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Finding the magnitude of an object's angular momentum about the origin.

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1.4 kg object at x = 2.00 m, y = 3.10 m moves at 4.9 m/s at an angle 45° north of east. Calculate the magnitude of the object's angular momentum about the origin.

    Answer is 5.3 kg m^2 /s

    2. Relevant equations
    These were the only three I could think of:
    L=Iω
    L=mvr
    ω=v/r




    3. The attempt at a solution

    First attempt:

    L=mvr

    L= (1.4)(4.9)sqrt(4+3.1^2)

    L=25.3077, not the right answer

    Second attempt:

    Diagram here.
    I tried to find the tangential velocity as shown:

    ∠2 = arctan (3.10/2.00) = 57.17°

    Vtan = 4.9cos(45) / cos(57.17) = 6.390914 m/s

    ω=v/r=6.390914/Sqrt(4+3.1^2) = 1.73234 rad/s

    Then I'm not really sure how to proceed after this because I'm not entirely sure what equation to use for I.

    Third attempt:

    Vtan=v ∴

    L=1.4(6.390914)(sqrt(4+3.1^2) = 8.94728 m/s, not the right answer

    Is my diagram wrong or am I missing something very important?
     
    Last edited: Nov 5, 2012
  2. jcsd
  3. Nov 5, 2012 #2

    ehild

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    Gold Member

    The magnitude of the angular momentum is L=mvr sin(theta), theta is the angle between the position vector and velocity.. Find theta. You almost have done.

    The angular momentum is the vector (cross-) product of the position vector with the linear momentum. [tex]\vec L = m\vec r \times\vec v [/tex].You have the position vector [itex]2\vec i +3.1 \vec j[/itex]. You can write up the velocity vector with its components, perform the product and determine the magnitude of the vector L



    ehild
     
  4. Nov 5, 2012 #3
    There was a response about finding the cross product of mr x v. I did this already, but I guess I made some small mistake the first time around. I have found the correct answer. Thank you whoever you are (I forgot your username, but I remember it was something like ehlib.
     
  5. Nov 5, 2012 #4

    ehild

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    It was me, but I changed my post after I saw your drawing. So all is good at last :smile:

    ehild
     
  6. Nov 5, 2012 #5
    Yes, I saw that earlier... I also tried doing that the first time around; however, I must have made a minor mistake somewhere because I found the correct answer using this method. Thank you very much! Here is my work for those looking for the solution:

    Position vector is:
    2.00i + 3.10 j

    Multiply that by m=1.4 kg, which gives:

    2.89i + 4.34 j

    Then you must find the x and y components of the velocity. Since it is at a 45-degree angle, both components are equal to each other:

    4.9(cos 45) = 3.4648 m/s

    Which gives us the velocity vector:

    3.4648i + 3.4648j

    You then find the cross product of mr x v which will give L = -5.335792.
    Since it's asking for magnitude, the answer is 5.335792.
     
  7. Nov 5, 2012 #6
    Thank you very much, sir. :)
     
  8. Nov 5, 2012 #7

    ehild

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    Gold Member

    You are welcome!
    You get the same result from L=mrvsin(theta)

    ehild
     

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    Last edited: Nov 6, 2012
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