(6)
If we put x, y, z = 0, the theorem is proved as regards axes parallel to Ox.
Next consider the kinetic energy of the system. If from a fixed point O we draw vectors OV1>, OV2> to represent the velocities of the several particles m1, m2 ..., and if we construct the vector
| OK> = | Σ ( m·OV> ) |
| Σ(m) |
(7)
this will represent the velocity of the mass-centre, by (3). We find, exactly as in the proof of Lagrange’s First Theorem (§ 11), that
1⁄2Σ (m·OV2) = 1⁄2Σ (m)·OK2 + 1⁄2Σ (m·KV2);
(8)
i.e. the total kinetic energy is equal to the kinetic energy of the whole mass supposed concentrated at G and moving with this point, together with the kinetic energy of the motion relative to G. The latter may be called the internal kinetic energy of the system. Analytically we have
| 1⁄2Σ { m (ẋ2 + ẏ2 + ż2) } = 1⁄2Σ(m)· { ( | dx | )2 + ( | dy | )2 + ( | dz | )2 } + 1⁄2Σ{ m(ζ2 + η̇2 + ζ̇2) }. |
| dt | dt | dt |