| Topic: |
Science > Physics |
| User: |
"Peter" |
| Date: |
21 Apr 2006 10:42:06 AM |
| Object: |
Work - impulse |
Hi, I understand that an impulsive force can do work; As a matter of
fact, I don't think there is any doubt about it. But, impulse = Ft,
whereas work = Fx. Isn't there an incompatibility here? What am I
missing? Could someone please explain? Thanks.
Peter
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| User: "Entropy" |
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| Title: Re: Work - impulse |
21 Apr 2006 05:24:26 PM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1145634126.902589.35250@i40g2000cwc.googlegroups.com...
Hi, I understand that an impulsive force can do work; As a matter of
fact, I don't think there is any doubt about it. But, impulse = Ft,
whereas work = Fx. Isn't there an incompatibility here? What am I
missing? Could someone please explain? Thanks.
Peter
Ok suppose you had no friction and a mass M on ice. The transfer function
that relates output position to input force is in Laplace notation
G(s) = 1/s^2
which has an impulse response t=t. If you put a step force into the system
you get an output position x(s) =1/s^3
which is aceleration. Now for an impulsive force input the output will be
defined as 1/s^2 since the Laplace of an impulse is unity.
Hence the output is just the impulse response x(t)=t. So the position vs
time graph is a linear slope of unity.
The acceleration is the derivative of this which is dx(t)/dt = 1 ie a
constant.
The work done is integral(delta(f).dx} ...how does this work?
Tom
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| User: "Gregory L. Hansen" |
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| Title: Re: Work - impulse |
22 Apr 2006 09:10:48 AM |
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In article <1145634126.902589.35250@i40g2000cwc.googlegroups.com>,
Peter <Poakfield@msn.com> wrote:
Hi, I understand that an impulsive force can do work; As a matter of
fact, I don't think there is any doubt about it. But, impulse = Ft,
whereas work = Fx. Isn't there an incompatibility here? What am I
missing? Could someone please explain? Thanks.
Peter
Work is energy, impulse is momentum. They can be related,
E = p^2/2m
Given an initial momentum p1 and a final momentum p2,
delta E = (p2^2 - p1^2)/2m
= (p2 - p1)(p1 + p2) / 2m
= delta p * avg(p)/m
When you push something, how much work you put into it depends on how fast
it's going, since
dW = F dx = F dx/dt dt = F v dt
Or, if forces and directions are nice and constant,
W = Ft * v
= impulse * v
--
"The average person, during a single day, deposits in his or her underwear
an amount of fecal bacteria equal to the weight of a quarter of a peanut."
-- Dr. Robert Buckman, Human Wildlife, p119.
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| User: "Gregory L. Hansen" |
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| Title: Re: Work - impulse |
22 Apr 2006 09:14:26 AM |
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In article <e2ddh8$13p$5@rainier.uits.indiana.edu>,
Gregory L. Hansen <glhansen@steel.ucs.indiana.edu> wrote:
In article <1145634126.902589.35250@i40g2000cwc.googlegroups.com>,
Peter <Poakfield@msn.com> wrote:
When you push something, how much work you put into it depends on how fast
it's going, since
dW = F dx = F dx/dt dt = F v dt
Or, if forces and directions are nice and constant,
W = Ft * v
= impulse * v
And here Greg starts by assuming that you're pushing a free particle, and
finished by assuming a particle with constant velocity, as something
sliding with friction on a surface.
--
"What's another word for thesaurus?" -- Steven Wright
"Let me look in my synonymicon." -- Thaddeus Stout
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| User: "Peter" |
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| Title: Re: Work - impulse |
22 Apr 2006 10:25:02 AM |
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Greg, my point is that the same amount of work = F dot x can produce
the same impulse = F t, that can result in different amounts of kinetic
energy (work) when it acts on an extended objects of different mass,
like levers, even when no energy losses occur. And this cannot be
right: something is wrong.
Peter
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| User: "Mike" |
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| Title: Re: Work - impulse |
22 Apr 2006 12:14:25 PM |
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Peter wrote:
Greg, my point is that the same amount of work = F dot x can produce
the same impulse = F t, that can result in different amounts of kinetic
energy (work) when it acts on an extended objects of different mass,
like levers, even when no energy losses occur. And this cannot be
right: something is wrong.
Peter
An example of different KEs for the same object:
You are sitting on a train travelling at a constant velocity v. The
windows are covered you cannnot see ouside. As fas as you are concerned
you are at rest. Your KE is a flat zero in your own frame, the train.
For Peter who stands by the tracks and measures the speed of the train,
your KE is mv^2/2.
These two frames are related by a Galilean transformation. KE is not
conserved under such transformations but rate of change of momentum and
force is as you can easily find out by differentiating such a
transformation twice. Thus, Work is the same under such transformations
although KE is not. Thus, if you get pushed by a force in the train,
you and Peter will calculate the same work.
Mike
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| User: "PD" |
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| Title: Re: Work - impulse |
22 Apr 2006 11:45:19 AM |
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Peter wrote:
Greg, my point is that the same amount of work = F dot x can produce
the same impulse = F t, that can result in different amounts of kinetic
energy (work) when it acts on an extended objects of different mass,
like levers, even when no energy losses occur. And this cannot be
right: something is wrong.
Peter
Specific example, please.
PD
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| User: "" |
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| Title: Re: Work - impulse |
21 Apr 2006 12:33:33 PM |
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In article <1145634126.902589.35250@i40g2000cwc.googlegroups.com>, "Peter" <Poakfield@msn.com> writes:
Hi, I understand that an impulsive force can do work; As a matter of
fact, I don't think there is any doubt about it. But, impulse = Ft,
whereas work = Fx. Isn't there an incompatibility here?
Nope. Impulse has units of momentum while work has units of energy.
No incompatibility.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "John C. Polasek" |
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| Title: Re: Work - impulse |
21 Apr 2006 12:24:52 PM |
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On 21 Apr 2006 08:42:06 -0700, "Peter" <Poakfield@msn.com> wrote:
Hi, I understand that an impulsive force can do work; As a matter of
fact, I don't think there is any doubt about it. But, impulse = Ft,
whereas work = Fx. Isn't there an incompatibility here? What am I
missing? Could someone please explain? Thanks.
Peter
Well of course an impulse does work. It's tricky stuff. The basic
formula is
Imp = FT = MV
where we can use square time-based force waveforms for integration, no
harm done. It is true that in most cases that you can't know F(t) when
you integrate F(t) from t = 0 to infinity, so how do you get Imp?
Simple, on the RHSide you can measure V and multiply by M giving you
Imp.
Well, it's quite obvious we now have mass M moving at V so W = .5MV^2.
or W = .5Imp*V.
The Dirac delta function is the formal method with the time span going
to zero.
John Polasek
http://www.dualspace.net
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| User: "" |
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| Title: Re: Work - impulse |
21 Apr 2006 01:54:21 PM |
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In article <dr4i42lbiadtobqonnik6obsv9trsvoat3@4ax.com>, John C. Polasek <jpolasek@cfl.rr.com> writes:
On 21 Apr 2006 08:42:06 -0700, "Peter" <Poakfield@msn.com> wrote:
Hi, I understand that an impulsive force can do work; As a matter of
fact, I don't think there is any doubt about it. But, impulse = Ft,
whereas work = Fx. Isn't there an incompatibility here? What am I
missing? Could someone please explain? Thanks.
Peter
Well of course an impulse does work. It's tricky stuff. The basic
formula is
Imp = FT = MV
There is no "of course" about it. Impulse need not do work.
It's not tricky. No waveforms or integration need be involved.
where we can use square time-based force waveforms for integration, no
harm done. It is true that in most cases that you can't know F(t) when
you integrate F(t) from t = 0 to infinity, so how do you get Imp?
Simple, on the RHSide you can measure V and multiply by M giving you
Imp.
The poster appears to be starting with Imp or Ft as the known, not MV.
Well, it's quite obvious we now have mass M moving at V so W = .5MV^2.
No. That is INCORRECT. You are conflating V with delta V.
The original poster didn't mention mass M.
The original poster didn't mention velocity V.
W = .5 M V(f)^2 - .5 M V(i)^2
V(f) = V(i) + V
To make your equations bear any resemblance to reality, let us
assume that initial velocity, V(i) = 0.
And then indeed, since initial energy is zero and final energy is
..5MV^2 we have W = .5MV^2.
But since V isn't a given, this formula isn't immediately useful.
or W = .5Imp*V.
And again, this is a formula for the variable we want (W) in terms of
a mix of knowns (Imp) and unknowns (V). Not very useful.
However...
From conservation of momentum and from the fact that we started with
zero momentum, we have M*V = Imp
Solving for V in terms of Imp, this gives V = Imp/M
Substituting into W = .5MV^2 we get:
W = .5MImp^2/M^2 = .5Imp^2/M
Substituting into W = .5Imp*V we also get
W = .5Imp*V = .5Imp^2/M
But what if the target is not initially at rest?
Assume we have a mass M moving at velocity V and subject to impulse Imp
Further assume that V is large compared with Imp/M
Then W ~= .5Imp dot V
where "dot" denotes the vector dot product.
And that equation, in my opinion, answers the stated question. The
way you relate impulse to work is by taking the dot product with
velocity.
If the target is moving slowly enough that the delta V imparted by
the impulse is significant compared to the target's initial velocity
then the relevant velocity is V+Imp/2M since that is the average
velocity over the duration of the impulsive event.
If the velocity is not aligned with the impulse, the work delivered
by a given impulse might be positive, negative or zero.
The Dirac delta function is the formal method with the time span going
to zero.
But there's no need to invoke the Dirac delta function in the context
of answering a high school physics question.
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| User: "" |
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| Title: Re: Work - impulse |
21 Apr 2006 02:04:03 PM |
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In article <01tIz8QFJPPM@eisner.encompasserve.org>, writes:
Assume we have a mass M moving at velocity V and subject to impulse Imp
Further assume that V is large compared with Imp/M
Then W ~= .5Imp dot V
Gah. Of course, that's W ~= Imp dot V.
The 0.5 has no place in that equation. It appears only in the
case of starting from rest where average velocity is half of final
velocity.
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| User: "John C. Polasek" |
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| Title: Re: Work - impulse |
21 Apr 2006 04:22:40 PM |
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On 21 Apr 2006 14:04:03 -0500, wrote:
In article <01tIz8QFJPPM@eisner.encompasserve.org>, writes:
Assume we have a mass M moving at velocity V and subject to impulse Imp
Further assume that V is large compared with Imp/M
Then W ~= .5Imp dot V
Gah. Of course, that's W ~= Imp dot V.
The 0.5 has no place in that equation. It appears only in the
case of starting from rest where average velocity is half of final
velocity.
What's all your posturing about? First, the above is wrong; you need
the 0.5.
And yes you admit if V0 = 0, I might have something.
I pointed out that in the majority of cases you only know the
resulting MV. Even with an accelerometer, all you'll get by
integrating over time is V, which we can regard as a simple
observable.
After FdT = Mdv or FT = MV you can obviously form the resulting energy
which is W = .5 M V^2 but not Imp dot V.
And yes, yes we are all acquainted with omega x V where any dot
product is zero.
The OP was asking for a relationship between Fdt and Fdx, and the
simple thing is to show that impulse has its energy.
Every material without exception has it compliance so the initial
impulse is first taken up in deflection after which springback
supplies the impetus. I invite you to demonstrate that the compressed
material as an intermediary has the same energy as the primitively
computable .5*Imp*V.
John Polasek
http://www.dualspace.net
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| User: "Peter" |
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| Title: Re: Work - impulse |
21 Apr 2006 02:42:59 PM |
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Thanks, John. It seems to me that if the force F = dp/dt, is not zero,
work is implied there, because work is needed to change the momentum of
an object. So, defining work as W = Fx is redundant. It is also clear
that an impulsive force is not fundamentally different from an ordinary
force. That a force be applied during a short or a long time does not
change the nature of the force. Something seems to be not quite right
with these definitions.
Peter
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| User: "" |
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| Title: Re: Work - impulse |
22 Apr 2006 01:43:09 AM |
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In article <1145648579.804020.10470@z34g2000cwc.googlegroups.com>, "Peter" <Poakfield@msn.com> writes:
Thanks, John. It seems to me that if the force F = dp/dt, is not zero,
work is implied there, because work is needed to change the momentum of
an object.
Nope. Consider a body moving in circle, lets say, in magnetic field.
Its momentum is changing continually but its energy remains constant.
So, defining work as W = Fx is reduntant.
It is *not* fx but F /dot x. Scalar product of vectors.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Mike" |
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| Title: Re: Work - impulse |
21 Apr 2006 03:03:22 PM |
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Peter wrote:
Thanks, John. It seems to me that if the force F = dp/dt, is not zero,
work is implied there, because work is needed to change the momentum of
an object.
Nope. Force changes the momentum of a body, not work, as implied by F =
dp/dt. A force can change the momentum and the work done by the force
can be zero, as in uniform circular motion.
So, defining work as W = Fx is redundant. It is also clear
that an impulsive force is not fundamentally different from an ordinary
force.
Where did you get the term "impulsive force" from? There is force F =
dp/dt and Impulse I = integral {F dt) over t.
That a force be applied during a short or a long time does not
change the nature of the force. Something seems to be not quite right
with these definitions.
Nothing is wrong with the definitions. These are 200 year old concepts
and tested over and over again. But if you think something is wrong, go
ahead and tell us what is it exactly and provide an experiment to
confirm it. I will be much interested in such discovery.
Mike
Peter
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| User: "Henning Makholm" |
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| Title: Re: Work - impulse |
21 Apr 2006 07:09:32 PM |
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Scripsit "Mike" <eleatis@yahoo.gr>
Where did you get the term "impulsive force" from? There is force F =
dp/dt and Impulse I = integral {F dt) over t.
There are authors who use the word "impulse" specifically to mean a
force whose time dependency is a Dirac delta function, i.e. an
infinite force that applies for an infinitely short times, and, in
doing so, transfers a particular finite amount of momentum.
Others use "impulse" to mean simply "change in momentum".
Adding to the confusion is that in some languages (e.g. Danish), the
word for momentum in general is very similar to the English word
"impulse".
--
Henning Makholm "Amanda, I'm a mad scientist!
Testing crazy things on myself and those
who are close to me is my job. It's what I do!"
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| User: "Peter" |
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| Title: Re: Work - impulse |
21 Apr 2006 06:45:23 PM |
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Thanks, Mike. When a point object -like a puck- collides with an arm
of an extended object of different mass -like a lever pivoted at its
center- at rest, or vice versa, angular momentum is conserved, but it
is impossible for kinetic energy to be conserved.
Peter
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| User: "Peter" |
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| Title: Re: Work - impulse |
21 Apr 2006 06:58:34 PM |
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Correction.
Thanks, Mike. When a point object -like a puck- collides with an arm
of an extended object of different mass -like a lever pivoted at its
center- at rest, or vice versa, and the incident object stops on
impact, angular momentum is conserved, but it is impossible for kinetic
energy to be conserved.
Peter
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| User: "Mike" |
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| Title: Re: Work - impulse |
22 Apr 2006 06:39:44 AM |
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Peter wrote:
Correction.
Thanks, Mike. When a point object -like a puck- collides with an arm
of an extended object of different mass -like a lever pivoted at its
center- at rest, or vice versa, and the incident object stops on
impact, angular momentum is conserved, but it is impossible for kinetic
energy to be conserved.
Kinetic energy is not always conserved. Only momentum is conserved in
collissions. I do not understand what this has to do with the
definitions of impulse and force and you alluding that there is
something wrong. If there is indeed something wrong you haven't
quantitatively described how the results of an experiment vary from the
theoretical results based on said definitions. For example, do you
disagree that a force orthogonal to dispacement vector does no work?
And if you do, can you provide a valid reason for such disagreement?
Mike
Peter
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| User: "" |
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| Title: Re: Work - impulse |
22 Apr 2006 01:46:30 PM |
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In article <1145710165.715899.189460@e56g2000cwe.googlegroups.com>, "Peter" <Poakfield@msn.com> writes:
Mike, I do not disagree that a force orthogonal to the displacement
vector does no work. What I say is that when one object collides with
another object of different mass at rest, and the incident object stops
on impact, momentum is conserved, but kinetic energy (work) cannot
possibly be conserved, even when no permanent deformation of the
objects occurs, and there is no friction or other causes of energy
loss. This is because momentum is a linear function of velocity, but
kinetic energy (work) is a quadratic function. This can occur when a
point mass, like a puck, collides with an extended object, like a
lever, where angular momentum is conserved. This means the same amount
of work = F dot x can result in different amounts of kinetic energy
(work), which shows there is something wrong with the definitions of
work.
Did it occur to you that, perhaps, there is something wrong with your
understanding here?
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Peter" |
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| Title: Re: Work - impulse |
22 Apr 2006 03:01:52 PM |
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"Did it occur to you that, perhaps, there is something wrong with your
understanding here?" Yes, it did, I checked, there is no
misunderstanding.
Peter
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| User: "Mike" |
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| Title: Re: Work - impulse |
22 Apr 2006 03:12:22 PM |
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Peter wrote:
"Did it occur to you that, perhaps, there is something wrong with your
understanding here?" Yes, it did, I checked, there is no
misunderstanding.
Peter
Peter why don't you write down the equations for the problem and show
what you mean. People here are used to that. Talk is cheap to them.
Mike
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| User: "Peter" |
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| Title: Re: Work - impulse |
22 Apr 2006 03:58:11 PM |
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Mike, I am questioning the standard equations for impulse Ft = mv and
work Fx = max = (1/2)mv^2. The former is an unnecessary and useless
manipulation of F = dp/dt, and the latter is only true for point
masses. It fails when applied to work done on extended objects.
Peter
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| User: "Mike" |
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| Title: Re: Work - impulse |
22 Apr 2006 06:58:30 PM |
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Peter wrote:
Mike, I am questioning the standard equations for impulse Ft = mv and
work Fx = max = (1/2)mv^2. The former is an unnecessary and useless
manipulation of F = dp/dt, and the latter is only true for point
masses. It fails when applied to work done on extended objects.
First of all, you were told before that these equations you present are
not the standard equations used ion physics.
I = int {f dt} over t
W = int (F dot ds) over path C
Actually, from the former the second law is derived and it is not a
useless manipulation of anything.
The "latter" is true for all objects not just for point masses.
When questioning physical principles you cannot say that you are just
questioning because this principle fails to apply in such and such
situation. You must prove why this is the case, eiter theoretically or
experimentally. Since you have not done so but you are asking us to
take your word for it, you are really wasting our time and yours. If
you cannot present valid reasons as to why your belief is true you are
going to be labelled as another crank by many people here who are fed
up with some idiots that visit frequently in sci.physics and declare
some concepts invalid without at least pointing to a result oir an
experiment, even wrong it does not matter, that corroborates their
statements.
So if you want to avoid such situation, either present valid arguments
based on theoretical of experimental results that corroborates your
statement above or I sugegst we drop this conversation because it is
pointless to talk about beliefs.
Mike
Peter
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| User: "Peter" |
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| Title: Re: Work - impulse |
23 Apr 2006 07:25:05 PM |
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Mike, I agree. Specifically, for instance, when a puck collides
(perpendicularly) with one arm of a lever of different mass at rest
(lever is a thin rod pivoted at its center), and the puck stops on
impact, even if no energy losses occur, it is impossible for angular
momentum and kinetic energy to be conserved at the same time. They can
only be conserved if it is true that m r1^2 w1 = M r^2/3 w2, and
(1/2)m(r1*w1)^2 = (1/2)M(r*w2)^2/3, where m and M are the masses of the
puck and lever, respectively, r1 the distance from the pivot the puck
strikes, r the length of each arm of the lever, and w1 and w2 the
angular velocities of the puck an instance before it hits the lever and
that of the lever an instance after the collision, respectively. But
this can only happen if the puck and the lever have the same moment of
inertia, and the puck stops on impact. However, empirical tests show
that if the puck and lever have the same moment of inertia, it is
impossible for the puck to stop on impact (although there are many
pucks of different masses that will stop on impact with the lever). Of
course I know why this happens, and I can prove it, but since it points
to a little problem with some currently cherished concepts, it is
probably not prudent to mention it now. though I will, in time. I hope
this is clear. Incidentally, all I am posting is copyrighted material.
Peter
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| User: "Mike" |
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| Title: Re: Work - impulse |
24 Apr 2006 04:19:04 AM |
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Peter wrote:
... of course I know why this happens, and I can prove it, but since it points
to a little problem with some currently cherished concepts, it is
probably not prudent to mention it now. though I will, in time. I hope
this is clear. Incidentally, all I am posting is copyrighted material.
Peter
I am afraid we cannot continue this discussion. I have heard such
claims before but there was never a proof presented, just promises. I
think people may have serious doubts you got something since you did
not even present he correct equations for work and impulse and you
still used the wrong ones even when your were pointed to the correct
ones. This MAY be an indications that you do not understand the
concepts. Nevertheless, even if you do, a promise of a proof is no
proof in physics. So nothing is clear with you and if you want to claim
the copyright to W = FX go ahead and do it. So long
Mike
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| User: "" |
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| Title: Re: Work - impulse |
22 Apr 2006 03:25:21 PM |
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In article <1145736112.529905.226610@t31g2000cwb.googlegroups.com>, "Peter" <Poakfield@msn.com> writes:
"Did it occur to you that, perhaps, there is something wrong with your
understanding here?" Yes, it did, I checked, there is no
misunderstanding.
:-)))) In this case, certainly, there is no reason for me to waste
even one additional second on this thread.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "" |
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| Title: Re: Work - impulse |
23 Apr 2006 06:53:11 AM |
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In article <Riw2g.35$25.2029@news.uchicago.edu>,
wrote:
In article <1145736112.529905.226610@t31g2000cwb.googlegroups.com>, "Peter"
<Poakfield@msn.com> writes:
"Did it occur to you that, perhaps, there is something wrong with your
understanding here?" Yes, it did, I checked, there is no
misunderstanding.
:-)))) In this case, certainly, there is no reason for me to waste
even one additional second on this thread.
Oh, shoot. I was just going to hand you 20 boxes of Kleenix.
/BAH
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| User: "" |
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| Title: Re: Work - impulse |
23 Apr 2006 02:07:40 PM |
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In article <e2fpr7$8ps_002@s1011.apx1.sbo.ma.dialup.rcn.com>, writes:
In article <Riw2g.35$25.2029@news.uchicago.edu>,
mmeron@cars3.uchicago.edu wrote:
In article <1145736112.529905.226610@t31g2000cwb.googlegroups.com>, "Peter"
<Poakfield@msn.com> writes:
"Did it occur to you that, perhaps, there is something wrong with your
understanding here?" Yes, it did, I checked, there is no
misunderstanding.
:-)))) In this case, certainly, there is no reason for me to waste
even one additional second on this thread.
Oh, shoot. I was just going to hand you 20 boxes of Kleenix.
Save them for a better use. This isn't it.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Peter" |
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| Title: Re: Work - impulse |
22 Apr 2006 07:49:25 AM |
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Mike, I do not disagree that a force orthogonal to the displacement
vector does no work. What I say is that when one object collides with
another object of different mass at rest, and the incident object stops
on impact, momentum is conserved, but kinetic energy (work) cannot
possibly be conserved, even when no permanent deformation of the
objects occurs, and there is no friction or other causes of energy
loss. This is because momentum is a linear function of velocity, but
kinetic energy (work) is a quadratic function. This can occur when a
point mass, like a puck, collides with an extended object, like a
lever, where angular momentum is conserved. This means the same amount
of work = F dot x can result in different amounts of kinetic energy
(work), which shows there is something wrong with the definitions of
work.
Peter
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| User: "Herman Trivilino" |
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| Title: Re: Work - impulse |
22 Apr 2006 08:30:05 AM |
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"Peter" <Poakfield@msn.com> wrote ...
point mass, like a puck, collides with an extended object, like a
lever, where angular momentum is conserved. This means the same amount
of work = F dot x can result in different amounts of kinetic energy
(work), which shows there is something wrong with the definitions of
work.
There is nothing "wrong" with the definition of work.
Work does not, in general, equal the change in kinetic energy. The
work-energy theorem is valid for particle-like objects, objects that cannot
change their internal energy.
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