Page:EB1911 - Volume 17.djvu/1001

Again, if the instantaneous position of G be taken as base, the angular momentum of the absolute motion is the same as the angular momentum of the motion relative to G. For the velocity of a particle m at P may be replaced by two components one of which (v̄) is identical in magnitude and direction with the velocity of G, whilst the other (v) is the velocity relative to G.

Fig. 70.

The aggregate of the components mv̄ of momentum is equivalent to a single localized vector Σ(m)·v̄ in a line through G, and has therefore zero moment about any axis through G; hence in taking moments about such an axis we need only regard the velocities relative to G. In symbols, we have

Σ { m(yż − zẏ) } = Σ(m)· ${\Big (}$ ȳ	dz̄	− z̄	dȳ	${\Big )}$ + Σ { m (ηζ − ζ̇η̇) }.
	dt		dt

(5)

Fig. 71.

since Σ(mξ) = 0, Σ(mξ̇) = 0, and so on, the notation being as in § 11. This expresses that the moment of momentum about any fixed axis (e.g. Ox) is equal to the moment of momentum of the motion relative to G about a parallel axis through G, together with the moment of momentum of the whole mass supposed concentrated at G and moving with this point. If in (5) we make O coincide with the instantaneous position of G, we have x̄, z̄, z = 0, and the theorem follows.

Finally, the rates of change of the components of the angular momentum of the motion relative to G referred to G as a moving base, are equal to the rates of change of the corresponding components of angular momentum relative to a fixed base coincident with the instantaneous position of G. For let G′ be a consecutive position of G. At the instant t + δt the momenta of the system are equivalent to a linear momentum represented by a localized vector Σ(m)·(v̄ + δv̄) in a line through G′ tangential to the path of G′, together with a certain angular momentum. Now the moment of this localized vector with respect to any axis through G is zero, to the first order of δt, since the perpendicular distance of G from the tangent line at G′ is of the order (δt)². Analytically we have from (5),

d	Σ { m (yż − zẏ) } = Σ(m)· ( ȳ	dz̄2	− z̄	d 2ȳ	) +	d	Σ { m(ηζ − ζη̇) }
dt		dt 2		dt 2		dt

(6)

If we put x̄, ȳ, z̄ = 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)· { (	dx̄	)2 + (	dȳ	)2 + (	dz̄	)2 } + 1/2Σ{ m(ζ2 + η̇2 + ζ̇2) }.
	dt		dt		dt

(9)

There is also an analogue to Lagrange’s Second Theorem, viz.

1/2Σ (m·KV2) = 1/2	ΣΣ (mpmq · VpVq2)	,
	Σm

(10)

which expresses the internal kinetic energy in terms of the relative velocities of the several pairs of particles. This formula is due to Möbius.

The preceding theorems are purely kinematical. We have now to consider the effect of the forces acting on the particles. These may be divided into two categories; we have first, the extraneous forces exerted on the various particles from without, and, secondly, the mutual or internal forces between the various pairs of particles. It is assumed that these latter are subject to the law of equality of action and reaction. If the equations of motion of each particle be formed separately, each such internal force will appear twice over, with opposite signs for its components, viz. as affecting the motion of each of the two particles between which it acts. The full working out is in general difficult, the comparatively simple problem of “three bodies,” for instance, in gravitational astronomy being still unsolved, but some general theorems can be formulated.

The first of these may be called the Principle of Linear Momentum. If there are no extraneous forces, the resultant linear momentum is constant in every respect. For consider any two particles at P and Q, acting on one another with equal and opposite forces in the line PQ. In the time δt a certain impulse is given to the first particle in the direction (say) from P to Q, whilst an equal and opposite impulse is given to the second in the direction from Q to P. Since these impulses produce equal and opposite momenta in the two particles, the resultant linear momentum of the system is unaltered. If extraneous forces act, it is seen in like manner that the resultant linear momentum of the system is in any given time modified by the geometric addition of the total impulse of the extraneous forces. It follows, by the preceding kinematic theory, that the mass-centre G of the system will move exactly as if the whole mass were concentrated there and were acted on by the extraneous forces applied parallel to their original directions. For example, the mass-centre of a system free from extraneous force will describe a straight line with constant velocity. Again, the mass-centre of a chain of