## Is angular momentum conserved here?

 Quote by tiny-tim yes no, there's plenty of internal forces, they just all add to zero
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.

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 Quote by aaaa202 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? :)
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*)
 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?
i'm not sure i understand your question

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
 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? Attached Thumbnails

 Quote by aaaa202 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?
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.

http://en.wikipedia.org/wiki/Screw_theory
“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.
http://helix.gatech.edu/ball2000/CD/...all2000-28.pdf
“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.”