Tensor of Inertia for Half Disk: Calc Angular Momentum

[springo]

Homework Statement

Find the tensor of inertia for a half disk with mass M and then use that to get the angular momentum along the axis in the figure.
http://img138.imageshack.us/img138/5312/problemr.png

Homework Equations

Moment of inertia for the whole disk (with mass 2M)
\begin{bmatrix}<br /> \frac{MR^{2}}{2} &amp; 0 &amp; 0 \\ <br /> 0 &amp; \frac{MR^{2}}{2} &amp; 0 \\ <br /> 0 &amp; 0 &amp; MR^{2}<br /> \end{bmatrix}

The Attempt at a Solution

For the tensor:
\begin{bmatrix}<br /> \frac{MR^{2}}{2} &amp; 0 &amp; 0 \\ <br /> 0 &amp; \frac{MR^{2}}{4} &amp; 0 \\ <br /> 0 &amp; 0 &amp; \frac{MR^{2}}{2}<br /> \end{bmatrix}
And the moment of inertia... I think I should apply Steiner's theorem, but I'm not quite sure how to apply it on a tensor.

Thanks for your help.

 

[jambaugh]

See:
http://en.wikipedia.org/wiki/Parallel_axis_theorem"

You take the displacement vector \mathbf{a}

(written as a column vector) and form the matrix \mathbf{a}\mathbf{a}^\top

The new moment of inertia tensor is then:
\mathbf{I}&#039;=\mathbf{I} + M(|a|^2\mathbf{1} -\mathbf{a}\mathbf{a}^\top)

[edited above, forgot the mass!]

 

[springo]

Thanks.
OK, so assuming my "attempt at a solution" tensor was right, I get to the following tensor:

<br /> \begin{bmatrix}<br /> \frac{MR^{2}}{2} &amp; 0 &amp; 0 \\ <br /> 0 &amp; \frac{MR^{2}}{4} &amp; 0 \\ <br /> 0 &amp; 0 &amp; \frac{MR^{2}}{2}<br /> \end{bmatrix}<br /> +\begin{bmatrix}<br /> 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; MR^{2} &amp; 0 \\ <br /> 0 &amp; 0 &amp; MR^{2}<br /> \end{bmatrix}<br /> =\begin{bmatrix}<br /> \frac{MR^{2}}{2} &amp; 0 &amp; 0 \\ <br /> 0 &amp; \frac{5MR^{2}}{4} &amp; 0 \\ <br /> 0 &amp; 0 &amp; \frac{3MR^{2}}{2}<br /> \end{bmatrix}<br />

I think this can't be right because then when I try to find variation in the angular momentum (at constant rotation speed), I get 0.