Mathematics of Physics and Engineering

Rigid Systems of Points 75

and it is therefore natural to introduce the quantity ICM = 2 mjll*j||^2!

which is called the moment of inertia of S around the line that passes
through the center of mass and is parallel to Q. Then

dLcM T. ~ ,„ „ „_*

dt = ICMUU. (2.2.27)

Our main goal is to extend equality (2.2.27) to a more general situation;

the moment of inertia thus becomes the main object to investigate. To

carry out this investigation, we backtrack a bit and look closely at the

angular momentum of the rigid system around the center of mass LCM =

Z3"=i mjVj x ij. By (2.2.22), we have Xj x ij = Xj x (w x Xj). Similar to

(2.2.25) we find Xj x (w x Xj) = ||tj||^2 w — (XJ • U>)XJ and

VCM = \ J2m^x^^2 u -J2mi(xi ' w)rJ- (2.2.28)

^' = 1 / 3 = 1

As written, equality (2.2.28) does not depend on the basis in the frame

OCM- To calculate LCM, we now choose a cartesian coordinate system

(i, j, k) in the frame OCM- Let Xj(t) = Xj(t)i + Vj(t)j+ Zj(t)k and

w(t) = wx(t) i + ojy(t)j + LJz(t) k. From (2.2.28) above,

LCM = ( J2mi(x2i + $ + **) I w - I ^2mixi vi I w*

* Vj=1 I (2.2.29)

u=i / \j=i

rrijZjXj wz,

where we omitted the time dependence notation to simplify the formula.

Since LCM = LCMX * + LCMV 3 + LCMZ «•, to compute the ^-component

LCMX of LCM, we replace the vector Xj in (2.2.29) with Xj, and the vector

u>, with wx:

n n n

LCMX =^x^2 rrij [y] + zf) - uy ^ rrijXjyj - wz ^ rrijXjZj;

3 = 1 3 = 1 3 = 1

similar representations hold for LcMy and LCMZ-