The body PCDQS vibrates in a vertical plane, being hung on a thread ABD, which as the body vibrates, winds about the curve ABX; and it is here proposed to determine the circumstances relating to the motion, proper data being supposed given. The part BD below the point where the thread leaves the curve, being supposed a right line passing through the centre of gravity C of the body, the resistance of the air not being taken into consideration ?

Let v be the velocity at a certain time of a particle P of the body, in the direction PM parallel to the horizon; let u be the velocity at the same time of the same particle P, in the direction parallel to BM, which is perpendicular to the hori zon; and let 2g be the absolute force of gravity, consequently Po is the fluxion of the momentum generated in P by the inertia of the system in the direction PM; and representing the fluxion of the time i P-2gPt will represent the fluxion of the momentum generated in P in the direction BM by the inertia of the system; but because the system by its inertia cannot put itself in motion or destroy any motion which it might have, it follows that the whole of these forces in the system must destroy each other. Furthermore, supposing B

the point at this instant where the thread ABX leaves the cheek; this point B will be the momentary centre of motion about which all the particles revolve, and through the medium of which, as a fulcrum, the effect of any particle P may, by the power of the lever, be transferred to any point C of the body; and consequently by the property of the lever, because the above named forces destroy each other, we have I, the sum of (ù P.

(v. P. BM) + the sum of (i P-2 g P. t.MP)= o. Draw PQI BC cutting & in Q, produce PM to cut BD in S, and draw Qn SM, cutting SM in n; therefore PM = BQ. cos of the angle BSM – PQ X sine of the angle BSM and if V be the absolute velocity of the centre of gravity C of the body, it is evident we should have, whilst B is the momentary centre, BC:V::BM: the velocity of M or its equal the velocity of P in the direction PM, that is, v; and, therefore,

BM =

PM =

vi.BC.

V

BC

V.; also we have BC:V::PM:u; and consequently

BC. น

V

consequently equation I becomes sum of

P)+sum of f(uù.BC)

P-2g. P. i. MP)=0; and.. II;

sum of vʊ P + sum of uù P= sum of 2g P i V. MP

BC

but sum

BC2

(because C is the centre of gravity of the body, and supposing

a be put for the sum of

CP
BC2

2g. Pi. V.

BC

P in the body, and the body

α

(1 + BC=

:)

V2; also sum of

=(by substituting for PM its value BQ. cos

i

of BSM-PQ. sine of BSM found above) 2g t cos of BSM . V:

because C is the centre of gravity and in consequence sum of BC, the body being 1 and the sum of PQ = 0.

BQ. P

Hence equation H

a V2

becomes III, fluxion of 1 + BC2 4g i. V cos of S. Let AE cutting the path CE of the centre of gravity C of the body, in E, be a line given in position to the horizon; draw CG parallel to the horizon cutting AË in G, put EG = GC y EC2 R = BC; and consequently because BC is perpendicular to the curve CE at C, the cosine

of the <S or its equal the cosine of BCG = and t is = whilst C descends; consequently the equation III becomes

4g.b-x, b being the value of x when Vo, consequently

We might here proceed in the subject in hand; but as I am not aware that any one has gone before me in this speculation, I consider it more satisfactory to have the corroborative evidence of a second investigation, and for this object I shall propose the following Lemma, which will be found worthy of consideration for other purposes as well as for this now in view.

Lemma. If a body OPD, fig. 2. after having been put in motion, move in virtue only of its inertia, and by some mechanism is continually changing its centre of motion, by continued and gradual, or, in other words, not sudden change, then I say the velocity of the centre of gyration of the body, corresponding to the point about which it may then be revolving, will be a constant quantity; that is the same as the velocity of any other point in the body, at the time when it is the centre of gyration, corresponding to the point about which it is at this other time revolving.

Fig. 2.

P

B

=

Let B be the centre about which the body is, revolving at a certain instant, B' infinitely near B be the centre about which it revolves the next instant, bisect BB' in r, and from every point P of the body, draw PB, Pr and PB', let w be the angu lar velocity of the body in the first instant, measured at the distance 1, and the momentum of P will be w. P. PB when it revolves about B, which is divisible into two, the one in the direction B'P, and the other perpendicular thereto, and which is w. P. PB. cos of the angle BPB', or omitting quantities infinitely small of the second degree simply w.P. PB the, former of these forces when B' becomes the centre is destroyed by the reaction of B', and the other gives to P the angular momentum w. P. PB. PB' = w. P. Pr2, neglecting infinitely small quantities of the second degree, consequently the whole angular momentum of the body = w sum of (P. Pr2;) Let Y be the centre of gyration of the body when it revolves about B, and Y' the centre of gyration of the body when it revolves about B', consequently, sum of BP2.P BY2= =, neglecting infinitely small quantibody

:

ties of the second degree, (Sum of Pr2. P+ sum of (2 Br. Pr. P cos of r)) ÷ body = (sum of (Pr2. P) + 2 Br. sum of Pr. P. cos of r)) body in the same way we find B' Y'2= (sum of (Pr2. P)—2. Br. sum of Pr.'P. of r)) body; .. B'Y'2. BY (sum of Pr. P4 Br2. (sum of Pr. P. cos of r2)÷body; or neglecting the term of which Br2 is the coefficient, it being infinitely small sum of Pr2P2; of the second degree, we find B Y'2 . BY2 = C

VOL. III.

body2

.. B' Y. BY

=

sum of Pr2. P
body

; consequently the angular

momentum of the body which was above shown to be w. sum of P Pr2 is equal to w. body X B'Y', BY, when the body revolves about B'; and Y' being the centre of gyration corresponding, the velocity of Y' will be the said angular force divided by B'Y'. into the body, and is consequently w. BY or the velocity which Y had when the body revolved about B, and therefore the fluxion of the velocity being equal to nothing, the velocity is constant QED. The above is not the only proof the lemma admits of, and the truth is likewise corroborated by the principal of vis-viva.

Hence to a second solution of the Problem. See Fig. 1. Let O be the centre of oscillation, and Y the centre of gyration of the body corresponding to the centre of suspension B; .. the velocity of Y V.

BY

BC

=

by the known ex

pression for the distance of the centre of gyration from the

✓ BO

point of suspension) V. = V. Q, Q being put for✔BO).

BC

BC

.. from the lemma were it not for gravity we should have V. Q a constant quantity, or VQ + VQ-o; that is V

QV
Q'

and consequently the excess of the real value of V above this' value must be generated by gravity; but gravity would gene

And consequently when the curve described by C is given, the time describing any part of it may be formed; but it may be more convenient in case the cheek ABX fig. 1, is