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Rolling Hoop - Angular Momentum

  1. Apr 21, 2007 #1
    1. The problem statement, all variables and given/known data

    A hoop is launched so that it slides with backspin across the floor. Due to friction on the floor, the hoop eventually reverses direction. A little after that, it stops slipping and rolls back to the point where it was launched. If the initial speed of the hoop was v and its initial angular velocity was omega, what is its speed as it rolls back?

    2. Relevant equations

    L = Iw + mvr
    Iw = Iw for conservation

    3. The attempt at a solution

    L = Iw - mvr (I thought "-" because of friction). I need to find the total angular momentum of the hoop with respect to a stationary point on the floor, but I am not sure how to do this mathematically. The angular momentum with respect to the hoop rolling back has to equal the angular momentum with friction making it slip away. Is the point of slip/roll significant?
  2. jcsd
  3. Apr 21, 2007 #2


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    I don't follow your description. I would except the hoop to stop moving forward, then reverse direction because of the backspin. In that case, the hoop would lose no angular momentum until the forward motion stopped.
    So it would begin its return trip with the same angular velocity it started.

    Of course, I might be misreading the situation.
  4. Apr 21, 2007 #3
    What you say makes sense I think... I went out and found a hoop, and tried this on a flat road, and the hoop does not roll back with the same angular or translational motion... so I'm not sure
  5. Apr 21, 2007 #4


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    I think you could be right. If there's friction slowing the translational movement, it must be slowing the angular movement also.

    I tried to work it out but it's not specific enough, or I'm too tired.
  6. Apr 21, 2007 #5
    Well the problem is asking for a symbolic answer.. and that's all I have
  7. Apr 22, 2007 #6
    still need help if anyone is interested,

    also, thank you for replying Mentz
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