about a vertical axis half-way between the points of attachment of the upper string. The equations of small motion are then of the tYPe

55—2w])-w2x= -P2x, j) -j-Zwii -w2y= -gzy. (35) This is satished by

d d:5é=A cos (at+e), y=B sin (¢1t+e), (36) provi e

(024-wz-P2)A-l-2awB =O, ()

2UwA+(a2-j-wz-q2)B =0. 37

Eliminating the ratio A:B we have (, ,2+w2 p2) (, ,2+w2 g2) 4a2w2 =0 (33) It is easily proved that the roots of this quadratic in U2 are always real, and that they are moreover both positive unless m2 lies between p” and qi. The ratio B/A is determined in each case by either of the equations (37); hence each root of the quadratic gives a solution of the type (36), with two arbitrary constants A, e. Since the equations (33) are linear, these two solutions are to be superposed. If the qua ratic (38) has a negative root, the trigonometrical functions in (36) are to be replaced by real exponential, and the sition x=o, y=o is unstable. This occurs only when the periodp((21r/w) of revolution of the arm lies between the two periods (21r/p, 21r/g) of oscillation when the arm is fixed. § 14. Central F orces. H odagraph.-The motion of a particle subject to a force which passes always through a fixed point O is necessarily in a plane orbit. For its investigation we require two equations; these may be obtained in a variety of forms. Since the impulse of the force in any element of time 6t has zero moment about O, the same will be true of the additional momentum generated. Hence the moment of the momentum (considered as a localized vector) about O will be constant. In symbols, if 'v be the velocity and p the perpendicular from O to the tangent to the path,

Pv=h» (I)

where h is a constant. If 5s be an element of the path, {>5s is twice the area enclosed by 55 and the radii drawn to its extremities from O. Hence if 5A be this area, we have 5A={, ~ p5sé hét, or

dA

5=ih- (2)

Hence equal areas are swept over by the radius vector in equal times.

If P be the acceleration towards O, we have dv dr

va; = -' Pa? (3)

since dr/ds is the cosine of the angle between the directions of r and 5s. We will suppose that P is a function of r only; then integrating (3) we find

é zz' = -fPdr-I-const., (4)

which is recognized as the equation of energy. Combining this with (1) we have

112 C ~

g= -2 jPdf, (5)

which completely determines the path except as to its orientation with respect to O.

If the law of attraction be that of the inverse square of the distance, we have P = p/rf, and hz zu

g 52 = C +7- (5)

Now in a conic whose focus is at O we have ?Ei=§ :F 5. 1 (7)

where l is half the latus-rectum, a is half the major axis, and the upper or lower sign is 'to be taken according as the conic is an ellipse or hyperbola. In the intermediate case of the parabola we have a=oo and the last term disappears. The equations (6) and (7) are identified by putting Since

l=h'/p, a=¢;4/C. (8)

h2

v“='§ 2=/1(§ =f=, § >. (9)

it appears that the orbit is an ellipse, parabola or hyperbola, according as 11” is less than, equal to, or greater than za/r. Now it appears from (6) that zp/r is the square of the velocity which would be acquired by a particle falling from rest at infinity to the distance r. Hence the character of the orbit depends on whether the velocity at any point is less than, equal to, or greater than the velocity from infinity, as it is called. In an elliptic orbit the area rrab is swept over in the time r =@ =Elr-gig (IO)

Eh i//1

since h=, u.*li=/t5ba'i by (8). The converse problem, to determine the law of force under which a given orbit can be described about a given pole, is solved by differentiating (5) with respect to 1; thus

map

P ==Fi;. (I I)

In the case of an ellipse described about the centre as pole welhave l%;)E=a2+b2»-r2; (12)

hence P=/ar, if, i=h2/a2b2. This merely shows that a particular ellipse may be described under the law of the direct distance provided the circumstances of projection be suitably adjusted. But since an ellipse can always be constructed with a given centre so as to touch a given line at a given point, and to have a given value of ab(=h// ju) we infer that the orbit will be elliptic whatever the initial circuénstances. Also the period is 21rab/h=21r// p, as previously foun . A

Again, in the equiangular spiral we have 19 =r sin n., and therefore P =y/V3, if ju =h2/sin' a.. But since an equiangular spiral having a given pole is completely determined by a given point and a given tangent, this type of orbit is not a general one for the law of the inverse cube. In order that the spiral may be described it is necessary that the velocity of projection should be adjusted to make h=/pt. sin u.. Similarly, in the case of a circle with the pole on the circumference we have p2 =r'/2a, P =;t/15, if a=8h2a,2; but this orbit is not a general one for the law of the inverse fifth power. In astronomical and other investigations relating to central forces it is often convenient to use polar co-ordinates with the centre of force as pole.

Let P, Q be the positions of a

moving point at times t, 15+ 513, and write OP=r, OQ=r+6r, 14

LPOQ=50, O being any fixed

origin. If u, -v be the component velocities at P along »

and perpendicular to OP (in

-v--6-u

I .U Q ' u-1-814

P

60

the direction of 0 increasing), FIG- 63we have

u=lim'Z=§ flv=lim @=r@' A (13)

M dl-' ' 5¢ dl

Again, the velocities parallel and perpendicular to OP change in the time 5t from u, 'v to u-1160, 11-|~u56, ultimately. The component accelerations at

du -116 d"'r dt? 2

af-va we -da) y

dv d6 1 if, d9 V (14)

Zi'&'l'“Z1?“i at (f a1)»

P in these directions are therefore respectively. q q

In the case of a central force, with O as pole, the transverse acceleration vanishes, so that p 12d0/dt=h, Y (15)

where h is constant; this shows (again) that the radius vector sweeps over equal areas in equal times. The radial resolution gives

§§ -f (1%) i= -P, no

where P, as before, denotes the acceleration towards O. If in this we put r= I/u, and eliminate t by means of (r 5), we obtain the general differential equation of central orbits, viz. A, dau P

W +14 =}jr, }§ - r (17)

If, for example, the law be that of the inverse square, we have P =;4u2, and the solution is of the form u=f§ {I+6 cos(0-u)}, (18)

where e, a. are arbitrary constants. This is recognized as the polar Zcpuation of a conic referred to the focus, the half latus-rectum being 2 ' .

M- ~