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Angular momentum of a rod rotating about an axis

  1. Apr 15, 2014 #1
    1. The problem statement, all variables and given/known data

    I'll provide a picture for clearer understanding. The problem is to calculate the angular momentum of the rod rotating about the z-axis. I have serious difficulties in deriving the inertia matrix, that's all I need help with.



    19Nguqj.jpg

    Progress

    Since the rod is rotating about the z-axis ##\Rightarrow \omega _{x}=\omega _{y}=0## the angular momentum simplifies to

    ## \boldsymbol{L}_{O}=-I_{xz}\omega _{z}\hat{\mathbf{i}}-I_{yz}\omega _{z}\boldsymbol{\hat{j}}+I_{zz}\omega_{z}\boldsymbol{\hat{k}} ##

    Split the rod in three parts (SEE FIGURE) and calculate for each body.

    I have correct terms for body A.

    For B:

    ##I_{xz}=\overline{I}_{xz}+md_{x}d_{z} = 0 +\rho b(0)(\frac{b}{2})=0##
    ##I_{yz}=\overline{I}_{yz}+md_{y}d_{z} = 0 +\rho b(b)(\frac{b}{2})=\frac{1}{2}\rho b^3##
    (Wrong)##I_{zz}=\overline{I}_{zz}+md^2 = 0 + (\rho b)(b^2+(\frac{b}{2})^2)=\frac{5}{4}\rho b^3## (d is distance from midpoint of B to origin O)

    For C:

    ##I_{xz}=\overline{I}_{xz}+md_{x}d_{z} = 0 +\rho b(\frac{b}{2})(b)=\frac{1}{2}\rho b^3##
    ##I_{yz}=\overline{I}_{yz}+md_{y}d_{z} = 0 +\rho b(b)(b)=\rho b^3##
    (Wrong)##I_{zz}=\overline{I}_{zz}+md^2 = \frac{1}{3}(\rho b)(b^2)+(\rho b)((2b^2)^2+(\frac{b}{2})^2)=\frac{55}{12}\rho b^3## (distance d is from midpoint of C to origin O)
     
  2. jcsd
  3. Apr 15, 2014 #2

    BiGyElLoWhAt

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    Curious, why is Ibar_zz = 0? are you treating the radius to be zero? if that's the case then your md^2 term should be rho(b)(a^2) as a is the distance from the z axis. If you aren't taking the segment to have 0 radius, then Ibar_zz should be non zero, and your term for md^2 should be rho(b)(a+radius_b)^2, not (b^2 + b/2 ^2)

    Follow that logic through with C. (how far is it from the z axis?)
    Also I believe you want 1/12 mr^2 for that rod, maybe not, I'm not up to par on moments of inertia, but I'm pretty sure a rod pivoted about one end is 1/12 mr^2.

    But using the parallel axis theorem is the right place to start.
     
  4. Apr 15, 2014 #3
    I think that since it is not mentioned about any radius we can safely assume it to be zero. What distance? The perpendicular distance? Or using Phytagorean ?

    I think this is slighly better but still wrong..

    For B:

    ##I_{zz}=\overline{I}_{zz}+md^2 = \frac{1}{3}(\rho b)(b^2)+(\rho b)(b^2+(\frac{b}{2})^2)=\frac{19}{12}\rho b^3 ##. I am not sure whether distance d is from the midpoint of B and or just the vertical distance because it is still wrong!

    For C:

    Correct it should be ##\frac{1}{12}(\rho b)(b^2)## since the distance of the mass particles to the axis varies. Still wrong..
     
  5. Apr 15, 2014 #4

    BiGyElLoWhAt

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    Oh wait, I didn't read the picture carefully enough. Replace all the a's with b's in the md^2 term
     
  6. Apr 15, 2014 #5

    BiGyElLoWhAt

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    Ok, so for A, it's cut and dry, [itex]\frac{1}{3} \rho b^3[/itex]because[itex] [(mb^2)=(\rho b)b^2][/itex]
    B, you have no moment before the axis shift, which leaves [itex]\rho b^3[/itex] after the shift
    C you have [itex]\frac{1}{3} mb^2 + md^2 = \frac{4}{3} \rho b^3[/itex]

    I don't know where you're getting this [itex]\frac{b}{2}[/itex] term from.

    Your total I about the z axis should be [itex]\frac{1}{3} \rho b^3 + \rho b^3 +\frac{4}{3} \rho b^3 = \frac{8}{3} \rho b^3[/itex]

    The only thing that's throwing me for a loop is the C chunk, it's not rotating like a bar with one end on the axis... do you see what I'm saying? So this solution may or may not be good...
     
  7. Apr 15, 2014 #6

    BiGyElLoWhAt

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