from a classical perspective, what is it about angular momentum fundamentally that means it has to be conserved? Surely if I have a rod about a fixed axis and a moving particle hits the end it will cause the rod to spin and therefore create angular momentum? (The particle isn't spinning around a point after the collision!!)

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    $\begingroup$ Just as a note about your second point. An object moving in a straight line can have non-zero angular momentum about a fixed point as long as the path of motion does not intersect that point. This is how the rod would start rotation, and therefore angular momentum would be conserved. $\endgroup$ – tmwilson26 Dec 2 '15 at 14:02
  • $\begingroup$ You need to learn to draw vector momentum diagrams. Typical textbook explanations include a rolling wheel, for example. $\endgroup$ – Carl Witthoft Dec 2 '15 at 14:59
  • $\begingroup$ Angular momentum needs not be conserved, in general. It does only when the Lagrangian is rotational invariant, i. e. when the system has isotropic symmetry in space. $\endgroup$ – gented Dec 2 '15 at 16:52

Surely if I have a rod about a fixed axis and a moving particle hits the end it will cause the rod to spin and therefore create angular momentum?

First off, there is no reason to expect that any of the conservation laws apply to the rod. A moving particle collides with the rod, and the rod has constraints that act on it to keep one end fixed. The collision and those constraint forces are external forces, some of which result in external torques. The conservation laws don't apply to the rod. They apply to the rod+particle+Earth system.

In general,

  • A system conserves energy if there is no transfer of energy between the system and the surrounding environment.
  • A system conserves linear momentum if no external forces act on the system and if all forces internal to the system obey the weak form of Newton's third law.
  • A system conserves angular momentum if no external torques act on the system and if all forces internal to the system obey the weak form of Newton's third law.

Secondly, you are ignoring that even point masses can have non-zero angular momentum. Angular momentum is always measured with respect to a point, not an axis. The angular momentum of a point mass is easily computed: It is $\vec L = m \vec r \times \vec v$, where $m$ is the mass of the point mass, $\vec r$ is the displacement vector from the central point to the point mass, and $\vec v$ is the velocity of the point mass. When viewed from the right perspective, the rod+particle system does conserve angular momentum. This "right perspective" is one in which the constraint forces on the rod exert zero torque.


Angular momentum is conserved in any system which has no external torques exerted upon it (much as linear momentum is conserved in any system which has not external forces exerted upon it).

The example of a moving particle hitting something and changing that somethings angular momentum is not relevant, since the something doesn't constitute a closed system in that case.

Your example fails to hold for the same reason that "hitting a golf ball means that linear momentum isn't conserved" does.

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    $\begingroup$ but what about if we include the particle in the system, then the particle would continue moving in a straight line? $\endgroup$ – lucky-guess Dec 2 '15 at 14:40
  • $\begingroup$ @EhabSyed If the particle stuck to the rod, then the rod/particle system would both rotate and translate, which conserves both linear and angular momentum. $\endgroup$ – tmwilson26 Dec 2 '15 at 14:59
  • $\begingroup$ what if it continued to move in a straight line or bounced off the rod? It would not be spinning around a point but the rod would. Then the angular momentum of the system before and after the collision will be different.. $\endgroup$ – lucky-guess Dec 2 '15 at 22:45

The contexts I have seen conserved quantities in is for holonomic systems whose legendre transformed variables of the lagrangian have some kind of nice property when it comes to the time derivative. Sometimes they look like some kind of lagrangian multiplier, but in the context of variational calculus. Usually Noether's theorem is quoted but I have never seen enough details of it to know what that means.

  • $\begingroup$ ok i get the implications of noether's theorems. but in this specific example, before the collision, the angular momentum of the particle and rod is 0, afterwards the rod is spinning. How is it then still conserved? $\endgroup$ – lucky-guess Dec 2 '15 at 23:25
  • $\begingroup$ @EhabSyed - The angular momentum of the rod+particle is not zero before the collision. Point masses have non-zero angular momentum if the radial vector and its derivative are not parallel or anti-parallel. $\endgroup$ – David Hammen Dec 3 '15 at 0:36

Angular momentum is conserved if the system behaves the same way regardless of it's orientation is space i.e its lagrangian is rotationally symmetric.

So if a system isn't sensitive to any rotation, then the angular momentum will be conserved.


protected by Qmechanic Dec 2 '15 at 23:38

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