Mathematics of Physics and Engineering

70 Systems of Point Masses

where TJE) = rj x Ff] is the external torque of FJ£) and T{ 0 E) is the
total torque on the system S by the external forces.

Next, we look at the ANGULAR MOMENTUM OF A SYSTEM RELATIVE
TO THE CENTER OP MASS. Again, let O be an arbitrary frame of reference,
let r j (t) be the position of mass rrij, 1 < j < n, and let TCM (*) De the
position of the center of mass. Define by Xj the position of rrij relative to
the center of mass:

tj = ^ - rCM. (2.2.10)

Then r-j = feu + ij and, according to (2.2.7),

n n

3 = 1 3 = 1

n n n

= ^2 rrij rCM x rcM + ^ rrij Xj x rCM + ^ rrij TCM X ij (2.2.11)

3 = 1 3 = 1 3 = 1

n

+ ^mjrj xij.

3 = 1

EXERCISE 2.2.3.c (a) Verify that

n

Y^mjXj = 0, (2.2.12)

j=i

where Xj is the position vector of the point mass rrij relative to the center

n n n n

of mass. Hint: ]>2 rrij tj = ^2 rrijrj — ^2 rrij rcM = ^2 mi ri ~ Mr CM — 0.

3 = 1 3=1 3 = 1 3=1

(b) Use (2.2.11) and (2.2.12) to conclude that

n

L 0 = M rCM x rc'M + ^ rrij Xj x ij. (2.2.13)

3=1

If we select the origin O of the frame at the center of mass of the sys-

tem, then TQM = 0, and we get the expression for the angular momentum

around the center of mass:

n

LCM = J^2mjXj xij, (2.2.14)

3 = 1