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Angular Momentum of a body after it loses mass

  1. Mar 13, 2012 #1
    Hey guys.

    If a body is spinning around the z - axis and an element of that body detaches on its own and flies off tangentially, does the rest of the remaining rotating body experience a change in angular velocity???

    My thought is that the angular velocity of the remaining rotating body remains constant (same as body+elements' combined angular velocity at the start), just like velocity of a man holding a bowling ball on a moving skateboard remains constant before and after he drops it because both bodies keep the same momentum before and after the release, so its all good.

    Now for the rotating issue.......

    The body (without the element) has a certain angular momentum as it spins. The element itself has a moment of inertia mR^2 and also has its own angular momentum, which is exactly mvR or in terms of w, mwR^2. Now, after the element flies off on a tangent, it has the same angular momentum with respect to the axis of rotation as when it was attached, no matter how far it flies right, which is ---- m(r x v), which always works out to be of magnitude mvR, which is the same as before???

    So does this mean that the body (without the element) would just keep rotating at the same angular velocity that the entire body + element had at the start of the problem because the angular momentum of the element never changes, so the same would need to be true for the remaining body as well.


    Please only send me a reply to this question if your willing to give me somewhat of a detailed response, because I'm really struggling with this idea...
     
  2. jcsd
  3. Mar 13, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hey there SmoothCat! Welcome to PF! :smile:
    If it flies off at the same speed as its tangential speed just before … so its angular momentum (about the centre) stays the same … then no, there's no change in angular velocity of the rest.

    That's because if the angular momentum of the detached part stays the same, then so must the angular momentum of the rest. :smile:

    (I'm sorry, but any more detailed response would be superfluous! :wink:)
     
  4. Mar 13, 2012 #3
    Thanks man, that answer was plenty. I just wanted to get more than a yes or no answer type thing. But, I'm glad that the answer matches what I was thinking and things didn't get messy!!!
     
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