Mathematics of Physics and Engineering

(^78) Systems of Point Masses
EXERCISE 2.2.11. Consider four identical point masses m at the vertices
of a square with side a. Denote by O the center of the square. Let (i, j, k)
be a cartesian system, at O, with the vectors i and j along the diagonals
of the square (draw a picture!). Verify the following statements: (a) The
center of mass of the system is at O. (b) The vectors i, j, k define the
three principal axis of the system, (c) The diagonal elements of the matric
IQM in {%, j, k) are ma^2 , ma^2 , 2ma^2. (d) If the system (i, j, k) is rotated
by 7r/4 around k (does not matter clock- or counterclockwise), the result is
again a principal axis frame for the system, and the matrix IQM does not
change.
EXERCISE 2.2.12/^1 When are some of the diagonal entries of IQM equal to
zero? Hint: not very often.
We now summarize our excursion into linear algebra: for every rigid
system S of point masses, there exists a special frame, called the principal
axes frame, in which the matrix of inertia IQM of the system is diagonal
and does not depend on time. This special frame is centered at the center
of mass and is fixed (not rotating) relative to the system S.
Let us go back to the analysis of the motion of the system 5. With
CM = LCMXI + LCMxz +LCMxi, U = OJXZ +u>yj +UJZK ,
and after multiplying by U on the left and by Uj on the right, equation
(2.2.31) becomes
£CM = ^CM^I (2.2.33)
where CCM = U^CCMUJ is the column vector {LCMx, LCMx, LCMz)T
and ft = C/fJ[/J is the column-vector (w, u>, w)T.
By construction, the principal axes frame rotates relative to the frame
O, and the corresponding rotation vector is w. Denoting the time derivative
in the frame O by Do, and in the the principal axes frame, by £>, and using
the relation (2.1.43) on page 64, we find
D 0 LCM(t) = D. LCM{t) + u(t) x LCM(t). (2.2.34)
To proceed, let us assume that the underlying frame O is inertial. Then
relation (2.2.17) applies, and we find D 0 LcM(t) = TcM(t): the change of
LCM (t) in the inertial frame is equal to the torque of all forces about the
center of mass. In the principal axes frame, we have TcM{t) = TQMx{t) i +
TcMy{t) j +T(XMz(i) k. Also, since the basis vectors i, f, k* are fixed in