Angular momentum of a solid cylinder rotating around axis

[zeralda21]

Homework Statement

I'll provide picture for clearer understanding. The solid cylinder of mass $m$ and radius $r$ revolves about its geometric axis at an angular rate $p$ rad/s. Simultaneously, the bracket and shaft revolve about x-axis at the rate $\omega$ rad/s. Determine the angular momentum about O.

The Attempt at a Solution

Due to symmetry it follows that $I_{xy}=I_{xz}=I_{yz}=0$ so the angular momentum reduces to $\boldsymbol{L_{O}}=I_{xx}\omega \boldsymbol{\hat{i}}+I_{yy}p\boldsymbol{\hat{j}}$.

Further thoughts: If the $mh^2$ is removed, does not that calculate the rotation of the cylinder IF the geometrical axis coincide with the $y$-axis ??

I have the calculated $I_{xx}$ correctly but $I_{yy}$ is incorrect.

$I_{yy}=\overline{I}_{yy}+md^2=\frac{1}{2}mr^2+mh^2$ but there shound NOT be an $mh^2$ ...and I want to understand why

 

[BiGyElLoWhAt]

say what? why shouldn't there be? It's a parallel axis shift What I see is 1/2 mb^2 + mh^2 for your moment in the y plane

 

[zeralda21]

Yes, exactly. It could be wrong in the key, because I got $I_{xx}$ correct. I'll double check.

 

[haruspex]

I don't see why there would be an mh²p term. The rotation p is not about an axis through O, so the parallel axis theorem does not apply.

 

[anchonee]

Why is I_yz = 0?

When looking at the YZ plane the geometry is not symmetrical about either axis?

 

[haruspex]

Why is I_yz = 0?

When looking at the YZ plane the geometry is not symmetrical about either axis?

It's hard to tell because the exact expression for the $\omega$ term was never posted, but I suspect the OP is a bit careless about whether $I_{xx}$ refers to moment about the mass centre or moment about O.