| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
30 Mar 2007 12:05:37 AM |
| Object: |
moment of inertia of a cube |
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
Thank you.
.
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| User: "Uncle Al" |
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| Title: Re: moment of inertia of a cube |
30 Mar 2007 11:06:34 AM |
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wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
Thank you.
We trivially locate the center of mass through which all relevant
rotation axes must pass. By what means can we distinguish any of the
eight corners or six faces to create a unique rotation axis? A point,
planes, and rotation axes of symmetry of the cube show the corners and
faces are inertially indistinguishable. Small stuff.
Does a rotation axis through a corner and the center of mass give
three identical orthogonal moments of inertia as does a rotation axis
through a face center and the center of mass? Do all 0.5(pi)
steradians of unique rotation axes give identical triplets of moments
of inertia? (Any axis through the top hemisphere goes through the
bottom hemisphere - mirror plane. Then chop out redundancy from the
other two mirror planes.)
How 'bout a tetrahedron with rotation axes passing through its center
of mass?
How 'bout a right circular cylinder whose height is (radius)[sqrt(3)],
with rotation axes passing through its center of mass? Do all unique
axes of rotation give orthogonal identical triplet moments of
inertia? The cylinder is not a Platonic solid nor does it have a
minimum surface/volume ratio (that would be height equal to
diameter). Symmetry again severely cuts down the possible unique
orientations of the first rotation axis.
Are we having fun yet? "8^)
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
.
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| User: "Douglas Eagleson" |
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| Title: Re: moment of inertia of a cube |
31 Mar 2007 07:55:39 AM |
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On Mar 30, 12:06 pm, Uncle Al <Uncle...@hate.spam.net> wrote:
b...@coolgroups.com wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
Thank you.
We trivially locate the center of mass through which all relevant
rotation axes must pass. By what means can we distinguish any of the
eight corners or six faces to create a unique rotation axis? A point,
planes, and rotation axes of symmetry of the cube show the corners and
faces are inertially indistinguishable. Small stuff.
Does a rotation axis through a corner and the center of mass give
three identical orthogonal moments of inertia as does a rotation axis
through a face center and the center of mass? Do all 0.5(pi)
steradians of unique rotation axes give identical triplets of moments
of inertia? (Any axis through the top hemisphere goes through the
bottom hemisphere - mirror plane. Then chop out redundancy from the
other two mirror planes.)
How 'bout a tetrahedron with rotation axes passing through its center
of mass?
How 'bout a right circular cylinder whose height is (radius)[sqrt(3)],
with rotation axes passing through its center of mass? Do all unique
axes of rotation give orthogonal identical triplet moments of
inertia? The cylinder is not a Platonic solid nor does it have a
minimum surface/volume ratio (that would be height equal to
diameter). Symmetry again severely cuts down the possible unique
orientations of the first rotation axis.
Are we having fun yet? "8^)
--
Uncle Alhttp://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2
You are learning to use third transform in solution. Very good.
"Axes" as third location. Very good.
.
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| User: "Uncle Al" |
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| Title: Re: moment of inertia of a cube |
31 Mar 2007 12:04:43 PM |
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Douglas Eagleson wrote:
On Mar 30, 12:06 pm, Uncle Al <Uncle...@hate.spam.net> wrote:
b...@coolgroups.com wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
Thank you.
We trivially locate the center of mass through which all relevant
rotation axes must pass. By what means can we distinguish any of the
eight corners or six faces to create a unique rotation axis? A point,
planes, and rotation axes of symmetry of the cube show the corners and
faces are inertially indistinguishable. Small stuff.
Does a rotation axis through a corner and the center of mass give
three identical orthogonal moments of inertia as does a rotation axis
through a face center and the center of mass? Do all 0.5(pi)
steradians of unique rotation axes give identical triplets of moments
of inertia? (Any axis through the top hemisphere goes through the
bottom hemisphere - mirror plane. Then chop out redundancy from the
other two mirror planes.)
How 'bout a tetrahedron with rotation axes passing through its center
of mass?
How 'bout a right circular cylinder whose height is (radius)[sqrt(3)],
with rotation axes passing through its center of mass? Do all unique
axes of rotation give orthogonal identical triplet moments of
inertia? The cylinder is not a Platonic solid nor does it have a
minimum surface/volume ratio (that would be height equal to
diameter). Symmetry again severely cuts down the possible unique
orientations of the first rotation axis.
Are we having fun yet? "8^)
--
Uncle Alhttp://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2
You are learning to use third transform in solution. Very good.
"Axes" as third location. Very good.
Idiot.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
.
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| User: "Sam Wormley" |
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| Title: Re: moment of inertia of a cube |
30 Mar 2007 12:07:23 AM |
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wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
Thank you.
Do your calculations extend to all the platonic solids?
.
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| User: "Y.Porat" |
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| Title: Re: moment of inertia of a cube |
30 Mar 2007 03:21:36 AM |
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On Mar 30, 8:07 am, Sam Wormley <sworml...@mchsi.com> wrote:
b...@coolgroups.com wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
Thank you.
Do your calculations extend to all the platonic solids?
----------since moment if inertia
the moment of inertia of a body is roattion
dependant
so if you take it at the center of garvity
it is unequivocally defined
but if you take it relative to another axis that isnot across the
center of gravity
you have endless solusions that atre dependant
on the distanance of your chosen axis
and the one through the center of gravity
now
a btw remark
in many cases it is more **convenient** to claculate it rather not at
the axis of the center of gravity (just a matetr of economising
calculations)
and then you can transform it to the
COG axis
by the Steiner law
ATB
Y.Porat
---------------------
on
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| User: "Andy Resnick" |
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| Title: Re: moment of inertia of a cube |
30 Mar 2007 01:13:54 PM |
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wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
I've read a few of the responses, but I am wondering what question you
are really asking- the moment of inertia of *any object*, being defined
as: I = Integral(r^2 dm) is always defined in terms of a coordinate
origin, which is taken to be on the axis of rotation (except for the
parallel axis theorem). So, unless your mass distribution falls off as
1/r^2 irrespective of your coordinate origin (something clearly
unphysical), there will be a dependence on where the axis of rotation is.
Or am I missing something? After all, it's friday afternoon, and I spent
the morning fighting with my microscope...
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
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| User: "" |
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| Title: Re: moment of inertia of a cube |
31 Mar 2007 06:18:14 AM |
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In article <eujgdo$k7i$1@eeyore.INS.cwru.edu>,
Andy Resnick <andy.resnick@op.case.edu> wrote:
Or am I missing something? After all, it's friday afternoon, and I spent
the morning fighting with my microscope...
<grin> Who won?
/BAH
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| User: "Andy Resnick" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 09:00:43 AM |
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wrote:
In article <eujgdo$k7i$1@eeyore.INS.cwru.edu>,
Andy Resnick <andy.resnick@op.case.edu> wrote:
Or am I missing something? After all, it's friday afternoon, and I spent
the morning fighting with my microscope...
<grin> Who won?
Me, of course :) . I maintain absolute dominion over my my stuff.
That, and the ability to fix broken parts....
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
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| User: "Edward Green" |
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| Title: Re: moment of inertia of a cube |
01 Apr 2007 01:31:40 PM |
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On Mar 30, 2:13 pm, Andy Resnick <andy.resn...@op.case.edu> wrote:
b...@coolgroups.com wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
I've read a few of the responses, but I am wondering what question you
are really asking- the moment of inertia of *any object*, being defined
as: I = Integral(r^2 dm) is always defined in terms of a coordinate
origin, which is taken to be on the axis of rotation (except for the
parallel axis theorem). So, unless your mass distribution falls off as
1/r^2 irrespective of your coordinate origin (something clearly
unphysical), there will be a dependence on where the axis of rotation is.
Or am I missing something? After all, it's friday afternoon, and I spent
the morning fighting with my microscope...
I think he is asserting is that for axes passing through the center of
the cube the moment of inertia is independent of the orientation of
the axis. I don't know if I ever knew this, but it's certainly
plausible: for some purposes cubic symmetry is as good as spherical
symmetry.
For any substance with cubic symmetry, for example, any property
represented by a rank-2 tensor is represented by a tensor proportional
to the unit tensor. Here would seem to be an application of this
general principle, with the tensor identified with the moment of
inertia tensor -- though I have to think about this more clearly.
It's probably possible to understand the property of the moment of
inertia by breaking the cube up into a union of sets of points, each
of which independently has the desired property. No doubt this idea
will gel in a moment or two. ;-)
.
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| User: "Andy Resnick" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 09:08:44 AM |
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Edward Green wrote:
On Mar 30, 2:13 pm, Andy Resnick <andy.resn...@op.case.edu> wrote:
<snip>
I think he is asserting is that for axes passing through the center of
the cube the moment of inertia is independent of the orientation of
the axis.
<snip>
Aha- the missing piece. Thanks.
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
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| User: "" |
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| Title: Re: moment of inertia of a cube |
01 Apr 2007 01:56:06 PM |
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In article <1175452300.078233.82830@n76g2000hsh.googlegroups.com>, "Edward Green" <spamspamspam3@netzero.com> writes:
On Mar 30, 2:13 pm, Andy Resnick <andy.resn...@op.case.edu> wrote:
b...@coolgroups.com wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
I've read a few of the responses, but I am wondering what question you
are really asking- the moment of inertia of *any object*, being defined
as: I = Integral(r^2 dm) is always defined in terms of a coordinate
origin, which is taken to be on the axis of rotation (except for the
parallel axis theorem). So, unless your mass distribution falls off as
1/r^2 irrespective of your coordinate origin (something clearly
unphysical), there will be a dependence on where the axis of rotation is.
Or am I missing something? After all, it's friday afternoon, and I spent
the morning fighting with my microscope...
I think he is asserting is that for axes passing through the center of
the cube the moment of inertia is independent of the orientation of
the axis. I don't know if I ever knew this, but it's certainly
plausible: for some purposes cubic symmetry is as good as spherical
symmetry.
Yes.
For any substance with cubic symmetry, for example, any property
represented by a rank-2 tensor
A symmetric rank-2 tensor, to be exact.
is represented by a tensor proportional
to the unit tensor. Here would seem to be an application of this
general principle, with the tensor identified with the moment of
inertia tensor -- though I have to think about this more clearly.
It is rather straightforward. A symmetric rank 2 tensor can be
diagonalized, by choosing an appropriate orthogonal base (principla
axes). In the case of cubic symmetry, the three principal axes are
identical and so are the three eigenvalues, meaning, once
diagonalised, the tensor is simply proportional to the unit tensor.
But, since the unit tensor doesn't change under orthogonal similarity
transformation, if a tensor is proportional to the unit tensor in one
base, it remains so in all bases.
It's probably possible to understand the property of the moment of
inertia by breaking the cube up into a union of sets of points, each
of which independently has the desired property. No doubt this idea
will gel in a moment or two. ;-)
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Edward Green" |
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| Title: Re: moment of inertia of a cube |
01 Apr 2007 05:26:02 PM |
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On Apr 1, 2:56 pm, wrote:
In article <1175452300.078233.82...@n76g2000hsh.googlegroups.com>, "Edward Green" <spamspamsp...@netzero.com> writes:
On Mar 30, 2:13 pm, Andy Resnick <andy.resn...@op.case.edu> wrote:
b...@coolgroups.com wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
I've read a few of the responses, but I am wondering what question you
are really asking- the moment of inertia of *any object*, being defined
as: I = Integral(r^2 dm) is always defined in terms of a coordinate
origin, which is taken to be on the axis of rotation (except for the
parallel axis theorem). So, unless your mass distribution falls off as
1/r^2 irrespective of your coordinate origin (something clearly
unphysical), there will be a dependence on where the axis of rotation is.
Or am I missing something? After all, it's friday afternoon, and I spent
the morning fighting with my microscope...
I think he is asserting is that for axes passing through the center of
the cube the moment of inertia is independent of the orientation of
the axis. I don't know if I ever knew this, but it's certainly
plausible: for some purposes cubic symmetry is as good as spherical
symmetry.
Yes.
For any substance with cubic symmetry, for example, any property
represented by a rank-2 tensor
A symmetric rank-2 tensor, to be exact.
is represented by a tensor proportional
to the unit tensor. Here would seem to be an application of this
general principle, with the tensor identified with the moment of
inertia tensor -- though I have to think about this more clearly.
It is rather straightforward. A symmetric rank 2 tensor can be
diagonalized, by choosing an appropriate orthogonal base (principla
axes). In the case of cubic symmetry, the three principal axes are
identical and so are the three eigenvalues, meaning, once
diagonalised, the tensor is simply proportional to the unit tensor.
But, since the unit tensor doesn't change under orthogonal similarity
transformation, if a tensor is proportional to the unit tensor in one
base, it remains so in all bases.
Well, yes... it is simple if you put it that way. :-)
The OP asked for a "real intuitive" way to see this, whether that
argument is "intuitive" must depend on how well one has internalized
the intermediates: you make it simple by bringing a lot to the table.
When you think about a cube and an arbitrary axis passing through the
center, it seems less obvious -- even miraculous -- that the moment of
inertia is independent of the orientation of the axis.
.
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| User: "" |
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| Title: Re: moment of inertia of a cube |
01 Apr 2007 05:43:45 PM |
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In article <1175466362.139896.106320@q75g2000hsh.googlegroups.com>, "Edward Green" <spamspamspam3@netzero.com> writes:
On Apr 1, 2:56 pm, wrote:
In article <1175452300.078233.82...@n76g2000hsh.googlegroups.com>, "Edward Green" <spamspamsp...@netzero.com> writes:
On Mar 30, 2:13 pm, Andy Resnick <andy.resn...@op.case.edu> wrote:
b...@coolgroups.com wrote:
I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?
I've read a few of the responses, but I am wondering what question you
are really asking- the moment of inertia of *any object*, being defined
as: I = Integral(r^2 dm) is always defined in terms of a coordinate
origin, which is taken to be on the axis of rotation (except for the
parallel axis theorem). So, unless your mass distribution falls off as
1/r^2 irrespective of your coordinate origin (something clearly
unphysical), there will be a dependence on where the axis of rotation is.
Or am I missing something? After all, it's friday afternoon, and I spent
the morning fighting with my microscope...
I think he is asserting is that for axes passing through the center of
the cube the moment of inertia is independent of the orientation of
the axis. I don't know if I ever knew this, but it's certainly
plausible: for some purposes cubic symmetry is as good as spherical
symmetry.
Yes.
For any substance with cubic symmetry, for example, any property
represented by a rank-2 tensor
A symmetric rank-2 tensor, to be exact.
is represented by a tensor proportional
to the unit tensor. Here would seem to be an application of this
general principle, with the tensor identified with the moment of
inertia tensor -- though I have to think about this more clearly.
It is rather straightforward. A symmetric rank 2 tensor can be
diagonalized, by choosing an appropriate orthogonal base (principla
axes). In the case of cubic symmetry, the three principal axes are
identical and so are the three eigenvalues, meaning, once
diagonalised, the tensor is simply proportional to the unit tensor.
But, since the unit tensor doesn't change under orthogonal similarity
transformation, if a tensor is proportional to the unit tensor in one
base, it remains so in all bases.
Well, yes... it is simple if you put it that way. :-)
The OP asked for a "real intuitive" way to see this, whether that
argument is "intuitive" must depend on how well one has internalized
the intermediates: you make it simple by bringing a lot to the table.
Yes, quite right. "Intuitive" is a relative, not absolute term.
When you think about a cube and an arbitrary axis passing through the
center, it seems less obvious -- even miraculous -- that the moment of
inertia is independent of the orientation of the axis.
I agree, my first reaction was "no way". But Polasek's argument was
convincing (even if not "intuitive").
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Timo Nieminen" |
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| Title: Re: moment of inertia of a cube |
01 Apr 2007 05:53:17 PM |
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On Sun, 1 Apr 2007 wrote:
"Edward Green" <spamspamspam3@netzero.com> writes:
When you think about a cube and an arbitrary axis passing through the
center, it seems less obvious -- even miraculous -- that the moment of
inertia is independent of the orientation of the axis.
I agree, my first reaction was "no way". But Polasek's argument was
convincing (even if not "intuitive").
Since the viscous drag on a spinning cube is orientation-independent in
the Stokes limit, I was not surprised. The explanation is, of course,
identical.
Should one expect the drag to be independent of orientation at high
Reynolds numbers?
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "" |
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| Title: Re: moment of inertia of a cube |
01 Apr 2007 07:24:46 PM |
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In article <Pine.LNX.4.50.0704020850040.13163-100000@localhost>, Timo Nieminen <timo@physics.uq.edu.au> writes:
On Sun, 1 Apr 2007 wrote:
"Edward Green" <spamspamspam3@netzero.com> writes:
When you think about a cube and an arbitrary axis passing through the
center, it seems less obvious -- even miraculous -- that the moment of
inertia is independent of the orientation of the axis.
I agree, my first reaction was "no way". But Polasek's argument was
convincing (even if not "intuitive").
Since the viscous drag on a spinning cube is orientation-independent in
the Stokes limit, I was not surprised. The explanation is, of course,
identical.
Indeed. And same goes for thermal expansion coefficients of cubic
crystals. That's where it did sound familiar to me.
Should one expect the drag to be independent of orientation at high
Reynolds numbers?
Hmm, you got me there. Intuitively, I don't think so.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 01:14:06 AM |
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In article <Pine.LNX.4.50.0704021523220.13534-100000@localhost>, Timo Nieminen <timo@physics.uq.edu.au> writes:
On Mon, 2 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
On Sun, 1 Apr 2007 wrote:
"Edward Green" <spamspamspam3@netzero.com> writes:
When you think about a cube and an arbitrary axis passing through the
center, it seems less obvious -- even miraculous -- that the moment of
inertia is independent of the orientation of the axis.
I agree, my first reaction was "no way". But Polasek's argument was
convincing (even if not "intuitive").
Since the viscous drag on a spinning cube is orientation-independent in
the Stokes limit, I was not surprised. The explanation is, of course,
identical.
Indeed. And same goes for thermal expansion coefficients of cubic
crystals. That's where it did sound familiar to me.
Should one expect the drag to be independent of orientation at high
Reynolds numbers?
Hmm, you got me there. Intuitively, I don't think so.
In the Stokes limit, it's all linear. Thus, the angular velocity and
rotational drag torque are related by a drag coefficient tensor D such
that torque = D omega. Lose the linearity, and what have you got? So I
don't really expect orientation independence at high Rn.
Yes, makes sense.
OTOH, the cube is still highly symmetric. From analogy with translational
motion, I'd expect turbulent rotational drag to be proportional to
|omega|^2. What can map one to the other? Perhaps there is still some
magic of symmetry?
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Uncle Al" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 09:42:11 AM |
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wrote:
In article <Pine.LNX.4.50.0704021523220.13534-100000@localhost>, Timo Nieminen <timo@physics.uq.edu.au> writes:
On Mon, 2 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
On Sun, 1 Apr 2007 wrote:
"Edward Green" <spamspamspam3@netzero.com> writes:
When you think about a cube and an arbitrary axis passing through the
center, it seems less obvious -- even miraculous -- that the moment of
inertia is independent of the orientation of the axis.
I agree, my first reaction was "no way". But Polasek's argument was
convincing (even if not "intuitive").
Since the viscous drag on a spinning cube is orientation-independent in
the Stokes limit, I was not surprised. The explanation is, of course,
identical.
Indeed. And same goes for thermal expansion coefficients of cubic
crystals. That's where it did sound familiar to me.
Should one expect the drag to be independent of orientation at high
Reynolds numbers?
Hmm, you got me there. Intuitively, I don't think so.
In the Stokes limit, it's all linear. Thus, the angular velocity and
rotational drag torque are related by a drag coefficient tensor D such
that torque = D omega. Lose the linearity, and what have you got? So I
don't really expect orientation independence at high Rn.
Yes, makes sense.
OTOH, the cube is still highly symmetric. From analogy with translational
motion, I'd expect turbulent rotational drag to be proportional to
|omega|^2. What can map one to the other? Perhaps there is still some
magic of symmetry?
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
It seems unlikely that the ballistic coefficient of a cube translating
through a medium in a direction normal to the plane of one face is
anywhere near the ballistic coefficient of a cube translating with a
centered corner head on.
The difference would be more extreme at high Reynolds numbers. A
subsonic bullet can be round nose (ogive), hollow point, or wad
cutter. It hardly makes a difference. A supersonic bullet is
tapered. A quality supersonic bullet is pointed. (The leading
asperity in a cheap rifle bullet causes boundary layer hoo-hah that
eases its passage through the shock while lessening the effects of
poor symmetry. A quality bullet is one shot, one kill.)
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
.
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| User: "Timo Nieminen" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 05:29:05 PM |
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On Mon, 2 Apr 2007, Uncle Al wrote:
mmeron@cars3.uchicago.edu wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
It seems unlikely that the ballistic coefficient of a cube translating
through a medium in a direction normal to the plane of one face is
anywhere near the ballistic coefficient of a cube translating with a
centered corner head on.
However, in the Stokes limit, it is.
The difference would be more extreme at high Reynolds numbers.
Yes, but I already said that.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Edward Green" |
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| Title: Re: moment of inertia of a cube |
08 Apr 2007 10:10:52 AM |
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On Apr 2, 6:29 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
On Mon, 2 Apr 2007, Uncle Al wrote:
mme...@cars3.uchicago.edu wrote:
Timo Nieminen <t...@physics.uq.edu.au> writes:
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
It seems unlikely that the ballistic coefficient of a cube translating
through a medium in a direction normal to the plane of one face is
anywhere near the ballistic coefficient of a cube translating with a
centered corner head on.
However, in the Stokes limit, it is.
I surmise that in the Stokes limit it's _all_ parasitic drag -- no
effective inertia, no lift or other dynamic effect. I wonder just
what the drag would depend on then? Maybe it doesn't depend on the
solid shape at all, but only the cross-sectional area in the direction
of motion?
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| User: "Timo A. Nieminen" |
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| Title: Re: moment of inertia of a cube |
08 Apr 2007 01:56:40 PM |
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On Mon, 8 Apr 2007, Edward Green wrote:
On Apr 2, 6:29 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
On Mon, 2 Apr 2007, Uncle Al wrote:
mme...@cars3.uchicago.edu wrote:
Timo Nieminen <t...@physics.uq.edu.au> writes:
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
It seems unlikely that the ballistic coefficient of a cube translating
through a medium in a direction normal to the plane of one face is
anywhere near the ballistic coefficient of a cube translating with a
centered corner head on.
However, in the Stokes limit, it is.
I surmise that in the Stokes limit it's _all_ parasitic drag -- no
effective inertia, no lift or other dynamic effect. I wonder just
what the drag would depend on then? Maybe it doesn't depend on the
solid shape at all, but only the cross-sectional area in the direction
of motion?
If only it were that simple! Compare a long needle with a sphere of
equal cross-sectional area and think about how the length affects the
drag.
The cube is also a counter-example - the cross-sectional area depends on
the orientation, but the drag does not.
The joy of the Stokes limit is that the equations are linear, and the
vector drag force is related to the vector velocity by a rank-2 tensor,
which brings us back to the starting point ...
Stokes flow is nice in that you can characterise the flow by a velocity
vector field (compare with turbulent flow, where that isn't enough), the
equations are linear, and sometimes you can even write down analytical
solutions. It's simple enough so that you can make it really complicated.
Highly turbulent flow, OTOH, you just give up on anything beyond crude
approximation, because it's so complicated you have to keep it really
simple.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Edward Green" |
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| Title: Re: moment of inertia of a cube |
15 Apr 2007 08:33:44 AM |
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On Apr 8, 2:56 pm, "Timo A. Nieminen" <t...@physics.uq.edu.au> wrote:
On Mon, 8 Apr 2007, Edward Green wrote:
On Apr 2, 6:29 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
On Mon, 2 Apr 2007, Uncle Al wrote:
mme...@cars3.uchicago.edu wrote:
Timo Nieminen <t...@physics.uq.edu.au> writes:
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
It seems unlikely that the ballistic coefficient of a cube translating
through a medium in a direction normal to the plane of one face is
anywhere near the ballistic coefficient of a cube translating with a
centered corner head on.
However, in the Stokes limit, it is.
I surmise that in the Stokes limit it's _all_ parasitic drag -- no
effective inertia, no lift or other dynamic effect. I wonder just
what the drag would depend on then? Maybe it doesn't depend on the
solid shape at all, but only the cross-sectional area in the direction
of motion?
If only it were that simple! Compare a long needle with a sphere of
equal cross-sectional area and think about how the length affects the
drag.
It's not obvious to me which way the effect goes. A long needle of
cross-sectional area equal to a sphere would have more surface area
presented to the flow: maybe that's important. OTOH we would think
of the needle as "streamlined".
That's an interesting word!
The cube is also a counter-example - the cross-sectional area depends on
the orientation, but the drag does not.
Right. I realized that later.
The joy of the Stokes limit is that the equations are linear, and the
vector drag force is related to the vector velocity by a rank-2 tensor,
which brings us back to the starting point ...
Stokes flow is nice in that you can characterise the flow by a velocity
vector field (compare with turbulent flow, where that isn't enough), the
equations are linear, and sometimes you can even write down analytical
solutions. It's simple enough so that you can make it really complicated.
Highly turbulent flow, OTOH, you just give up on anything beyond crude
approximation, because it's so complicated you have to keep it really
simple.
I've thought of one peculiar property of Stokes flow: if it's true
that the model is equivalent to a massless fluid, then no momentum can
be transferred to the fluid: yet the fluid can exert force on an
object, with the object of course exerting equal and opposite force on
the fluid. So where does the momentum go? I suppose to the
boundaries -- the Stokes fluid behaves in someways like an ideal
massles rigid body.
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| User: "Timo A. Nieminen" |
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| Title: Re: moment of inertia of a cube |
21 Apr 2007 03:49:13 PM |
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On Sun, 15 Apr 2007, Edward Green wrote:
On Apr 8, 2:56 pm, "Timo A. Nieminen" <t...@physics.uq.edu.au> wrote:
If only it were that simple! Compare a long needle with a sphere of
equal cross-sectional area and think about how the length affects the
drag.
It's not obvious to me which way the effect goes. A long needle of
cross-sectional area equal to a sphere would have more surface area
presented to the flow: maybe that's important.
Think: slowly moving through syrup.
I've thought of one peculiar property of Stokes flow: if it's true
that the model is equivalent to a massless fluid, then no momentum can
be transferred to the fluid: yet the fluid can exert force on an
object, with the object of course exerting equal and opposite force on
the fluid. So where does the momentum go? I suppose to the
boundaries -- the Stokes fluid behaves in someways like an ideal
massles rigid body.
Yes, the momentum goes to the distant boundaries (or infinity, if you
prefer). For translation, if the object is stationary, and the infinite
fluid is moving past, there's an infinite amount of momentum available, so
no problem. For a stationary fluid and translating object, the flow isn't
steady-state, so a little care might be needed.
The perfect steady state case is a sphere rotating in a (distantly)
stationary fluid - "easy" to explicitly show that the outward angular
momentum flux through a spherical surface concentric with the sphere is
the same for all such surfaces (external to the sphere).
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 12:31:46 PM |
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In article <46111643.EC47C9CD@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
wrote:
In article <Pine.LNX.4.50.0704021523220.13534-100000@localhost>, Timo Nieminen <timo@physics.uq.edu.au> writes:
On Mon, 2 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
On Sun, 1 Apr 2007 wrote:
"Edward Green" <spamspamspam3@netzero.com> writes:
When you think about a cube and an arbitrary axis passing through the
center, it seems less obvious -- even miraculous -- that the moment of
inertia is independent of the orientation of the axis.
I agree, my first reaction was "no way". But Polasek's argument was
convincing (even if not "intuitive").
Since the viscous drag on a spinning cube is orientation-independent in
the Stokes limit, I was not surprised. The explanation is, of course,
identical.
Indeed. And same goes for thermal expansion coefficients of cubic
crystals. That's where it did sound familiar to me.
Should one expect the drag to be independent of orientation at high
Reynolds numbers?
Hmm, you got me there. Intuitively, I don't think so.
In the Stokes limit, it's all linear. Thus, the angular velocity and
rotational drag torque are related by a drag coefficient tensor D such
that torque = D omega. Lose the linearity, and what have you got? So I
don't really expect orientation independence at high Rn.
Yes, makes sense.
OTOH, the cube is still highly symmetric. From analogy with translational
motion, I'd expect turbulent rotational drag to be proportional to
|omega|^2. What can map one to the other? Perhaps there is still some
magic of symmetry?
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
It seems unlikely that the ballistic coefficient of a cube translating
through a medium in a direction normal to the plane of one face is
anywhere near the ballistic coefficient of a cube translating with a
centered corner head on.
I agree, it seems very unlikely.
The difference would be more extreme at high Reynolds numbers. A
subsonic bullet can be round nose (ogive), hollow point, or wad
cutter. It hardly makes a difference. A supersonic bullet is
tapered. A quality supersonic bullet is pointed. (The leading
asperity in a cheap rifle bullet causes boundary layer hoo-hah that
eases its passage through the shock while lessening the effects of
poor symmetry. A quality bullet is one shot, one kill.)
It still helps to aim right, as well.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 05:44:04 PM |
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In article <Pine.LNX.4.50.0704030829170.20881-100000@localhost>, Timo Nieminen <timo@physics.uq.edu.au> writes:
On Mon, 2 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
Fairly trivially, it is independent. With cubic symmetry and every
equation concerned being linear, I suspected this might be the case.
Landau saved me from going through the details, with equation (20.15)
giving the Stokes drag force on an arbitrary finite body,
F = eta A v, where A is a symmetric rank-2 tensor. Cubic symmetry means
that for a cube, A = a I, where d is a scalar. QED.
Lovely, same thing again. So, in anything that involves nothing
beyond rank-2 tensors, cube is equivalent to a sphere.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Timo Nieminen" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 05:33:00 PM |
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On Mon, 2 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
Fairly trivially, it is independent. With cubic symmetry and every
equation concerned being linear, I suspected this might be the case.
Landau saved me from going through the details, with equation (20.15)
giving the Stokes drag force on an arbitrary finite body,
F = eta A v, where A is a symmetric rank-2 tensor. Cubic symmetry means
that for a cube, A = a I, where d is a scalar. QED.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 11:14:54 PM |
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In article <Pine.LNX.4.50.0704030844330.20982-100000@localhost>, Timo Nieminen <timo@physics.uq.edu.au> writes:
On Tue, 3 Apr 2007, Timo Nieminen wrote:
On Mon, 2 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
Fairly trivially, it is independent. With cubic symmetry and every
equation concerned being linear, I suspected this might be the case.
Landau saved me from going through the details, with equation (20.15)
giving the Stokes drag force on an arbitrary finite body,
F = eta A v, where A is a symmetric rank-2 tensor. Cubic symmetry means
that for a cube, A = a I, where d is a scalar. QED.
H. K. Moffat, Six lectures on general fluid dynamics and two on
hydromagnetic dynamo theory, pp 149-234 in R. Balian & J.-L. Peube
(eds), Fluid Dynamics (Gordon and Breach, 1977) notes this result
explicitly, and also states "A moment's reflection (perhaps more than a
moment!) will convince you that the same is true for a particle exhibiting
the same degree of freedom as any of the other regular solids
(tetrahedron, octahedron etc.)"
Yes, for the same reason. Three orhtogonal and equivalent axes.
Octahedron? I'll have to think about that one.
Moffat next gives a very nice result: for a translating non-rotating
particle to experience a torque, the particle must be chiral ie lack
mirror symmetry. Thus, the translating cube is torque-free!
You mean "translating in fluid", I gather.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Uncle Al" |
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| Title: Re: moment of inertia of a cube |
03 Apr 2007 02:03:55 PM |
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wrote:
In article <Pine.LNX.4.50.0704030844330.20982-100000@localhost>, Timo Nieminen <timo@physics.uq.edu.au> writes:
On Tue, 3 Apr 2007, Timo Nieminen wrote:
On Mon, 2 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
The turbulent drag on a translating cube depends of the orientation. I'll
have to see if the Stokes drag on a translating cube does.
An interesting problem. My hands are full right now, so I'll resist the
temptation, but let us know if you've any results.
Fairly trivially, it is independent. With cubic symmetry and every
equation concerned being linear, I suspected this might be the case.
Landau saved me from going through the details, with equation (20.15)
giving the Stokes drag force on an arbitrary finite body,
F = eta A v, where A is a symmetric rank-2 tensor. Cubic symmetry means
that for a cube, A = a I, where d is a scalar. QED.
H. K. Moffat, Six lectures on general fluid dynamics and two on
hydromagnetic dynamo theory, pp 149-234 in R. Balian & J.-L. Peube
(eds), Fluid Dynamics (Gordon and Breach, 1977) notes this result
explicitly, and also states "A moment's reflection (perhaps more than a
moment!) will convince you that the same is true for a particle exhibiting
the same degree of freedom as any of the other regular solids
(tetrahedron, octahedron etc.)"
Yes, for the same reason. Three orhtogonal and equivalent axes.
Octahedron? I'll have to think about that one.
Moffat next gives a very nice result: for a translating non-rotating
particle to experience a torque, the particle must be chiral ie lack
mirror symmetry. Thus, the translating cube is torque-free!
You mean "translating in fluid", I gather.
For the same reason, two opposite parity mass distributions in a
vacuum background will violate the Equivalence Principle. In this
case angular momentum must be imparted so everything cannot cancel
versus the fixed stars. That the Earth both spins on its axis and
orbits the sun is convenient - the combination cannot continuously
cancel over a 24-hour span.
One then wants the smallest possible self-similar chiral emergent
scale, hence atomic lattices in a periodic crysal lattice -
enantiomorphic space groups P3(1)21 and P3(2)21 - benzil (parity
calorimetry experiment) and alpha-quartz (parity Eotvos experiment).
http://www.arxiv.org/abs/physics/0702045
<Timo Nieminen>
Thank you.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
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| User: "" |
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| Title: Re: moment of inertia of a cube |
03 Apr 2007 12:42:36 AM |
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In article <Pine.LNX.4.50.0704031452310.21331-100000@localhost>, Timo Nieminen <timo@physics.uq.edu.au> writes:
On Tue, 3 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
Moffat next gives a very nice result: for a translating non-rotating
particle to experience a torque, the particle must be chiral ie lack
mirror symmetry. Thus, the translating cube is torque-free!
You mean "translating in fluid", I gather.
Yes, and in the Stokes limit.
It's nice to see it's true - we assumed it was an adequate approximation
for modelling the optical trapping of a cube [1], and exactly correct is
certainly adequate.
For sure, can't ask for "more correct" than this.
[1] http://www.arxiv.org/abs/physics/0702045
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Timo Nieminen" |
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| Title: Re: moment of inertia of a cube |
02 Apr 2007 11:59:04 PM |
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On Tue, 3 Apr 2007 wrote:
Timo Nieminen <timo@physics.uq.edu.au> writes:
Moffat next gives a very nice result: for a translating non-rotating
particle to experience a torque, the particle must be chiral ie lack
mirror symmetry. Thus, the translating cube is torque-free!
You mean "translating in fluid", I gather.
Yes, and in the Stokes limit.
It's nice to see it's true - we assumed it was an adequate approximation
for modelling the optical trapping of a cube [1], and exactly correct is
certainly adequate.
[1] http://www.arxiv.org/abs/physics/0702045
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Edward Green" |
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| Title: Re: moment of inertia of a cube |
22 Apr 2007 11:59:01 AM |
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On Apr 3, 12:59 am, Timo Nieminen <t...@physics.uq.edu.au> wrote:
On Tue, 3 Apr 2007 wrote:
Timo Nieminen <t...@physics.uq.edu.au> writes:
Moffat next gives a very nice result: for a translating non-rotating
particle to experience a torque, the particle must be chiral ie lack
mirror symmetry. Thus, the translating cube is torque-free!
You mean "translating in fluid", I gather.
Yes, and in the Stokes limit.
It's nice to see it's true - we assumed it was an adequate approximation
for modelling the optical trapping of a cube [1], and exactly correct is
certainly adequate.
It appears to me to be "incorrect".
Uncle Al already has already as much as stated this for general flow
regimes, though he positioned himself as a defender with the
qualification "net non-zero torque over time": but the claim makes no
mention of long term time average, so any non-zero torque, however
transient, is sufficient to complete the counterexample : a
shuttlecock illustrates.
What about Stokes flows? Like relativistic motion they have the charm
of the counter-intuitive, so maybe by some magic they contrive to keep
non-chiral bodies torque-free about _any_ axis. This would require
not merely magic though, but miraculous intervention. A sufficient
counter example follows:
Consider a compound body consisting of two identical spheres separated
by a long strut of negligible cross section. The drag force is
concentrated on the spheres, in the direction of flow, and equal. So,
one might say, no torque. Or is there? As Andy Resnick implied in
reaction to the original question, moments of inertia (or torques)
must be evaluated about a particular axis or point: we can't without
qualification simply speak of the "torque".
But surely, one might say, we mean "torque about the center of mass":
that's what allows us to decompose the motion into a linear
acceleration in reaction to net force and angular acceleration in
response to torque. The center of mass is in the center of the
strut. Or is it? Nobody told us the spheres were of equal density!
By altering the masses while preserving the shape we can place the COM
anywhere along the strut without altering the forces: the body will
experience a torque about the COM and begin to rotate.
But maybe we consider altering the density cheating. Can a similar
theorem can be salvaged for bodies of uniform density? Consider again
a compound body, now consisting of spheres of masses m, m/2 and m/2.
The spheres are again connected by struts of negligible cross-section,
as shown by the (planar) diagram below:
O m
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___|___ _COM_
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o o m/2, m/2
The lighter spheres are equidistant from an axis passing through the
COM, the heavier sphere at the same distance on the opposite arm, the
separations sufficient so the perturbative effect on each other's flow
fields is negligible. (All spheres are of equal density, and flow
into the page).
Stokes drag on a sphere is proportional to the area, so if we halve
the mass of a sphere the drag is multiplied by 2^(-2/3). Adding the
areas of the smaller spheres we find the total area has increased by a
factor of 1.26. So by subdividing one sphere we have increased the
drag on that side of the moment arm by ~25% without moving the COM.
The body experiences a torque about the COM.
QED
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