Mathematics of Physics and Engineering

Non-Rigid Systems of Points 69

From (2.2.7) it follows that

AT n n

3 = 1 j=l 3 = 1

since rj x rj = 0. Now, we again assume that the frame O is inertial. Then
Fj = rrij Tj and dLo/dt = Y^j=i rj x Fj = !Cj=i Tj, where Tj = Tj X FJ
is the torque about O of the force Fj acting on rrij. We define the total
torque To = 2j=i Tj and conclude that

^ = r 0. ( 2. 2. 8 )

Equation (2.2.8) is an extension of (2.1.6), page 41, to finite systems of
point masses. If To = 0, then Lo is constant, that is, angular momentum
is conserved.
In general, unlike the equation for the linear momentum (2.2.5), the
torque To in (2.2.8) includes both the internal and external forces. If
the internal forces are central, then only external forces appear in (2.2.8).
Indeed, let us compute the total torque in the case of CENTRAL INTERNAL

FORCES. The internal force FJJ) acting on particle j is FJ7) = f^ F^. By
fc=i
k?j
Newton's Third Law, we have F$ = -F$. Then

, j- n n n n
-^ = J2rjX F> = £-, x Ff> + J> x X>g>
j=i j=i j'=i k=i
k?j
n n..
The terms in the product Yl rj x 2 -^k can ^e arranged as a sum of
J'=I fc=i

pairs rj x FJj/ + r^ x FJy for each (j, k) with j ^ /c. Also, TJ X FJfc' +

rk x FJL- = (r^ — rk) x Fjk- The vector rk — T"j is on the line joining rrij

and rrik- If the forces F-k are central, as in the cases of gravitational and

electrostatic forces, then the vector FL is parallel to the vector (rj —rk),

and (rj — rk) x Fk' = 0. In other words, the internal forces do not

contribute to the torque, and (2.2.8) becomes

d^ = ±rjxFf=±T^=T^, (2.2.9)

3 = 1 .7 = 1