have 58.86 newtons of force on both sided, and the system is again balanced, it will not accelerate. It was however 11 kg moving 1.3355 m/sec. and it had 14.69 units of momentum. If momentum is conserved it will now be 12 kg moving (v m/sec) with 14.69 units of momentum. Momentum = mv; 14.69/12 = v; v = 1.224 m/sec.
OK; according to those of you that want to cancel momentums that are going in different directions, the 6 kg going down and the 5 kg going up would cancel all but 1.3355 of the 14.69 units of momentum. This I presume would leave you with only 1.3355 units of momentum to share with the 12th kg that was picked up off of the floor. Now would that give the 11th and 12th kg 1.3355/2 = .6678 m/sec, or .6678units of momentum each? And if you had to share this paltry 1.3355 units of momentum with all the 12 kg you would be moving only .1113 m/sec. In your thinking the system would all but stop instead of slowing from 1.3355 m/sec to 1.224 m/sec as Newton would predict, which would be the same relationship of 12kg sharing the same momentum that was held by 11kg.
Even harder for the cancellation club to explain would be 5 kg going up and 5 going down (in an Atwood) at 1 m/sec., the Atwood is then arranged to pickup a 1 kilogram object off of the floor. The cancellation club concepts would have no available momentum that could be used to pick up the 1 kg mass. This would mean that the 10 kg moving one meter per second would be immediately jerked to a stop as soon as it tried to lift the 1 kg object.
Those of us familiar with experiments know that the one kilogram will be lifted. The ten units of momentum will be shared with the extra kilogram, giving the whole system an initial velocity of .9091 m/sec. This lifting will be stopped by the one extra kilogram (9.81 N) on the 6 kg side, but it will first rise .4634 m. d = ½ v²/a ; ½ .90909²/.8918 = .4634 m.
Vectors cancel when the two (or more) forces are pushing or pulling in opposite directions or working obliquely to each other. The moving masses in the Atwood will work together to lift the one kilogram object. Each point of mass in the spinning rim will work together to lift a one kilogram object as well. They work together they do not cancel.
Whenever this inappropriate cancellation of vectors began for Atwood’s machines and spinning objects, Newtonian physics was no longer used in evaluating there motion. No doubt the physics that replaced it is totally worthless.
This faulty concept of vector cancellation helps to leaves an empty gap in the field of physics that you can drive a bus through.
What motivates me is that free energy can be made from the gravitational field. These arguments about faulty concepts are an attempt to get people to assist me in this endeavored. Specifically I would like someone to build a cylinder and spheres machine, and see if you get the same results as I. For example; some think that a spinning rim has no momentum, well how much force * time did it take to make it spin, that is how much momentum it has. And that is how much momentum it will surrender even if the quantity of mass in motion is reduced. This is what the cylinder and spheres experiment does.
This is a real experiment, I have done most of it, the portion I haven’t done can be found in the literature.
An Atwood’s machine is a pulley wheel with masses suspended on both sides of the pulley. Let’s mount a pulley wheel near the ceiling and let’s drape an inelastic string over the pulley and hang one 5 kg mass on each end of the string. What happens?
Answer: Nothing happens the system is balanced with 49.05 N of force (5 kg each side) on both sides of the pulley, the forces cancel each other.
Now let’s add a 1 kg mass to the right side just below the 5 kg mass. What happens?
Answer: We now have 58.86 newtons of force on one side and 49.05 newtons on the other. So we have 9.81 newtons accelerating 11 kg. F = ma 9.81 = 11 * a, a = .8918 m/sec/sec. After the 1 extra kg side falls one meter the whole 11 kg has a velocity of: d = ½ v²/a , √(2*1*.8918) = 1.3355 m/sec. This is 6 kg going down and 5 kg going up.
After a 1 m drop, let’s pick up a 1 kg mass off of the floor by attaching a string to the left (ascending) side 5 kg mass. What happens?
Answer: We now have 58.86 newtons of force on both sided, and the system is again balanced, it will not accelerate. It was however 11 kg moving 1.3355 m/sec. and it had 14.69 units of momentum. If momentum is conserved it will now be 12 kg moving (v m/sec) with 14.69 units of momentum. Momentum = mv; 14.69/12 = v; v = 1.224 m/sec.
OK; according to those of you that want to cancel momentums that are going in different directions, the 6 kg going down and the 5 kg going up would cancel all but 1.3355 of the 14.69 units of momentum. This I presume would leave you with only 1.3355 units of momentum to share with the 12th kg that was picked up off of the floor. Now would that give the 11th and 12th kg 1.3355/2 = .6678 m/sec, or .6678units of momentum each? And if you had to share this paltry 1.3355 units of momentum with all the 12 kg you would be moving only .1113 m/sec. In your thinking the system would all but stop instead of slowing from 1.3355 m/sec to 1.224 m/sec as Newton would predict, which would be the same relationship of 12kg sharing the same momentum that was held by 11kg.
Even harder for the cancellation club to explain would be 5 kg going up and 5 going down (in an Atwood) at 1 m/sec., the Atwood is then arranged to pickup a 1 kilogram object off of the floor. The cancellation club concepts would have no available momentum that could be used to pick up the 1 kg mass. This would mean that the 10 kg moving one meter per second would be immediately jerked to a stop as soon as it tried to lift the 1 kg object.
Those of us familiar with experiments know that the one kilogram will be lifted. The ten units of momentum will be shared with the extra kilogram, giving the whole system an initial velocity of .9091 m/sec. This lifting will be stopped by the one extra kilogram (9.81 N) on the 6 kg side, but it will first rise .4634 m. d = ½ v²/a ; ½ .90909²/.8918 = .4634 m.
Vectors cancel when the two (or more) forces are pushing or pulling in opposite directions or working obliquely to each other. The moving masses in the Atwood will work together to lift the one kilogram object. Each point of mass in the spinning rim will work together to lift a one kilogram object as well. They work together they do not cancel.
Whenever this inappropriate cancellation of vectors began for Atwood’s machines and spinning objects, Newtonian physics was no longer used in evaluating there motion. No doubt the physics that replaced it is totally worthless.
This faulty concept of vector cancellation helps to leaves an empty gap in the field of physics that you can drive a bus through.
What motivates me is that free energy can be made from the gravitational field. These arguments about faulty concepts are an attempt to get people to assist me in this endeavored. Specifically I would like someone to build a cylinder and spheres machine, and see if you get the same results as I. For example; some think that a spinning rim has no momentum, well how much force * time did it take to make it spin, that is how much momentum it has. And that is how much momentum it will surrender even if the quantity of mass in motion is reduced. This is what the cylinder and spheres experiment does.
QUOTE (Limon+Dec 23 2006, 12:42 AM)
What motivates me is that free energy can be made from the gravitational field.
... but before I get into that I'm not ducking the vector cancellation deal because it's particularly subtle and I haven't dealt with it in decades ...
Anyway, yes, energy is free for the taking in a gravitational field so long as you don't try to return to where you started from
. That is, things like waterfalls and tides are great for grabbing "free" energy.
But, Limon, most if not all of your ideas seem to center around "returning to the starting point" and continuing the cycle. One fundamental barrier would seem to be the theorem that any closed path in a inverse square field has zero work. Or something along those lines ... I'm sure rpenner could fill in the gaps of my memory ... something about the line integral being zero.
In any case, it would seem that whatever you gain from the field you must ultimately give back to the field in order to return to your starting point. If you are familiar with this theorem, Limon, it would be interesting to hear your opinion of how it is not an obstacle to your ideas.
And just skimming your tirade against the unfortunate vector, it seems you're lumping them all in together: vectors measuring force, those measuring constant motion, and the innocent momentum vector. They all have to be playing on the same side before the cancelling can begin. I really can't see what you have against them, unless you hate the idea of "net force", but then you'd have to hate most of Newtons laws, then, too.
... but before I get into that I'm not ducking the vector cancellation deal because it's particularly subtle and I haven't dealt with it in decades ...
Anyway, yes, energy is free for the taking in a gravitational field so long as you don't try to return to where you started from
. That is, things like waterfalls and tides are great for grabbing "free" energy.But, Limon, most if not all of your ideas seem to center around "returning to the starting point" and continuing the cycle. One fundamental barrier would seem to be the theorem that any closed path in a inverse square field has zero work. Or something along those lines ... I'm sure rpenner could fill in the gaps of my memory ... something about the line integral being zero.
In any case, it would seem that whatever you gain from the field you must ultimately give back to the field in order to return to your starting point. If you are familiar with this theorem, Limon, it would be interesting to hear your opinion of how it is not an obstacle to your ideas.
And just skimming your tirade against the unfortunate vector, it seems you're lumping them all in together: vectors measuring force, those measuring constant motion, and the innocent momentum vector. They all have to be playing on the same side before the cancelling can begin. I really can't see what you have against them, unless you hate the idea of "net force", but then you'd have to hate most of Newtons laws, then, too.
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