iMechanica - Comments for "Are Linear and Angular Momenta Interconvertible?"
https://imechanica.org/node/8288
Comments for "Are Linear and Angular Momenta Interconvertible?"enLinear Momentum could be a special case only for rigid bodies
https://imechanica.org/comment/15596#comment-15596
<a id="comment-15596"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/14825#comment-14825">Reply to Ajit:</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear All,
</p>
<p>
After so much time, I am convinced of some flaws in my arguments. So, though the interest in blog has long been over, I thought it would be prudent for me to write this comment.
</p>
<p>
My argument that the linear momentum balance principle could be a special case of angular momentum balance principle, implicitly assumes like concepts like transmissibility of forces, and moments being free vectors. However, these concepts are valid only for rigid bodies. Hence, linear momentum balance principle and angular momentum balance principle are independent principles for deformable bodies.
</p>
<p>
However, for rigid bodies, the linear momentum balance principle could be replaced by angular momentum balance principle with respect to different axes or reference points. Of course, we should have enough number of independent equations (six in 3D). Hence, the angular momentum balance principle can not be avaoided, as there can only be 3 (in 3D) independent equations from linear momentum balance principle.
</p>
<p>
Regards,
</p>
<p>
Jayadeep
</p>
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</ul>Wed, 13 Oct 2010 05:20:20 +0000Jayadeep U. B.comment 15596 at https://imechanica.orgthe Green-Naghdi-Rivlin theorem
https://imechanica.org/comment/14868#comment-14868
<a id="comment-14868"></a>
<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Dear Ajit, Although I didn't follow up all the threads, I think it is useful to mention the Green-Naghdi-Rivlin theorem (or principle?). The GNR theorem shows that the integral forms of conservation of mass, and balance of linear and angular momentum can be deduced from the integral form of energy balance and the requirement of invariance under the superposed rigid body motion. I think it is relevant to the "single physical situation" you asked. Sincerely yours,Sangyul Ha</p>
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</ul>Tue, 01 Jun 2010 12:27:19 +0000dubukingcomment 14868 at https://imechanica.orgAjit:
https://imechanica.org/comment/14792#comment-14792
<a id="comment-14792"></a>
<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Ajit: i do not read all of your note (sorry, you wrote very lenghty), but just a clue:
</p>
<p>
If we assume a deformable body, under a nonuniform initial velocity field; conversion is possible, e.g. assume an incompressible inviside fluid; writing enstrophy (magnitude of vorticity) conservation equation, there is a vortex streching therm which does this job. In 2D cases, as vortex streching is equal to zero, enstrophy will be remained fixed as time proceeds.
</p>
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</ul>Sun, 30 May 2010 09:46:58 +0000resacomment 14792 at https://imechanica.orgReply to Jaydeep
https://imechanica.org/comment/14832#comment-14832
<a id="comment-14832"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/14825#comment-14825">Reply to Ajit:</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Hi Jaydeep,
</p>
<p>
Please see another post I made regarding this issue here [<a href="http://imechanica.org/node/8313" target="_blank">^</a>].
</p>
<p>
--Ajit
</p>
<p>
- - - - - <br />
[E&OE]
</p>
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</ul>Thu, 27 May 2010 16:10:37 +0000Ajit R. Jadhavcomment 14832 at https://imechanica.orgReply to Ajit:
https://imechanica.org/comment/14825#comment-14825
<a id="comment-14825"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/14809#comment-14809">Reply to Jaydeep</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Ajit,
</p>
<p>
Your example above considers a situation in which the particle will accelerate along the path (and not at constant speed). Definitely the angular momentum will change, but we can relate the rate of change of angular momentum to the applied moment (of the force about the reference point). In the example you considered above, there is a constant moment acting, which means that the angular momentum will change at a constant rate. Removing the moment arms from the equation, you can get the linear equation of motion, <em>F</em> = <em>ma</em>.
</p>
<p>
In case of planetary motion, Kepler considered the planet to be a particle. When the object is not having size (a particle) rotation is not a meaningful concept. That is what I meant to convey.
</p>
<p>
Thanks for suggesting that there is some merit in my way of thinking. Let me know, if there is any flaw in it...
</p>
<p>
Regards,
</p>
<p>
Jayadeep
</p>
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</ul>Thu, 27 May 2010 12:14:13 +0000Jayadeep U. B.comment 14825 at https://imechanica.orgTo Ajit,
https://imechanica.org/comment/14812#comment-14812
<a id="comment-14812"></a>
<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Hi Ajit,
</p>
<p>
I mentioned stress tensor symmetry as an example of something you get from the conservation of angular momentum that you can not get from linear momentum conservation. To me an independent principle gives you extra information that you can not get otherwise. I was giving such an example.
</p>
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</ul>Wed, 26 May 2010 17:45:26 +0000Temesgen Markoscomment 14812 at https://imechanica.orgReplies to comments on Linear and Angular Momenta--2
https://imechanica.org/comment/14808#comment-14808
<a id="comment-14808"></a>
<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Hi all,
</p>
<p>
Thanks.
</p>
<p>
0. Here is the link to Jim Barber's excellent clarification that I had alluded to in the post above: [<a href="http://imechanica.org/node/319#comment-1144" target="_blank">^</a>], and while we are at it, don't miss in the same thread Prof. Ballarini's brief comment about why a 1/r singularity can be permitted in one context but not another [<a href="http://imechanica.org/node/319#comment-939" target="_blank">^</a>].
</p>
<p>
1. Temesgen, may be I didn't bring it out right, but yes, I could see that you had agreed with me in your first comment too. Now, the remark "just a different way of looking at things" is especially appreciated.
</p>
<p>
I still am not sure what is that you wish to motivate by referring to the symmetry of the stress tensor.
</p>
<p>
2. This point just jots down the thoughts that I occurred as I read Temesgen and rohtav's comments.
</p>
<p>
There *is* a difference between a single particle (i.e. an infinitesimally small element of matter), and a finite body. The issue of the adjacent particles (or differential elements) is absent in the first case whereas it is present in the second.
</p>
<p>
For a single particle, I hope that we can consider the independence of the two momentum conservation principles as settled.
</p>
<p>
Now, consider the <em>geometry</em> of a spinning disk. Split this domain into elements of finite sizes. (This isn't FE modeling though.) Two cases are possible: (i) the elements are not glued to each other, and (ii) they are. Now, increase the number of elements. In the limit of ever increasing number of elements, the elements themselvs become infinitesimal, and so, we call them particles again.
</p>
<p>
In the first case (no glue), we can obtain the total linear and angular momenta by simply adding (superimposing) the respective momenta due to the single particles.
</p>
<p>
In the second case (glue), forces (i.e. momenta) can be exchanged between adjacent particles within the disk. Stress arises precisely out of this reason. The angular/linear momentum of a given particle may lead, via the glue, to a change in the value of the angular/linear momentum for the adjacent particle---this change being measured from the value it would have if the first case (no glue) were to apply. The nature of this change is such that the <em>angular</em> momentum for one particle may influence the <em>linear </em>momentum for the adjacent particle, and <em>vice versa</em>.
</p>
<p>
However, note that in my formulation in the top post, I was talking of an object that is unconstrained and unacted upon by a force---which is not the case here.
</p>
<p>
Notice that the volume integral of all the linear and angular momenta of all the particles would still come out to be precisely the same as in the first case. (The geometry and mass distribution being identical in both the cases.)
</p>
<p>
In short, no matter how weirdly the deformable body may move in space and what be the nature of transients in momenta that get set up in it, for the object taken as a whole, each of the linear and angular momenta will continue to carry the same value (so long it doesn't run into a physical constraint and isn't acted upon by an external force or a torque).
</p>
<p>
The principle of impossibility of conversion from one form to another will still apply to the body taken as a whole.
</p>
<p>3. rohtav: No, you do not have to read my writing in its entirety... Just a question: Since I can't/don't/won't make it compulsory, an interesting issue arises: does someone at SUT make it compulsory for you to read my posts and reply to them?</p>
<p>
As to your "clue:" The particular nature of the displacement field gradient tensor really doesn't matter, and so, the point 2. above again applies to the fluid you consider.
</p>
<p>
To see that it does apply, consider an isolated volume of a fluid with a certain initial angular and linear momentum. If you take a finite material volume inside a fluid, the point will not apply because this volume will continue exchanging momenta with the surrounding fluid. So, you have to take a complete "droplet" of sorts, and then think of the momenta for this entire mass of fluid vs. those for the particles making it up.
</p>
<p>
4. For rigid body, thinking on the fly, I guess that once again the point 3 above applies. The displacement gradient now contains only the rotation tensor but no strain tensor. Am I right?
</p>
<p>
5. Sure let me know if I am making mistakes. It is easy to do so. (For instance, somewhere above, I said that you first define the linear and angular momenta and then interrelate them. After Temesgen's reply, I now think that no, first you really need to define the linear one and only then can you define the angular one. .... Or is it? (LOL!))
</p>
<p>
6. Since there doesn't seem to be anything wrong with the principle (of impossibility of conversion), the issue which now becomes more interesting (to me!) is the possible reason why mechanicians might not have thought about it. I gave some thought to this idea, and will share my thinking in a separate post sometime later. (Hint: Ask what are the comparative advantages of the momentum approach and the energy approach, and then think why, even if true, this principle wouldn't have much practical utility and hence little theoretical value.)
</p>
<p>
</p>
<p>
- - - - - <br />
[E&OE]
</p>
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</ul>Wed, 26 May 2010 16:03:14 +0000Ajit R. Jadhavcomment 14808 at https://imechanica.orgReply to Jaydeep
https://imechanica.org/comment/14809#comment-14809
<a id="comment-14809"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/14791#comment-14791">Replies to Ajit's Comments</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Jaydeep,
</p>
<p>
2. If you try to calculate the angular momentum of a particle moving<br />
with constant speed in a straight line with respect to an axis at a<br />
finite distance (axis perpendicular to the line of motion), it will be a<br />
constant
</p>
<p>
No, it won't be a constant. Here is a simple way to look at it. Take a graph paper and set up an xy coordinate system. A particle exists at point P(0, 10) cm. It is acted upon by an oblique force: 100 i + 50 j N. Find out the torque of the force at the origin point 0(0,0). The simplest procedure is to take the moment arm as 10 cm and the force as 100 i N.
</p>
<p>
Perhaps an even simpler exampple: take a cantilever and apply an oblique force at the free end and compute its moment at the fixed end.
</p>
<p>
So, it will not be constant...
</p>
<p>
... Or, Jaydeep, is it that the way that the component of the force changes and the angle changes, in some way, they cancel each other out? ... Let me think about it and come back..
</p>
<p>
BTW, the term rotation still applies to planetary motion; it's just that the orbit is not circular but elliptic. ... We speak of the instantaneous radius of curvature for paths that are neither circular nor elliptic.
</p>
<p>
3. Why can't we extend the same logic and use six equations in terms of<br />
angular momentum with respect to the properly chosen reference points? </p>
<p>
You can, and if it helps simplify analysis, you should. But I would like to see a concrete example before I try it myself.
</p>
<p>
Regards,
</p>
<p>
--Ajit
</p>
<p>
- - - - <br />
[E&OE]
</p>
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</ul>Wed, 26 May 2010 15:53:53 +0000Ajit R. Jadhavcomment 14809 at https://imechanica.orgReplies to Ajit's Comments
https://imechanica.org/comment/14791#comment-14791
<a id="comment-14791"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/14781#comment-14781">Replies to comments on linear and angular momenta</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Ajit,
</p>
<p>
My replies to Ajit's comments above. Item numbers are same as those in Ajit's post.
</p>
<p>
1. I also feel there is something wrong with my argument, though I can not come to a conclusion on what it is... May be I will think more, as well as wait for some more comments.
</p>
<p>
2. If you try to calculate the angular momentum of a particle moving with constant speed in a straight line with respect to an axis at a finite distance (axis perpendicular to the line of motion), it will be a constant, not varying as mentioned in the post above. The angular momentum, being obtained as a cross-product, depends only on the shortest distance between the line of motion and the axis (constant) and velocity (constant). Interestingly, in this case the angular momentum equation directly gives the linear momentum equation. A more general case of application of angular momentum conservation to a particle (where rotation is a meaningless concept) is the <a href="http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion">Kepler's Second Law of Planetary Motion</a>.
</p>
<p>
3. Okey, my example above is an argument based on mathematical similarity. We have six equilibrium equations in 3D: 3 force equations and 3 equations in terms of first moment of forces. (2D has three equations). However, if we prefer to do so, we can think of six equations in terms of moments of forces, by choosing proper reference points. Similarly we have three equations of conservation linear momentum and three equations of conservation of first moment of linear momentum. Why can't we extend the same logic and use six equations in terms of angular momentum with respect to the properly chosen reference points?
</p>
<p>
Regards,
</p>
<p>
Jayadeep
</p>
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</ul>Tue, 25 May 2010 14:30:47 +0000Jayadeep U. B.comment 14791 at https://imechanica.orgAjit, what I mean by
https://imechanica.org/comment/14782#comment-14782
<a id="comment-14782"></a>
<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
<span class="Apple-style-span">Ajit,</span>
</p>
<p>
<span class="Apple-style-span">what I mean by independent is one can not derive one from the other. In the dynamics of systems of particles, if you start with the definitions of angular momentum and torque, and then take the time derivative of angular momentum and plug in the conservation of linear momentum, you get the equation for conservation of angular momentum. If you have one, the other one does not give you any extra equation, just a different way of looking at things.</span>
</p>
<p>
<span class="Apple-style-span">For deformable bodies, you can not derive one from the other and each one counts as an extra equation. Remember in continuum mechanics the conservation of angular momentum gives symmetry of the stress tensor in the absence of body torques. </span>
</p>
<p>
<span class="Apple-style-span">My understanding is that for rigid bodies the two principles are independent.</span>
</p>
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</ul>Mon, 24 May 2010 19:16:59 +0000Temesgen Markoscomment 14782 at https://imechanica.orgReplies to comments on linear and angular momenta
https://imechanica.org/comment/14781#comment-14781
<a id="comment-14781"></a>
<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Thanks for your comments. Here I bundle together my replies.
</p>
<p>
1. Jaydeep: re. translation as a special case of rotation with the axis at infinity. I do not agree.
</p>
<p>
Geometrically (which is not the same even as "kinematically" let alone "kinetically"), the straight line may be seen as the limiting case of an arc in the limit that the radius of curvature exceeds all bounds. But note, when we are speaking of the geometry in this way, we haven't yet introduced any physical object on the path.
</p>
<p>
The geometrical statement may be meaningful. But this does not mean that it is meaningful to speak of an angular momentum of a body around an axis whose location is farther than any great distance. The reason it is not meaningful to speak this way is that in this process, the torque of the body also increases without bound. As such, it fails to be a proper procedure. Note that a similar procedure is OK in certain other contexts---the physical situations wherein as the reference point is taken infinitely away, the variable under question also continually diminishes to attain a finite limit. For an example of this latter category, consider the Somerfeld radiation condition. As to elasticity, I also vaguely recall a clarification on similar lines once given here at iMechanica by Prof. Jim Barber.
</p>
<p>
Actually, it is easier to understand this issue by appeal to its physics. If you hold in mind the context that the only way angular momentum can be imparted to a body is via a torque, and that a torque can come into being only if there is a <em>couple</em> of forces (and not just a single force), then it's immediately clear that some kind of a physical connection must exist between the translating body <em>and</em> the axis---else, two forces cannot be applied. Now as this axis goes infinitely away, the size of the body itself approaches infinity, which leads to all sorts of contradictions---e.g., that it would require an infinite torque, but since the orders are similar, that it would not be a definite amount, etc. Notice that the the blown up size is not an issue for field theoretical problems.
</p>
<p>
2. Continuing on Jaydeep's suggestion: It's easier to think of calculating instantaneous angular momenta for a translating particle if the axis is a <em>finite </em>distance away. For instance, if you stand at a railway station, even if the rail-track is straight, you can always associate an angular momentum to a passing train with respect to your position. However, note that this angular momentum would be continually changing, and so, you would have to include (I guess virtual kind of) torques in your analysis, thereby hopelessly complicating the whole situation.
</p>
<p>
Once again, the example serves to highlight the importance of first fixing up the physical ideas before attempting mathematical analysis. Really speaking, there is no causual physical connection between you and the train. In all such situations, you either get contradictions or hopelessly complex analysis.
</p>
<p>
The reason people do think along such complicated lines is because the relativist theorists of the 20th have helped wipe out the distinction between the physical and the mathematical, and most physicists even today militate against the idea of drawing the line.
</p>
<p>
3. Jaydeep, I am not clear what you mean to bring out in giving the example of the simple bending. ... I mean, static equilibrium requires both force and momentum balance; the simple bending model abstracts the couple actions as point moments (even though in reality there are no point-moments) and it involves no additional force. Note, we are not replacing forces by moments. We are only representing torques involving finite distances as if they were applied at a point. Thus, I don't think that this example takes forward the point you were trying to make.
</p>
<p>
4. Temesgen: In dynamics of particles, we first define the two quantities, and later on, we can establish a mathematical relationship between them. For example, we establish infinitesimal rotations as vectors, and then relate the instantaneous change in linear quantities to the instantaneous angular quantities.
</p>
<p>
However, this does not mean that one set of quantities physically gets converted into the other set, thereby. Physical reality is independent of our calculations. If you burn petrol (gasoline), you get heat. There is a real, physical conversion, not just a recalcuation. Such is not the case in merely mathematically relating the two momenta.
</p>
<p>
And, as far as I know/think, the two principles remain independent also for dynamics of rigid bodies. (Come to think of it, this is easier to accept the their independence for rigid bodies than for deformable bodies).
</p>
<p>
5. Hi G., could you please tell us your first name? (Would be a convenience :) )
</p>
<p>
The key question is not whether you can use your knowledge of the total energy and one of its parts to calculate the remaining part. The key question is: does the angular energy change into the linear form or vice versa.
</p>
<p>
6. The basic issue is this: For an isolated object, is there any way that the angular momentum gets converted to the linear momentum, or vice versa. And, if not, why not make a principle out of it, esp. because the total energy principle would prove to be necessary but not sufficient in the absence of such a principle.
</p>
<p>
So, if sufficiency can only be had by having this additional principle, why not state it explicitly.
</p>
<p>
And, if the proposal is theoretically wrong, what is the reason that it is wrong.
</p>
<p>
</p>
<p>
- - - - -
</p>
<p>
[E&OE]
</p>
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</ul>Mon, 24 May 2010 15:27:48 +0000Ajit R. Jadhavcomment 14781 at https://imechanica.orgLinear momentum could be a special case of angular momentum
https://imechanica.org/comment/14769#comment-14769
<a id="comment-14769"></a>
<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
I am sure that everyone here knows that the translation is a special case of rotation, with the axis at infinity. So extending this concept, it should be possible to replace the force equations of motion by moment equations of motion, by suitably choosing the axes. The conservation of momemtum is only a special case.
</p>
<p>
I can see some similarity with the case of not using force equilibrium equation, instead using moment equilibrium equations with respect to different axes, in static problems. Recall the simple 2D beam problems for finding support reactions (shear force and bending moment diagrams).
</p>
<p>
Regards,
</p>
<p>
Jayadeep U.B.
</p>
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</ul>Mon, 24 May 2010 07:35:44 +0000Jayadeep U. B.comment 14769 at https://imechanica.orgI think the two types of
https://imechanica.org/comment/14760#comment-14760
<a id="comment-14760"></a>
<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>I think the two types of momentum are interconvertible. The key for the conversion is that the energy of the body considered (the sum of the linear and angular energy) has to remain constant during the conversion from linear to angular momentum and vice versa.</p>
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</ul>Sun, 23 May 2010 16:54:38 +0000George Papazafeiropouloscomment 14760 at https://imechanica.orgLinear and Angular Momentum
https://imechanica.org/comment/14757#comment-14757
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<p><em>In reply to <a href="https://imechanica.org/node/8288">Are Linear and Angular Momenta Interconvertible?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>As far as I know, conservation of linear and angular momentum are independent principles when it comes to deformable bodies. On the other hand, in the dynamics of systems of particles one can derive the conservation of angular momentum from the conservation of linear momentum.</p>
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</ul>Sun, 23 May 2010 14:27:11 +0000Temesgen Markoscomment 14757 at https://imechanica.org