Work - impulse
Peter
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
Mike
Actually, for a constant force F, I = F x delta(t), the time interval
the force applies.
In general, I = integral{Fx dt} over time t
On the other hand, Work is the integral of force F over the path:
W = integral{ F dot ds) over C (path)
Some people confuse the two definitions.
An impulse does not do work, it measures change in momentum p. Force
does work. As a matter of fact, momentum can be changing with no work
performed, as in the case of uniform circular motion.
Mike
srp
You could define impulse as being "a brief acceleration", if this
makes it clearer for you.
André Michaud
Peter
object has uniform circular motion, no work is being done on it.
Peter
Peter
was done on it?
Peter
John C. Polasek
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
PD
Precisely. The centripetal force, which acts perpendicular to the
tangential motion, does no work but does deliver an impulse.
An impulse is an external change in momentum, which can mean either a
change in the magnitude of the momentum (then there's work, too) or a
change in the direction only (no work), or both.
PD
mme...@cars3.uchicago.edu
>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,
me...@cars.uchicago.edu | chances are he is doing just the same"
Mike
Because the change of momentum is orthogonal to the path. It gives
rise to a centripetal force directed towards the center of rotation. F
dot ds is zero everywhere along the path.
The confusion often arises from the following: in order to set a body
in uniform circular motion you must do work, of course. Example: using
you hand or using an electric motor. But the force on the body does
not do any work. This is the point.
You can look at it in another way. Energy is the ability to do work.
The energy of a body rotating uniformly equals mv^2/2 for
non-relativistic conditions. In uniform circular motion v is constant
and thus dE/dt = 0. No energy change along the path means no work is
done anywhere along the path since there is no flow of energy from the
outside.
Mike
bri...@encompasserve.org
> On 21 Apr 2006 08:42:06 -0700, "Peter" <Poak...@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.
bri...@encompasserve.org
> 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.
Peter
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
Igor
Apparently, you're still not getting it. First of all, there is no
such thing as an impulsive force. There is just force (vector) defined
as you already did as dp/dt. There's also impulse (vector), which is a
change in momentum, defined as int F dt. Then there's work (scalar),
which is int F*dx (where * represents the dot product). And work is
not necessary to change an object's momentum. If dx is perpendicular
to F, then work is zero, but impulse is still a finite number. This is
what happens in uniform circular motion. No work done, but momentum is
constantly changing, if only in its direction. But this still
constitutes non-zero impulse.
Sam Wormley
Mike
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
mme...@cars3.uchicago.edu
>Thanks Andre. Doesn't the brief acceleration of an object imply work
>was done on it?
>
already had, and its direction, relative to the impulse.
John C. Polasek
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
Entropy
"Peter" <Poak...@msn.com> wrote in message
news:1145634126....@i40g2000cwc.googlegroups.com...
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
Peter
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
Peter
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
Henning Makholm
> 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!"
srp
> Thanks Andre. Doesn't the brief acceleration of an object imply work
> was done on it?
I must say that in a classical mechanics context a short definition
of "work" does not come to me as easily as one for impulse.
For example, the whole body of experimental data is in agreement
with Newton’s principle of inertia that any particle possessing
inertia will resist having its direction of motion changed from
straight line trajectory by a force acting transversally and in
all such cases energy has to be expended to force the change in
direction, which means that work has to be performed to induce
the change in direction. Here, impulse is deemed to imply, and
rightly so, that work has been done.
This is conform to the 2nd Principle of Thermodynamics that implies
that it is impossible that any change of state of motion could occur
without some expenditure of energy.
However, there is a centuries old assumption in physics circles
that angular momentum does not imply a change of state, and that
perfectly circular motion (which fundamentally is "constant change
in direction") does not involve that work is being performed. In this
case impulse is apparently not deemed to imply that work is being
done.
I have my own idea on this issue, not shared by most, which I will
refrain from stating, since it would conflict with classical mechanics
definition and could only mix you up at this point, which is not what
I intend.
André Michaud
mme...@cars3.uchicago.edu
>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.
Mike
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
Mike
srp wrote:
> Peter a écrit :
> > Thanks Andre. Doesn't the brief acceleration of an object imply work
> > was done on it?
>
> I must say that in a classical mechanics context a short definition
> of "work" does not come to me as easily as one for impulse.
>
> For example, the whole body of experimental data is in agreement
> with Newton's principle of inertia that any particle possessing
> inertia will resist having its direction of motion changed from
> straight line trajectory by a force acting transversally and in
> all such cases energy has to be expended to force the change in
> direction, which means that work has to be performed to induce
> the change in direction. Here, impulse is deemed to imply, and
> rightly so, that work has been done.
>
> This is conform to the 2nd Principle of Thermodynamics that implies
> that it is impossible that any change of state of motion could occur
> without some expenditure of energy.
>
> However, there is a centuries old assumption in physics circles
> that angular momentum does not imply a change of state, and that
> perfectly circular motion (which fundamentally is "constant change
> in direction") does not involve that work is being performed. In this
> case impulse is apparently not deemed to imply that work is being
> done.
Impulse results in a change of momentum: I = F dt = m dv for a constant
force. From this you get Newton's seconf law: F = mdv/dt for a constant
mass.
Work is performed by a force F that displaces a mass m along a path C
according to:
W = int {F dot ds) over C
If there is a change in direction of momentum only in such a way that
the change is orthogonal to the path C but not change in magnitude then
work is not being done. It is easy to see that since the energy content
of the mass m remains the same: KE=mv^2/2 = constant. If work were to
be done then the energy content would not remain the same.
In contrast, in elliptical orbits, since the radial dispancement r
changes, there is work being done by the gravity force. However, the
mechanical energy remains constant, there is just an exchange between
kinetic and potential energy.
>
> I have my own idea on this issue, not shared by most, which I will
> refrain from stating, since it would conflict with classical mechanics
> definition and could only mix you up at this point, which is not what
> I intend.
You refrain because you are afraid to be called a crank. Why should you
be afraid? If you tell us now you do not run the risk of someone else
making the same discovery and claiming it before you. I think that if
you know something you should go ahead and say it. Just place a note
that you hold the copyright to the material. The added benefit is that
you may end up improving the concept from the feedaback you will get.
On the other hand, if you say nothing it means 99% of the time that you
have nothing.
Mike
>
> André Michaud
Peter
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
Herman Trivilino
> 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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Peter
objects that cannot change their internal energy. But shouldn't it also
be valid for extended objects that cannot change their internal energy?
Otherwise, we have cases where the same amount of work can result in
different amounts of kinetic energy, which should not be.
Peter
Gregory L. Hansen
Peter <Poak...@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.
Gregory L. Hansen
Gregory L. Hansen <glha...@steel.ucs.indiana.edu> wrote:
>In article <1145634126....@i40g2000cwc.googlegroups.com>,
>Peter <Poak...@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
Peter
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
PD
srp wrote:
> Peter a écrit :
> > Thanks Andre. Doesn't the brief acceleration of an object imply work
> > was done on it?
>
> I must say that in a classical mechanics context a short definition
> of "work" does not come to me as easily as one for impulse.
>
> For example, the whole body of experimental data is in agreement
> with Newton's principle of inertia that any particle possessing
> inertia will resist having its direction of motion changed from
> straight line trajectory by a force acting transversally and in
> all such cases energy has to be expended to force the change in
> direction,
That is incorrect. No energy is expended in forcing Earth to change its
direction so to make its path an ellipse around the sun.
> which means that work has to be performed to induce
> the change in direction. Here, impulse is deemed to imply, and
> rightly so, that work has been done.
>
> This is conform to the 2nd Principle of Thermodynamics that implies
> that it is impossible that any change of state of motion could occur
> without some expenditure of energy.
I don't know where the heck you got that idea.
>
> However, there is a centuries old assumption in physics circles
> that angular momentum does not imply a change of state, and that
> perfectly circular motion (which fundamentally is "constant change
> in direction") does not involve that work is being performed. In this
> case impulse is apparently not deemed to imply that work is being
> done.
That's correct.
Mike
Peter wrote:
> Mike, I do not disagree that a force orthogonal to the displacement
> vector does no work.
Good we got past that.
> 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.
Kinetic energy is conserved if the collision is perfectly elastic
without any losses to to heat dissipation or stress energy release.
> This is because momentum is a linear function of velocity, but
> kinetic energy (work) is a quadratic function.
Momentum is a vector. KE is a scalar. They are both frame dependent
quantities with the exception that momentum is always conserved in the
absence of external forces.
> 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.
No it does not. You are equating KE with Work. That is not the case.
More importantly, Work is not equal to F dot x but to integral of {F
dor dx) over the path C.
Do a favor to yourself and get an introductory College level physics
book and study these concepts. Try to solve the problems at the end of
the chapter. Be sure that if your are skeptical about the concepts and
how they work it is because you have not been exposed to them in their
strict form, just as an introductory level.
I am sure if you do that you will have no questions. Everything works
just fine in Mechanics. No magic, no bogus concepts, no violations. If
there was a bit of violation of the defintions you mentioned then
equivalence of inertial to gravitational mass would not have been
experimentally confirmed with an accuracy of at least 1 part in a
trillion. Some people have done us some favors.
Mike
>
> Peter
Mike
PD wrote:
> srp wrote:
> > Peter a écrit :
> > > Thanks Andre. Doesn't the brief acceleration of an object imply work
> > > was done on it?
> >
> > I must say that in a classical mechanics context a short definition
> > of "work" does not come to me as easily as one for impulse.
> >
> > For example, the whole body of experimental data is in agreement
> > with Newton's principle of inertia that any particle possessing
> > inertia will resist having its direction of motion changed from
> > straight line trajectory by a force acting transversally and in
> > all such cases energy has to be expended to force the change in
> > direction,
>
> That is incorrect. No energy is expended in forcing Earth to change its
> direction so to make its path an ellipse around the sun.
But you agree that work is done by the gravity force since the radial
distance from the force center changes slightly (in the case of the
earth eccenticity is only 0.02). At the same time however, Potential
energy changes by the same amount and thus total mechanical energy
remains constant. i.e.e there is no need to pump energy to have the
earth revolve around the sun and no energy is released by that motion.
However, since the earth is not perfectly round and there is the tidal
effect, there is energy dissipation in the earth-moon system with the
result being the moon distance from the eartg increases by 3cm every
year, or so.
Mike
PD
No, actually this is not right. It is certainly possible for a
collision between two unequal masses, one of them being initially at
rest, to conserve both momentum and kinetic energy.
Let's take a simple 1D example.
m1 = 3 kg
m2 = 1 kg
v1i = 4 m/s
v2i = 0 m/s
v1f = 2 m/s
v2f = 6 m/s
The initial momentum is (3)(4) + (1)(0) = 12 kg*m/s
The final momentum is (3)(2) + (1)(6) = 12 kg*m/s, which is the same.
The initial KE is (1/2)(3)(4)^2 + (1/2)(1)(0)^2 = 24 joules
The final KE is (1/2)(3)(2)^2 + (1/2)(1)(6)^2 = 24 joules, which is the
same.
Since the initial and final energies are the same, we know that the
collision was elastic and that no energy was dissipated to heat or
deformation.
Seeing is believing!
PD
PD
Mike wrote:
> PD wrote:
> > srp wrote:
> > > Peter a écrit :
> > > > Thanks Andre. Doesn't the brief acceleration of an object imply work
> > > > was done on it?
> > >
> > > I must say that in a classical mechanics context a short definition
> > > of "work" does not come to me as easily as one for impulse.
> > >
> > > For example, the whole body of experimental data is in agreement
> > > with Newton's principle of inertia that any particle possessing
> > > inertia will resist having its direction of motion changed from
> > > straight line trajectory by a force acting transversally and in
> > > all such cases energy has to be expended to force the change in
> > > direction,
> >
> > That is incorrect. No energy is expended in forcing Earth to change its
> > direction so to make its path an ellipse around the sun.
>
> But you agree that work is done by the gravity force since the radial
> distance from the force center changes slightly (in the case of the
> earth eccenticity is only 0.02). At the same time however, Potential
> energy changes by the same amount and thus total mechanical energy
> remains constant. i.e.e there is no need to pump energy to have the
> earth revolve around the sun and no energy is released by that motion.
My mistake. There is no energy expended over an entire orbit, except
for the deformation losses you meantion below.
In a circular orbit, there is no energy expended at any point in the
orbit, except for the deformation losses you mention below.
PD
PD
Specific example, please.
PD
Mike
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
Mike
Agree, in the case of the elliptic closed path, net work done is zero
since of forces are conservative (holonomic system). In the case of the
circular closed orbit no work is done at any point on the orbit.
Mike
mme...@cars3.uchicago.edu
>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.
>
understanding here?
Peter
misunderstanding.
Peter
Mike
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
mme...@cars3.uchicago.edu
>"Did it occur to you that, perhaps, there is something wrong with your
>misunderstanding.
>
even one additional second on this thread.
Peter
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
Dirk Van de moortel
"PD" <TheDrap...@gmail.com> wrote in message news:1145724062....@t31g2000cwb.googlegroups.com...
Careful, he demanded v1f = 0:
"and the incident object stops on impact"
The only ways to have v2i = 0 and v1f = 0, are
if m1 = m2 or if v1i = v2f = 0.
Dirk Vdm
Mike
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
PD
Ah, yes, I missed that. That doesn't happen in real life unless there
is some energy dissipated somewhere.
srp
Not necessarily. In electromagnetism, energy is induced as a function
of distance (and only distance) between charged particles, Irrespective
of whether work is done the force induced energy as a function of a
given distance can simply not vary. If work should consume some
energy for whatever reason, if the distance remains the same, the
energy level is maintained (constantly being replenished and
maintained by the acting force, in accordance with the fundamental
law)
> In contrast, in elliptical orbits, since the radial dispancement r
> changes, there is work being done by the gravity force. However, the
> mechanical energy remains constant, there is just an exchange between
> kinetic and potential energy.
In CM yes.
>>I have my own idea on this issue, not shared by most, which I will
>>refrain from stating, since it would conflict with classical mechanics
>>definition and could only mix you up at this point, which is not what
>>I intend.
>
>
> You refrain because you are afraid to be called a crank. Why should you
> be afraid?
I am not afraid. I simply have no chip on my shoulder.
> If you tell us now you do not run the risk of someone else
> making the same discovery and claiming it before you.
I have had this discussion years ago, on this very forum. Not
really interested in a repeat.
You simply need to dig into electromagnetism until you understand.
> I think that if you know something you should go ahead and say it.
I have said it in the past.
> Just place a note that you hold the copyright to the material.
No need for a note. The material is copyrited to the hilt and has
been spreading into institution for years.
> The added benefit is that you may end up improving the concept from
> the feedaback you will get.
I am here to learn.
> On the other hand, if you say nothing it means 99% of the time that
> you have nothing.
Apparently nothing that would mean anything to you, no doubt.
André Michaud
srp
> srp wrote:
>
>>Peter a écrit :
>>
>>>Thanks Andre. Doesn't the brief acceleration of an object imply work
>>>was done on it?
>>
>>I must say that in a classical mechanics context a short definition
>>of "work" does not come to me as easily as one for impulse.
>>
>>For example, the whole body of experimental data is in agreement
>>with Newton's principle of inertia that any particle possessing
>>inertia will resist having its direction of motion changed from
>>straight line trajectory by a force acting transversally and in
>>all such cases energy has to be expended to force the change in
>>direction,
>
>
> That is incorrect. No energy is expended in forcing Earth to change its
> direction so to make its path an ellipse around the sun.
That is the classical assumption, as I stated.
I suggest you try forcing a baseball to change its trajectory
without any work being expended.
>>which means that work has to be performed to induce
>>the change in direction. Here, impulse is deemed to imply, and
>>rightly so, that work has been done.
>>
>>This is conform to the 2nd Principle of Thermodynamics that implies
>>that it is impossible that any change of state of motion could occur
>>without some expenditure of energy.
>
>
> I don't know where the heck you got that idea.
As I said, 2nd principle of thermodynamics.
No change of state of motion possible without energy being expended.
André Michaud
jmfb...@aol.com
mme...@cars3.uchicago.edu wrote:
>In article <1145736112.5...@t31g2000cwb.googlegroups.com>, "Peter"
<Poak...@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
Herman Trivilino
> Herman, I agree that the work-energy theorem is valid for particle-like
> objects that cannot change their internal energy. But shouldn't it also
> be valid for extended objects that cannot change their internal energy?
No, it is not. Take, for example, the work done on a gas during an
isothermal process.
> Otherwise, we have cases where the same amount of work can result in
> different amounts of kinetic energy, which should not be.
I'm not sure what criteria you are using to establish what "should not be".
Using the definitions of work and kinetic energy, you find that the change
in kinetic energy doesn't always equal the work done. Theory predicts, and
experiments confirm. In this sense, they "should not be" equal.
If you are using your own sense of what "should not be" then perhaps your
sense is wrong. Wrong, that is, because the way you think Nature ought to
behave is not matching the way Nature does behave.
Or, perhaps you think that the definition of work needs to modified so that
the work done always equals the change in kinetic energy. You could simply
define work to be equal to the change in kinetic energy. Of course, your
definition will not match the definition that others use. You'll have
trouble, too, when you try to state things like the First Law of
Thermodynamics. The way it's currently stated is based on a definition of
work that doesn't coincide with yours. You could then come up with a new
statement of the First Law. Etcetera.
In the end, though, the definitions that get accepted are the definitions
that demonstrate themselves to be useful. If your definition of work is not
useful, it'll never get accepted.
PD
srp wrote:
> PD a écrit :
> > srp wrote:
> >
> >>Peter a écrit :
> >>
> >>>Thanks Andre. Doesn't the brief acceleration of an object imply work
> >>>was done on it?
> >>
> >>I must say that in a classical mechanics context a short definition
> >>of "work" does not come to me as easily as one for impulse.
> >>
> >>For example, the whole body of experimental data is in agreement
> >>with Newton's principle of inertia that any particle possessing
> >>inertia will resist having its direction of motion changed from
> >>straight line trajectory by a force acting transversally and in
> >>all such cases energy has to be expended to force the change in
> >>direction,
> >
> >
> > That is incorrect. No energy is expended in forcing Earth to change its
> > direction so to make its path an ellipse around the sun.
>
> That is the classical assumption, as I stated.
>
> I suggest you try forcing a baseball to change its trajectory
> without any work being expended.
Sure. Just fire it at 25000 mph so that the force of gravity is always
perpendicular to its velocity. The fact that you can find examples
where work is done in a trajectory does not imply that work is always
done in a trajectory.
>
> >>which means that work has to be performed to induce
> >>the change in direction. Here, impulse is deemed to imply, and
> >>rightly so, that work has been done.
> >>
> >>This is conform to the 2nd Principle of Thermodynamics that implies
> >>that it is impossible that any change of state of motion could occur
> >>without some expenditure of energy.
> >
> >
> > I don't know where the heck you got that idea.
>
> As I said, 2nd principle of thermodynamics.
>
> No change of state of motion possible without energy being expended.
Please explain in some detail where the "2nd principle of
thermodynamics" says this.
PD
PD
Dirk Van de moortel wrote:
I thought about this a little more in the shower, and there is indeed a
way to have this happen, even in 1D. Have the larger puck be incident
with an angular velocity w1i, and say the two moments of inertia are I1
and I2 (equal to c1*m1 and c2*m2, respectively, where c1 and c2 are
geometry dependent).
Now we'll have more constraints:
(m1)(v1i) + (m2)(0) = (m1)(0) + (m2)(v2f)
(I1)(w1i) + (I2)(0) = (I1)(w1f) + (I2)(w2f)
(1/2)(m1)(v1i)^2 + (1/2)(I1)(w1i)^2 + (1/2)(m2)(0)^2 + (1/2)(m2)(0)^2 =
(1/2)(m1)(0)^2 + (1/2)(I1)(w1f)^2 + (1/2)(m2)(v2f)^2 + (1/2)(I2)(w2f)^2
Now we have three equations and three unknowns: v2f, w1f, w2f.
Looks solvable to me.
This doesn't mean that Peter is write about the formula for work being
wrong. He just hasn't accounted for where that work can go (into
rotational kinetic energy).
PD
mme...@cars3.uchicago.edu
>In article <Riw2g.35$25....@news.uchicago.edu>,
> mme...@cars3.uchicago.edu wrote:
><Poak...@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.
>
>
Dirk Van de moortel
"PD" <TheDrap...@gmail.com> wrote in message news:1145818191.2...@e56g2000cwe.googlegroups.com...
Yes, absolutely, even if w1i = 0.
>
> This doesn't mean that Peter is write about the formula for work being
> wrong. He just hasn't accounted for where that work can go (into
> rotational kinetic energy).
If for example w1i = 0, there will obviously be 2 solutions:
one with clock/anticlock rotation
w1f < 0 < w2f
and another with anticlock/clock rotation
w2f < 0 < w1f
I guess that, if for instance I1 = I2, you must have m1 > m2.
You have an inspiring shower :-)
Dirk Vdm
PD
Except if w1i and w2i were both zero, and they were symmetric pucks,
then the only way that they could induce spin is if there were a
contact force that generated torque, which means that they made contact
off-center, in which case the problem would no longer be 1D.
However, if the two objects were not symmetric but were instead, say,
dog-bone-shaped, then there very well could be a contact force that
would induce a torque even though the force is still along the
direction of motion of the center of mass. In this case, you could get
rotation where there was none before and still have 1D motion.
PD
Peter
(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
srp
> srp wrote:
>
>>PD a écrit :
>>
>>>srp wrote:
>>>
>>>
>>>>Peter a écrit :
>>>>
>>>>
>>>>>Thanks Andre. Doesn't the brief acceleration of an object imply work
>>>>>was done on it?
>>>>
>>>>I must say that in a classical mechanics context a short definition
>>>>of "work" does not come to me as easily as one for impulse.
>>>>
>>>>For example, the whole body of experimental data is in agreement
>>>>with Newton's principle of inertia that any particle possessing
>>>>inertia will resist having its direction of motion changed from
>>>>straight line trajectory by a force acting transversally and in
>>>>all such cases energy has to be expended to force the change in
>>>>direction,
>>>
>>>
>>>That is incorrect. No energy is expended in forcing Earth to change its
>>>direction so to make its path an ellipse around the sun.
>>
>>That is the classical assumption, as I stated.
>>
>>I suggest you try forcing a baseball to change its trajectory
>>without any work being expended.
>
> Sure. Just fire it at 25000 mph so that the force of gravity is always
> perpendicular to its velocity.
No. Don't let gravity do the job for you.
I challenge you to cause a baseball to change its trajectory without
any work being expended.
Just do it and prove to us that no work was needed to do it.
> The fact that you can find examples
> where work is done in a trajectory does not imply that work is always
> done in a trajectory.
Wow! So now you admit that work is needed to deflect a trajectory,
but then by the same token, you assert that work is not needed to
deflect a trajectory.
So the fundamental laws of nature sometimes act and sometimes don't
for identical problems.
Howfully clear thinking.
>>>>which means that work has to be performed to induce
>>>>the change in direction. Here, impulse is deemed to imply, and
>>>>rightly so, that work has been done.
>>>>
>>>>This is conform to the 2nd Principle of Thermodynamics that implies
>>>>that it is impossible that any change of state of motion could occur
>>>>without some expenditure of energy.
>>>
>>>
>>>I don't know where the heck you got that idea.
>>
>>As I said, 2nd principle of thermodynamics.
>>
>>No change of state of motion possible without energy being expended.
>
>
> Please explain in some detail where the "2nd principle of
> thermodynamics" says this.
The second principle of thermodynamics says that in an isolated
system, a spontaneous change of state cannot be produced, for any
physical event possessing a certain level of energy, unless
circumstances allow this event to reach a lower state of energy.
Hence _no_ change in the state of motion of a body can occur without
an expenditure of energy (work).
Even if a stabilizing force constantly replenishes the energy can't
prevent such expenditure to occur.
André Michaud
Mike
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
PD
srp wrote:
> PD a écrit :
> > srp wrote:
> >
> >>PD a écrit :
> >>
> >>>srp wrote:
> >>>
> >>>
> >>>>Peter a écrit :
> >>>>
> >>>>
> >>>>>Thanks Andre. Doesn't the brief acceleration of an object imply work
> >>>>>was done on it?
> >>>>
> >>>>I must say that in a classical mechanics context a short definition
> >>>>of "work" does not come to me as easily as one for impulse.
> >>>>
> >>>>For example, the whole body of experimental data is in agreement
> >>>>with Newton's principle of inertia that any particle possessing
> >>>>inertia will resist having its direction of motion changed from
> >>>>straight line trajectory by a force acting transversally and in
> >>>>all such cases energy has to be expended to force the change in
> >>>>direction,
> >>>
> >>>
> >>>That is incorrect. No energy is expended in forcing Earth to change its
> >>>direction so to make its path an ellipse around the sun.
> >>
> >>That is the classical assumption, as I stated.
> >>
> >>I suggest you try forcing a baseball to change its trajectory
> >>without any work being expended.
> >
> > Sure. Just fire it at 25000 mph so that the force of gravity is always
> > perpendicular to its velocity.
>
> No. Don't let gravity do the job for you.
??
>
> I challenge you to cause a baseball to change its trajectory without
> any work being expended.
Put a baseball in orbit.
Or put a small satellite in orbit. Will that do?
>
> Just do it and prove to us that no work was needed to do it.
>
> > The fact that you can find examples
> > where work is done in a trajectory does not imply that work is always
> > done in a trajectory.
>
> Wow! So now you admit that work is needed to deflect a trajectory,
> but then by the same token, you assert that work is not needed to
> deflect a trajectory.
No. What I said is that there are *instances* of work being involved in
changing a trajectory. This does not mean that work is *required* in
changing a trajectory. Get a grip.
>
> So the fundamental laws of nature sometimes act and sometimes don't
> for identical problems.
>
> Howfully clear thinking.
>
> >>>>which means that work has to be performed to induce
> >>>>the change in direction. Here, impulse is deemed to imply, and
> >>>>rightly so, that work has been done.
> >>>>
> >>>>This is conform to the 2nd Principle of Thermodynamics that implies
> >>>>that it is impossible that any change of state of motion could occur
> >>>>without some expenditure of energy.
> >>>
> >>>
> >>>I don't know where the heck you got that idea.
> >>
> >>As I said, 2nd principle of thermodynamics.
> >>
> >>No change of state of motion possible without energy being expended.
> >
> >
> > Please explain in some detail where the "2nd principle of
> > thermodynamics" says this.
>
> The second principle of thermodynamics says that in an isolated
> system, a spontaneous change of state cannot be produced, for any
> physical event possessing a certain level of energy, unless
> circumstances allow this event to reach a lower state of energy.
>
> Hence _no_ change in the state of motion of a body can occur without
> an expenditure of energy (work).
That is, of course, crap. If you have degenerate states of motion with
the same energy level, there is no work required to go from one state
to another. Your understanding of the 2nd law of thermodynamics is as
shallow as your understanding of simple Newtonian mechanics. And to
think I once tried to explain special relativity to you.
PD
Peter
point-like objects of different mass, it is impossible for the incident
object to stop on impact. By the way, I derived from conservation on
momentum only, a simple equation to find the final velocity of either
of the two objects in such collisions. It is vf = (2P/M)-vi, where vf
and vi are the final and initial velocities of either object, P the
total momentum, and M the total mass of the two objects.
Peter
Peter
(perpendicularly) with one arm of a lever at rest (the lever is a thin
rod pivoted at its longitudinal center) with different mass, and the
puck stops completely on impact. I say that in this case, it is
impossible for the final kinetic energy of the lever, to be equal to
the initial kinetic energy of the puck, even if no energy losses occur.
Moreover, if the same puck, with the same kinetic energy collides with
one arm of a heavier lever and stops on impact, the kinetic energy of
the heavier lever would be different from that of the first lever. I
hope this is clear.
Peter
PD
It is a standard exercise to show that in a 1D *elastic* collision
between a big, moving object (mass M, velocity V) and a stationary,
smaller object (mass m), the final velocities are
(2m/(M+m))*V and ((M-m)/(M+m))*V.
It is left as a smaller exercise to the reader to determine which
velocity belongs to which mass by deriving these expressions, and to
show that (in general) the relative velocity is a conserved quantity in
elastic collisions.
PD
PD
Keep in mind that the kinetic energy of the lever consists of TWO
terms, the linear term and the angular term. Unlike momentum, the
linear and angular kinetic energies are not conserved separately but
are mixed.
KE = (1/2)mv^2 + (1/2)Iw^2
PD
tj Frazir
Impulsive force is POTENTUAL energy.
You cant move the brick untill motion i conducted to the other end of
the brick at c.
You cant move the brick untill you overcome RESISTANCE to impulsive
FORCE
tj Frazir
One end of the brick stops or starts then the other end starts or stops
.
If you hit the brick one end is moving and the other end must wait to
move till c gets there .
IF the brick was 1 light year long ,,the other end would not move for a
year.
Peter
All I say is that you always do the same work to impart to a given puck
a certain velocity, but if the puck collides (perpendicularly) with an
arm of a lever pivoted at its center with different mass, and the puck
stops on impact, it is impossible for the lever to acquire the same
kinetic energy the puck had, even if no energy is lost to friction or
other causes.
Peter
PD
Peter, as an exercise, consider the following simple problem:
A dinner plate is placed on its edge and is rolled down a hill. The
vertical height of the hill is 2.5 m. The mass of the plate is 0.43 kg,
and its radius is 11 cm.
Calculate the linear speed of the dinner plate by the time it gets to
the bottom of the hill. Assume the plate rolls without slipping, and
that there is no substantial slowing due to drag, and that the plate is
essentially a thin disk of uniform thickness, and that the plate's
initial speed is very small and can be ignored.
Show your work. (Hint: The energy supplied to the plate comes
completely from the work done by gravity.)
PD
srp
No. You are still using gravitation.
_you_ try to deflect a baseball from its trajectory without
doing any work, and come back to impress us with the Nobel
grade result.
>
>
>>Just do it and prove to us that no work was needed to do it.
>>The fact that you can find examples
>>>where work is done in a trajectory does not imply that work is always
>>>done in a trajectory.
>>
>>Wow! So now you admit that work is needed to deflect a trajectory,
>>but then by the same token, you assert that work is not needed to
>>deflect a trajectory.
>
> No. What I said is that there are *instances* of work being involved in
> changing a trajectory. This does not mean that work is *required* in
> changing a trajectory. Get a grip.
Exactly what I understond. Double standard.
In physical reality, either work is needed to change a trajectory
or it is not. Period. The fact that gravitational force or electrostatic
force can continuously replenish the spent energy has nothing to
do with it.
Deflecting the trajectory of a particle or larger body requires an
expenditure of energy. That's all there is to it.
Dont waste your time. You obviously identify disagreement with
"not understanding". Its your problem, not mine.
I found that very few physicists had any real understanding of
electromagnetism and all of them are to be found at accelerator
sites, and even then they are few even among experimentalists
in high energy sites.
You definitely do not belong to that tiny group. Anyone can
resolve equations on a sheet of paper without having any idea
of what they really are about.
I suggest you read and _understand_ the following books before
you come snapping at me about stuff you have no knowledge
whatsoever about but what was known in Newton's time.
"Principle of Charged Particle Acceleration" by Stanley Humphries
"Electromag netic Theory" by Julius Adams Stratton
André Michaud
Peter
consisted of two terms. I thought a lever had only rotational kinetic
energy equal to (1/2)Iw^2. Could you please explain? Thanks.
Peter
Peter
which occurs on a horizontal plane.
Peter
Dirk Van de moortel
"Peter" <Poak...@msn.com> wrote in message news:1145890770.4...@u72g2000cwu.googlegroups.com...
> PD, I was not aware that the kinetic energy of a rotating lever
> consisted of two terms. I thought a lever had only rotational kinetic
> energy equal to (1/2)Iw^2. Could you please explain? Thanks.
Not familiar with kinetic energy?
Perhaps you could read some kind of introductory physics text.
That usually helps :-)
Dirk Vdm
Peter
in case my memory is failing me, I checked, and all say the total
rotational kinetic energy of spinning object with a fixed axis is
(1/2)Iw^2. Are you joshing me?
Peter
Dirk Van de moortel
"Peter" <Poak...@msn.com> wrote in message news:1145893828.6...@e56g2000cwe.googlegroups.com...
> Dirk, I have several excellent textbooks on introductory physics. Just
> in case my memory is failing me, I checked, and all say the total
> rotational kinetic energy of spinning object with a fixed axis is
> (1/2)Iw^2.
And what do they say about the kinetic energy of a rotating
object that is moving with a velocity v?
> Are you joshing me?
Are you?
Dirk Vdm
Randy Poe
That is true. That is the rotational part of kinetic energy.
It also has translational kinetic energy.
- Randy
Peter
axis is fixed.
Peter
Peter
relative to its frame of reference, that would be different, but that
is not the case.
Peter
PD
Peter wrote:
> PD, Yes, but I don't see the connection with the puck-lever collision,
> which occurs on a horizontal plane.
Work the problem, and you will.
PD
PD
The kinetic energy of a macroscopic body which is rotating about its
center of mass and whose center of mass is also translating is
KE = (1/2)mv^2 + (1/2)Iw^2.
Now, you have proposed a scenario where a puck that is initially moving
strikes off-center a lever that is initially stationary, and this
collision is elastic.
It is not possible for the puck and the center of mass of the lever to
be both stationary after the collision. That would violate conservation
of linear momentum. It is possible for the puck to be stationary after
the collision, but it is not possible for the center of mass of the
lever to be stationary.
PD
Dirk Van de moortel
"Peter" <Poak...@msn.com> wrote in message news:1145897947.6...@v46g2000cwv.googlegroups.com...
> Dirk, if the rotating lever had, besides, some translational velocity,
> relative to its frame of reference, that would be different, but that
> is not the case.
Then the collision can't be 100% elastic.
Dirk Vdm
Dirk Van de moortel
"PD" <TheDrap...@gmail.com> wrote in message news:1145904757.1...@u72g2000cwu.googlegroups.com...
I think Peter need a shower :-)
Dirk Vdm
PD
Peter wrote:
> Randy, the lever I am talking about has no translational motion: its
> axis is fixed.
If the axis is *fixed* then the puck and the lever do not represent an
isolated system, and some momentum and energy is transferred out of the
system to the body to which the axis is fixed.
This is sort of like the ball bouncing off a brick wall scenario. Since
the total momentum is *forward* before the collision, then the total
momentum is *still* forward after the collision, even though the ball
is going *backward* after the collision. "But," you say, "the wall is
*fixed*. It doesn't go anywhere." If the wall is attached to the Earth,
then it is the Earth *and* the wall that go forward (though very, very
slowly) after the collision.
PD
Peter
there is absolutely no doubt that the puck stops completely on impact
with the lever, and since, after the collision, the lever is rotating
about an axis anchored on a large table, there is absolutely no doubt
the axis stays put. Relative to the axis, the puck has an initial
angular momentum, which is transferred to the lever during the
collision. Looking at it this way, there is no violation of
conservation of angular momentum. I know this is correct because I took
three years of classical mechanics, and this is what I learned. If the
initial angular momentum of the puck -relative to the axis- were zero,
since the lever does have an angular momentum after the collision,
conservation of angular momentum would be violated. In the opposite
case, when an arm of a rotating lever collides with a stationary puck
-if the puck has the right mass and position- the lever will stop
completely on impact, and the puck will acquire its angular momentum,
relative to the axis. Personally I think linear-angular momentum is
conserved, because they cannot be conserved separately. But I need help
with this problem, if you are interested, it might be worth your while,
because I know it is important.
Peter
PD
Peter wrote:
> PD, I have performed this kind of experiments dozens of times, and
> there is absolutely no doubt that the puck stops completely on impact
> with the lever, and since, after the collision, the lever is rotating
> about an axis anchored on a large table,
Then it is not an isolated system and some energy and momentum is
transferred to the large table.
> there is absolutely no doubt
> the axis stays put. Relative to the axis, the puck has an initial
> angular momentum, which is transferred to the lever during the
> collision. Looking at it this way, there is no violation of
> conservation of angular momentum.
That is correct. An object moving in a straight line has angular
momentum with respect to an axis not on that line.
> I know this is correct because I took
> three years of classical mechanics, and this is what I learned. If the
> initial angular momentum of the puck -relative to the axis- were zero,
> since the lever does have an angular momentum after the collision,
> conservation of angular momentum would be violated. In the opposite
> case, when an arm of a rotating lever collides with a stationary puck
> -if the puck has the right mass and position- the lever will stop
> completely on impact, and the puck will acquire its angular momentum,
> relative to the axis.
That is also correct, for the same reason.
> Personally I think linear-angular momentum is
> conserved, because they cannot be conserved separately.
But they ARE conserved separately. (In your case, however, you need to
also consider the linear momentum of the table, since the lever is
anchored to it.)
> But I need help
> with this problem, if you are interested, it might be worth your while,
> because I know it is important.
>
Well, having a correct understanding of the problem is important.
Also, making sure that you recover some of what you learned in three
years of classical mechanics is important.
PD
Peter
losses whatsoever, one can see that kinetic energy cannot possibly be
conserved; although, of course, this does not imply a violation of
conservation of energy, and it has a very logical explanation, which I
could give you, if you are interested and want to help me, because I
need help with this problem.
Peter
Dirk Van de moortel
"Peter" <Poak...@msn.com> wrote in message news:1145912409.8...@i39g2000cwa.googlegroups.com...
Sorry, but I don't think I can help anymore.
Cheers and good luck,
Dirk Vdm
Peter
in the ship, will that change the momentum of the ship?
Peter
PD
Peter wrote:
> PD, suppose you are in a spaceship, and you throw a ball toward a wall
> in the ship, will that change the momentum of the ship?
>
If you are anchored to the floor of the spaceship, yes. If course, once
the ball hits the wall of the ship, the momentum of the ship will be
altered again. And then if you catch the ball again after the bounce
off the wall, then the momentum of the ship will be altered again. Can
you guess what the final state will be?
If not, no. You will recoil from throwing the ball. Of course, once the
ball hits the wall of the spaceship, then yes, the momentum of the
spaceship will be altered. And once your recoiling body hits the
opposite wall of the spaceship, the momentum of the ship will be
altered once again. If you catch the ball again, can you guess what the
final state will be?
And if you throw the ball out of the back end of the spaceship while
leaning against the front end of the ship, the ship's momentum will
definitely be changed. This is in fact how rocket engines work.
PD
Mike
The derivative of angular momentum L wrt to time is the torque T. For
conservation of angular momentum you must have no external torque on
your system. Thus, it may depend on how you rotate the lever. In
practice, even if you have high quality ball bearings you will get a
torque reaction on the system when the collision takes place. Of
course, if you use some kind of motor to rotate the lever you should
know that this will produce a reactive torque also. These experiments
are tricky to perform and a reasonable setup may cost anywhere from 50K
to 200k or even more. You need to measure angular velocity carefully
using high definition encoders and distances using laser
interforometers because any violation in the order of 1 part in a
billion would send Newtonian mechanics to the heuristics compartment.
In short, you did not perform any experiments. All you have done is
thinking about them and for someone who thinks W = kx it is very hard
to think correctly about them. And ven if you have done any, you have
no done them correctly since you cannot control the conditions
necessary for reliable resutls.
of course, if you have read the Final theory you do not need to do
anything because according to the book heavily advertized by google the
"work function" [sic] is incorrect, energy is never conserved, momentum
is not conserved and you need to do work to keep a planer on a circular
orbit around a start.
Yihhhhhhhhhaaaaaa
Mike
>
> Peter
Peter
the axis, and according to my understanding, it is not possible to
change the angular momentum of a system with a force exerted on the
axis, because the lever arm is zero.
Also, according to my understanding, the lever exerts a force on the
axis, but according to Newton's third law, the axis exerts an equal and
opposite force on the lever, and these forces constitute an
action-reaction force pair that cancels, because neither the lever nor
the axis can accelerate, for the physical constraints. It is as if I am
tied to a pole, and I push on the pole, and the pole pushes back on me,
but neither can move, because we are tied.
I think I have a correct understanding of the problem, but the help I
need is to make other people understand. There are important and
far-reaching implications in this matter.
Peter
Peter
gas particles that it ejects out of the ship; an equal and opposite
reaction force accelerates the ship, according to Newton's third law.
You would have to throw the ball out of the spaceship to change its
momentum.
Peter
Herman Trivilino
Peter, I am stating that the work done does NOT, in genearl, equal the
change in kinetic energy.
It is therefore no surprise (to anyone who understands) to learn that you
have found an example where the work done does not equal the change in
kinetic energy.
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Dirk Van de moortel
"Peter" <Poak...@msn.com> wrote in message news:1145921505.2...@j33g2000cwa.googlegroups.com...
Alarm bell.
Dirk Vdm
Mike
Herman Trivilino wrote:
> "Peter" <Poak...@msn.com> wrote ...
>
> > Herman, I am no suggesting the definition of work should be changed.
> > All I say is that you always do the same work to impart to a given puck
> > a certain velocity, but if the puck collides (perpendicularly) with an
> > arm of a lever pivoted at its center with different mass, and the puck
> > stops on impact, it is impossible for the lever to acquire the same
> > kinetic energy the puck had, even if no energy is lost to friction or
> > other causes.
>
>
> Peter, I am stating that the work done does NOT, in genearl, equal the
> change in kinetic energy.
He (and others) also seem not to understand that the work (energy)
needed to put an object in rotation in the first place is not related
to the work done by the centripetal force on the object, which for
uniform circular motion is zero.
Along with the other problem, like for instance that W = Fx and I = Ft,
it is easy to see why he can reach the conclusions he does.
Mike
Peter
this problem as a thought experiment, which is perfectly licit in the
field of classical mechanics, because it has well defined laws. I did
tests to determine the mass a puck should have, and the distance from
the pivot the point it strikes (on the lever) should have, to stop on
impact. I challenge you to do similar thought experiments and come up
with a different result. No, I have not read the book you mention, nor
do I intend to read it.
Peter
Peter
kinetic, it can also be potential, or heat, or whatever. But it has to
be there. Where is it in this case?
Peter
Peter
Peter
Peter
related to the centripetal force? Of course, the centripetal force does
no work, because its point of application does not move.
Peter
PD
Permanently, yes.
But in the other cases, where you are throwing the ball inside the
spaceship, the spaceship's momentum can also be changed, at least until
the ball makes contact again.
Look at it this way. Suppose you are looking at the spaceship, you, and
the ball in a frame of reference where the center of mass of this
system is not moving. If you throw the ball inside the spaceship, the
center of mass of the whole system must still remain in the same place.
So if part of the mass (the ball) is moving, what happens to the rest
of the mass in the system?
PD
Dirk Van de moortel
"Peter" <Poak...@msn.com> wrote in message news:1145971108....@y43g2000cwc.googlegroups.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in
> message news:yLk3g.390688$jT5.11...@phobos.telenet-ops.be...
> Dirk, could you please elucidate? Thanks.
| " ... but the help I
| need is to make other people understand. There are important and
| far-reaching implications in this matter."
http://www.apa.org/journals/features/psp7761121.pdf
Dirk Vdm