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What is the direction of the red ball's angular momentum

  1. Apr 12, 2015 #1
    1. The problem statement, all variables and given/known data

    What is the direction of the red ball's angular momentum about the point A?
    (see figure)

    Same as the momentum.
    Out of the page.
    Zero magnitude.
    Opposite to the momentum.
    Same as rA.
    Into the page.


    2. Relevant equations
    Angular Momentum


    3. The attempt at a solution

    I believe the answer is same as the momentum because the angular momentum is 'the moment of momentum'
     

    Attached Files:

  2. jcsd
  3. Apr 12, 2015 #2

    rcgldr

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    Is the ball rolling up the plane or down the plane? If this is during an instant where the ball is stopped, then it has zero momentum.
     
  4. Apr 12, 2015 #3
    Unfortunately that info is not given, but if we assumed it was down wouldn't the answer just be same as momentum?
     
  5. Apr 12, 2015 #4

    mfb

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    How do you determine angular momentum of a point-mass?
     
  6. Apr 12, 2015 #5

    SammyS

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    The figure helps tremendously.

    screen-shot-2015-04-12-at-2-25-48-pm-png.81898.png
    It looks to me like the ball is at the position of the red dot and is moving horizontally.

    How do one calculate angular momentum ?
     
  7. Apr 12, 2015 #6

    L = r x p
    L = same as momentum then ?
     
    Last edited: Apr 12, 2015
  8. Apr 12, 2015 #7

    mfb

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    r x p and p are not the same, and do not point in the same direction.
    What do you know about the cross-product?
     
  9. Apr 12, 2015 #8
    Cross product is r⃗A × F⃗

    This helps you find torque
     
  10. Apr 12, 2015 #9

    mfb

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    And what is your result?
     
  11. Apr 12, 2015 #10
    Opposite to the momentum?
     
  12. Apr 12, 2015 #11

    haruspex

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    If you have two vectors and take their cross product, ##\vec c = \vec a \times \vec b##, what is the angle between ##\vec a## and ##\vec c##? It is always the same answer.
     
  13. Apr 12, 2015 #12

    it is 180 degrees, I'm still a little lost of what the answer could be though..
     
  14. Apr 12, 2015 #13

    mfb

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    It is not 180 degrees. Check the definition and basic properties of the cross-product.
     
  15. Apr 12, 2015 #14

    SammyS

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    Do you know the Right Hand Rule for cross-product ?
     
  16. Apr 12, 2015 #15
    Vector C will be perpendicular to both A & B. The angle will be between 0 and 180 degrees


    • A x B = A B sin([PLAIN]https://www.physics.uoguelph.ca/tutorials/torque/theta.gif) [Broken]

    not familiar with right hand rule but I'm trying to learn it right now
     
    Last edited by a moderator: May 7, 2017
  17. Apr 12, 2015 #16

    mfb

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    Right.
    You can know the exact angle between a and c. This is not the angle between a and b!
     
  18. Apr 12, 2015 #17
    It would be going out of the page then!?!?!?

    because it will be pointing upwards
     
    Last edited: Apr 12, 2015
  19. Apr 12, 2015 #18

    mfb

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    Right.
     
  20. Apr 12, 2015 #19

    rcgldr

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    If the ball is rolling down hill (left to right), then the angular momentum vector points into the page (right hand rule).
     
  21. Apr 12, 2015 #20
    So who is right here? I only have one try
     
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