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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 ?

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2 Answers 2

up vote 3 down vote accepted

That's correct; there's no change to the angular momentum here with respect to the center of mass. Remember that $L = r \times p$. If $p$ is constant (as it is once they both let go of the rope), then you can see that $r_\perp$ is constant also, and $r_\parallel \times p$ is guaranteed to be zero.

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and w.r.t. any other frame on the rope , angular momentum is conserved , however , it'll be relative for those points. –  nonagon Apr 27 '13 at 18:07

I see you have ticked Muphrid's answer but I still think a less abstract reply could be useful.
1. Yes, angular momentum is conserved. 2. Yes, They will move off at constant velocity at a tangent to their circular motion. 3. Yes, their angular momentum will be of equal magnitude. 4. "but opposite sign." - No - they are travelling in opposite directions, but also on opposite sides of their centre of mass. For example, they both have clockwise motions. Therefore their angular momenta have the same sign - and add to the same as its original value - no matter how far away they get.

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