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Is angular momentum conserved here?

  1. Aug 17, 2012 #1
    There is no net torque in this system so angular momentum should be conserved. I can't see how that is however.

    Our system is: A stick with a massless spring attached at one end. The spring is compressed and at the other end there is a point particle of mass m.

    The spring is then released. Causing a torque on the particle aswell as on the stick relative to the center of mass of the stick, the torquevectors being of same length but opposite sign. The pole will rotate and its angular momentum will remain the same (right?). But the particle will move away from the pole and thus change its location relative to the center of mass of the stick. But does that then not mean that the angular momentum will change? Or is it wrong to use the center of mass of the stick?
  2. jcsd
  3. Aug 17, 2012 #2


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    hi aaaa202! :smile:

    the angular momentum of the particle is the perpendicular distance from the centre of mass of the stick to the particle's line of motion "cross" the velocity of the particle

    that is a constant, and is equal and opposite to the angular momentum of the stick :wink:

    (oh, and you can use any point as your centre of angular momentum …

    the only time you have to use the centre of mass, or the centre of rotation, is when there's an external torque and so you need to differentiate the angular momentum …

    you get extra awkward terms then, if you don't use the centre of mass, or the centre of rotation :yuck:)
  4. Aug 17, 2012 #3
    okay but I am not quite sure: Why is it that the stick will rotate around its center of mass rather than any other point on it?
  5. Aug 17, 2012 #4


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    the centre of mass is the only point on the stick that moves in a straight line

    the stick as a whole rotates uniformly about that moving point :smile:

    (it also rotates uniformly about any other point on the stick, but such other point would be moving along a loopy line, so that's not very convenient :wink:)
  6. Aug 17, 2012 #5
    How do you see that any other point moves in a loopy line? I do know that the center of mass moves as though it was only acted upon by external forces. But if you push a stick at its center of mass surely also the rest of the stick will move in a straight line like the com?
  7. Aug 17, 2012 #6


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    because the centre of mass moves in a straight line, and any other point on the stick rotates around it (if it's rotating) :wink:
    yes, that's correct :smile:
  8. Aug 17, 2012 #7
    Now I am getting confused.

    You said before that the toruqe caues the stick to rotate uniformly about any point - that was how I understood it. But we just discovered that without any torque all the point move in a straight line. So shouldn't the stick then rotate about every point? hmm okay that sounds weird too.
    Say there was no translation of the center of mass when you pushed the stick. I know that is not possible since every force will both tend to translate and rotate. Then the stick would still look like a stick rotating about its center of mass, but what about the other points?
  9. Aug 17, 2012 #8


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    yes, the stick does rotate about every point, but the angular speed is zero :wink:
    as you say, that's not possible
  10. Aug 17, 2012 #9
    what? :confused:

    The angular speed zero? How is it then a rotation?

    This is getting weirder and weirder for me. I know that you can mathematically show that the torques around the center of mass are only caused by external torques. Is this a part of the explanation.
    But I'm still lost. Relative to every point on the stick there is a torque, yet when you picture it in your mind, I can only see the stick rotate around the center of mass. That is weird..
  11. Aug 17, 2012 #10


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    if you can have a zero speed, you can have a zero angular speed
    you've lost me :confused:

    if the angular speed is constant, there is no torque

    (there was an impulsive torque to get it to that angular speed, but then that stopped)
  12. Aug 17, 2012 #11
    Okay let me describe how I see it, and you can correct me where I go wrong.
    The force from the spring on the stick makes all of the atoms in the stick translate with a velocity given by equations of linear momentum conservation. However, the force also generates a torque around the center of mass making it rotate around it.
    What I don't understand is this: Why does it rotate around the center of mass? What makes that point so special? Relative to an arbitrary point on the stick there is generated a torque from the force but surely when you look at a spinning wheel it is only spinning around the center, i.e. the center of mass.
  13. Aug 17, 2012 #12
    You two might not be on the same page here.
    We choose the center of mass as the origin because, as seen by an inertial observer, the center of mass is not rotating about any point and moves in a line. In this inertial frame, we observe all other points on be rotating about the CM, and thus they are kept at a constant radial distance from it, allowing us to specify the position of the point with only the angle (s) it makes with the origin.
  14. Aug 17, 2012 #13
    Okay, but WHY is it then that the center of mass has this property? That's more or less what I have been trying to get at.
  15. Aug 17, 2012 #14
    Oh yes sorry, right now I am thinking about the instant where the spring exerts a force. At this instant there is a torque from its force relative to any point on the stick. Yet when we observe the stick, it only seems to rotate around the center of mass.
  16. Aug 17, 2012 #15


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    (a different velocity for each atom, of course)

    if the wheel is moving (eg rolling along the ground), it is only our prejudice that makes us say that it is rotating about its centre

    in reality, as you probably know, a rolling wheel is rotating about the point of contact with the ground! :wink:
    there is always an instantaneous centre of rotation, which can be anywhere

    (and is "at infinity" if the body is not rotating)

    it only seems to rotate about the centre of mass, because we tend to follow the centre of mass with our eyes :smile:
  17. Aug 17, 2012 #16
    So you could also view the movement of a stick translating and rotating as a rotation around for instance the blue point indicated on the attached figure?
    What complications would arise with this approach? Would you be able to use the equation:
    Ʃτ = Iα
    to find the angular acceleration of the stick?
    And since only the center of mass moves as if only acted on by external forces... Would this then not mean that there would be internal forces at this point messing up my head? :(

    Attached Files:

  18. Aug 17, 2012 #17


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    yes :smile:

    but that would only be the instantaneous centre of rotation
    yes, that will give you the instantaneous angular acceleration about the centre of rotation

    (I of course has to be about the centre of rotation)
    no, there's plenty of internal forces, they just all add to zero
  19. Aug 17, 2012 #18
    Does this not only hold for the centre of mass or is this a case where it actually counts for all points? If so, can you name an example where it only works for the center of mass, so I can distinguish? :)

    And more: Why is it only instantaneous? There is a torque around that point all the time? Do you mean that because the stick moves all the time the rotation around it is only instantaneous? But then is the rotation around the center of mass not also instantaneous? Surely that moves too - what distingiushes those two coordinate frames?

    Or said in another way:
    I don't understand why it is much easier to work with the center of mass. I don't understand why this point is so special. I understand it translates as though only acted on by external forces but it seems that all points on the stick do this in this situation.

    Sorry I ask so much, but I actually feel that I get a little closer to a complete understanding every time. I will consider your answer to this reply final for now and try to go back and get a better understanding by thinking everything over. Thanks so much for your help for now.
    Last edited: Aug 17, 2012
  20. Aug 17, 2012 #19


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    the internal forces at a particular point (on a body with no external force at that point) will sum to zero only if that point has zero acceleration

    if the body is rotating, then the only point that isn't accelerating is the centre of mass

    (and the only point that isn't moving is the centre of rotation*)
    i'm not sure i understand your question :confused:

    if the body is moving freely, there is no torque about any point

    * this is in 2D … for a 3D body, there will be an instantaneous axis with zero velocity
  21. Aug 18, 2012 #20
    Okay ill try to explain my problem more pictorially. Why is the stick moving as in scenario 2 and not as in scenario 1 on the attached picture?

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  22. Aug 18, 2012 #21
    You are picturing the motion of a rigid body, which includes both rotation and displacement. What is probably bothering you is that the mathematical description of such motion is not unique. This is because any motion can be pictured by a screw transformation about any arbitrary axis. The arbitrary axis does not have to correspond to any physically relevant parameter of the rigid body.
    Graduate mechanics courses present the mathematics of rigid bodies in several representations. Although the physics is “classical”, some of the mathematics seems anti-intuitive at first sight because the descriptions aren’t unique.
    One learns in introductory physics to choose the leverage point that is most “convenient” for the calculations. However, “convenient” isn’t the same as “unique”. Very often the center of mass is used as a leverage point because it is most “convenient”. Again, convenient isn’t unique.
    Both the leverage point and the axes of rotation are arbitrary. It is possible that your left cerebral hemisphere has chosen one axis and your right cerebral hemisphere has chosen another axis of rotation. Both hemispheres are correct. However, solving a physics involves working with one axis consistently.

    “Screw theory refers to the algebra and calculus of pairs of vectors, such as forces and moments and angular and linear velocity, that arise in the kinematics and dynamics of rigid bodies.
    The conceptual framework was developed by Sir Robert Stawell Ball in 1876 for application in kinematics and statics of mechanisms (rigid body mechanics).[3]
    The value of screw theory derives from the central role that the geometry of lines plays in three dimensional mechanics, where lines form the screw axes of spatial movement and the lines of action of forces. The pair of vectors that form the Plücker coordinates of a line define a unit screw, and general screws are obtained by multiplication by a pair of real numbers and addition of vectors. A remarkable result of screw theory is that geometric calculations for points using vectors have parallel geometric calculations for lines obtained by replacing vectors with screws. This is termed the transfer principle.

    A remarkable result of screw theory is that geometric calculations for points using vectors have parallel geometric calculations for lines obtained by replacing vectors with screws. This is termed the transfer principle.[4]
    Screw theory notes that all rigid-body motion can be represented as rotation about an axis along with translation along the same axis; this axis need not be coincident with the object or particle undergoing displacement. In this framework, screw theory expresses displacements, velocities, forces, and torques in three dimensional space.”

    Sometimes the best way to learn a topic is from a graduate student’s thesis. This thesis covers the moment of inertial in rigid body theory.
    “Rigid-Body Inertia and Screw Geometry
    This paper reviews the geometric properties of the inertia of rigid bodies in the light
    of screw theory. The seventh chapter of Ball’s treatise [1] defines principal screws of inertia for a general rigid body based on Ball’s co-reciprocal basis of screws. However, the application of that work to the important cases of planar- and spherical-motion is not satisfactory. The following paper proposes a new formulation of the screws of inertia which is more easily applicable, and compares it with common mathematical devices for treating rigid body inertia such as the inertia tensor [6, 9]. This brings to light a geometric perspective of inertia that does not often accompany this topic.”
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