| 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: "Peter" |
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| Title: Re: Work - impulse |
22 Apr 2006 08:53:57 AM |
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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?
Otherwise, we have cases where the same amount of work can result in
different amounts of kinetic energy, which should not be.
Peter
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| User: "Herman Trivilino" |
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| Title: Re: Work - impulse |
23 Apr 2006 10:31:42 AM |
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"Peter" <Poakfield@msn.com> wrote ...
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.
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| User: "Peter" |
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| Title: Re: Work - impulse |
24 Apr 2006 08:13:56 AM |
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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
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| User: "Herman Trivilino" |
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| Title: Re: Work - impulse |
24 Apr 2006 10:35:12 PM |
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"Peter" <Poakfield@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.
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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| User: "Mike" |
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| Title: Re: Work - impulse |
25 Apr 2006 07:08:39 AM |
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Herman Trivilino wrote:
"Peter" <Poakfield@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
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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| User: "Peter" |
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| Title: Re: Work - impulse |
25 Apr 2006 08:42:53 AM |
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Mike, and who said the work required to put an object in rotation is
related to the centripetal force? Of course, the centripetal force does
no work, because its point of application does not move.
Peter
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| User: "Peter" |
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| Title: Re: Work - impulse |
25 Apr 2006 08:16:00 AM |
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Herman, doesn't work transfer energy? Of course it does not have to be
kinetic, it can also be potential, or heat, or whatever. But it has to
be there. Where is it in this case?
Peter
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| User: "PD" |
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| Title: Re: Work - impulse |
24 Apr 2006 08:52:22 AM |
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Peter 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
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
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| User: "Peter" |
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| Title: Re: Work - impulse |
24 Apr 2006 10:07:09 AM |
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PD, Yes, but I don't see the connection with the puck-lever collision,
which occurs on a horizontal plane.
Peter
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| User: "PD" |
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| Title: Re: Work - impulse |
24 Apr 2006 01:45:06 PM |
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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
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| User: "Mike" |
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| Title: Re: Work - impulse |
22 Apr 2006 11:30:17 AM |
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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
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| User: "PD" |
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| Title: Re: Work - impulse |
22 Apr 2006 11:41:02 AM |
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Peter wrote:
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
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
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| User: "Dirk Van de moortel" |
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| Title: Re: Work - impulse |
22 Apr 2006 05:22:13 PM |
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"PD" <TheDraperFamily@gmail.com> wrote in message news:1145724062.540909.19140@t31g2000cwb.googlegroups.com...
Peter wrote:
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
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!
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
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| User: "PD" |
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| Title: Re: Work - impulse |
23 Apr 2006 01:49:51 PM |
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Dirk Van de moortel wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message news:1145724062.540909.19140@t31g2000cwb.googlegroups.com...
Peter wrote:
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
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!
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
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
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| User: "Dirk Van de moortel" |
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| Title: Re: Work - impulse |
23 Apr 2006 02:15:18 PM |
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"PD" <TheDraperFamily@gmail.com> wrote in message news:1145818191.205725.229590@e56g2000cwe.googlegroups.com...
Dirk Van de moortel wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message news:1145724062.540909.19140@t31g2000cwb.googlegroups.com...
Peter wrote:
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
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!
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
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.
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
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| User: "PD" |
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| Title: Re: Work - impulse |
23 Apr 2006 03:29:52 PM |
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Dirk Van de moortel wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message news:1145818191.205725.229590@e56g2000cwe.googlegroups.com...
Dirk Van de moortel wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message news:1145724062.540909.19140@t31g2000cwb.googlegroups.com...
Peter wrote:
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
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!
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
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.
Yes, absolutely, even if w1i = 0.
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
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
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| User: "Peter" |
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| Title: Re: Work - impulse |
24 Apr 2006 07:58:54 AM |
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Dirk, in the problem I am talking about, a puck collides
(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
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| User: "PD" |
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| Title: Re: Work - impulse |
24 Apr 2006 08:04:46 AM |
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Peter wrote:
Dirk, in the problem I am talking about, a puck collides
(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
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
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| User: "Peter" |
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| Title: Re: Work - impulse |
24 Apr 2006 09:59:30 AM |
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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.
Peter
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| User: "Dirk Van de moortel" |
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| Title: Re: Work - impulse |
24 Apr 2006 10:22:30 AM |
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"Peter" <Poakfield@msn.com> wrote in message news:1145890770.468707.308080@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
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| User: "Peter" |
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| Title: Re: Work - impulse |
24 Apr 2006 10:50:28 AM |
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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. Are you joshing me?
Peter
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| User: "Dirk Van de moortel" |
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| Title: Re: Work - impulse |
24 Apr 2006 10:55:55 AM |
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"Peter" <Poakfield@msn.com> wrote in message news:1145893828.654566.277030@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
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| User: "Peter" |
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| Title: Re: Work - impulse |
24 Apr 2006 11:59:07 AM |
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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.
Peter
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| User: "Dirk Van de moortel" |
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| Title: Re: Work - impulse |
24 Apr 2006 01:54:36 PM |
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"Peter" <Poakfield@msn.com> wrote in message news:1145897947.614546.279390@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
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| User: "Peter" |
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| Title: Re: Work - impulse |
24 Apr 2006 04:00:09 PM |
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Dirk, working it as a though experiment, assuming there are no energy
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
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| User: "Dirk Van de moortel" |
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| Title: Re: Work - impulse |
24 Apr 2006 04:06:46 PM |
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"Peter" <Poakfield@msn.com> wrote in message news:1145912409.889641.111240@i39g2000cwa.googlegroups.com...
Dirk, working it as a though experiment, assuming there are no energy
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.
Sorry, but I don't think I can help anymore.
Cheers and good luck,
Dirk Vdm
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| User: "PD" |
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| Title: Re: Work - impulse |
24 Apr 2006 01:52:37 PM |
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Peter wrote:
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.
Peter
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
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| User: "Dirk Van de moortel" |
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| Title: Re: Work - impulse |
24 Apr 2006 01:57:06 PM |
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"PD" <TheDraperFamily@gmail.com> wrote in message news:1145904757.198376.184590@u72g2000cwu.googlegroups.com...
Peter wrote:
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.
Peter
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.
I think Peter need a shower :-)
Dirk Vdm
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| User: "Peter" |
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| Title: Re: Work - impulse |
24 Apr 2006 03:43:40 PM |
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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, 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
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| User: "PD" |
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| Title: Re: Work - impulse |
24 Apr 2006 03:52:57 PM |
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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
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