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Conservation of spin, energy and momentum

  1. Jul 22, 2010 #1
    I have some problems understanding when a system is isolated or not. I know the definition:
    Well, my confiusion is better shown by an example.

    Consider an Merry-go-round (MGR):
    149-FunGoRound.jpg

    A person runs towards the MGR while the MGR stand still. Call this situation 1). He the jumps on it. This obviously makes the MGR rotate. Call this situation 2). Define that for the spin, the center of rotation is in the middle of the MGR.

    Im confused regarding what properties is conserved. Angular momentum, energy and (linear momentum). If you dont care about my ramblings, you can stop read now ;p

    I would think that momentum (linear, p=mv) is not conserved. In 1) the person running makes the system have a p>0, but in 2) the system will only have angular momentum, i.e p=0. Is this correct?

    I would say that energy is not conserved due to the fact that it is friction that makes the person stick to the MGR. But im not sure about this. I can think of situations where one would not need friction to stick to the MGR, for example if one run into a wall placed on the MGR perpedicular to it. This makes me confused.

    And then you have angular momentum (spin). The person has angular momentum around the axis of rotation just before he hits the MGR in 1). in 2) the MGR and person both have Angular momentum. I would say that the spin is conserved, because there is no external net torque acting on the person-MGR system. However, you have that friction again.. One can perhaps look at the friction as a torque.
     
  2. jcsd
  3. Jul 22, 2010 #2

    Doc Al

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    Linear momentum of the system is not conserved. The MGR is fixed by its axis, which exerts an external force on it.

    However the person 'sticks' to the MGR, the person essentially makes an inelastic collision with the MGR. Mechanical energy is not conserved. (The same would be true if the person just ran and jumped onto a wagon.)

    There are no external torques acting on the system. (Presumably, the axle of the MGR is frictionless and exerts no torque.) Whatever forces the MGR and person exert on each other (whether friction or other), those forces produce no net torque on the system. Angular momentum is conserved.
     
  4. Jul 22, 2010 #3
    Thanks for the quick reply!
    Can you explain to me why energy is not conserved in inelastic collisions?
    And why can you neglect the force acting from the axis on the system in the angular momentum case, but not in the linear momentum case?

    Do I got it right:
    -Isolated system for angular momentum <=> no external net *torque* action on the system
    -Isolated system for linear momentum <=> no external net *force* acting on the system

    Why is the force from the axis considered external anyway? It is a part of the carousel, and thus a part of the system, and thus internal, not external.
     
  5. Jul 22, 2010 #4

    Doc Al

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    Whenever things collide and stick together, mechanical energy is 'lost'. You can prove this for yourself by setting up a few simple examples of colliding objects and applying conservation of momentum to find the final speed after the collision.
    The reason is just as you state next:

    Right!

    It's perfectly OK to consider the axle as part of the system if you like, but then realize that something must be exerting a force (an external force) to hold the axle in place.
     
  6. Jul 22, 2010 #5
    What if I consider the axis AND the ground that holds it AND the earth that carries the ground as a part of the system? Will there still be an external force acting on the system?
     
  7. Jul 22, 2010 #6

    Doc Al

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    No. You can always expand your system so that any force is then part of the system. Of course, that's not always a helpful thing to do.

    Considering the entire earth and all its parts as a system, then sure, linear momentum of that system is conserved when the person jumps onto the MGR.
     
  8. Jul 22, 2010 #7
    Interesting... Yet confusing. =)

    Thank you for your help, Doc!
     
  9. Jul 22, 2010 #8
    Btw, what happens to the "excess" energy after an inelastic collision?
    Edit:
    I know the basic answer probably is "heat, deforming the colliding bodies and vibrational waves (sound)". However, why is there is no heat loss in elastic collisions? And would you hear a sound when you are standing close to an elastic collision? If yes, then energy has been lost.. I can't understand why there should be more heat loss in an inelastic collision than in an elastic one..
     
    Last edited: Jul 22, 2010
  10. Jul 22, 2010 #9

    Doc Al

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    It's a matter of degree. All real collisions of macroscopic bodies lose energy to some degree. But some collisions are approximately elastic to a high degree--for example, a collision between steel balls may well be close to elastic. (But you're right, you'll still lose a bit of energy.) On the other hand, a collision between two lumps of clay that stick together is far from elastic.

    What makes the difference is the ability of the bodies to store and release elastic energy. (Thus the name elastic collision.) Steel is highly elastic--it springs back after you deform it; clay is not--it just gets mushed and stays that way.
     
  11. Jul 28, 2010 #10
    If you are running in a st. line that does not passthrough the center of MGR, yourself and MGR together (system) has an angular momentum. When you jump on to the MGR and come to a dead stop, MGR and yourself together rotate with an angular momentum equal to that before you jumped. Angular momentum is conserved. No friction is assumed in this case. The kinetic energy is conserved. If friction is assumed to be present, heat is produced to the extent of loss of kinetic energy.

    If the line along which you run passes through the center of MGR, then when you jump on to MGR, you will simply slide along the surface and land on the ground . If you come to a stop on the ground, KE is transformed into heat. If you don't come to a stop and keep moving, KE is lost to the extent that your speed decreases. If your speed does not decrease, both linear momentum, angular momentum and KE are conserved. If your speed decreases, the linear momentum of yourself and the earth together is conserved, so is the angular momentum wrt MGR.
     
  12. Jul 28, 2010 #11

    Doc Al

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    Kinetic energy is not conserved. (You and the MGR exert force on each other, else how can you end up rotating together.)
     
  13. Jul 28, 2010 #12
    In kinematics, you don't invoke the concept of force at all. You rotate together because of the angular momentum.
     
  14. Jul 29, 2010 #13

    Delta²

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    Isnt it possible to construct a device or make an experiment that converts linear momentum to angular momentum or vice versa?
    I guess the answer is no, since conservation of linear and angular momentum hold together. It is somehow confusing because conservation of momentum holds but not as total (angular+linear) but only for each kind separately.
     
    Last edited: Jul 29, 2010
  15. Jul 29, 2010 #14

    Doc Al

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    Using conservation of angular momentum alone you can show that kinetic energy is not conserved. (Nonetheless, you and the MGR do exert forces on each other!)
     
  16. Jul 29, 2010 #15
    You are right. Linear momentum and angular momentum are not inter convertible.
    Since you know that each of them is converted individually, it goes without saying that they are conserved together too!
     
  17. Jul 29, 2010 #16
    I don't understand exactly what you are saying. But, I can perhaps say something more if you can show (as you say) that KE is not conserved, from conservation of angular momentum alone.
     
  18. Jul 30, 2010 #17

    Doc Al

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    Linear and angular momentum are conserved (or not) independently. They are different physical quantities with different units, so adding them in a single expression is not physically meaningful.
     
  19. Jul 30, 2010 #18

    Doc Al

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    Here you go:

    m ≡ mass of person
    p ≡ initial momentum of person
    I ≡ rotational inertia of MGR

    Assume that the person is moving tangentially to the MGR.

    Conservation of angular momentum:
    Rp = (I + mR2

    Initial KE: ½p2/m

    Final KE: ½(I + mR22 = ½R2p2/(I + mR2)

    As long as I > 0, the Final KE will be less than the Initial KE.
     
  20. Jul 30, 2010 #19

    Delta²

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    Yes you are right about this, however conceptually they are both momentums and one could argue mathematically that [tex]P_{total}=P_{angular}+P_{linear}=C[/tex] holds but only because [tex]P_{angular}=C_1, P_{linear}=C_2[/tex] hold.
     
    Last edited: Jul 30, 2010
  21. Jul 30, 2010 #20

    Doc Al

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    I'd say that was a meaningless expression. What would be the units of Ptotal? How do you add quantities with different dimensions?
     
  22. Jul 30, 2010 #21

    Delta²

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    You add them as real numbers :smile:

    Btw, on your other post u treat the inertia of the person as that of a point particle having mass m, doesnt it depends on the inertia of the person if [tex]KE_{final}<KE_{initial}[/tex] (for example if the inertia of the person is taken as [tex]\frac{1}{2}mR^2[/tex] it doesnt seem to hold for every [tex]I>0[/tex]
     
  23. Jul 30, 2010 #22

    Doc Al

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    :rofl: As long as you're willing to ignore units, you can make all sorts of silly expressions.

    I do treat the person as a point particle with mass m. The rotational inertia of the person is mR2, assuming they jump onto the edge of the MGR.

    If the person jumps onto the MGR at some other point, the rotational inertia will be less. But the final KE will still be less than the initial for any perfectly inelastic collision.
     
  24. Jul 30, 2010 #23

    Delta²

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    I dont think they are silly, at least not mathematically. But anyway.

    I guess it only makes the math more complex to consider different rotational inertia for the person. Since the collision is inelastic the conclusion holds, still it is kind of surprising that without making any assumptions about friction forces or other internal forces, not even a reference to them, just from conservation of momentum it follows that the final KE will be reduced.
     
  25. Jul 31, 2010 #24
    You are quite right. We cannot and should not add linear and angular momenta for the simple reason, as you said, that they have different units.

    More interesting is your derivation. I stand corrected; the person and MGR do not rotate together after collision, as I stated earlier.

    Assume masses of the person and MGR to be equal. In this case, the person comes to a stop and the MGR starts rotating with an angular velocity that gives angular momentum equal to the initial angular momentum of the person, thus satisfying the conservation of angular momentum. Because we assumed no friction, the person keeps slipping at the same location where he landed on MGR, as MGR keeps rotating. This is similar to 1-D elastic collision, in linear motion.

    When the masses are unequal, and MGR has greater mass, the person transfers a portion of his momentum to MGR and himself reverses his direction of motion and move with a slower speed. MGR rotates with angular momentum received from the person. When the mass of MGR is so high that the mass of the person is negligible in comparison, the person simply reverses his velocity, i.e. retraces his path with same speed as he had before jumping; MGR remains still - does not rotate.

    The essential point is that the sign of the relative angular momentum changes its sign after collision.

    If, on the other hand, the mass of MGR is lower, then the person continues his motion after the collision, along the same line but with a lower value of speed. MGR rotates with an angular momentum that has same sign as the angular momentum of the person, and a speed that gives the reversal of angular velocity. This satisfies the conservation of angular momentum.

    If you represent angular momentum vector by an arrow, then those vectors corresponding to the person and MGR can be manipulated just as in the case of elastic collisions in linear motion.

    Reversal of relative angular velocity due to elastic collision is the key to the solution of the problem, just as in the case of elastic collisions in linear motion.
     
    Last edited: Jul 31, 2010
  26. Jul 31, 2010 #25
    When the masses are comparable, the axis of rotation exerts a force on the MGR under the collision, so this is not similar to 1-D linear elastic collision, as you state.

    But when the mass of the MGR is so big that the mass of the "person" bouncing off the MGR again is negligible, then the axis of rotation does not exert a force on the MGR, and thus the collision is similar to 1-D linear elastic collision.

    .. I think :)
     
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