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Conservation of Angular Momentum of a stick

  1. Feb 21, 2013 #1
    Here's the scenario:

    A uniform stick of length L and mass M lies on a horizontal surface. A point particles of mass m approaches the stick with velocity v on a straight line perpendicular to the stick that intersects with the stick at one end. After the collision, the particle is at rest.

    My question is, the angular momentum should be conserved in this system right? After the collision, the stick will gain angular momentum as it will be rotating (as well as translating) after the collision. But initially, the particle could either have or do not have angular momentum depending on whether the particle is rotating. Assuming that the particle does not rotate, then the particle should not have any angular momentum. Then in this case how is the angular momentum conserved? The initial angular momentum would be zero initially and non-zero after the collision right?

    "The law of conservation of angular momentum states that when no external torque acts on an object or a closed system of objects." In this case the torque is an internal torque in the particle-stick system right?

    Please help thanks.
     
  2. jcsd
  3. Feb 21, 2013 #2

    cepheid

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    A particle can't "rotate" because it is considered to have all of its mass located at a single point in space. That's what physicists mean by "particle": a point mass. It is not an extended body.

    However, the particle can have angular momentum around a point. Angular momentum is not associated only with the rotation (spin) of extended bodies. It is also associated with the motion of a point particle relative to some reference point. For example, for a system consisting of a planet orbiting a star, there is some angular momentum associated with the orbit of that planet around the star (not just the spin of the planet around its axis). In other words, even if the planet weren't spinning, just orbiting in a circle, the system would still have angular momentum.

    For a point particle moving with linear momentum p, the angular momentum L of that particle around some reference point is given by the equation$$\mathbf{L} = \mathbf{r}\times\mathbf{p}$$where r is a position vector going from the reference point to the particle.

    So, even if the particle isn't spinning (which it can't), there is still non-zero angular momentum initially.
     
    Last edited: Feb 21, 2013
  4. Feb 21, 2013 #3
    From which point you are calculating the angular momentum, and as well stating its conservation?
     
  5. Feb 21, 2013 #4
    Ok thanks!

    As the particle moves, the angular momentum of the particle will vary as it moves right? So in this problem we calculate the angular momentum of the stick from the pivot of the stick which is at the end of the stick opposite to where the particle collides with it. So I should also use that pivot as a reference point to calculate the angular momentum of the particle right.

    So the initial momentum before collision should be:
    L=r x p = L x mv = Lmv.

    Correct?
     
  6. Feb 21, 2013 #5
    Hi, I think I understood the problem now thanks to cepheid. I was assuming that the angular momentum of the particle is zero at all point without using any reference point.
     
  7. Feb 21, 2013 #6

    cepheid

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    In general, yes, but you can just consider angular momentum at the instant just before the collision, and then again at the instant just after it.

    It might be more convenient to use the centre of mass of the stick as your reference point, since, after the collision, all of the angular momentum of the system will be due to the rotation of the stick around its centre of mass, so it makes sense to consider conservation of angular momentum around that point.

    I think you're using the letter 'L' for two different things here: the angular momentum, and the length of the stick. So that's confusing. Maybe use a different letter for the length of the stick, like ##\ell##. But basically this is correct, except there will be a slight change in the magnitude of r because of my comment above about using the centre of mass as the reference point. So the distance won't be ##\ell##, but it will be related to ##\ell##.
     
  8. Feb 21, 2013 #7
    Oh right, I was thinking the pivot is at the end of the stick. Ok you're right, the pivot should be at the center of mass.


    Good catch, I didn't notice I was using 'L' for two different things. Thanks for pointing that out.

    Thanks again!
     
  9. Feb 21, 2013 #8
    If you consider the end of the stick where the collision occurs, in this point the resultant torque is 0. This is because all forces pass trough this point. So there is conservation of angular momentum in this point. The initial velocity of the ball also passes trough this point, so the angular momentum is zero. After the collision the stick gets a both translational and rotational motion. The velocity of the translational motion can be calculated by the conservation of linear momentum. Then we decompose the two motions and we sum vectorially the angular momentum of each one. Note the translational angular momentum will be negative (if we consider the positive axis going out the page). The rotational angular momentum will be positive. If we sum both and equal 0, we find the angular velocity. If I didn't missed anything w=mv/6ML, where w=angular velocity, v=initial velocity of the ball, m = mass of the ball, M = mass of the stick

    Try to calculate by yourself and see if you get the same answer
     
  10. Mar 14, 2013 #9
    What about rotational angular momentum, e.g a disc that is rotating. Does the angular momentum depends on where my point of reference is?
     
  11. Mar 15, 2013 #10

    cepheid

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    Yes. The fact is: L = Iω for a rigid body spinning around an axis, where I is the moment of inertia of the body around that axis. I think that if you start with the definition of angular momentum I gave before, you can show that you get the same answer independently of what reference point you choose.
     
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