Physics Help Forum Angular momentum conservation violation?
 User Name Remember Me? Password

 Advanced Mechanics Advanced Mechanics Physics Help Forum

 Feb 10th 2009, 02:45 AM #1 Physics Team   Join Date: Feb 2009 Posts: 1,425 Angular momentum conservation violation? Hi All, A thick disc spinning clockwise on a vertical axis is brought into contact edgewise with another identical disc and then moved back. Now both of them spin. Since both discs are identical and angular momentum is conserved, both spin slower than the original disc at half the speed. But the second disc obviously spins anticlockwise and hence its angular momentum,is in the opposite direction. Thus the sum of angular momenta ,( which is a vector sum) works out to zero or less than the original .Is conservation violated ? Someone help please!
 Feb 10th 2009, 03:22 PM #2 Senior Member   Join Date: Dec 2008 Location: Las Cruces NM Posts: 256 This is an interesting question. The conservation of angular momentum applies to the angular momentum measured about a given point in a closed system. The two angular momentum vectors of the disks mentioned in the question are not measured about a common point so there is no reason to expect their sum to be constant. Angular momentum can be measured with reference to a point without having a situation where an object rotates around that point. It is the cross product of one vector that is the radius vector from the point to a mass with another vector that is the mass-times-the-velocity of the mass. For example, suppose mass m1 is on the y-axis moving along it with velocity v1. Object m2 is also on the y-axis moving along it with velocity v2. Measure angular momentum about the point (-1.0). It is a vector in the z direction that has magnitude is (1)(m1 v1) + (1)(m2 v2). If the objects collide, after the collision they have new velocities v1' and v2'. By (linear) momentum conservation m1 v1' + m2 v2' = m1 v1 + m2 v2. So the angular momentum after the collision, which is (1) (m1 v1') + (1)(m2 v2') is conserved. We can also measure torques about the point (+1,0). They would also be conserved before and after the collision. The magnitude of this torque is (-1) m1 v1 + (-1) m2 v2.. However their is no reason to expect that the angular momentum of mass m1 measured about point (-1,0) plus the angular momentum of mass m2 measured about the point (+1,0) to be conserved in collisions. If we take the center of the first disk as the reference point, how would we calculate the momentum of the second disk about that point after the collision? Is this related to the "parallel axis" theorem for the moment of inertia computations?
 Feb 12th 2009, 10:11 AM #3 Physics Team   Join Date: Feb 2009 Posts: 1,425 Angular Momentum Thanks tashirosgt. Actually i stumbled on this when i was trying to think up a classical analog to spin in quantum mech. Sort of 'classical pauli exclusion principle' ; two spinning objects in contact necessarily have opposite spins. The example you gave about objects on Y axis was more reminiscent of orbital angular momentum in atomic physics (though there is no actual revolution around a centre)this reminds me more of spin. Since spin is also conserved what if we look at these two as forming a spin sytem? One experiences the contact as a torque which slows it down while the other as a torque which causes it to spin.However, the total spin angular momentum also seems to be in trouble here. I dont think there is any relation to parallel axis theorem though i havent thought about it much. I have another query about torques which i will post seperately.
 Apr 28th 2009, 10:35 PM #4 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 Kindly see the attachment. Edit : Sorry. Could not attach the file. Last edited by Parvez; Apr 28th 2009 at 10:39 PM.
 Jun 11th 2009, 12:57 AM #5 Junior Member   Join Date: Jun 2009 Posts: 4 Still don't understand after reading... Guys, could we revisit this please? I'm not sure I understand how angular momentum is conserved here. I've read tashirosgt's explanation and I see that you can't expect conservation when measuring angular momentum about two different points. Nevertheless if this is the explanation can you (or anyone) construct an example based on measuring the system before and after from one single point (maybe one that's not on either discs axes)? The fact that the angular velocity vector of the second disc is negative baffles me in terms of conservation of momentum. Please be as detailed and as slow in explaining as possible! Thanks all!, mgri
 Jun 11th 2009, 03:13 AM #6 Physics Team   Join Date: Feb 2009 Posts: 1,425 I don’t think I understand it well either. If we were to consider an analogy of the earth orbiting the sun, its total angular momentum should be the sum of its orbital angular momentum and its spin angular momentum. At least this is what is done in atomic physics (the spin angular momentum there is not due to actual spin of the electron, but is something intrinsic to it). So if we take two discs very close to each other, (one spinning fast and the other stationary,) but equidistant from a point O , the orbital angular momentum about O is zero. Let one disc be gently pushed (very slowly). Now this has an angular momentum about O. After they touch, they in all probability will move in such a manner that the orbital angular momentum part is conserved. This can be easily visualized. But now they spin in opposite directions at the same rate and this is a drastic change. It is here that I can’t understand how angular momentum is conserved. It might be that we can’t sum orbital and spin momentum here, but that is done in atomic physics.
 Jun 11th 2009, 10:04 PM #7 Junior Member   Join Date: Jun 2009 Posts: 4 How about this....? Ok I think I may understand what's happening here, although I'm not sure! The image I've made and attached should illustrate what I'm talking about. Basically I think that when the discs are brought into contact there would be a force transmitted to the axles that support these two spinning discs. This force would then be transmitted to the earth (or in the case of the image i've drawn the force is transmitted to the frame/gimbal that is supporting the spinning discs). Ultimately what i'm saying is that we can't be looking at the spinning discs in isolation, we need the entire system. In the image I've drawn I think the entire frame supporting the discs would begin to rotate and thus account for the apparently missing angular momentum. Initially I had a hard time figuring out the source of the force ( rhyme!) I just mentioned. Drawing a free body diagram didn't reveal it to me, but thinking of the following situation seems to satisfy me:Imagine a 1m x 10cm x 10cm block lying with it's long end ointing along the x axis ontop of a frictionless surface in the xy plane. If you were to smack one end of this block perpendicular to it's long axis (i.e. in the y dirn) what would happen? You'd obviously get rotation around the block's centre of mass but in addition you'll get translation motion of the centre of mass in the y direction. If you now think of the frictional force existing at the edge of the discs when they're put in contact as analogous to the force smacking the block you might see where I'm coming from: There must be a force applied to the disc's axis that is analogous to the force causing translation of the block's centre of mass. Since the axis isn't free to move the force is transmitted elsewhere thus distributing the angular momentum elsewhere. What do you think??? Does anyone know how to quantify the linear momentum of my block's centre of mass if it's given the kind smack I described? What is this general area of physics called? Attached Thumbnails
 Jun 12th 2009, 10:27 AM #8 Member   Join Date: Apr 2009 Posts: 71 Hi, this is what I think, Initially the angular momentum (measured from a point distance d from the centre of mass of the disk) is given by. L= $\displaystyle 2 \omega(I_{cm}+Md^2)$ (Paralell axis theorem) Finally L= $\displaystyle \omega(I_{cm}+Md^2)$ for each individual disk. Hence the final L is given by $\displaystyle \omega(I_{cm}+Md^2)$ + $\displaystyle \omega(I_{cm}+Md^2)$= $\displaystyle 2 \omega(I_{cm}+Md^2)$ (and we have conservation). Note $\displaystyle \omega$ has the same sign each time as the disks move in the same direction relative to our reference point (not to their centres). How is this? Last edited by C.E; Jun 12th 2009 at 01:08 PM.
 Jun 12th 2009, 10:42 PM #9 Junior Member   Join Date: Jun 2009 Posts: 4 C.E, I think there's a couple of things not quite right about what you've said below. First of all, I'm not convinced the parallel axis theorum should be applied to the initial position as there is no motion of the centre of mass of the spinning disk. The parallel axis theorum applies only to motion of a centre of mass as far as I know. What I'm saying is that a system's angular momentum due to a spinning disk is constant no matter where it's measured from... please someone correct me if i'm wrong on this . Secondly, I disaggree with this statement: " has the same sign each time as the disks move in the same direction relative to our reference point (not to their centres). When you say "the disks move" could you be more specific... do you mean (1) the disks rotate in the same direction, or their (2) centre of masses move in the same direction? Either way I don't think its a correct statement for the following reasons: 1 - , describes the angular velocity of either of the disks about their CoMs.. This is clearly opposite. 2 - the centre of masses of both disks can't spontaneously move in the same direction after being brought in contact. What actually happens is as I described previously: the centre of masses of the disks begin to rotate about each other (or if they're not free to do so the axles transmit the angular momentum to whatever they're attached to). It's definitley and interesting one but I'm pretty sure I understand it now. C.E. let me know if you don't agree... mgri Last edited by mgri; Jun 13th 2009 at 09:45 AM.
 Jun 12th 2009, 11:30 PM #10 Physics Team   Join Date: Feb 2009 Posts: 1,425 Thanks. The more i think about it, the more iam convinced that you are right. Even if we were to change the problem slightly and have two spheres on a frictionless surface, i feel they will orbit each other to conserve angular momentum. Stretching things a bit, maybe this is how the earth got its initial orbital momentum when it was ejected from the sun! So we may conclude that we have to consider both spin angular momentum and orbital angular momentum for conservation to hold which is what they do in atomic physics. Though i had mentioned this before, i did not see the connection till i saw your image which is excellent.How did you make it? But what do you mean by this? The parallel axis theorum applies only to linear momentum as far as I know.

 Tags angular, conservation, momentum, violation

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post assaftolko Advanced Mechanics 1 Sep 2nd 2012 08:26 AM amirshrestha Kinematics and Dynamics 2 Apr 28th 2012 11:49 AM hapflir Advanced Mechanics 5 Oct 13th 2009 10:38 PM monomoco Advanced Mechanics 1 Mar 15th 2009 10:45 PM orgoistough Advanced Mechanics 1 Oct 31st 2008 09:14 PM