Displaced Axes Theorem- quick question.

[binbagsss]

I am using the displaced axis theorem:

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

where $\hat{A}$can be represented as a matrix, the elements of which are determined by:

$A$$_{\alpha\beta}$=|Rc$^{2}$|δ$_{\alpha\beta}$ - Rc$_{\alpha}$Rc$_{\beta}$


I know that it is derived from substituting in rk=rk'+Rc into the definition of the moment of inertia tensor, where rk is the position vector of the of the kth particle of a rigid body from the point at which we are calculating I, Rc is the position vector of the centre of mass of the rigid body.

My question is, that if Rc only has a , x component , say, s.t Rc$_{y}$ and Rc$_{z}$=0, then , looking at A, I attain 0 for all my matrix elements.

Picturing this, surely this can not be correct? - That $\hat{I}$=$\hat{I}$com when computing the moment of inertia about an axis displaced solely in x,y,z from an axis located at the bodies com.

Quick question on a similar note, looking at the expression of A, I am struggling to see how the last term yields a scalar, like the |Rc$^{2}|$δ$_{\alpha\beta}$. I know that $\alpha$ and $\beta$ =x,y,z. But aren't Rc,$\alpha$ and Rc,$\beta$, rank one tensors, so that they are vectors.

Many thanks to anyone who can shed some light on this, greatly appreciated !

 

[binbagsss]

anyone?

 

[BruceW]

if Rc only has a , x component , say, s.t Rc$_{y}$ and Rc$_{z}$=0, then , looking at A, I attain 0 for all my matrix elements.

that's not true. not all the elements will be zero. it is true that A$_{xx}$ is zero, and any non-diagonal elements are zero.

 

[binbagsss]

Ohhhh ! I think I see, so considering Ayy and Azz, Ayy=Azz=|Rc^2| ?

 

[BruceW]

yep, that's it!

 

[binbagsss]

thanks a lot.

 

[BruceW]

ah, no problem! it's interesting stuff.