| 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: "CWatters" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 03:09:14 PM |
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"Mark Teller" <markteller2000@yahoo.com> wrote in message
news:1180168104.656062.322530@n15g2000prd.googlegroups.com...
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?
The ball has zero angular momentum with respect to it's own center of mass
BUT it does have angular momentum with respect to the center of mass of the
dumbell.
.
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| User: "Dirk Van de moortel" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 05:08:32 AM |
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"Mark Teller" <markteller2000@yahoo.com> wrote in message news:1180168104.656062.322530@n15g2000prd.googlegroups.com...
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?
Think about 1 point object going in a straight line. It has no
angular momentum around any point of its path, but it *does*
have angular momentum around any other point in space.
Now, can you imagine 2 point objects colliding without
angular momentum around any point? Yes, *only* if their
paths have a common point. Will these point objects show
"rotation" after the collision? No.
Now, can you imagine 3 objects colliding without angular
momentum around any point? Yes, only if their paths have
a common point. But be careful, imagine that 2 out of those
3 objects have a fixed between them distance (and form a
dumbbell). Can these 3 objects have a common point on
their paths? No. So there is no point in space around which
the ball and the dumbbell have zero angular momentum - not
before the collision, and therefore not after the collision.
See?
Dirk Vdm
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| User: "Mark Teller" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 07:38:46 AM |
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On May 26, 3:08 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Mark Teller" <markteller2...@yahoo.com> wrote in messagenews:1180168104.656062.322530@n15g2000prd.googlegroups.com...
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?
Think about 1 point object going in a straight line. It has no
angular momentum around any point of its path, but it *does*
have angular momentum around any other point in space.
Now, can you imagine 2 point objects colliding without
angular momentum around any point? Yes, *only* if their
paths have a common point. Will these point objects show
"rotation" after the collision? No.
Now, can you imagine 3 objects colliding without angular
momentum around any point? Yes, only if their paths have
a common point. But be careful, imagine that 2 out of those
3 objects have a fixed between them distance (and form a
dumbbell). Can these 3 objects have a common point on
their paths? No. So there is no point in space around which
the ball and the dumbbell have zero angular momentum - not
before the collision, and therefore not after the collision.
See?
Dirk Vdm
Thanks for the answer. It is a perspective I didnt think of.
So it appears that in theory curvilinear motion analysis is just a
question of perspective and can be completely derived from linear
motion analysis by imposing the condition of fixed distance between
particles in a body.
Am I right or does curvilinear motion / analysis have an identity by
itself, not derived from linear motion analysis?
Thanks again
Mark
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| User: "Dirk Van de moortel" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 10:25:25 AM |
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"Mark Teller" <markteller2000@yahoo.com> wrote in message news:1180355926.558455.245550@g37g2000prf.googlegroups.com...
On May 26, 3:08 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Mark Teller" <markteller2...@yahoo.com> wrote in messagenews:1180168104.656062.322530@n15g2000prd.googlegroups.com...
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?
Think about 1 point object going in a straight line. It has no
angular momentum around any point of its path, but it *does*
have angular momentum around any other point in space.
Now, can you imagine 2 point objects colliding without
angular momentum around any point? Yes, *only* if their
paths have a common point. Will these point objects show
"rotation" after the collision? No.
Now, can you imagine 3 objects colliding without angular
momentum around any point? Yes, only if their paths have
a common point. But be careful, imagine that 2 out of those
3 objects have a fixed between them distance (and form a
dumbbell). Can these 3 objects have a common point on
their paths? No. So there is no point in space around which
the ball and the dumbbell have zero angular momentum - not
before the collision, and therefore not after the collision.
See?
Dirk Vdm
Thanks for the answer. It is a perspective I didnt think of.
So it appears that in theory curvilinear motion analysis is just a
question of perspective and can be completely derived from linear
motion analysis by imposing the condition of fixed distance between
particles in a body.
Am I right or does curvilinear motion / analysis have an identity by
itself, not derived from linear motion analysis?
Curvilinear motion is curvilinear from every perspective,
just like linear motion is linear from every perspective.
In your example, the only effect of imposing the condition
of fixed distance between the 2 masses, is that, still in your
example, it just is not possible to have any point around
which the total angular momentum before the collision is zero.
If the ball hits the dumbbell and you demand that the dumbbell
starts spinning, even if the ball is stopped dead, it must have
hit the dumbbell off its center of mass. Therefore that centre
of mass cannot lie on the ball's path, and the ball must have
had angular momentum around that point. If the ball stops
dead on impact, the dumbbell will take over that angular
momentum about its centre of mass.
Dirk Vdm
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| User: "Mark Teller" |
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| Title: Re: Conservation of angular momentum |
29 May 2007 07:25:47 AM |
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On May 28, 8:25 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Mark Teller" <markteller2...@yahoo.com> wrote in messagenews:1180355926.558455.245550@g37g2000prf.googlegroups.com...
On May 26, 3:08 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Mark Teller" <markteller2...@yahoo.com> wrote in messagenews:1180168104.656062.322530@n15g2000prd.googlegroups.com...
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?
Think about 1 point object going in a straight line. It has no
angular momentum around any point of its path, but it *does*
have angular momentum around any other point in space.
Now, can you imagine 2 point objects colliding without
angular momentum around any point? Yes, *only* if their
paths have a common point. Will these point objects show
"rotation" after the collision? No.
Now, can you imagine 3 objects colliding without angular
momentum around any point? Yes, only if their paths have
a common point. But be careful, imagine that 2 out of those
3 objects have a fixed between them distance (and form a
dumbbell). Can these 3 objects have a common point on
their paths? No. So there is no point in space around which
the ball and the dumbbell have zero angular momentum - not
before the collision, and therefore not after the collision.
See?
Dirk Vdm
Thanks for the answer. It is a perspective I didnt think of.
So it appears that in theory curvilinear motion analysis is just a
question of perspective and can be completely derived from linear
motion analysis by imposing the condition of fixed distance between
particles in a body.
Am I right or does curvilinear motion / analysis have an identity by
itself, not derived from linear motion analysis?
Curvilinear motion is curvilinear from every perspective,
just like linear motion is linear from every perspective.
Any reason why curvilinear motion has an identity of its own and
cannot be derived as a special case of linear motion analysis?
I tried looking up Wikipedia and did find that curvilinear motion is
always treated separately.
Why is this so?
Thanks
Mark
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
29 May 2007 07:50:39 AM |
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"Mark Teller" <markteller2000@yahoo.com> wrote in message
news:1180441547.383878.124790@g37g2000prf.googlegroups.com...
Any reason why curvilinear motion has an identity of its own and
cannot be derived as a special case of linear motion analysis?
I tried looking up Wikipedia and did find that curvilinear motion is
always treated separately.
Why is this so?
No doubt it can be traced back to some symmetry of the
space in which we are embedded. Conservations laws all
result from such symmetries via Noether's Theorem.
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| User: "Dirk Van de moortel" |
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| Title: Re: Conservation of angular momentum |
29 May 2007 10:18:38 AM |
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"Mark Teller" <markteller2000@yahoo.com> wrote in message news:1180441547.383878.124790@g37g2000prf.googlegroups.com...
On May 28, 8:25 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Mark Teller" <markteller2...@yahoo.com> wrote in messagenews:1180355926.558455.245550@g37g2000prf.googlegroups.com...
On May 26, 3:08 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Mark Teller" <markteller2...@yahoo.com> wrote in messagenews:1180168104.656062.322530@n15g2000prd.googlegroups.com...
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?
Think about 1 point object going in a straight line. It has no
angular momentum around any point of its path, but it *does*
have angular momentum around any other point in space.
Now, can you imagine 2 point objects colliding without
angular momentum around any point? Yes, *only* if their
paths have a common point. Will these point objects show
"rotation" after the collision? No.
Now, can you imagine 3 objects colliding without angular
momentum around any point? Yes, only if their paths have
a common point. But be careful, imagine that 2 out of those
3 objects have a fixed between them distance (and form a
dumbbell). Can these 3 objects have a common point on
their paths? No. So there is no point in space around which
the ball and the dumbbell have zero angular momentum - not
before the collision, and therefore not after the collision.
See?
Dirk Vdm
Thanks for the answer. It is a perspective I didnt think of.
So it appears that in theory curvilinear motion analysis is just a
question of perspective and can be completely derived from linear
motion analysis by imposing the condition of fixed distance between
particles in a body.
Am I right or does curvilinear motion / analysis have an identity by
itself, not derived from linear motion analysis?
Curvilinear motion is curvilinear from every perspective,
just like linear motion is linear from every perspective.
Any reason why curvilinear motion has an identity of its own and
cannot be derived as a special case of linear motion analysis?
I guess because it is the other way around: linear motion is a
special case of arbitrary (curvilinear) motion :-)
I tried looking up Wikipedia and did find that curvilinear motion is
always treated separately.
Why is this so?
Probably because it is less simple, and from a pedagogical viewpoint
it is better to start simple and gradually add more complexity.
I really don't understand why you insist on this...
Perhaps these in-turn-questions help:
Why is Galilean reltivity treated before special relativity
which is in turn treated before general relativity?
Why is electrostatics treated before electrodynamics?
Why is 2D geometry treated before 3D?
Why are linear equations treated before quadratic equations?
Why is addition treated before multiplication?
Why do they teach us to talk before they teach to read and write?
Does that help?
Dirk Vdm
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 10:18:02 AM |
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On May 28, 8:38 am, Mark Teller <markteller2...@yahoo.com> wrote:
On May 26, 3:08 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Mark Teller" <markteller2...@yahoo.com> wrote in messagenews:1180168104.656062.322530@n15g2000prd.googlegroups.com...
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 netangular
momentumbefore collision, while there is after the collision?
Think about 1 point object going in a straight line. It has no
angularmomentumaround any point of its path, but it *does*
haveangularmomentumaround any other point in space.
Now, can you imagine 2 point objects colliding without
angularmomentumaround any point? Yes, *only* if their
paths have a common point. Will these point objects show
"rotation" after the collision? No.
Now, can you imagine 3 objects colliding withoutangular
momentumaround any point? Yes, only if their paths have
a common point. But be careful, imagine that 2 out of those
3 objects have a fixed between them distance (and form a
dumbbell). Can these 3 objects have a common point on
their paths? No. So there is no point in space around which
the ball and the dumbbell have zeroangularmomentum- not
before the collision, and therefore not after the collision.
See?
Dirk Vdm
Thanks for the answer. It is a perspective I didnt think of.
So it appears that in theory curvilinear motion analysis is just a
question of perspective and can be completely derived from linear
motion analysis by imposing the condition of fixed distance between
particles in a body.
Am I right or does curvilinear motion / analysis have an identity by
itself, not derived from linear motion analysis?
Thanks again
Mark- Hide quoted text -
- Show quoted text -
To the best of my knowledge, only Newton's motion laws govern all
motion, and no other. Newton's laws can, and should, be used to
analyze all motion. To simplify things, I think it is best to deal
with point particles, with the understanding that all extended objects
are made of point particles. To calculate the total momentum of an
extended object, one only needs to sum the momenta of its individual
particles.
Peter
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| User: "Sam Wormley" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 10:28:31 AM |
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Peter wrote:
To the best of my knowledge, only Newton's motion laws govern all
motion, and no other. Newton's laws can, and should, be used to
analyze all motion. To simplify things, I think it is best to deal
with point particles, with the understanding that all extended objects
are made of point particles. To calculate the total momentum of an
extended object, one only needs to sum the momenta of its individual
particles.
Peter
To deal with high gravitation or velocities, one needs the tools of
relativity.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
http://scienceworld.wolfram.com/physics/MomentumFour-Vector.html
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 11:24:53 AM |
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On May 28, 11:28 am, Sam Wormley <sworml...@mchsi.com> wrote:
Peter wrote:
To the best of my knowledge, only Newton's motion laws govern all
motion, and no other. Newton's laws can, and should, be used to
analyze all motion. To simplify things, I think it is best to deal
with point particles, with the understanding that all extended objects
are made of point particles. To calculate the totalmomentumof an
extended object, one only needs to sum the momenta of its individual
particles.
Peter
To deal with high gravitation or velocities, one needs the tools of
relativity.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
http://scienceworld.wolfram.com/physics/MomentumFour-Vector.html
You are right, of course. I should have added that this is true only
in the macroscopic world.
Peter
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 11:45:16 AM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180369493.176025.322550@q66g2000hsg.googlegroups.com...
On May 28, 11:28 am, Sam Wormley <sworml...@mchsi.com> wrote:
To deal with high gravitation or velocities, one needs the tools of
relativity.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
http://scienceworld.wolfram.com/physics/MomentumFour-Vector.html
You are right, of course. I should have added that this is true only
in the macroscopic world.
What makes you think that relativity doesn't apply to the
macroscopic world?
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| User: "The Ghost In The Machine" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 12:09:46 PM |
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In sci.physics, Greg Neill
<gneillREM@OVEsympatico.ca>
wrote
on Mon, 28 May 2007 12:45:16 -0400
<465b0663$0$15553$9a6e19ea@news.newshosting.com>:
"Peter" <Poakfield@msn.com> wrote in message
news:1180369493.176025.322550@q66g2000hsg.googlegroups.com...
On May 28, 11:28 am, Sam Wormley <sworml...@mchsi.com> wrote:
To deal with high gravitation or velocities, one needs the tools of
relativity.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
http://scienceworld.wolfram.com/physics/MomentumFour-Vector.html
You are right, of course. I should have added that this is true only
in the macroscopic world.
What makes you think that relativity doesn't apply to the
macroscopic world?
It does, but at the level of two colliding automobiles (30 m/s each with
respect to the roadway) one gets an error of at most 2 * 10^-14 in
comparison with Newton's calculation methods. ;-)
Of course, at velocities such as 0.99 c the error is a lot higher.
--
#191,
Useless C++ Programming Idea #992381111:
while(bit&BITMASK) ;
--
Posted via a free Usenet account from http://www.teranews.com
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 12:35:52 PM |
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On May 28, 12:45 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180369493.176025.322550@q66g2000hsg.googlegroups.com...
On May 28, 11:28 am, Sam Wormley <sworml...@mchsi.com> wrote:
To deal with high gravitation or velocities, one needs the tools of
relativity.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
http://scienceworld.wolfram.com/physics/MomentumFour-Vector.html
You are right, of course. I should have added that this is true only
in the macroscopic world.
What makes you think that relativity doesn't apply to the
macroscopic world?
It does, but insignificantly.
Peter
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| User: "Sam Wormley" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 12:51:00 PM |
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Peter wrote:
On May 28, 12:45 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180369493.176025.322550@q66g2000hsg.googlegroups.com...
On May 28, 11:28 am, Sam Wormley <sworml...@mchsi.com> wrote:
To deal with high gravitation or velocities, one needs the tools of
relativity.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
http://scienceworld.wolfram.com/physics/MomentumFour-Vector.html
You are right, of course. I should have added that this is true only
in the macroscopic world.
What makes you think that relativity doesn't apply to the
macroscopic world?
It does, but insignificantly.
Peter
Except with high gravitation or velocities, the relativistic effects are
not insignificant. The Global Positioning System is an excellent example
of a technology where relativistic effects cannot be ignored.
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| User: "Sam Wormley" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 01:55:54 PM |
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Mark Teller wrote:
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
If you pick any frame and include both the ball and the dumbbell
in that system--that system has some nonzero angular momentum that
will be conserved whether the collision is just gravitational or
involves physical contact.
.
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| User: "Eric Gisse" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 03:48:29 AM |
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On May 26, 1:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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?
The answer lies within the _proper_ answering of the following
question:
How do you define angular momentum?
Mark
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| User: "Mark Teller" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 03:58:37 AM |
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On May 26, 1:48 pm, Eric Gisse <jowr...@gmail.com> wrote:
On May 26, 1:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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?
The answer lies within the _proper_ answering of the following
question:
How do you define angular momentum?
Mark- Hide quoted text -
- Show quoted text -
I would imagine in both the cases around their centres of mass.
Obviously the centre is not the same, but I guess it may not make a
difference.
Am I right?
Are you suggesting the net angular momentum would be zero considering
that the ball spin around its axis (in the opposite direction) would
offset that of the dumbbell?
.
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| User: "Ben C" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 06:53:05 AM |
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On 2007-05-26, Mark Teller <markteller2000@yahoo.com> wrote:
On May 26, 1:48 pm, Eric Gisse <jowr...@gmail.com> wrote:
On May 26, 1:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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?
The answer lies within the _proper_ answering of the following
question:
How do you define angular momentum?
[...]
I would imagine in both the cases around their centres of mass.
Obviously the centre is not the same, but I guess it may not make a
difference.
Am I right?
I think you use the same rotation origin for all the objects involved,
before and after the collision.
It doesn't matter what origin, so long as it's the same. In this case,
the centre of mass of the dumbbell at the time of the collision is one
choice. At the instant of collision, the dumbbell starts to rotate
around this origin, and the ball rotates around it in the opposite
direction; but faster of course if the ball has less mass.
Are you suggesting the net angular momentum would be zero considering
that the ball spin around its axis (in the opposite direction) would
offset that of the dumbbell?
The ball definitely doesn't spin around its own axis after the
collision. Because it's a ball, the force of the collision acts through
its centre of mass, which means it doesn't pick up any spin. After the
collision it departs in a straight line with no spin. The dumbbell
departs in another straight line, but rotating around its own centre of
mass.
This is where it gets trickier, but I think if you calculate all angular
momenta with respect to the same origin you should find it all still
adds up to zero at any point in time before or after the collision.
Even though the ball is moving away from the collision in a straight
line, it still has an angular momentum about any origin provided v x r
is not zero where r is the vector from the ball's position to that
origin.
Another way of thinking of this is that if you draw the ball's path as a
straight line through space, the ball has an angular momentum about any
point not intersected by that line. It's that which offsets the angular
momentum of the dumbbell.
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| User: "Ben C" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 07:02:33 AM |
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On 2007-05-26, Ben C <spamspam@spam.eggs> wrote:
On 2007-05-26, Mark Teller <markteller2000@yahoo.com> wrote:
[...]
This is where it gets trickier, but I think if you calculate all angular
momenta with respect to the same origin you should find it all still
adds up to zero at any point in time before or after the collision.
Correction: as Dirk Van de moortel just explained, the angular momentum
in this system never adds up to zero, whatever origin you choose, before
or after the collision. But whatever origin you do choose, it should add
up to the _same_ amount of total angular momentum before and after.
It can add up to zero if the ball hits the dumbbell head-on, on a line
through its centre of mass, but of course in that case the dumbbell
won't rotate.
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| User: "Eric Gisse" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 03:18:05 PM |
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On May 26, 1:58 am, Mark Teller <markteller2...@yahoo.com> wrote:
[...]
Read the responses of the others. I was hoping you'd look up the
definition of angular momentum and see something..
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| User: "z" |
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| Title: Re: Conservation of angular momentum |
29 May 2007 03:01:29 PM |
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On May 26, 4:58 am, Mark Teller <markteller2...@yahoo.com> wrote:
On May 26, 1:48 pm, Eric Gisse <jowr...@gmail.com> wrote:
On May 26, 1:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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?
The answer lies within the _proper_ answering of the following
question:
How do you define angular momentum?
Mark- Hide quoted text -
- Show quoted text -
I would imagine in both the cases around their centres of mass.
Obviously the centre is not the same, but I guess it may not make a
difference.
Am I right?
Are you suggesting the net angular momentum would be zero considering
that the ball spin around its axis (in the opposite direction) would
offset that of the dumbbell?- Hide quoted text -
- Show quoted text -
you're partway there but you forgot:
the angular momentum resulting from the movement of the two centers
of mass around their joint center of mass.
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| User: "Igor" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 12:33:50 PM |
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On May 26, 4:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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?
Who says that there is a net angular momentum after the collision? If
angular momentum if conserved, then there can't be.
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 01:30:51 PM |
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"Igor" <thoovler@excite.com> wrote in message
news:1180200830.642806.88310@o5g2000hsb.googlegroups.com...
On May 26, 4:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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?
Who says that there is a net angular momentum after the collision? If
angular momentum if conserved, then there can't be.
Choose a convenient coordinate system, say the center of
momentum of the system. If the ball is going to hit the
dumbbell off-center, then already the system has a net
angular momentum. This value will be conserved.
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| User: "Dirk Van de moortel" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 01:23:33 PM |
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"Igor" <thoovler@excite.com> wrote in message news:1180200830.642806.88310@o5g2000hsb.googlegroups.com...
On May 26, 4:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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?
Who says that there is a net angular momentum after the collision? If
angular momentum if conserved, then there can't be.
If angular momentum was present before, then there will be
after as well.
And it *is* present before.
Dirk Vdm
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| User: "Peter" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 04:48:40 PM |
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On May 26, 4:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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
Your problems with angular momentum are due to the fact that it is now
generally (incorrectly) considered to be something different from
linear momentum (different units, different dimensions, one is a
vector, the other is not). In fact, they are the same thing, and, for
a particle, it is equal to mv; linear and angular momentum readily
convert into each other (there are many examples of this). The
direction of the velocity of an object in rectilinear motion does not
change, of course, and the direction of the velocity of a point
object in curvilinear motion changes continuously. In either case, if
no net external force acts in the direction of its motion, the
magnitude of its momentum |mv| does not change. In the case of
extended objects, since they are all made of point particles, their
total momentum (linear or angular is the sum of the momenta of their
constituent particles. I hope this helps. If you have any questions,
please do not hesitate to ask.
Peter
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
26 May 2007 06:21:36 PM |
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"Peter" <Poakfield@msn.com> wrote in message
news:1180216120.806455.97030@m36g2000hse.googlegroups.com...
On May 26, 4:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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
Your problems with angular momentum are due to the fact that it is now
generally (incorrectly) considered to be something different from
linear momentum (different units, different dimensions, one is a
vector, the other is not). In fact, they are the same thing, and, for
a particle, it is equal to mv; linear and angular momentum readily
convert into each other (there are many examples of this).
No. That is wrong. Linear and angular momentum are two
separate beasts, and are conserved independantly. If you
think otherwise, then your understanding of basic physics
is faulty.
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| User: "Ditto" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 03:11:43 PM |
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On 27 May, 00:21, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Peter" <Poakfi...@msn.com> wrote in message
news:1180216120.806455.97030@m36g2000hse.googlegroups.com...
On May 26, 4:28 am, Mark Teller <markteller2...@yahoo.com> wrote:
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
Your problems with angular momentum are due to the fact that it is now
generally (incorrectly) considered to be something different from
linear momentum (different units, different dimensions, one is a
vector, the other is not). In fact, they are the same thing, and, for
a particle, it is equal to mv; linear and angular momentum readily
convert into each other (there are many examples of this).
No. That is wrong. Linear and angular momentum are two
separate beasts, and are conserved independantly.
Are you sure? I can't think of a situation where AM is conserved and
LM isn't, but can where LM is and AM isn't.
If you
think otherwise, then your understanding of basic physics
is faulty.- Hide quoted text -
- Show quoted text -
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 03:45:55 PM |
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"Ditto" <jmc8197@softhome.net> wrote in message
news:1180383103.420338.223010@g4g2000hsf.googlegroups.com...
On 27 May, 00:21, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
No. That is wrong. Linear and angular momentum are two
separate beasts, and are conserved independantly.
Are you sure? I can't think of a situation where AM is conserved and
LM isn't, but can where LM is and AM isn't.
Really? Care to share?
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| User: "Ditto" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 04:53:53 PM |
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On 28 May, 21:45, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Ditto" <jmc8...@softhome.net> wrote in message
news:1180383103.420338.223010@g4g2000hsf.googlegroups.com...
On 27 May, 00:21, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
No. That is wrong. Linear and angular momentum are two
separate beasts, and are conserved independantly.
Are you sure? I can't think of a situation where AM is conserved and
LM isn't, but can where LM is and AM isn't.
Really? Care to share?
A couple acting on a rod. The sum of the two forces is zero so linear
momentum is conserved, but angular momentum isn't.
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| User: "Greg Neill" |
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| Title: Re: Conservation of angular momentum |
28 May 2007 05:04:08 PM |
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"Ditto" <jmc8197@softhome.net> wrote in message
news:1180389233.421503.244990@u30g2000hsc.googlegroups.com...
On 28 May, 21:45, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"Ditto" <jmc8...@softhome.net> wrote in message
news:1180383103.420338.223010@g4g2000hsf.googlegroups.com...
On 27 May, 00:21, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
No. That is wrong. Linear and angular momentum are two
separate beasts, and are conserved independantly.
Are you sure? I can't think of a situation where AM is conserved and
LM isn't, but can where LM is and AM isn't.
Really? Care to share?
A couple acting on a rod. The sum of the two forces is zero so linear
momentum is conserved, but angular momentum isn't.
If the forces are generated by something external to the
system, then neither linear nor angular momentum need be
conserved in the system under consideration; the system
is not closed.
If the couple is produced by elements of the system, then
both linear and angular momentum must be conserved.
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