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Conservation of Angular Momentum vs. Conservation of Energy

  1. Jan 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi all! I have a very simple problem, which seems to get two different answers depending on whether you use conservation of angular momentum, or energy. Both quantities seem to be conserved:

    Initially we have a disk of radius a spinning about its center of mass at known angular velocity ω. Suddenly a random point along the disks diameter becomes fixed in place and the disk begins spinning about that point with angular velocity Ω. Find the final angular velocity Ω.

    2. Relevant equations
    Using conservation of angular momentum:
    ##L_i=I_{CM}*\omega=\frac{1}{2}ma^2*\omega##
    ##L_f=I_{edge}*\Omega=(I_{CM}+ma^2)*\Omega=\frac{3}{2}ma^2*\Omega##
    ⇒##L_i=L_f##
    ⇒##\Omega=\frac{1}{3}*\omega##

    Using Conservation of energy:
    ##E_i=\frac{1}{2}I_{CM}\omega^2=\frac{1}{4}ma^2\omega^2##
    ##E_f=\frac{1}{2}I_{edge}\Omega^2=\frac{3}{4}ma^2\Omega^2##
    ⇒##E_i=E_f##
    ⇒##\Omega=\frac{1}{\sqrt{3}}*\omega##

    3. The attempt at a solution

    As you can see, both methods give similar results, but not the same. Why would this be? If there some reason why Energy or Angular momentum would not be conserved? Is it something to do with the fact that angular momentum is only conserved about the center of mass, which it is no longer spinning about in the final scenario?

    Thank you in advance!
     
  2. jcsd
  3. Jan 18, 2015 #2

    Orodruin

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    The pinching of one point of the disk is essentially an inelastic collision that does not conserve energy. Angular momentum is conserved in both cases (for a particular point, which is not the CoM, so think long and hard about which).
     
  4. Jan 18, 2015 #3

    BvU

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    Hello BH, welcome to PF :)

    I have questions here: You state you are "Using conservation of angular momentum" , but isn't that conservation of angular momentum about an axis ? It looks to me as if your ##L_i## and ##L_j## are around different axes.

    @Orodruin: where would the energy go ?
     
  5. Jan 18, 2015 #4

    Stephen Tashi

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    Neither is conserved in a system if external forces act on the system. Can you suddenly freeze a pont on the edge of the disk without using external forces? I, myself, don't know if freezing a point on the edge of the disk without using force is a standard "let's pretend" device used in mechanics problems. ( I wonder if you are supposed to do a calculation with a rotating reference frame.)

    In the absence of external forces, angular momentum computed about any fixed point is conserved. Angular momentum doesn't have to be computed about the center of mass, although it's often convenient to compute there. A spinning object has an angular momentum about any point in space, even if it's not the point the object is "spinning around".

    You can't apply conservation of momentum by equating angular momentum computed about one point in the "before" scenario and about a different point in the "after" scenario. To assert equality, you must compute it about the same point "before" and "after".
     
  6. Jan 19, 2015 #5

    Orodruin

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    Where does energy usually go in inelastic collisions?

    Edit: And indeed the OP should be made aware of the fact that the angular momentum wrt different points can be different and in particular, in this problem, there is the underlying idea that fixing a point only gives forces acting at that point (and thus zero torque, which gives conservation of angular momentum wrt that point).

    However, the original disk was only rotating with no CoM velocity. Thus the original angular momentum is the same wrt any point is the same. Let us not completely disregard the possibility that the OP already knew this.
     
    Last edited: Jan 19, 2015
  7. Jan 19, 2015 #6

    BvU

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    Friction, deformation and that kind of violent and difficult stuff.
    So I build a nice little low-friction catch H that snaps shut when it hooks on pin A. (Mind the little red snapper :) )!
    Strong enough to not deform in the process. Add a bit of lubricant to satisfy the theorists.
    Pin A can exercise torque, force, what have you. But it does not contribute or take away kinetic energy: it's fixed.

    My picture is that H catches on A. Kinetic energy is conserved (where would it go?) and A becomes the center of rotation. I vote for the ##1\over\sqrt 3##, but I am well aware that physics is not a popularity contest. I haven't sorted out what to do with the angular momentum conservation part (*) yet, so I'll keep some egg at hand in case someone proves me all wrong.
    Disc.jpg
    (*) Is it supposed to be easy to go from ##L_i## around C to ##L_i## around H ?
     
  8. Jan 19, 2015 #7

    Orodruin

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    The point is that you have to have some deformation in order to provide the force. The transfer of energy into heat can also be due to sound dampening inside the materials. Even in the case where you have a very very large spring constant meaning essentially no deformation, it is the dampening of the resulting sound waves that will in the end turn out as heat in the materials.

    As long as the point A is sufficiently point-like, there are going to be no torque around that point and thus no change of angular momentum. As you say, without the external torque, where would the angular momentum go?

    I would like to make a similar example for linear motion: You have a non-deformable spherical bullet that just fits into a non-deformable spherical shell and assume the collision is elastic. When the bullet reaches the shell, it snaps shut so that they have to travel together. Where did the energy go? Where did the momentum go? (In this case there are not even any external forces to consider.) No external forces means no change in total momentum and I would like to claim that invariably the excess energy must be transfered to internal degrees of freedom of the compound object such as vibrations (or in the case of damped vibrations, heat in the end).
     
  9. Jan 19, 2015 #8

    BvU

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    The egg at hand is all shattered o:) In fact my physics sense had already convinced me I was erring about the catching fairy tale. Same issue as with the shell/bullet. Kudos to Oro!

    And supermassive knew from the beginning that it's a very simple problem ! :)
     
  10. Jan 19, 2015 #9

    BruceW

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    we should still try to explain (mathematically) why the pin and catch experiment cannot conserve kinetic energy.
     
  11. Jan 19, 2015 #10

    Orodruin

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    I would say the mathematical reply would involve non-invariance under time translations (there is a difference before and after the pin has caught) and Noether's theorem. I think this is above the level we can assume that the OP is familiar with.
     
  12. Jan 19, 2015 #11

    BruceW

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    uh.. but it's a pretty simple system. I feel like there should be a fairly easy answer. We could even replace the catch with a small rod pointing outward from the disc, and consider what will happen in the first few moments as the rod strikes the pin.

    edit: maybe this is too off-topic though.
     
  13. Jan 19, 2015 #12

    Orodruin

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    What is going to happen is that you are going to excite vibrational modes inside the system. Whether or not it will continue to vibrate depends on the damping of the vibrations. You could technically take the limit where the system becomes stiffer and stiffer, but the energy will eventually end up in the same way. You could not replace it with a rod as a simple rod would most likely lead to a rebound and thus not a perfectly inelastic collision. This would transfer less energy to vibrational modes but conserve more of the kinetic energy in terms of how it is considered in the given problem. Again, I think it is overkill for the OP's purposes.
     
  14. Jan 19, 2015 #13

    BruceW

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    I was thinking about it for a while, and it doesn't matter what shape the object has. The kinetic energy will never be conserved in this situation. Is this a particular example of some more general theorem? Maybe it's related to non-invariance under time translation, like you say.
     
  15. Jan 19, 2015 #14

    BvU

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    Excellent idea. Then we probably come to the conclusion that fully elastic would mean ##\vec \omega \rightarrow -\vec\omega##

    But that's different from the situation in the OP
     
  16. Jan 19, 2015 #15

    Orodruin

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    This situation would require no net force and a net torque on the disk. Unless you have (at least) two pins I do not see how this could be arranged. I would say the CoM will start moving and the disk would spin at a different angular velocity.
     
  17. Jan 19, 2015 #16

    BvU

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    More eggs please o:)
    I'm running out of them for today :)
     
  18. Jan 19, 2015 #17
    Wow thank you guys! I forgot to turn on email notifications so I didn't realize you all replied. A lot of food for thought, I'm going to have to think about this one more.

    Thank you!
     
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