Mathematics of Physics and Engineering

74 Systems of Point Masses

system. Define the corresponding rotation vector u> according to (2.1.39)
on page 63. Let us apply equation (2.1.46), page 66, to the motions of the
point mass rrij relative to frames O and 0. Denote by rj the position
vector of rrij in the frame O (draw a picture!) Then Xj = rj — TCM is
the position vector of rrij in OCM • Because of the rigidity condition, the
position vector rij of rrij in the frame Oi does not change in time, and so
f\j = 0 and rij = 0. By (2.1.42) on page 64, rj = rcM + w x Xj, and
then, by (2.1.46), 'fj = TTCM + u x (u> XXJ) + U; x Xj. Since the frame OCM
is not rotating relative to O, we have x = f j — rcM, and

Xj=uxxj, (2.2.22)

Xj = u> x (u> x Xj) + u> x Xj. (2.2.23)

Next, we use identity (1.2.27) on page 22 for the cross product:

w x (u> x Xj) = (w • Xj) OJ - (a; • u>) Xj = (u • Xj) LJ - J^1 Xj, (2.2.24)

where u) = ||u>||. To compute the rate of change of the angular momentum,

we will need the cross product Xj x Xj. Applying (1.2.27) one more time,

Xj X(UXXJ) = \\xj\\^2 u-{xj-uj)xj. (2.2.25)

Putting everything together, we get

Xj x ij = (u • Xj) XjXu> + \\XJf cj - (XJ • u) Xj. (2.2.26)

After summation over all j, the right-hand side of the last equality does
not look very promising, and to proceed we need some new ideas. Let
us look at (2.2.26) in the most simple yet non-trivial situation, when the
rotation axis is fixed in space, and all the point masses rrij are in the plane
perpendicular to that axis. Then w • tj = 0 for all j, and <ii = Co Q, where
Q is the unit vector in the direction of u>; note that since the rotation axis
is fixed, the vectors w and w are parallel. Accordingly, equality (2.2.24)
becomes U}X(LJ xxj) = —co^2 Xj, and since d(toCj)/dt = u u), equality (2.2.25)
becomes Xj X(UJXXJ) = ||tj ||^2 wQ. Summing over all j in (2.2.26) and taking
into account these simplifications, we find from (2.2.16) on page 71

U1U),