CHAPTER XVII.

THE LAGRANGE EQUATIONS.

308. IN any motion of a system there are a certain number of independent variables, which can be chosen in different ways, and which completely represent the position and configuration of the system.

Such a set of variables is called a set of generalised coordinates, and the number of them represents the number of degrees of freedom of the system.

Let them be 0, 4, 4,... so that the coordinates of any particle of the system will be functions of t, 0, &, ¥,...

i.e.

x= f (t, 0, &, ¥,...), etc.

It must be understood however that this method is only adapted to those cases in which x, y, and z are independent of 0, 4, 4, etc.

The system of time-fluxes of momenta, or of effective forces, is the exact equivalent of the system of acting forces, and therefore if, at any instant, a slight displacement of the configuration be imagined, the virtual work of the effective forces will be equal to that of the acting forces, that is to the loss of potential energy of the system.

Expressed in mathematical symbols this statement gives the equation

Em {28+jôy+28z}=-8V,

where V is the potential energy of the system.

Therefore, observing that 80, 84, are arbitrary quantities,

Do dx dx dx
8+
+

=

Dt dt de аф

$+...

representing the partial t-flux of x, and

dV

de

D

=

Dt &m ( x

D

Dt

di

dė

+

dx

Ση

D (dx

Dt (aa).

=

d2x

dodt

dx

de'

dz dy +2 do

Ση

...)

D di

-Σm (à Dt do

D dx -Em & Dt de (i

Dt Im (&
Ση
Em (à

+

de

+

...)

...).

+

d2x

$+...

.) – Σm (à

Dx

Dt

Now let T be the kinetic energy, so that

2T=Σm (x2 + ÿ3 + ¿2);

309. We now proceed to illustrate the use of these equations by the solution of some examples.

Ex. 1. Two heavy rods AB, BC, jointed at B, swinging in a vertical plane about the fixed end A.

and

If and are the inclinations of AB and BC to the vertical, these are a set of generalised coordinates.

If m, m' are the masses, 2a, 2b the lengths, and x, y, the coordinates of G, the centre of gravity of BC,

where

− V = mga cos 0 + m'g (2a cos 0 + b cos 4),

OC= 2a cos 0 + b cos, y = 2a sin 0 + b sin 4.

.. 2T=({ma2 + 4m'a2) Ò2 ÷ §m′b2¿2 + 4m'abÒ¿ cos († — 0), and Lagrange's equations become

+m') 4a3Ò + 2m'ab¿ cos (p − 0)} — 2m'abė¿ sin (4 – 0)

=-mga sin 0-2m'ga sin 0.........(a),

d

1

dt {§m'b3¿ + 2m'abÒ cos (Þ — 0)} + 2m'ab0$ sin (4 – 0)

A solution by means of the time-fluxes of momenta is given in art. 251.

The equation (B) is the same as that obtained by taking moments about B, and, if we add together (a) and (B), we get the equation of moments about A.

Small oscillations of these two rods.

If we neglect small quantities of the second order, the equations (a) and (B) take the forms,

M)

+m') 4a20 + 2m'abö

2m'abÄ + ‡m'b2Ï

(m + 2m') gal,

m'gbp.

==

Writing these equations in the forms,

(D2 + a2) 0 + ß3D2¢ = 0, (D2+λ2) $ + μ2D20 = 0, we obtain {(D2 + a2) (D2 + x2) — ẞ3μ3D1} 0 = 0.

It will be found that the values of p2 obtained from the auxiliary equation

(p2 + a2) (p2 + λ2) — B2μ2p1 = 0,

are real and negative.

It follows therefore that if these values are - n2, -n', the solution of the equation is of the form

0 = A cos (nt + e) + B cos (n't + €).

and

Or, if we multiply the second of these equations by k, and add it to the first, and assume that

B2 + kλ2 = k (a2 + kμ2),

we have two equations of the form,

D2 (0 +kp)+(a2 + kμ2) (0+kp) = 0,

leading to the same result.

Ex. 2. To deduce the equations of motion of a particle in polar coordinates.

In the figure of Art. 30, let mP, mQ, mR be the forces acting on the particle in the directions OP, PT, and perpendicular to the plane ZOP; then

also

2T = m {ř2 + r2ẻ2 + r2 sin2 0¿2},

where r, 0, p, are the generalised coordinates.

We then obtain, from the Lagrange equations,

rö sin 0 + 2řò̟ sin 0 + 2rẻ¿ cos 0 = R.

The left-hand members of these equations are, it will be seen, identical with the expressions obtained in Art. 30 for the accelerations.

Ex. 3. Motion of a heavy rod inside a smooth spherical shell.

For the simplification of formulæ we will take the case in which the rod subtends an angle of 120° at the centre of the shell.

If PQ is the rod, take for axes of (1), (2) and (3) the radius OA parallel to the rod, the radius OB bisecting it at right angles, and the radius OC perpendicular to the plane OPQ.

If A, B, C are the principal moments of inertia at G the centre of the rod, and if a is the radius of the inner surface of the shell,

Let u, v, w be the component velocities, parallel to the moving axes, of G, the centre of gravity of the rod.