Parallel Axis Theorem and interia tensors

1. Nov 30, 2008

pimpalicous

Is the parallel axis theorem always valid for inertia tensors? We have only seen examples with flat (2d) objects and was wondering if it would also be valid for 3d objects, like a h emisphere, for example. Thanks.

2. Nov 30, 2008

aerospaceut10

Yep, just use the distance projected from the axis that it's being measured around.

3. Nov 30, 2008

D H

Staff Emeritus
The general form for the parallel axis theorem is

$$\mathbf{J} = \mathbf{I} + m(r^2\mathbf{1} - \boldsymbol{r}\boldsymbol{r}^T)$$

where

$\mathbf{J}$ is the inertia tensor of some object about some point removed from the center of mass,
$\mathbf{I}$ is the inertia tensor of the object about its center of mass
$m$ is the object's mass
$\boldsymbol{r}$ is the displacement of the point in question from the center of mass, expressed as a column vector
$\mathbf{1}$ is the identity matrix.

Last edited: Dec 1, 2008
4. Dec 1, 2008

I think you meant $$\mathbf{J} = \mathbf{I} + m(r^2\mathbf{1} - \mathbf{r}\mathbf{r}^T)$$, because $$\mathbf{r}^T \mathbf{r} = r^2$$, which is not even the right type of tensor.

To show that this gives the correct moment of inertia about a certain axis, let e be a unit vector giving the direction of this axis. Then the moment of inertia about this axis is
$$\mathbf{e}^T\mathbf{Je} = \mathbf{e}^T \mathbf{Ie} + m(r^2 \mathbf{e}^T\mathbf{1e} - \mathbf{e}^T\mathbf{rr}^T\mathbf{e})$$
$$= \mathbf{e}^T \mathbf{Ie} + mr^2 - m(\mathbf{r} \cdot \mathbf{e})^2$$
$$= \mathbf{e}^T \mathbf{Ie} + mr_\perp^2$$, where $$r_\perp$$ is the perpendicular distance from the centre of mass to the axis.

Last edited: Dec 1, 2008
5. Dec 1, 2008

D H

Staff Emeritus

Oops. I corrected my post, thanks.

6. Dec 1, 2008

pimpalicous

so does that mean if we find the inertia tensor at the center of the base of a hemisphere we can use the parallel axis theorem to find the tensor for the center of mass

7. Dec 1, 2008

Yes, but in reverse. You know J, and you must find $$\mathbf{I} = \mathbf{J} - m(r^2 \mathbf{1} - \mathbf{rr}^T)$$.