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Linear and Angular Momentum 
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#1
Aug3004, 10:06 PM

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Hi, I have a question about momentum. Is linear and angular momentum just special cases of a more general total momentum equation, or are each two independent equations. Is it possible for angular momentum to be converted into linear momentum, or are they two fundamentally different and non related ideas. The reason I ask is because I came acorss a paper from a somewhat shady website that said it was possible for there to be equal but not opposite reaction. The reason he state was becuase of what he called a general conservation of momentum, where he said the initial translational momnetum = the final translational momentum + the linear equivalent of the final angular momentum. He did this by stating that the angular momentum L is mv*l , where m is the mass, v is the velocity and l is the distance. He said if he divided the angular momentum by the lever arm l, he could get its linear equivelant, and use that in the momentum equation that i typed earlier. He uses this to show that linear momentum equation is just a special case of a more general momentum equation, and that its possible to make a machine that can propell itself forward with no external force outside, I think he claims so by some form of a rotatingwheel that converts angular momentum to linear momentum, but this sounds like a load of crap to me, maybe because he writes most people dismiss this arugment, but thats just me.
http://montalk.net/newton/newtonmath.html this is his paper if you want to see what im talking about. 


#2
Aug3104, 07:16 AM

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The paper you cite has a ball striking a "stick" that pivots about its middle. All of his calculations assume that stick only rotates and its center of mass does not rotate. That means the stick has to be attached to something at its pivot point. It is not a closed system and "conservation of linear momentum" does not hold.
Yes, linear momentum can be converted into angular momentum. If the stick in the article were not attached, then the ball striking it would cause it to both rotate and move away from its original position. You would have to use both conservation of momentum and conservation of energy (including the rotation) to find the actual values. IF you calculated the momentum vector of every point in the figure and summed (integrated) them all, the result would be a constant momentum vector angular momentum for a rigid body is just an easy way of handling that. 


#3
Aug3104, 07:55 AM

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But I talked with my physics professor on this and he said that angular momentum and linear momentum are irespective of eachother. He said that the system will have some intial angular momentum equal to m*v *l, which is the mass times the velocity times the distance of the billiard ball from the center of mass of the dumbell situation. He said that this would equal the angular momentum after the collision, which should equal the sum of how the dumbell rotates plus the product of what speed the billiard ball moves and its distance from the center of mass of the dumbell. He also said you must conserve linear momentum independly, so that you will have three equations with three unknowns, angular momentum, linear momentum, and conservation of energy. He said it is only in the conservation of energy equation that you can include the energy to cause both translation and rotation. Cyrus 


#4
Aug3104, 01:01 PM

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Linear and Angular Momentum
Your professor is correct. Under the appropriate conditions, angular momentum, linear momentum, and energy (rotational plus translational) are separately conserved.
I looked over that site you referenced in your first post. In my opinion, it's a pure crackpot site. Complete nonsense. 


#5
Aug3104, 01:20 PM

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Doc Al is right, that site is covered in pot shards.
I'll just post a link to an earlier thread of yours with "a rather lengthy" comment from me http://www.physicsforums.com/showthread.php?t=39505 


#6
Aug3104, 02:07 PM

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hey arildno,
But now you see what my question was when I posted that dumbell question. According to my phyics professor I should have three equations with three unknowns. But after I read your post you showed me a system of 12 equations! yikes. Where did the rest of these come from. I would assume that your way would give the same anwser. I also asked my professor if its possible to determing collisions involving three bodies or more. He said there is no way under any circumstance to determine the effects of the collision becuase there is not enough information and equations to handel a problem of that type. Do you know of any way or do you concur with him on that issue? Thanks for your help. Cyrus 


#7
Sep104, 02:46 AM

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Actually, I use 13 equations
This should come as no surprise, since: A rigid body has 6 degrees of freedom. In addition, the force (or impulse) is unknown as well, giving 3 extra unknowns. Hence, you should expect to need 6+6+3=15 unknowns. The "normal impact" condition specifies the tangential impulses to be zero, hence you're left with 13 unknowns. The equations of linear and angular momentum of both objects give you 12 equations; the last is given with equal normal velocity at the contact point at the end of the first impact period. This uniquely determines the motion state at the end of the first impact period. To bring the analysis to finish, we use a simple relation between the secondary impulse and the primary impulse in order to gain the final motion states. Your professor is incorrect. Let M objects make N normal impacts with each other (that is, there exist N points of collisions, distibuted among the M objects. The location of these N points are ASSUMED KNOWN) We therefore have 6M+N unknowns to determine! In the first impact period, 6M independent equations are gained from linear and angular momentum equations. The remaining N equations are the equal normal velocity demands at the N contact points. Specifying the secondary impulses in terms of the primary impulses enable us to determin the final motion state. Hope this clears up a bit; if you'd like a detailed discussion of your dumbbell problem, just say so EDIT: I would, however stress that your professor has identified a very important point, although he had dismissed the possibility of solving it too soon: In order to solve general impact problems, you have to use some assumptions regarding the basically unknown impulse. The normal impact assumption is by far the most useful one (it implies the "normal" twocollision equations). To give one subtle example: Assume that you have solved a general impact problem up to and including the first period of impact. In particular, we assume that we've solved an oblique impact problem, where the primary tangential impulse(s) have not been zero. What now of modelling the secondary impulse? Now, we know this is associated with the reversal of elastic deformations; but we cannot in general conclude that the coefficients of elasticity is equal in the normal and tangential directions!! Reversals in the normal directions arre typically responses to compression deformations, whereas reversals in the tangential directions are responses to shear deformations. Most materials have completely different attitudes to these deformation types. Hence, it is not generally true that the primary and secondary impulses are collinear.. 2.EDIT: Finally, I'll explain why your professor is correct in saying you'll need only 3 equations in your case. This is not because my approach is wrong (it is not; it is the most general approach of impact analysis if we want to avoid a differential equations approach). You'll strictly need only 3 equations because you've got lots of information in this particular case yet unused!! A) 2dimensional normal impact problem: This means each rigid body has only 3 degrees of freedom; hence, in general, you'll need only 3+3+1=7 equations (in the first period). B) One of the objects is a sphere/circle. Since we have normal/direct impact, this means that the sphere cannot gain rotational velocity as a result of the impact (i.e, it experience a centered impact) This reduces our number of fundamental unknowns by one. C)Unidirectional motion: Impacts are along the horizontal, hence there will be generated no vertical C.M. velocities; our fundamental unknown velocities are therefore reduced to 3: New velocity of the sphere, C.M velocity of dumbbell, angular velocity of dumbbell. Under our assumptions, we see that for the system sphere+dumbbell, we gain by conservation of linear and angular momentum, we have 2 of the necessary 3 equations. (Note that the impulse itself falls out in the system!) Do we need the impulse here (that is, a total of 4 unknowns), or can we find a 3rd equation relating the velocities? This is given by Newton's condition, which for an elastic collision says that the final separation velocity at the contact point must be the negative of the initial collision velocity at the contact point. (Collision velocity is one object's relative velocity "into" the other object) Newton's condition is certainly implied in my approach; occasionally (as in your case), Newton's condition plus conservation of the system's angular and linear momenta are sufficient to solve the problem. That is, we may bystep the admittedly cumbersome procedure I've sketched; just remember that for general problems, this remains the technique to solve them (in contrast to simplified versions which merely hold in certain cases). 


#8
Sep104, 01:03 PM

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Which course did you learn this stuff in? I dident do it in physics one, and I have not seen it in any engineering dynamics books. I assume its a grad course or upper level junior/senior course?



#9
Sep104, 01:12 PM

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In the now defunct "Introductory Classical Mechanics" course at the University of Oslo.
It was taught with excellent rigour, but rather "oldfashioned" in that "obsolete" theories like classical impact theory was taught. You'll be hard put to find a textbook on this published after 1945. 


#10
Sep104, 01:14 PM

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Just one question:
Did my explanation make sense to you? 


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