Angular momentum of a rod rotating about an axis

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.

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)

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