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Angular Momentum Incorrect Graph?

  1. May 13, 2017 #1

    gv3

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    1. The problem statement, all variables and given/known data
    Determine the total magnitude of angular momentum Ho of the particle about point O. The velocity of the particle is 5.5 m/s.


    engineering 4.png


    2. Relevant equations
    Ho= r x mv

    3. The attempt at a solution
    The answer is 43.04. My question is, isnt the graph wrong? If you take the magnitude of the vector shown in the graph, it doesnt equal 5.5 it equals 6. Without the proper vector from the graph isn't the problem unsolvable? I only knew the answer because i was asked this question twice.
     
  2. jcsd
  3. May 14, 2017 #2

    gneill

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    Staff: Mentor

    The only vector I can see indicated in the figure is V, which is shown to be directed along the path AB. AB itself is not the velocity vector, it just indicates its direction.
     
  4. May 14, 2017 #3

    gv3

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    But isn't that vector 2i-4j-4k?
     
  5. May 14, 2017 #4

    gneill

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    Staff: Mentor

    A position vector from A to B might be that, and it would have the same direction as V. But it is not V.
     
  6. May 14, 2017 #5

    gv3

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    My mistake you are right. How would the velocity vector be found then? My professor in his lecture notes looks like he just took the direction of V shown.
     
  7. May 14, 2017 #6

    gneill

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    Staff: Mentor

    All you need to take from the vector AB is its direction. Are you familiar with unit vectors? Can you form a unit vector that has the same direction as AB?
     
  8. May 14, 2017 #7

    gv3

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    yeah it would just be the position vector divided by the magnitude.
     
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