Tensor of inertia

1. Jan 23, 2010

springo

1. The problem statement, all variables and given/known data
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 [Broken]

2. Relevant equations
Moment of inertia for the whole disk (with mass 2M)
$$\begin{bmatrix} \frac{MR^{2}}{2} & 0 & 0 \\ 0 & \frac{MR^{2}}{2} & 0 \\ 0 & 0 & MR^{2} \end{bmatrix}$$

3. The attempt at a solution
For the tensor:
$$\begin{bmatrix} \frac{MR^{2}}{2} & 0 & 0 \\ 0 & \frac{MR^{2}}{4} & 0 \\ 0 & 0 & \frac{MR^{2}}{2} \end{bmatrix}$$
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.

Last edited by a moderator: May 4, 2017
2. Jan 23, 2010

jambaugh

See:
http://en.wikipedia.org/wiki/Parallel_axis_theorem" [Broken]

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:
$$\mathbf{I}'=\mathbf{I} + M(|a|^2\mathbf{1} -\mathbf{a}\mathbf{a}^\top)$$

[edited above, forgot the mass!]

Last edited by a moderator: May 4, 2017
3. Jan 23, 2010

springo

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

$$\begin{bmatrix} \frac{MR^{2}}{2} & 0 & 0 \\ 0 & \frac{MR^{2}}{4} & 0 \\ 0 & 0 & \frac{MR^{2}}{2} \end{bmatrix} +\begin{bmatrix} 0 & 0 & 0 \\ 0 & MR^{2} & 0 \\ 0 & 0 & MR^{2} \end{bmatrix} =\begin{bmatrix} \frac{MR^{2}}{2} & 0 & 0 \\ 0 & \frac{5MR^{2}}{4} & 0 \\ 0 & 0 & \frac{3MR^{2}}{2} \end{bmatrix}$$

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.

Last edited: Jan 23, 2010