| ξ = M0u = | ∂T | , η = M0v = | ∂T | , ζ = M0w = | ∂T | |
| ∂u | ∂v | ∂w |
(14)
whilst those of the relative angular momentum are given by (7). The preceding formulae are sufficient for the treatment of instantaneous impulses. Thus if an impulse (ξ, η, ζ, λ, μ, ν) change the motion from (u, v, w, p, q, r) to (u′, v′, w′, p′, q′, r′) we have
| M0 (u′ − u) = ξ, | M0 (v′ − v) = η, | M0(w′ − w) = ζ, |
| A (p′ − p) = λ, | B (q′ − q) = μ, | C (r′ − r) = ν, |
(15)
where, for simplicity, the co-ordinate axes are supposed to coincide with the principal axes at the mass-centre. Hence the change of kinetic energy is
T′ − T = ξ · 1⁄2 (u + u′) + η · 1⁄2 (v + v′) + ζ · 1⁄2 (w + w′),
+ λ · 1⁄2 (p + p′) + μ · 1⁄2 (q + q′) + ν · 1⁄2 (r + r′).
(16)
The factors of ξ, η, ζ, λ, μ, ν on the right-hand side are proportional to the constituents of a possible infinitesimal displacement of the solid, and the whole expression is proportional (on the same scale) to the work done by the given system of impulsive forces in such a displacement. As in § 9 this must be equal to the total work done in such a displacement by the several forces, whatever they are, which make up the impulse. We are thus led to the following statement: the change of kinetic energy due to any system of impulsive forces is equal to the sum of the products of the several forces into the semi-sum of the initial and final velocities of their respective points of application, resolved in the directions of the forces. Thus in the problem of fig. 77 the kinetic energy generated is 1⁄2M (κ2 + Cq2)ω′2, if C be the instantaneous centre; this is seen to be equal to 1⁄2F·ω′·CP, where ω′·CP represents the initial velocity of P.
The equations of continuous motion of a solid are obtained by substituting the values of ξ, η, ζ, λ, μ, ν from (14) and (7) in the general equations