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Homework Help: Conservation of angular momentum

  1. Oct 22, 2009 #1


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    Bear with me, I am mathematically challenged chemist.

    I am trying to understand following situation:

    There are two masses connected with a line. They rotate about an axis perpendicular to the line. System have some easy to calculate angular momentum. In a closed system angular momentum is constant, so as long as we don't touch it, angular momentum is not changing.

    Line breaks.

    Now we have two masses going in straight lines. Obviously at the moment line broke each mass had some linear momentum and it still has the same linear momentum, tangential to the previous trajectory.

    However, I can't see what have happened to the angular momentum. L=rxp - r and p vectors are getting parallel, so the cross product becomes zero.

    I suppose I am mising something simple, but I can't see it :grumpy: The only explanation I can think of is that angular momentum conservation is only kind of emergent property of the rotating system, while linear momentum conservation is the 'real' principle, but I feel like that's rather a bold statement.
  2. jcsd
  3. Oct 22, 2009 #2
    A similar problem. Just one mass here. A planet revloving around a highly massive sun. There is some initial angular momentum thats unchanged as gravity is central. Now what would happen if gravity is 'switched off'?
  4. Oct 22, 2009 #3

    Doc Al

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    The cross product does not become zero! L = r X p = rp sinθ; rsinθ remains constant. (Even though r and θ both change, rsinθ remains constant.)
  5. Oct 22, 2009 #4


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    I have no back of the envelope at hand, so I had to try on the postit note card. Assuming r0 is the distance between rotation axis and the line on which mass moves now

    [tex]r \sin \theta = r_0[/tex]

    You learn all your life and you die stupid :rofl:
  6. Oct 22, 2009 #5


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    The component of r which is perpendicular to p is constant so r x p is constant.
  7. Oct 22, 2009 #6
    Maybe try considering first a simpler problem.

    I have a single massive object moving in space. There are no forces acting on it, and it remains moving at a constant velocity. There is no torque on it, and it is not rotating in any way.

    What is its angular momentum with respect to the origin? Does it even have angular momentum? Go back to your formula and figure out what the angular momentum is, given the velocity, mass, and position of the object.
  8. Oct 22, 2009 #7
    Angular momentum is defined by product of distance and linear momentum,
    it doesn't necessarily has to imply rotational movement

    in your problem, even when the mass is moving in a straight line, you can
    see that the product of distance and momentum remains as before.
  9. Oct 22, 2009 #8


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    I am also of the opinion that linear momentum is more fundamental than angular momentum.

    Still, as has been pointed out in earlier messages in this thread, even if there is no centripetal force, sustaining a rotating system, angular momentum can still be defined.

    I have attached an image, (I don't know how the approval process works here.) The image comes from the http://www.cleonis.nl/physics/phys256/angular_momentum.php" [Broken]

    You can see that a particle in uniform motion will sweep out equal areas in equal intervals of time, similar to Kepler's area law. Hence in the case of an object in uniform motion you can take any point in space that is in uniform motion and define the angular momentum of that object with respect to the chosen "pivot" point.

    If you have two objects in uniform motion, you can define the sum of their angular momentums with respect to their common center of mass. If the two object sudden start exerting a force upon each other the angular momentum will be conserved.

    On my website there is also a http://www.cleonis.nl/physics/ejs/angular_acceleration_simulation.php" [Broken]. Interestingly, conservation of angular momentum can be regarded as a logical consequence of the work/energy theorem.


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    Last edited by a moderator: May 4, 2017
  10. Oct 23, 2009 #9


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    Dearly Missed

    Note that what you've got there is the equation of a straight line, in POLAR coordinates.
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