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Is angular momentum conserved if you move off at a Tangent ?

  1. Apr 27, 2013 #1
    Lets imagine a binary system of two astronauts in space connected to one another via light rope.

    The rope is taut and they're spinning round and round with their axis of rotation being the the axis perpendicular to the their centre of mass.

    Now, my question is this. Lets say they each let go of the rope; they move off at tangents.

    Is angular momentum conserved ? And what is their subsequent motion ?

    My hypothesis : They will move off at constant velocity at a tangent to their circular motion. I believe that because they now under the effects of no external force by the rotational analogue of newton's second law (dL/dt = T) their angular momentum will be of equal magnitude but opposite sign.

    Does this make sense ?

    EDIT : If you can recommend some good reference websites on this topic that would be nice too. Thanks !
     
  2. jcsd
  3. Apr 27, 2013 #2

    Nugatory

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    yes.
    You're right about them moving off at a tangent, wrong about the angular momenta. It will be equal magnitude and same sign, so that angular momentum is conserved. You calculate the angular momentum of point masess like our drifting astronauts as
    L=r x mv
    where the 'x' is the vector cross-product operation and r is the position vector from the origin. Try it on your astronauts and you'll see that it doesn't change when they let go of the rope; as they drift apart the angle between v and r changes in a way that is exactly canceled by the change in the magnitude of r.

    (The moment of inertia of a non-point object is calculated by integrating the formula for point masses across the object. Any first-year mechanics text will be a good reference).
     
  4. Apr 27, 2013 #3
    I comprehend everything in your answer except the bit in bold. Can you explain how they change ?

    I understand that lLl = lrllpl*sinθ.... Does this have anything to do with your explanation ?
     
  5. Apr 27, 2013 #4
    Look at [itex] rsin\theta [/itex]. It is the perpendicular distance from the origin to an astronaut and will not change even when r does.
     
  6. Apr 27, 2013 #5
    I assume you are referring to : lLl = lrllpl*sinθ

    Yes, what you say make sense. Hence, I agree that lLl is proportional to lpl in the scenario.

    However, wouldn't v have different signs for each astronaut since they are moving off in opposite directions ?
     
  7. Apr 27, 2013 #6
    Yes, they do have opposite signs but don't you have an absolute value in your equation? If you are worried about the value of L and not just its magnitude, then you could use the right hand rule for both astronauts to see that the angular momentum vector is in the same direction for both.
     
  8. Apr 27, 2013 #7
    Yes, I know that the absolute value of angular momentum would be equal for each astronaut, but my initial argument was that L will be equal in each astronaut's case but have opposite signs.

    The reason for the opposite signs (in my mind) is the opposite signs for velocity (moving in opoosite directions).

    However, Nugatory says the magnitude will be same but they will have the SAME signs.

    This I do not understand.

    EDIT : I'm talking about the vector quantities of L.
     
  9. Apr 27, 2013 #8
    Do you know what a cross product is? If you construct the angular momentum vector for each astronaut, you will see that the angular momentum for both astronauts is in the same direction and same sign (before and after) by using the right hand rule.

    Edit: You can also think of it this way. Even though the momentum vectors are in opposite directions, so are the position vectors, so the cross product is the same in both cases.
     
    Last edited: Apr 27, 2013
  10. Apr 27, 2013 #9

    Nugatory

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    That's the rule for the magnitude of the angular momentum vector. The direction of the angular momentum vector is perpendicular to the plane in which r and p lie, and points either up or down according to the right-hand rule. In your example, they'll both point in the same direction so will add up to the original angular momentum instead of cancelling to zero.

    The components of the cross-product L of two vectors A and B are given by
    [tex]L_i = \sum\sum\epsilon_{ijk}A_{j}B_{k}[/tex]
    where the summations are across j and k, and the epsilon symbol is defined here (no need to read past the first few lines). This turns out to be equivalent to the rule you gave above for calculating the magnitude of the vector plus using the right-hand rule to choose its direction.
     
    Last edited: Apr 27, 2013
  11. Apr 27, 2013 #10

    rcgldr

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    Note that the total angular momentum will also include the rate of rotation for each astronaut. This portion of the total angular momentum is also conserved.
     
  12. Apr 27, 2013 #11

    WannabeNewton

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    elemis maybe it would benefit to do this in terms of vectors and see that the result does indeed hold. Fix the origin to the center of the circular binary orbit of radius ##R## and assume that the rotation speed is a constant ##v = \omega R##. The position vectors of the two point masses at a given instant of time will be ##r_{1} = R\hat{r}, r_{2} = -R\hat{r}## and their tangential velocities will be ##\mathbf{v}_{1} = v\hat\theta, \mathbf{v}_{2} = -v\hat{\theta}## hence ##L_{1} = mr\times \mathbf{v} = mRv\hat{r}\times \hat{\theta} = mRv\hat{z} = L_{2}## so ##L = 2mRv\hat{z}##.

    Now let's say they both let go of the ropes. They each go off on a tangent with velocity ##v##. At this point it would be more convenient to use cartesian coordinates so let's do that and let's also orient our coordinate system (with the same fixed origin from above) so that the x-axis is aligned with the direction they go off in. Draw the position vector from the origin to any one of the point masses and you'll see that ##r_{1} = vt\hat{x} - R\hat{y}, r_{2} = -vt\hat{x} + R\hat{y}## and ##\mathbf{v}_{1} = v\hat{x}, \mathbf{v}_{2} = -v\hat{x}## thus ##L_{1} = m(vt\hat{x} - R\hat{y})\times (v\hat{x}) = mRv\hat{z}, L_{2} = m(-vt\hat{x} + R\hat{y})\times (-v\hat{x}) = mRv\hat{z}## so again we see that ##L = 2mRv\hat{z}##.
     
  13. Apr 27, 2013 #12

    WannabeNewton

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    This is a very good point but I think the OP was thinking of a situation where the astronauts could be treated as point masses, in which case they wouldn't have spin angular momentum. But yeah the distinction wouldn't really matter since, as you note, it would be conserved anyways.
     
  14. Apr 27, 2013 #13

    BobG

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    Which reference frame?

    If you're using a non-inertial rotating reference frame, then both astronauts are moving in a positive direction. If you're using an inertial non-rotating frame, then the signs of the components of your radius will also be changing. Using an easy example, if the first astronaut moved positive y when the radius lie on the positive x axis, and the second astronaut moved negative y when the radius lie on the negative x axis, then the sign of their angular momentum hasn't changed. It lies on the positive z axis for both.

    Likewise, linear momentum is also conserved, as you started with none and ended with the astronauts' linear momentum cancelling out.

    All of that makes sense mathematically, but, obviously, once the astronauts are moving off on their own path, it makes more sense to refer to their linear momentum and the angular momentum of the rope (plus whatever it was connected to). The angular momentum of the rope has obviously decreased once the astronauts have departed.

    But total momentum was still conserved regardless of how you look at it or refer to it.
     
    Last edited: Apr 27, 2013
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