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Angular momentum and impulse

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  1. Aug 21, 2015 #1
    True or false.jpg
    Hopefully the image is self-explanatory, if not:

    A cylinder is rotating around its central axis with angular momentum L1; an angular impulse, ΔL is then added to the cylinder, perpendicularly with respect to L1.

    The hypothetical result is: the cylinder has one angular mometum at the end, L-result. L-result is equivalent in magnitude to L1, and its direction is shifted counter-clockwise by a certain ammount so that a line parallel to L1 draw from the tip of L-result will meet the tip of ΔL, mathemetically, the angle has increased in the counter-clockwise direction by (90-arccos(ΔL / L1)) degrees. (<-- that stuff is got from basic trignometry)

    Is this true? If it is false, what is the right way to do it?

    Additional question:
    will the cylinder "follow" the angular momentum arrow and also point in the new direction, or will it remain in place and then rotate around the new arrow in a weird fashion?
     
  2. jcsd
  3. Aug 22, 2015 #2

    mfb

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    The vectors add, so the resulting angular momentum will have a larger magnitude.

    Neither. It will precess.
     
  4. Aug 22, 2015 #3
    I heard from various sources that if the two angular momentum vectors are perpendicular to one another, they will only change the direction of the final angular mometum. I understand, albeit to a shallow extent, precession, and I think precession only occurs when a torque is continuously exerted on a wheel. Also, it would be the most helpful of all if you can give the final angular momentum expressed in terms of L1 and ΔL.
     
  5. Aug 22, 2015 #4

    A.T.

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  6. Aug 23, 2015 #5

    mfb

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    As vector: L = L1 + ΔL
    This is conservation of angular momentum.
    As magnitude: ##|L|=\sqrt{L1^2 + ΔL^2}## - using the right angle between the two components.

    See the section "Torque-free" in the wikipedia article.
     
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