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Rotational Kinematics

  1. Apr 9, 2016 #1
    [Note: Post moved to homework forum by mentor]

    So I stumbled upon a reviewer for my physics exam tomorrow and I was wondering how the equation was formulated. Your help is very much appreciated :) ! Normally I would consult my professor for this but it's Sunday in my country today so I can't.

    Question:
    A uniform stick with length 3.00 [m] and mass
    5.00 [kg] is moving and rotating about its center of mass (CM) as
    shown in the figure. If the stick and point O both lie in the same
    xy-plane, what is the total angular momentum of the stick at point
    O at the instant shown?

    answer:
    L = (r )(p) +1/12mL^2(w) = -76.7kmm^2/s positive k hat

    Picture: http://postimg.org/image/u5eeo77el/96cf1d1b/
    p6.png
     
    Last edited by a moderator: Apr 9, 2016
  2. jcsd
  3. Apr 9, 2016 #2
    The first term ## r \times p## comes from the definition of angular momentum of any particle (or center of mass), while the second term comes from the fact that this object has a finite extent (i.e. moment of inertia). However, technically, the formula is incorrect since the first term is a vector while the second term seems to be a scalar (this doesn't disrupt the answer since you're given the direction of the rotation, just note that the formula isn't technically correct)
     
  4. Apr 9, 2016 #3

    haruspex

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    ω should be a vector. Are you saying it seems to be a scalar because it is not in bold? The value is shown in the diagram as a vector (##\hat k##).
     
  5. Apr 10, 2016 #4
    how did it yield a negative answer? What I got was letter D. my solution is: (5kg)(7.50m/s)(4m)sin(35) + 1/12(5kg)(3m)^2(2.50 rad) = 95.411
    Thanks for the reply, I really need to learn this before tomorrow. Cheers :)
     
  6. Apr 10, 2016 #5
    I got it already *ZOINKS* I just needed to use the right hand rule
     
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