where r1, r2 are the distances of m1, m2 from their mass-centre G. Hence if the extraneous forces (e.g. gravity) have zero moment about G, ω will be constant. Again, the tension R of the string is given by
| R = m1ω2r1 = | m1m2 | ω2a, |
| m1 + m2 |
where a = r1 + r2. Also by (10) the internal kinetic energy is
| 1⁄2 | m1m2 | ω2a2 |
| m1 + m2 |
The increase of the kinetic energy of the system in any interval of time will of course be equal to the total work done by all the forces acting on the particles. In many questions relating to systems of discrete particles the internal force Rpq (which we will reckon positive when attractive) between any two particles mp, mq is a function only of the distance rpq between them. In this case the work done by the internal forces will be represented by
−Σ ∫ Rpg drpq,
when the summation includes every pair of particles, and each integral is to be taken between the proper limits. If we write
V = Σ ∫ Rpq drpq,
(11)
when rpq ranges from its value in some standard configuration A of the system to its value in any other configuration P, it is plain that V represents the work which would have to be done in order to bring the system from rest in the configuration A to rest in the configuration P. Hence V is a definite function of the configuration P; it is called the internal potential energy. If T denote the kinetic energy, we may say then that the sum T + V is in any interval of time increased by an amount equal to the work done by the extraneous forces. In particular, if there are no extraneous forces T + V is constant. Again, if some of the extraneous forces are due to a conservative field of force, the work which they do may be reckoned as a diminution of the potential energy relative to the field as in § 13.