| Topic: |
Science > Physics |
| User: |
"Mark Teller" |
| Date: |
26 May 2007 03:28:24 AM |
| Object: |
Conservation of angular momentum |
I guess this must be a trivial question.
Nevertheless I am having trouble with it.
Say there are two bodies in motion approaching each other. One of them
is a ball and another of a dumbbell shape. Both of them are in pure
translation with no rotation about any axis.
As the ball hits one end of the dumbbell it (dumbbell) starts
spinning.
How would you account for the fact that there was no net angular
momentum before collision, while there is after the collision?
Mark
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 10:51:15 AM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180531776.698460.235360@m36g2000hse.googlegroups.com...
On May 29, 6:17 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
No. The lever is not constrained at all: it can rotate freely; there
is very little friction at the pivot.
Crap. It is pinned at its center to a large mass (the
table and building and Earth,...).
Yes, but the Earth is a spaceship that will not accelerate, except
very briefly, while the ball is being accelerated, but when the ball
hits the lever, decelerates and stops, so does the spaceship. The net
effect on the spaceship is zero.
More crap. Conservation of momentum says that if the
ball stops, whatever stops it takes on (at least some) of
its momentum. So the Earth accelerates briefly, yes, but
then continues with its new velocity indefinitely.
[snip]
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 03:16:50 PM |
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On May 30, 11:51 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180531776.698460.235360@m36g2000hse.googlegroups.com...
On May 29, 6:17 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
No. The lever is not constrained at all: it can rotate freely; there
is very little friction at the pivot.
Crap. It is pinned at its center to a large mass (the
table and building and Earth,...).
Yes, but the Earth is a spaceship that will not accelerate, except
very briefly, while the ball is being accelerated, but when the ball
hits the lever, decelerates and stops, so does the spaceship. The net
effect on the spaceship is zero.
More crap. Conservationofmomentumsays that if the
ball stops, whatever stops it takes on (at least some) of
itsmomentum. So the Earth accelerates briefly, yes, but
then continues with its new velocity indefinitely.
I think that point was elucidated by someone in this thread.
Peter
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| User: "PD" |
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| Title: Re: Conservation of angular momentum |
29 May 2007 03:15:04 PM |
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On May 29, 1:51 pm, Peter <Poakfi...@msn.com> wrote:
On May 29, 1:53 pm, PD <TheDraperFam...@gmail.com> wrote:
On May 29, 12:34 pm, Peter <Poakfi...@msn.com> wrote:
On May 29, 12:50 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180453813.561517.235110@q75g2000hsh.googlegroups.com...
Do you mean that what I saw was a mirage?
No, you simply didn't notice all of the momenta transfers
going on. The track was connected to the Earth which
also participated. Just because the Earth is really,
really big and didn't move much doesn't mean it wasn't
there.
Besides, what about the
collision between a point object of mass m and an equal-arm lever of
mass 6m. Calculation using r x p shows it is impossible for both
angularmomentumand kinetic energy to be transferred, even under
ideal conditions. Nobody offered a satisfactory explanation.
Show your calculation and we'll show where you're wrong
wither in the mechanics of your math or your assumptions
about the analysis. If you can't understand the corrections,
that's not our problem.
As Samuel Johnson once said to a particularly thick-headed
individual, "Sir, I have found you an explanation, but I
am not obliged to find you an understanding."
Do you agree that a radial force on an object in circular motion can
only change the direction of its velocity but not the magnitude of its
velocity?
A steel point object withangularmomentumm(L/4)^2w collides
perpendicularly with the middle of the right arm of an steel, equal-
arm lever (at rest) on a vertical pivot, and stops on impact. The
moment of inertia of the lever is 6mL^2/12. Assuming no energy losses
due to friction or other causes, how canangularmomentumand kinetic
energy of the system point object-lever be conserved?
You can't. The stopping on impact flags the fact that the collision is
not elastic, and therefore kinetic energy is not conserved. It is only
in *elastic* collisions that kinetic energy is conserved, and you
cannot arbitrarily create an elastic collision where one body comes to
rest.
PD- Hide quoted text -
- Show quoted text -
When a steel ball of mass m collides head-on (without rolling) with
another steel ball of the same mass at rest, it stops dead on impact.
You know this.
Yes, when it impacts a steel ball of the *same mass*. When it impacts
a steel ball of *different* mass or impacts a lever, it generally will
NOT come to rest. You know this. Or perhaps you don't.
The same thing happens when the point object collides
with the center of one arm of the lever. Of course, if it hits any
other spot, it either rebounds or keeps going.
Peter- Hide quoted text -
- Show quoted text -
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| User: "Ben C" |
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| Title: Re: Conservation of angular momentum |
29 May 2007 03:57:42 PM |
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On 2007-05-29, PD <TheDraperFamily@gmail.com> wrote:
On May 29, 1:51 pm, Peter <Poakfi...@msn.com> wrote:
[...]
When a steel ball of mass m collides head-on (without rolling) with
another steel ball of the same mass at rest, it stops dead on impact.
You know this.
Yes, when it impacts a steel ball of the *same mass*. When it impacts
a steel ball of *different* mass or impacts a lever, it generally will
NOT come to rest. You know this. Or perhaps you don't.
It is possible to choose a rod, a ball and a place on the rod for the
ball to hit such that the ball stops dead and the rod rotates. That
isn't the specific problem here.
I don't know what the specific problem is though. It seems from earlier
postings that Peter might not be taking into account the transfer of
linear momentum to the rod and that's why he's getting an apparent
violation of conservation of energy.
The total energy of the rod after the collision is a combination of
linear kinetic energy and rotational kinetic energy. If the rod is on a
spindle attached to something then obviously the linear k.e. is just
transferred to that something. It's still conserved.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
29 May 2007 05:28:34 PM |
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On May 29, 4:57 pm, Ben C <spams...@spam.eggs> wrote:
On 2007-05-29, PD <TheDraperFam...@gmail.com> wrote:
On May 29, 1:51 pm, Peter <Poakfi...@msn.com> wrote:
[...]
When a steel ball of mass m collides head-on (without rolling) with
another steel ball of the same mass at rest, it stops dead on impact.
You know this.
Yes, when it impacts a steel ball of the *same mass*. When it impacts
a steel ball of *different* mass or impacts a lever, it generally will
NOT come to rest. You know this. Or perhaps you don't.
It is true that when an object impacts on a lever, it usually does not
stop on impact. The combination of masses of the point object and
lever, and the point the object must strike, to stop on impact, are
unique, and only through trial and error one can find them. Once it is
found, one realizes the system is very sensitive to variations of any
kind. If everything is not right, the point object either rebounds or
keeps going, but when all is O. K. the incident object stops dead on
impact.
It is possible to choose a rod, a ball and a place on the rod for the
ball to hit such that the ball stops dead and the rod rotates. That
isn't the specific problem here.
That is exactly the problem here.
I don't know what the specific problem is though. It seems from earlier
postings that Peter might not be taking into account the transfer of
linearmomentumto the rod and that's why he's getting an apparent
violation of conservation of energy.
I do believe energy is always conserved. This in not an example of
violation of conservation of energy.
If that were the case, it would be impossible (with this kind of
setup) to obtain conservation of both angular momentum and kinetic
energy, at least analytically, as it is.
The total energy of the rod after the collision is a combination of
linear kinetic energy and rotational kinetic energy. If the rod is on a
spindle attached to something then obviously the linear k.e. is just
transferred to that something. It's still conserved.
If this test were carried out on a spaceship (and the Earth is a
spaceship) one realizes clearly that it is impossible to give the
spaceship some linear momentum from actions inside the spaceship. To
give the spaceship some linear momentum, an outside force is required.
Peter
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
29 May 2007 05:46:56 PM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180477713.948024.195170@m36g2000hse.googlegroups.com...
If that were the case, it would be impossible (with this kind of
setup) to obtain conservation of both angular momentum and kinetic
energy, at least analytically, as it is.
Momentum is always conserved, regardless of the setup.
The total energy of the rod after the collision is a combination of
linear kinetic energy and rotational kinetic energy. If the rod is on a
spindle attached to something then obviously the linear k.e. is just
transferred to that something. It's still conserved.
If this test were carried out on a spaceship (and the Earth is a
spaceship) one realizes clearly that it is impossible to give the
spaceship some linear momentum from actions inside the spaceship. To
give the spaceship some linear momentum, an outside force is required.
Continuous motion is not possible without opening the system
(such as by ejecting mass), but oscillatory motion is possible.
The center of momentum will not change.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 08:34:03 AM |
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On May 29, 6:46 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180477713.948024.195170@m36g2000hse.googlegroups.com...
If that were the case, it would be impossible (with this kind of
setup) to obtainconservationof bothangularmomentumand kinetic
energy, at least analytically, as it is.
Momentumis always conserved, regardless of the setup.
I agree.
The total energy of the rod after the collision is a combination of
linear kinetic energy and rotational kinetic energy. If the rod is on a
spindle attached to something then obviously the linear k.e. is just
transferred to that something. It's still conserved.
If this test were carried out on a spaceship (and the Earth is a
spaceship) one realizes clearly that it is impossible to give the
spaceship some linearmomentumfrom actions inside the spaceship. To
give the spaceship some linearmomentum, an outside force is required.
Continuous motion is not possible without opening the system
(such as by ejecting mass), but oscillatory motion is possible.
The center ofmomentumwill not change.
I already answered that point.
Peter
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| User: "Ben C" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 02:09:38 AM |
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On 2007-05-29, Peter <Poakfield@msn.com> wrote:
On May 29, 4:57 pm, Ben C <spams...@spam.eggs> wrote:
[...]
The total energy of the rod after the collision is a combination of
linear kinetic energy and rotational kinetic energy. If the rod is on a
spindle attached to something then obviously the linear k.e. is just
transferred to that something. It's still conserved.
If this test were carried out on a spaceship (and the Earth is a
spaceship) one realizes clearly that it is impossible to give the
spaceship some linear momentum from actions inside the spaceship. To
give the spaceship some linear momentum, an outside force is required.
The spaceship as a whole yes, but you can move things around in the
spaceship, and if you do, the spaceship hull may have to move the other
way in order to keep the centre of mass of the whole lot in the same
place.
Here's my view of what happens: you shoot the ball out, the spaceship
reacts by moving backwards, and, provided the ball's path does not
intersect its centre of mass, also by rotating (A). When the ball hits
the lever and stops, the spaceship decelerates again until it has no
linear momentum. It also decelerates rotationally, but not by enough to
zero all of the angular velocity it gained due to (A). It remains
spinning about its COM in the opposite direction to the lever, but much
more slowly than the lever since it has a higher MoI.
Note that at no point does the COM of the whole lot move relative to the
stars. You shoot the ball and the spaceship moves backwards, but the
ball moves forwards. The distribution of mass in the whole assembly is
changing, but the COM doesn't move.
When the ball hits the lever, the linear momentum transfer to the table
is what stops the spaceship in its backwards slide. If it didn't then
the spaceship would keep on going and you then you _would_ have the
absurdity of the spaceship pulling itself along by its own bootstraps
without an external force.
Consider the reaction on the spaceship as the ball is fired. If you're
happy with the idea that this causes the spaceship to move backwards,
then something's got to put a stop to that before the COM of the whole
assembly moves relative to the stars.
.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 08:39:28 AM |
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On May 30, 3:09 am, Ben C <spams...@spam.eggs> wrote:
On 2007-05-29, Peter <Poakfi...@msn.com> wrote:
On May 29, 4:57 pm, Ben C <spams...@spam.eggs> wrote:
[...]
The total energy of the rod after the collision is a combination of
linear kinetic energy and rotational kinetic energy. If the rod is on a
spindle attached to something then obviously the linear k.e. is just
transferred to that something. It's still conserved.
If this test were carried out on a spaceship (and the Earth is a
spaceship) one realizes clearly that it is impossible to give the
spaceship some linearmomentumfrom actions inside the spaceship. To
give the spaceship some linearmomentum, an outside force is required.
The spaceship as a whole yes, but you can move things around in the
spaceship, and if you do, the spaceship hull may have to move the other
way in order to keep the centre of mass of the whole lot in the same
place.
Here's my view of what happens: you shoot the ball out, the spaceship
reacts by moving backwards, and, provided the ball's path does not
intersect its centre of mass, also by rotating (A). When the ball hits
the lever and stops, the spaceship decelerates again until it has no
linearmomentum. It also decelerates rotationally, but not by enough to
zero all of theangularvelocity it gained due to (A). It remains
spinning about its COM in the opposite direction to the lever, but much
more slowly than the lever since it has a higher MoI.
Note that at no point does the COM of the whole lot move relative to the
stars. You shoot the ball and the spaceship moves backwards, but the
ball moves forwards. The distribution of mass in the whole assembly is
changing, but the COM doesn't move.
When the ball hits the lever, the linearmomentumtransfer to the table
is what stops the spaceship in its backwards slide. If it didn't then
the spaceship would keep on going and you then you _would_ have the
absurdity of the spaceship pulling itself along by its own bootstraps
without an external force.
Consider the reaction on the spaceship as the ball is fired. If you're
happy with the idea that this causes the spaceship to move backwards,
then something's got to put a stop to that before the COM of the whole
assembly moves relative to the stars.
You are right. I agree.
Peter
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| User: "Ben C" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 11:36:19 AM |
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On 2007-05-30, Peter <Poakfield@msn.com> wrote:
On May 30, 3:09 am, Ben C <spams...@spam.eggs> wrote:
On 2007-05-29, Peter <Poakfi...@msn.com> wrote:
On May 29, 4:57 pm, Ben C <spams...@spam.eggs> wrote:
[...]
The total energy of the rod after the collision is a combination of
linear kinetic energy and rotational kinetic energy. If the rod is on a
spindle attached to something then obviously the linear k.e. is just
transferred to that something. It's still conserved.
If this test were carried out on a spaceship (and the Earth is a
spaceship) one realizes clearly that it is impossible to give the
spaceship some linearmomentumfrom actions inside the spaceship. To
give the spaceship some linearmomentum, an outside force is required.
The spaceship as a whole yes, but you can move things around in the
spaceship, and if you do, the spaceship hull may have to move the other
way in order to keep the centre of mass of the whole lot in the same
place.
Here's my view of what happens: you shoot the ball out, the spaceship
reacts by moving backwards, and, provided the ball's path does not
intersect its centre of mass, also by rotating (A). When the ball hits
the lever and stops, the spaceship decelerates again until it has no
linearmomentum. It also decelerates rotationally, but not by enough to
zero all of theangularvelocity it gained due to (A). It remains
spinning about its COM in the opposite direction to the lever, but much
more slowly than the lever since it has a higher MoI.
Note that at no point does the COM of the whole lot move relative to the
stars. You shoot the ball and the spaceship moves backwards, but the
ball moves forwards. The distribution of mass in the whole assembly is
changing, but the COM doesn't move.
When the ball hits the lever, the linearmomentumtransfer to the table
is what stops the spaceship in its backwards slide. If it didn't then
the spaceship would keep on going and you then you _would_ have the
absurdity of the spaceship pulling itself along by its own bootstraps
without an external force.
Consider the reaction on the spaceship as the ball is fired. If you're
happy with the idea that this causes the spaceship to move backwards,
then something's got to put a stop to that before the COM of the whole
assembly moves relative to the stars.
You are right. I agree.
So what's the problem? As far as I can see, linear and angular momentum
and energy are all conserved here.
.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 03:26:10 PM |
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On May 30, 12:36 pm, Ben C <spams...@spam.eggs> wrote:
On 2007-05-30, Peter <Poakfi...@msn.com> wrote:
On May 30, 3:09 am, Ben C <spams...@spam.eggs> wrote:
On 2007-05-29, Peter <Poakfi...@msn.com> wrote:
On May 29, 4:57 pm, Ben C <spams...@spam.eggs> wrote:
[...]
The total energy of the rod after the collision is a combination of
linear kinetic energy and rotational kinetic energy. If the rod is on a
spindle attached to something then obviously the linear k.e. is just
transferred to that something. It's still conserved.
If this test were carried out on a spaceship (and the Earth is a
spaceship) one realizes clearly that it is impossible to give the
spaceship some linearmomentumfrom actions inside the spaceship. To
give the spaceship some linearmomentum, an outside force is required.
The spaceship as a whole yes, but you can move things around in the
spaceship, and if you do, the spaceship hull may have to move the other
way in order to keep the centre of mass of the whole lot in the same
place.
Here's my view of what happens: you shoot the ball out, the spaceship
reacts by moving backwards, and, provided the ball's path does not
intersect its centre of mass, also by rotating (A). When the ball hits
the lever and stops, the spaceship decelerates again until it has no
linearmomentum. It also decelerates rotationally, but not by enough to
zero all of theangularvelocity it gained due to (A). It remains
spinning about its COM in the opposite direction to the lever, but much
more slowly than the lever since it has a higher MoI.
Note that at no point does the COM of the whole lot move relative to the
stars. You shoot the ball and the spaceship moves backwards, but the
ball moves forwards. The distribution of mass in the whole assembly is
changing, but the COM doesn't move.
When the ball hits the lever, the linearmomentumtransfer to the table
is what stops the spaceship in its backwards slide. If it didn't then
the spaceship would keep on going and you then you _would_ have the
absurdity of the spaceship pulling itself along by its own bootstraps
without an external force.
Consider the reaction on the spaceship as the ball is fired. If you're
happy with the idea that this causes the spaceship to move backwards,
then something's got to put a stop to that before the COM of the whole
assembly moves relative to the stars.
You are right. I agree.
So what's the problem? As far as I can see, linear andangularmomentum
and energy are all conserved here.- Hide quoted text -
- Show quoted text -
The problem is when a point mass of mass m collides with a lever of
mass 6m, and the point mass stops on impact. It is impossible for both
momentum and kinetic energy to be conserved at the same time, even in
theory.
Peter
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 04:02:25 PM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180556770.061867.105590@o5g2000hsb.googlegroups.com...
The problem is when a point mass of mass m collides with a lever of
mass 6m, and the point mass stops on impact. It is impossible for both
momentum and kinetic energy to be conserved at the same time, even in
theory.
Kinetic energy, in general, is not conserved. Total energy
is conserved, momentum is conserved.
.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 04:50:28 PM |
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On May 30, 5:02 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180556770.061867.105590@o5g2000hsb.googlegroups.com...
The problem is when a point mass of mass m collides with a lever of
mass 6m, and the point mass stops on impact. It is impossible for both
momentumand kinetic energy to be conserved at the same time, even in
theory.
Kinetic energy, in general, is not conserved. Total energy
is conserved,momentumis conserved.
I think I found the solution to my problem, but I need help to do
something with it. Would you want to help me? An open mind is
required, besides having a PhD and being connected to some university.
Peter
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
30 May 2007 07:44:58 PM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180561828.528929.239770@h2g2000hsg.googlegroups.com...
On May 30, 5:02 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180556770.061867.105590@o5g2000hsb.googlegroups.com...
The problem is when a point mass of mass m collides with a lever of
mass 6m, and the point mass stops on impact. It is impossible for both
momentumand kinetic energy to be conserved at the same time, even in
theory.
Kinetic energy, in general, is not conserved. Total energy
is conserved,momentumis conserved.
I think I found the solution to my problem, but I need help to do
something with it. Would you want to help me? An open mind is
required, besides having a PhD and being connected to some university.
State your problem clearly and allow for clarifying or
simplifying assumptions to be discussed. Once the
actual problem is formalized, I'm sure that there are
folks here who could help. I don't think that degrees
or university associations need be a stipulation, as
correct mathematics and physics remain correct regardless
of their source. Incorrect mathematics and physics will
be spotted in short order.
.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 09:00:08 AM |
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On May 30, 8:44 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180561828.528929.239770@h2g2000hsg.googlegroups.com...
On May 30, 5:02 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180556770.061867.105590@o5g2000hsb.googlegroups.com...
The problem is when a point mass of mass m collides with a lever of
mass 6m, and the point mass stops on impact. It is impossible for both
momentumand kinetic energy to be conserved at the same time, even in
theory.
Kinetic energy, in general, is not conserved. Total energy
is conserved,momentumis conserved.
I think I found the solution to my problem, but I need help to do
something with it. Would you want to help me? An open mind is
required, besides having a PhD and being connected to some university.
State your problem clearly and allow for clarifying or
simplifying assumptions to be discussed. Once the
actual problem is formalized, I'm sure that there are
folks here who could help. I don't think that degrees
or university associations need be a stipulation, as
correct mathematics and physics remain correct regardless
of their source. Incorrect mathematics and physics will
be spotted in short order.- Hide quoted text -
The problem is that in actual tests (carried out many times with much
care) it is found that when a point object, say it is a steel puck, of
mass m, collides with the middle of an arm of an equal-arm steel lever
of mass 6m, it stops dead on impact. However, analytically, even
assuming there is no friction or other causes of energy loss, there is
no way that both angular momentum and kinetic energy be conserved.
Peter
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 09:13:15 AM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180620008.647131.105960@h2g2000hsg.googlegroups.com...
The problem is that in actual tests (carried out many times with much
care) it is found that when a point object, say it is a steel puck, of
mass m, collides with the middle of an arm of an equal-arm steel lever
of mass 6m, it stops dead on impact. However, analytically, even
assuming there is no friction or other causes of energy loss, there is
no way that both angular momentum and kinetic energy be conserved.
Another poster already outlined the math of the situation
where it can happen. He made the assumption that the
lever was unsupported (so could translate after impact),
but that assumption can be mitigated by presuming that the
lever is fixed to a mass that is so large that its change
in motion would be utterly negligible from the point of view
of an inertial observer. In that case, from the point of
view of the observer, essentially all the kinetic energy of
the puck would end up as rotational energy in the lever.
.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 09:40:29 AM |
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On May 31, 10:13 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180620008.647131.105960@h2g2000hsg.googlegroups.com...
The problem is that in actual tests (carried out many times with much
care) it is found that when a point object, say it is a steel puck, of
mass m, collides with the middle of an arm of an equal-arm steel lever
of mass 6m, it stops dead on impact. However, analytically, even
assuming there is no friction or other causes of energy loss, there is
no way that bothangularmomentumand kinetic energy be conserved.
Another poster already outlined the math of the situation
where it can happen. He made the assumption that the
lever was unsupported (so could translate after impact),
but that assumption can be mitigated by presuming that the
lever is fixed to a mass that is so large that its change
in motion would be utterly negligible from the point of view
of an inertial observer. In that case, from the point of
view of the observer, essentially all the kinetic energy of
the puck would end up as rotational energy in the lever.
The equal-arm lever rotates about a vertical pivot secured to a large
table. Of course, the table does not move at all in this test.
Peter
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 10:23:14 AM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180622429.487502.53490@u30g2000hsc.googlegroups.com...
On May 31, 10:13 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180620008.647131.105960@h2g2000hsg.googlegroups.com...
The problem is that in actual tests (carried out many times with much
care) it is found that when a point object, say it is a steel puck, of
mass m, collides with the middle of an arm of an equal-arm steel lever
of mass 6m, it stops dead on impact. However, analytically, even
assuming there is no friction or other causes of energy loss, there is
no way that bothangularmomentumand kinetic energy be conserved.
Another poster already outlined the math of the situation
where it can happen. He made the assumption that the
lever was unsupported (so could translate after impact),
but that assumption can be mitigated by presuming that the
lever is fixed to a mass that is so large that its change
in motion would be utterly negligible from the point of view
of an inertial observer. In that case, from the point of
view of the observer, essentially all the kinetic energy of
the puck would end up as rotational energy in the lever.
The equal-arm lever rotates about a vertical pivot secured to a large
table. Of course, the table does not move at all in this test.
Peter
Don't say it doesn't move at all. That's simply not
possible on so many levels. You *can* say that its
movement is negligible within experimental accuracy.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 11:24:52 AM |
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On May 31, 11:23 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180622429.487502.53490@u30g2000hsc.googlegroups.com...
On May 31, 10:13 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180620008.647131.105960@h2g2000hsg.googlegroups.com...
The problem is that in actual tests (carried out many times with much
care) it is found that when a point object, say it is a steel puck, of
mass m, collides with the middle of an arm of an equal-arm steel lever
of mass 6m, it stops dead on impact. However, analytically, even
assuming there is no friction or other causes of energy loss, there is
no way that bothangularmomentumand kinetic energy be conserved.
Another poster already outlined the math of the situation
where it can happen. He made the assumption that the
lever was unsupported (so could translate after impact),
but that assumption can be mitigated by presuming that the
lever is fixed to a mass that is so large that its change
in motion would be utterly negligible from the point of view
of an inertial observer. In that case, from the point of
view of the observer, essentially all the kinetic energy of
the puck would end up as rotational energy in the lever.
The equal-arm lever rotates about a vertical pivot secured to a large
table. Of course, the table does not move at all in this test.
Peter
Don't say it doesn't move at all. That's simply not
possible on so many levels. You *can* say that its
movement is negligible within experimental accuracy.- Hide quoted text -
Yes, you are right, of course.
Peter
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| User: "PD" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 10:23:07 AM |
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On May 31, 9:40 am, Peter <Poakfi...@msn.com> wrote:
On May 31, 10:13 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180620008.647131.105960@h2g2000hsg.googlegroups.com...
The problem is that in actual tests (carried out many times with much
care) it is found that when a point object, say it is a steel puck, of
mass m, collides with the middle of an arm of an equal-arm steel lever
of mass 6m, it stops dead on impact. However, analytically, even
assuming there is no friction or other causes of energy loss, there is
no way that bothangularmomentumand kinetic energy be conserved.
Another poster already outlined the math of the situation
where it can happen. He made the assumption that the
lever was unsupported (so could translate after impact),
but that assumption can be mitigated by presuming that the
lever is fixed to a mass that is so large that its change
in motion would be utterly negligible from the point of view
of an inertial observer. In that case, from the point of
view of the observer, essentially all the kinetic energy of
the puck would end up as rotational energy in the lever.
The equal-arm lever rotates about a vertical pivot secured to a large
table. Of course, the table does not move at all in this test.
Peter
Ah, but it does, just not measurably enough for you to notice. When a
ripe peach falls to the ground, the ground falls up to the peach as
well. What's remarkable about this is that the change in the momentum
of the peach is *identical* to the change in the momentum of the
ground, even though the motion of the latter is imperpectible.
PD
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 11:21:08 AM |
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On May 31, 11:23 am, PD <TheDraperFam...@gmail.com> wrote:
On May 31, 9:40 am, Peter <Poakfi...@msn.com> wrote:
On May 31, 10:13 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180620008.647131.105960@h2g2000hsg.googlegroups.com...
The problem is that in actual tests (carried out many times with much
care) it is found that when a point object, say it is a steel puck, of
mass m, collides with the middle of an arm of an equal-arm steel lever
of mass 6m, it stops dead on impact. However, analytically, even
assuming there is no friction or other causes of energy loss, there is
no way that bothangularmomentumand kinetic energy be conserved.
Another poster already outlined the math of the situation
where it can happen. He made the assumption that the
lever was unsupported (so could translate after impact),
but that assumption can be mitigated by presuming that the
lever is fixed to a mass that is so large that its change
in motion would be utterly negligible from the point of view
of an inertial observer. In that case, from the point of
view of the observer, essentially all the kinetic energy of
the puck would end up as rotational energy in the lever.
The equal-arm lever rotates about a vertical pivot secured to a large
table. Of course, the table does not move at all in this test.
Peter
Ah, but it does, just not measurably enough for you to notice. When a
ripe peach falls to the ground, the ground falls up to the peach as
well. What's remarkable about this is that the change in themomentum
of the peach is *identical* to the change in themomentumof the
ground, even though the motion of the latter is imperpectible.
PD- Hide quoted text -
- Show quoted text -
You are right, of course, I know that. But the kinetic energy of the
table is negligible compared to the missing energy.
Peter
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| User: "Ben C" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 01:17:23 PM |
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On 2007-05-31, Peter <Poakfield@msn.com> wrote:
On May 31, 11:23 am, PD <TheDraperFam...@gmail.com> wrote:
On May 31, 9:40 am, Peter <Poakfi...@msn.com> wrote:
[...]
The equal-arm lever rotates about a vertical pivot secured to a large
table. Of course, the table does not move at all in this test.
Peter
Ah, but it does, just not measurably enough for you to notice. When a
ripe peach falls to the ground, the ground falls up to the peach as
well. What's remarkable about this is that the change in themomentum
of the peach is *identical* to the change in themomentumof the
ground, even though the motion of the latter is imperpectible.
PD- Hide quoted text -
- Show quoted text -
You are right, of course, I know that. But the kinetic energy of the
table is negligible compared to the missing energy.
If the table is attached to something very heavy (e.g. the Earth) then
there is a negligible transfer of linear momentum, and therefore if the
ball stops, as near as dammit all its energy should end up as rotational
kinetic energy of the lever.
This means the ball has to hit the lever further out. According to my
calculations, it should not be possible to stop the ball with an
anchored lever weighing 6m (where m is the mass of the ball), since you
would need to hit the lever at a radius of 0.7m, and it's only 1m long
(so the radius at which you can strike it is limited to 0.5m).
So anywhere the ball hits that lever, assuming it's attached rigidly to
the Earth, I think the ball should bounce back and not stop.
In a collision with an anchored lever, I make the radius at which the
ball must hit the rod sqrt((k / 12) * L^2) where L is the length of the
rod, and km = M, the mass of the rod. This is based on the formula for
moment of inertia of a rod of (M * L^2) / 12.
This means that k must be <=3 for it to be possible to stop the ball. If
k = 3, i.e. if the rod has 3x the mass of the ball, and is attached to a
very heavy table, then the ball will stop if it hits the lever right at
the end. Anywhere else the ball bounces back.
To stop the ball halfway down the lever, the lever should have 0.75 the
mass of the ball.
So if you really have an experiment in which the lever's mass is 6m, and
the ball stops, then there is a problem.
My guess is the most likely explanation is that my calculations are
wrong. The second most likely explanation is that energy _is_ being
transferred to the spindle, either by shuffling the table across the
floor a bit, or by flexing the spindle.
Post a link to photographs of your experimental setup if you can and we
can have a look. What sort of bearings is the spindle running on? If you
provide actual masses for the ball, lever, and where the ball strikes
the lever, we can work out how much energy is "missing" and see how
reasonable is it for it to have been absorbed by the spindle.
.
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 01:32:46 PM |
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"Ben C" <spamspam@spam.eggs> wrote in message
news:slrnf5u2q5.t9o.spamspam@bowser.marioworld...
If the table is attached to something very heavy (e.g. the Earth) then
there is a negligible transfer of linear momentum, and therefore if the
ball stops, as near as dammit all its energy should end up as rotational
kinetic energy of the lever.
The linear momentum is transfered to the lever+table+Earth,
it's just that the M component of the Mv is so large that the
v is negligible within observational capability.
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| User: "Ben C" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 03:22:54 PM |
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On 2007-05-31, Greg Neill <gneillREM@OVEsympatico.ca> wrote:
"Ben C" <spamspam@spam.eggs> wrote in message
news:slrnf5u2q5.t9o.spamspam@bowser.marioworld...
If the table is attached to something very heavy (e.g. the Earth) then
there is a negligible transfer of linear momentum, and therefore if the
ball stops, as near as dammit all its energy should end up as rotational
kinetic energy of the lever.
The linear momentum is transfered to the lever+table+Earth,
it's just that the M component of the Mv is so large that the
v is negligible within observational capability.
Exactly, and it's that tiny v which makes the energy transferred to
table+Earth negligible.
If you shoot a ball at a wall (attached to Earth etc.) at velocity v the
ball bounces back at you at very nearly -v. Almost all the kinetic
energy remains in the ball.
So if the spindle is well mounted and doesn't flex, and the table
doesn't move, etc., it shouldn't be possible to stop a ball with a lever
weighing 6x what the ball does. The lever has to weigh <=3x for that to
be possible. >3x and the ball will bounce back and not stop.
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 03:35:56 PM |
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"Ben C" <spamspam@spam.eggs> wrote in message
news:slrnf5ua5g.t9o.spamspam@bowser.marioworld...
On 2007-05-31, Greg Neill <gneillREM@OVEsympatico.ca> wrote:
"Ben C" <spamspam@spam.eggs> wrote in message
news:slrnf5u2q5.t9o.spamspam@bowser.marioworld...
If the table is attached to something very heavy (e.g. the Earth) then
there is a negligible transfer of linear momentum, and therefore if the
ball stops, as near as dammit all its energy should end up as
rotational
kinetic energy of the lever.
The linear momentum is transfered to the lever+table+Earth,
it's just that the M component of the Mv is so large that the
v is negligible within observational capability.
Exactly, and it's that tiny v which makes the energy transferred to
table+Earth negligible.
If you shoot a ball at a wall (attached to Earth etc.) at velocity v the
ball bounces back at you at very nearly -v. Almost all the kinetic
energy remains in the ball.
So if the spindle is well mounted and doesn't flex, and the table
doesn't move, etc., it shouldn't be possible to stop a ball with a lever
weighing 6x what the ball does. The lever has to weigh <=3x for that to
be possible. >3x and the ball will bounce back and not stop.
And this is a problem because...?
If your ball is stopping despite your lever massing <3x that
of the ball I'd start looking for frictional losses. When
the ball strikes the lever there's going to be a short term
high lateral force on any pivot. There will be acoustic
losses in the lever as it vibrates, too. Have you calculated
the vibration modes of the lever?
.
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| User: "Ben C" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 04:19:06 PM |
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On 2007-05-31, Greg Neill <gneillREM@OVEsympatico.ca> wrote:
"Ben C" <spamspam@spam.eggs> wrote in message
news:slrnf5ua5g.t9o.spamspam@bowser.marioworld...
On 2007-05-31, Greg Neill <gneillREM@OVEsympatico.ca> wrote:
"Ben C" <spamspam@spam.eggs> wrote in message
news:slrnf5u2q5.t9o.spamspam@bowser.marioworld...
If the table is attached to something very heavy (e.g. the Earth) then
there is a negligible transfer of linear momentum, and therefore if the
ball stops, as near as dammit all its energy should end up as
rotational
kinetic energy of the lever.
The linear momentum is transfered to the lever+table+Earth,
it's just that the M component of the Mv is so large that the
v is negligible within observational capability.
Exactly, and it's that tiny v which makes the energy transferred to
table+Earth negligible.
If you shoot a ball at a wall (attached to Earth etc.) at velocity v the
ball bounces back at you at very nearly -v. Almost all the kinetic
energy remains in the ball.
So if the spindle is well mounted and doesn't flex, and the table
doesn't move, etc., it shouldn't be possible to stop a ball with a lever
weighing 6x what the ball does. The lever has to weigh <=3x for that to
be possible. >3x and the ball will bounce back and not stop.
And this is a problem because...?
It's not a problem but we can conclude from this that some energy is
apparently "missing" just on the information that the lever weighs 6m
and the ball stops.
If your ball is stopping despite your lever massing <3x that
of the ball I'd start looking for frictional losses.
You mean >3x. Peter says his lever weighs 6x what the ball does.
When the ball strikes the lever there's going to be a short term high
lateral force on any pivot. There will be acoustic losses in the
lever as it vibrates, too. Have you calculated the vibration modes of
the lever?
Of course there must be energy losses somewhere. I don't draw the
conclusion that energy (or angular or linear momentum) is not conserved!
Acoustic losses are probably not very high since generally speaking it
takes very little power to make a lot of noise. But I don't know the
actual masses and sizes involved so I don't know how much energy we're
talking about.
.
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
31 May 2007 04:59:52 PM |
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"Ben C" <spamspam@spam.eggs> wrote in message
news:slrnf5udes.t9o.spamspam@bowser.marioworld...
On 2007-05-31, Greg Neill <gneillREM@OVEsympatico.ca> wrote:
If your ball is stopping despite your lever massing <3x that
of the ball I'd start looking for frictional losses.
You mean >3x. Peter says his lever weighs 6x what the ball does.
Oops. My bad. Yes, I should have said >3x.
When the ball strikes the lever there's going to be a short term high
lateral force on any pivot. There will be acoustic losses in the
lever as it vibrates, too. Have you calculated the vibration modes of
the lever?
Of course there must be energy losses somewhere. I don't draw the
conclusion that energy (or angular or linear momentum) is not conserved!
Acoustic losses are probably not very high since generally speaking it
takes very little power to make a lot of noise. But I don't know the
actual masses and sizes involved so I don't know how much energy we're
talking about.
By acoustical I meant to imply vibration modes in the
lever with stress/strain frictional losses in addition
to any tiny sound emissions.
If the lever arm is made of a material with significant
flexure, it could dissipate a lot of energy via vibration.
If it's thick enough to prevent these vibrations to a large
degree, we should take a closer look at the assumption that
its moment of inertia is that of the classic "thin rod".
Actual dimensions and composition would be helpful.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
01 Jun 2007 08:56:22 AM |
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On May 31, 5:59 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Ben C" <spams...@spam.eggs> wrote in message
news:slrnf5udes.t9o.spamspam@bowser.marioworld...
On 2007-05-31, Greg Neill <gneill...@OVEsympatico.ca> wrote:
If your ball is stopping despite your lever massing <3x that
of the ball I'd start looking for frictional losses.
You mean >3x. Peter says his lever weighs 6x what the ball does.
Oops. My bad. Yes, I should have said >3x.
When the ball strikes the lever there's going to be a short term high
lateral force on any pivot. There will be acoustic losses in the
lever as it vibrates, too. Have you calculated the vibration modes of
the lever?
Of course there must be energy losses somewhere. I don't draw the
conclusion that energy (orangularor linearmomentum) is not conserved!
Acoustic losses are probably not very high since generally speaking it
takes very little power to make a lot of noise. But I don't know the
actual masses and sizes involved so I don't know how much energy we're
talking about.
By acoustical I meant to imply vibration modes in the
lever with stress/strain frictional losses in addition
to any tiny sound emissions.
If the lever arm is made of a material with significant
flexure, it could dissipate a lot of energy via vibration.
If it's thick enough to prevent these vibrations to a large
degree, we should take a closer look at the assumption that
its moment of inertia is that of the classic "thin rod".
Actual dimensions and composition would be helpful.
The lever consists of a steel rod of 1/4 x 1/4 x 12 inches, with a
mass of about 100 grams.
Peter
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
01 Jun 2007 10:19:00 AM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180706182.052941.84600@p47g2000hsd.googlegroups.com...
The lever consists of a steel rod of 1/4 x 1/4 x 12 inches, with a
mass of about 100 grams.
Then I make the moment of inertia to be
I = M*(L^2/12 + d^2/6) = 7.75 x 10^-4 kg*m^2
where M = mass of lever
L = length of lever
d = width of lever = height of lever
Not too big a difference from the "thin rod" assumption.
Your mass looks to be a tad low. The density of steel
is on the order of 7.85 gm/cm^3, which would put your
lever at 96.5 grams.
.
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
01 Jun 2007 12:11:20 PM |
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On Jun 1, 11:19 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180706182.052941.84600@p47g2000hsd.googlegroups.com...
The lever consists of a steel rod of 1/4 x 1/4 x 12 inches, with a
mass of about 100 grams.
Then I make the moment of inertia to be
I = M*(L^2/12 + d^2/6) = 7.75 x 10^-4 kg*m^2
where M = mass of lever
L = length of lever
d = width of lever = height of lever
Not too big a difference from the "thin rod" assumption.
Your mass looks to be a tad low. The density of steel
is on the order of 7.85 gm/cm^3, which would put your
lever at 96.5 grams.
I said the mass of the lever is about 100 grams; it is exactly 96.6
grams.
Peter
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