# Parallel Axis Theorem derived from the Displaced Axes Theorem.

1. Dec 24, 2013

### binbagsss

I am trying to attain the parallel axis theorem from the displaced axes therom.

I have the displaced axes thorem stated in this form:

$\hat{I}$=$\hat{I}$com+M$\hat{A}$

-Where Rc is the position of the centre of mass position-
-Where $\hat{}$is the inertia tensor of a rigid body wrt to rotations about the origin, $\hat{}$com with respect to rotations about its COM.

-$\hat{}$can be represented as a matrix, the elements of which are determined by the elements of the COM position vector:

Aαβ=Rc^2δαβ - RcαRcβ

And the parallel axes theorem stated as :

I=Icom+Md^2 **

- Where d is the distance of parallel axis from the axis of roation passing through the COM.

I think I am having a hardtime linking these two statements due to my lack of understanding of index notation.

-First of all, $\hat{I}$ is a 3x3 matrix right? So $\hat{A}$ must be a 3x3 matrix?
-I am told α=1,2,3 = x,y,z. I assume then that β=1,2,3=x’,y’z’?; as we are linking different coordinate systems as we are liking inertia tensors representing axes of rotation passing through different origins.
-From the first term of Aαβ, I conclude that α=β or this term would be equal to zero. Looking at **, there is clearly no negative , so this must be the case.
-Most importantly, when we deduce this corollary – parallel axis theorem – from the displaced axes theorem, have we gone from $\hat{A}$which is 3x3, to a 1x1 matrix? So in this case, I am looking to attain A – in matrix form – as a 1x1 matrix.
-Trying to attain this expression –looking at what I need to get to – it appears that RcαRcβ =0. I’m not sure why this would be – probably again due to my lack of understanding – would it be something along the lines of generally looking at 6 axes in total in the displaced axes theorem – 3 in each coordinate system – but only 2 axes in the parallel axes theorem – 1 in each coordinate system?

Many thanks to anyone who can shed some light on this, really appreciate it !!

Last edited: Dec 24, 2013
2. Dec 25, 2013

### vanhees71

I can't read your math very well. Please use LaTeX. The derivation of the axis theorem (I suppose that's what's known as Steiner's theorem in the German literature) is straight forward, using the definition of the inertia tensor. Take $\vec{r}^{(k)}$ to be the position vectors wrt. to an arbitrary point and fixed in the body. Then the tensor of inertia is
$$\Theta_{ij}=\sum_{k} m_k \left (\vec{r}^{(k)} \cdot \vec{r}^{(k)} \delta_{ij}-r_i^{(k)} r_{j}^{(k)} \right ).$$
Then you define $\vec{x}^{(k)}$ as the position vectors wrt. to the center-mass frame and $\vec{R}$ the vector pointing from the center of mass to the original point in the body. Then you have
$$\vec{r}^{(k)}=\vec{x}^{(k)}-\vec{R}.$$
Plug this into the formula and use the CM property of $\vec{R}$ to get the axis theorem for the tensor.

For the moment of inertia for the body rotating around some fixed axis through the original point, you just have to multiply the inertial tensor from the left and from the right with the unit vector pointing in the direction of this fixed axis.

3. Jun 27, 2015

### fabatka

Hi there!
I have a problem properly understanding my textbook's derivation of the parallel axis theorem, and I've searced the forums for similar proofs. The proof in question starts out the same way as your post, vanhees71, so if you could enlighten me, I'd be really grateful.
So by writing in the $\vec r = \vec x - \vec R$ into the expression in parentheses of the moment of inertia (for simplicity, without the (k) index for the different points of mass):

$\delta _{ij} (\vec r \cdot \vec r) - r_i r_j \ = \delta _{ij} [(\vec x)^2 + (\vec R)^2 - 2(\vec x \cdot \vec R)] - x_i x_j - R_i R_j + x_i R_j + x_j R_i$

From this expression, the parts that contribute to the moment of inertia wrt the center of mass can be extracted, so without those what remains is:

$\delta _{ij} [ (\vec R)^2 - 2(\vec x \cdot \vec R)] - R_i R_j + x_i R_j + x_j R_i$

And according to Steiner's theorem the only parts that remain from this is

$\delta _{ij} (\vec R)^2 - R_i R_j$,

plus the parts that contribute to the moment of inertia wrt the center of mass.
So the following equation must be true:

$\delta _{ij} [- 2(\vec x \cdot \vec R)] + x_i R_j + x_j R_i = 0$,

but I can not see why.
The first part is a diagonal matrix, and the second and third parts are transposed matrices of each other, so the $x_i R_j$ matrix must be skew-symmetric, am I right? But why would it be skew-symmetric?