ROTATION — 5th ed. (1815)

ROTATION, is a term which expresses the motion of the different parts of a solid body round an axis, and distinct from the progressive motion which it may have in its revolution round a distant point. The earth has a rotation round its axis, which produces the vicissitudes of day and night; while its revolution round the sun, combined with the obliquity of the equator, produces the varieties of summer and winter.

The mechanism of this kind of motion, or the relation which subsists between the intensity of the moving forces, modified as it may be by the manner of application, and the velocity of rotation, is highly interesting, both to the speculative philosopher and to the practical engineer. The precession of the equinoxes, and many other astronomical problems of great importance and difficulty, receive their solutions from this quarter: and the actual performance of our most valuable machines cannot be ascertained by the mere principles of equilibrium, but require a previous acquaintance with certain general propositions of rotatory motion.

It is chiefly with the view of assisting the engineer that we propose to deliver in this place a few fundamental propositions; and we shall do it in as familiar and popular a manner as possible, although this may cause the application of them to the abstruse problems of astronomy to be greatly deficient in the elegance of which they are susceptible.

When a solid body turns round an axis, retaining its shape and dimensions, every particle is actually describing a circle round this axis, and the axis passes through the centre of the circle, and is perpendicular to its plane. Moreover, in any instant of the motion, the particle is moving at right angles with the radius vector, or line joining it with its centre of rotation. Therefore, in order to ascertain the direction of the motion of any particle P (fig. 1.), we may draw a straight line PC from the particle perpendicular to the axis AB of rotation.

rotation. This line will lie in the plane of the circle P m n of rotation of the particle, and will be its radius vector; and a line PQ drawn from the particle perpendicular to this radius vector will be a tangent to the circle of rotation, and will have the direction of the motion of this particle.

3 The whole body being supposed to turn together, it is evident, that when it has made a complete rotation, each particle has described a circumference of a circle, and the whole paths of the different particles will be in the ratio of these circumferences, and therefore of their radii; and this is true of any portion of a whole turn, such as $\frac{1}{2}$ , $\frac{1}{3}$ , or 20 degrees, or any arch whatever; therefore the velocities of the different particles are proportional to their radii vectors, or to their distances from the axis of rotation.

And, lastly, all these motions are in parallel planes, to which the axis of rotation is perpendicular.

5 For the rotation of different lines in respect of velocity may be compared. When we compare the rotations of different bodies in respect of velocity, it is plain that it cannot be done by directly comparing the velocity of any particle in one of the bodies with that of any particle of the other; for, as all the particles of each have different velocities, this comparison can establish no ratio. But we familiarly compare such motions by the number of complete turns which they make in equal times, and we say that the second hand of a clock turns 60 times faster than the minute hand; now this comparison is equally just in any part of a turn as in the whole. While the minute hand moves round one degree, the second-hand moves 60; therefore, as the length or number of feet in the line uniformly described by a body in its progressive motion is a proper measure of its progressive velocity, so the number of degrees described by any particle of a whirling body in the circumference of its circle of rotation, or the angle described by any radius vector of that body, is a proper measure of its velocity of rotation. And in this manner may the rotation of two bodies be compared; and the velocity is with propriety termed ANGULAR VELOCITY.

9 An angle is directly as the length of the circumference on which it stands, and inversely as the radius of the circle, and may be expressed by the fraction of which the numerator is the arch, and the denominator the radius. Thus the angle PC p may be expressed by $\frac{Pp}{PC}$ . This fraction expresses the portion of the radius which is equal to the arch which measures the angle; and it is converted into the usual denomination of degrees, by knowing that one degree, or the 360th part of the circumference, is $\frac{1}{57.296}$ of the radius, or that an arch of 57.296 degrees is equal to the radius.

243, &c. When a solid body receives an impulse on any one point, or when that point is anyhow urged by a moving force, it cannot move without the other points also moving. And whatever is the motion of any particle, that particle must be conceived as urged by a force precisely competent to the production of that motion, by acting immediately on the particle itself. If this is not the particle immediately acted on by the external force, the force which really impels it is a force arising from the cohesion of the body. The particle immediately impelled by the external force is pressed towards its neigh-

bouring particles, or is drawn away from them; and, by this change of place, the connecting forces are brought into action, or are excited; they act on the particles adjoining, and change, or tend to change, their distances from the particles immediately beyond them; and thus the forces which connect this next series of particles are also excited, and another series of particles are made to exert their forces; and this goes on through the body till we come to the remote particle, whose motion we are considering. The forces which connect it with the adjoining series of particles are excited, and the particle is moved. We frequently say that the external moving force is propagated through the body to the distant particle; but this is not accurate. The particle is really and immediately moved by the forces which connect it with those adjoining. It will greatly assist our conception of the manner in which motion is thus produced in a distant particle, if we consider the particles as so many little balls, connected with each other by slender spiral springs like cork-screws. This would compose a mass which would be compressible, or which could be stretched, &c. And if we give an impulse to one of these balls, we shall see the whole assemblage in motion round any axis which we may suppose to support it. Now any one of these balls is really and immediately moved by the elasticity of the spiral wires which join it to its neighbours.

8 We are but little acquainted with the nature of these connecting forces. It can be learned only by the phenomena which are their effects. These are various, almost beyond description; but the mechanical philosopher has little to do with this variety. The distinctions which are the immediate causes of fluidity, of hardness, of softness, elasticity, ductility, are not of very difficult conception. There is one general fact which is sufficient for our present purpose—the forces by which the particles of bodies act on each other are equal. This is a matter of unexceptional experience; and no other foundation can be given to it as a law of mechanical nature.

An immediate consequence of this law is, that when two external forces A and B are in equilibrium by the intervention of a solid body (or rather when a solid body is in equilibrium between two external forces), these forces are equal and opposite; for the force A is in fact in immediate equilibrium with the opposite forces exerted by the particle to which it is applied, and is therefore equal and opposite to the force resulting from the combination of all the forces which connect that particle with the series of particles immediately adjoining. This resulting force may with propriety be called the equivalent of the forces from the combination of which it results. The use of this term will greatly abbreviate language. This first set of connecting forces consists of a number of distinct forces corresponding to each particle of the series, and each force has an equal and opposite force corresponding to it: therefore the compound force by which the first series of particles acts on that to which the external force A is applied, is equal and opposite to the compound force which connects this first series with the next series. And the same thing must be said of each succeeding series of particles, till we come at last to the particle to which the external force B is immediately applied. The force exerted by this particle is equal and opposite to that external

Rotation. ternal force; and it is equal to the compound force exerted by the second series of particles on that side; therefore the forces A and B are equal and opposite.

10 It results from this proposition, that when any number of external forces are applied to a solid body, and it is in equilibrium between them, they are such as would be in equilibrium if they were all applied to one point. Let the forces $aA, bB, cC$ (fig. 2.), be applied to three particles of the solid body. Therefore $aA$ is immediately in equilibrium with an equal and opposite force $A\alpha$ , resulting from the composition of the force $AD$ , which connects the particles A and B, and the force $AE$ which connects A with C. In like manner $bB$ is immediately in equilibrium with $B\beta$ , the equivalent of the forces $BF$ and $BG$ ; and $cC$ is in immediate equilibrium with the equivalent $C\gamma$ of the forces $CH$ and $CI$ . We shall conceive it very clearly if we suppose the three forces $aA, bB, cC$ , to be exerted by means of threads pulling at the solid body. The connecting parts between A and B, as also between A and C, are stretched. The lines AB and AC may be considered as elastic threads. Each thread is equally stretched through its whole length; and therefore if we take AD to represent the force with which the particle A is held back by the particle B, and if we would also represent the force with which B is held back by A, we must make BF equal to AD. Now (No. 9.), the forces AD and BF are equal and opposite; so are the forces AE and CI; so are the forces CH and BG. Now it is evident, that if the six forces AD, BF, BG, CH, CI, AE, were applied to one particle, the particle would be in equilibrium; for each force is accompanied by an equal and opposite force: and if the force $A\alpha$ were applied in place of AD, AE, the equilibrium would remain, because $A\alpha$ is equivalent to AD and AE. The same is true of $B\beta$ and $C\gamma$ . Therefore if the three forces $A\alpha, B\beta, C\gamma$ , were applied to one point, they would be in equilibrium. Consequently if the three forces $aA, bB, cC$ , which are respectively equal and opposite to $A\alpha, B\beta, C\gamma$ , are so applied, they will be in equilibrium. It is plain that this demonstration may be extended to any number of forces.

We may just remark by the bye, that if three forces are thus in equilibrium, they are acting in one plane; and, if they are not parallel, they are really directed to one point: for any one of them must be equal and opposite to the equivalent of the other two; and this equivalent is the diagonal of a parallelogram, of which the other two are the sides, and the diagonal and sides of any parallelogram are in one plane; and since they are in one plane, and any one of them is in equilibrium with the equivalent of the other two, it must pass through the same point with that equivalent, that is, through the point of concurrence of the other two.

These very simple propositions are the foundation of the whole theory of statics, and render it a very simple branch of mechanical science. It has been made abstruse by our very attempts to simplify it. Many elaborate treatises have been written on the fundamental property of the lever, and in them all it has been thought next to an insuperable difficulty to demonstrate the equilibrium of a straight lever when the parallel forces are inversely as their distances from the fulcrum.

Rotation. We think the demonstrations of Archimedes, Fontenex, D'Alembert, and Hamilton, extremely ingenious; but they only bring the mind into such a state of conception that it cannot refuse the truth of the proposition; and, except Mr Hamilton's, they labour under the disadvantage of being applicable only to commensurable distances and forces. Mr Vince's, in the Philosophical Transactions for 1794, is the most ingenious of them all; and it is wonderful that it has not occurred long ago. The difficulty in them all has arisen from the attempt to simplify the matter by considering a lever as an inflexible straight line. Had it been taken out of this abstract form, and considered as what it really is, a natural body, of some size, having its particles connected by equal and opposite forces, all difficulty would have vanished.

11 Mechanical science has been rendered abstruse by attempts at simplification. That we may apply these propositions to explain the mode of motion of rotation, we must recollect an unquestionable proposition in dynamics, that the force which produces any motion is equal and opposite to the force which would prevent it, when applied in the same place and in the same line, or which would extinguish it in the same time in which we suppose it to be produced. Therefore the force which is excited and made to act on any particle of a body, by the action of an external force on another particle, so as to cause it to move round an axis, is equal and opposite to the force which, when applied to that particle in the opposite direction, would be in equilibrium with the external force.

12 The only distinct notion we can form of the magnitude of any moving force is the quantity of motion which it can produce by acting uniformly during some given time. This will be had by knowing the velocity which it will produce in a body of known bulk. Thus we know that the weight of ten pounds of matter acting on it for a second will cause it to fall 16 feet with an uniformly accelerated motion, and will leave it in a state such that it would move on for ever at the rate of 32 feet in a second; which we call communicating the velocity of 32 feet per second. In the same manner, the best way of acquiring a distinct conception of the rotatory effort of a moving force, is to determine the quantity of rotatory motion which it can produce by acting uniformly during some known time.

13 Let a solid body turn round an axis passing through the point C (fig. 3.) perpendicular to the plane of this figure. Let this rotation be supposed to be produced by an external force acting in the direction FP. Let this force be such, that if the body were free, that is, unconnected with any axis supported by fixed points, it would, by acting uniformly during a small moment of time, cause its centre of gravity G (A) to describe a line of a certain length parallel to FP. This we know to

14 And of the quantity and effect of rotatory motion. Fig. 3.

(A) We take this term in its usual sense, as expressing that point where the sum of the equal gravitations of each particle may be supposed united. It is by no means (though commonly supposed) the point where the equivalent of the real gravitations of the particles may be supposed to act, and to produce the same motion as when acting

Rotation. to be the effect of a moving force acting on any solid body in free space. The centre of gravity will always describe a straight line. Other particles may chance to move differently, if the body, besides its progressive motion, has also a motion of rotation, as is generally the case. Draw $GI$ parallel to $FP$ , and make $GI$ to $GC$ as the velocity which the external force would communicate to the centre of the body (if moving freely, unconnected with a supported axis), to the velocity which it communicates to it in the same time round the axis $Cc$ . Also let $m$ be the number of equal particles, or the quantity of matter in the body. Then $mGI$ will express the quantity of motion produced by this force, and is a proper measure of it as a moving force; for $GI$ is twice the space described during the given time with an uniformly accelerated motion.

15 But since the body cannot move any way but round the axis passing through $C$ , the centre $G$ will begin to move with the velocity and in the direction, $GH$ , perpendicular to the line $CG$ ( $N^o$ 2.). And any particle $A$ can only move in the direction $AL$ , perpendicular to $CA$ . Moreover, the velocities of the different particles are as their radii vectores; and $CG$ is actually equal to the line $GH$ , which expresses the velocity of a particle in $G$ . Therefore $CA$ will in like manner express the velocity of the particle $A$ . If $A$ expresses its quantity of matter, $A \cdot CA$ will express its quantity of motion, and will represent the force which would produce it by acting uniformly during the moment of time.

16 We expressed the external moving force by $m \cdot GI$ . Part of it is employed in exciting the force $A \cdot CA$ , which urges the particle $A$ . In order to discover what part of the external force is necessary for this purpose, draw $CP$ perpendicular to $FP$ . The preceding observations show us, that the force wanted at $A$ is equal to the force which, when applied at $P$ in the direction $FP$ , would balance the force $A \cdot CA$ applied to $A$ in the direction $LA$ . Therefore (by the propriety of the lever $ACP$ , which is impelled at right angles at $A$ and $P$ ) we must have $CP$ to $CA$ as the force $A \cdot CA$ to the balancing pressure, which must be exerted at $P$ , or at any point in the line $FP$ . This pressure is therefore $\frac{A \cdot CA \cdot CA}{CP}$ or $\frac{A \cdot CA^2}{CP}$ . As we took $m \cdot GI$ for the measure of the whole external force, $GI$ being the velocity which it would communicate to the whole body moving in free space, we may take $G \cdot i$ for the velocity which would be communicated to the whole body by

the pressure $\frac{A \cdot CA^2}{CP}$ , and then this pressure will be

VOL. XVIII. Part I

properly expressed by $m \cdot G \cdot i$ . In like manner, $m \cdot a \cdot k$ may express the portion of the external force employed in communicating to another particle $B$ the motion which it acquires; and so on with respect to all the particles of the body.

It must be desirable to see the manner in which the forces are really concerned in giving motion to the different particles.

Suppose the external force to act immediately on the external particle $F$ . The line $FC$ connecting this particle with the axis in $C$ is either stretched or compressed by the effort of giving motion to a remote particle $A$ . It is plain that, in the circumstances represented in the figure, the line $FC$ is compressed, and the axis is pushed by it against its supports in the direction $C \lambda$ ; and the body must, on this account, resist in the opposite direction $F f$ . The particle $A$ is dragged out of its position, and made to begin its motion in the direction $AL$ perpendicular to $AC$ . This cannot be, unless by the connexion of the two lines $AC$ , $AF$ . $A$ resists by its inertia, and therefore both $AC$ and $AF$ are stretched by dragging it into motion. By this resistance the line $AC$ tends to contract itself again, and it pulls $C$ in the direction $C c$ , and $A$ in the direction $A a$ ; and if we take $C c$ to represent the action on $C$ , $A a$ must be taken equal to it. In like manner $AF$ is stretched and tends to contract, pulling $F$ in the direction $F \phi$ and $A$ in the direction $A \alpha$ with equal forces. Thus the particle $A$ is pulled in the directions $A a$ and $A \alpha$ ; the particle $F$ is pulled in the direction $F \phi$ , and pushed in the direction $F f$ ; and $C$ is pulled in the direction $C c$ , and pushed in the direction $C x$ . $A a$ and $A \alpha$ have produced their equivalent $AL$ , by which $A$ is dragged into motion; $F f$ and $F \phi$ produce their equivalent $F g$ , by which the external force is resisted, and $F g$ is equal and opposite to $m \cdot G \cdot i$ ; the forces $C c$ and $C x$ produce their equivalent $C d$ by which the axis is pressed on its supports, and this is resisted by an equal and opposite reaction of the supports in the direction $dC$ . The forces therefore which excite in the body the motion $A \cdot AL$ are both external, viz. the impelling force $gF$ , and the supporting force $dC$ . $AL$ therefore is not only the immediate equivalent of $A a$ and $A \alpha$ , but also the remote equivalent of $gF$ and $dC$ . We may therefore ascertain the proportion of $gF$ (that is, of $m \cdot G \cdot i$ ) to $AL$ (that is, of $A \cdot AC$ ), independent of the property of the lever. $gF$ is to $AL$ in the ratio compounded of the ratios of $gF$ to $F \phi$ or $A \alpha$ , and of $A \alpha$ to $AL$ . But we shall obtain it more easily by considering $gF$ as the equivalent of $AL$ and $dC$ . By what has been demonstrated above, the

P p directions

acting on each particle separately. It is this point only when all the particles gravitate alike, and in parallel directions. If the body were near the centre of the earth, for instance, the gravitations of the different particles would neither be nearly equal nor in parallel lines; and the place of its real centre of gravity, on which the equivalent of its whole gravitation may be supposed to act, would be very different from $G$ . Were we to denominate the point $G$ , as usually determined, by its mathematical properties, we would call it the CENTRE OF POSITION; because its distance from any plane, or its position with respect to any plane, is the average distance and position of all the particles. The true designation of $G$ is "the point through which if any plane whatever be made to pass, and if perpendiculars to this plane be drawn from every particle, the sum of all the perpendiculars on one side of this plane is equal to the sum of all the perpendiculars on the other side."

If we were to denominate $G$ by its mechanical properties, we would call it the CENTRE OF INERTIA; for this is equal in every particle, and in the same direction: and it is not in consequence of gravity, but of inertia, that the body describes with the point $G$ a line parallel to $FP$ . We wish this remark to be kept in mind.

Rotation. directions of the three forces $gF$ , $AL$ , and $dC$ must meet in one point $E$ , and $gF$ must be equal to the diagonal $tE$ of the parallelogram $Eest$ , of which the sides $Ee$ , $Et$ are respectively equal to $AL$ and $dC$ . Now $tE$ is to $Ee$ as the sine of the angle $teE$ to the sine of the angle $Ete$ , that is, as the sine of $CEA$ to the sine of $CEP$ , that is, as $CA$ to $CP$ , as we have already demonstrated by the property of the lever. We preferred that demonstration as the shortest, and as abundantly familiar, and as congenial with the general mechanism of rotatory motions. And the intelligent reader will observe, that this other demonstration is nothing but the demonstration by the lever expanded into its own elements. Having once made our readers sensible of this internal process of the excitement and operation of the forces which connect the particles, we shall not again have recourse to it.

17 It is evident that the sum of all the forces $gF$ , or $m.GI$ , must be equal to the whole moving force $m.GI$ , that $m.P\rho$ may be $=m.GI$ . That is, we must have $m.GI = \int \frac{A.CA^2}{CP}$ ; or, because $CP$ is given when the position of the line $FP$ is given, we must have $m.GI = \int \frac{A.CA^2}{CP}$ , where both $A$ and $CA$ are variable quantities.

18 This equation gives us $m.GI.CP = \int A.CA^2$ . Now we learn in mechanics that the energy of any force applied to a lever, or its power of producing a motion round the fulcrum, in opposition to any resistance whatever, is expressed by the product of the force by the perpendicular drawn from the fulcrum on the line of its direction. Therefore we may call $m.GI.CP$ the momentum ( $B$ ), energy, or rotatory effort, of the force $m.GI$ . And in like manner $\int A.CA^2$ is the sum of the momentum of all the particles of the body in actual rotation; and as this rotation required the momentum $m.GI.CP$ to produce it, this momentum balances, and therefore may express the energy of all the resistances made by the inertia of the particles to this motion of rotation. Or $\int A.CA^2$ may express it. Or, take $\rho$ to represent the quantity of matter in any particle, and $r$ to represent its radius vector, or distance from the axis of rotation, $\int \rho.r^2$ will express the momentum of inertia, and the equilibrium between the momentum of the external force $m.GI$ , acting in the direction $FP$ , and the combined momenta of the inertia of all the particles of the whirling body, is expressed by the equation $m.GI.CP = \int A.CA^2 = \int \rho.r^2$ . The usual way of studying elementary mechanics gives us the habit of associating the word equilibrium with a state of rest; and this has made our knowledge so

imperfect. But there is the same equilibrium of the actual immediate pressures when motion ensues from the action. When a weight $A$ descending raises a smaller weight $B$ by means of a thread passing over a pulley, the thread is equally stretched between the acting and resisting weights. The strain on this thread is undoubtedly the immediate moving force acting on $B$ , and the immediate resisting force acting on $A$ .

The same equation gives us $GI = \frac{\int \rho.r^2}{m.CP}$ .

Now $GI : CG = \frac{\int \rho.r^2}{m.CP} : CG = \int \rho.r^2 : m.CP.CG$ ; 19

but $CG$ represents the velocity of the centre. Hence we derive this fundamental proposition $\int \rho.r^2 : m.CP.CG = GI : CG$ ; or, that $\int \rho.r^2$ is to $m.CP.CG$ as the velocity of the body moving freely to the velocity of the centre of gravity round the axis of rotation.

Therefore the velocity of the centre is $= \frac{m.GI.CP.CG}{\int \rho.r^2}$ 20

The velocity of any point $B$ is $= \frac{m.GI.CP.CB}{\int \rho.r^2}$ 21

This fraction represents the length of the arch described by the point $B$ in the same time that the body unconnected with any fixed points would have described $GI$ .

Therefore the angular velocity (the arch divided by the radius) common to the whole body is $= \frac{m.GI.CP}{\int \rho.r^2}$ .

It may be here asked, how this fraction can express an angle? It evidently expresses a number; for both the numerator and denominator are of the same dimensions, namely, surfaces. It therefore expresses the portion of the radius which is equal to the arch measuring the angle, such as $\frac{1}{4}$ , $\frac{1}{3}$ , $\frac{1}{2}$ , &c. And to have this angle in degrees, we have only to recollect that the radius is 57,2958.

This angular velocity will be a maximum when the axis of rotation passes through the centre of gravity $G$ . For draw from any particle $A$ the line $Aa'$ perpendicular to $CG$ , and join $AG$ . Then $CA^2 = GA^2 + CG^2 \pm 2CG \times Ga$ . Therefore $\int CA^2 = \int GA^2 + \int CG^2 \pm \int 2CG \times Ga$ , $= \int GA^2 + m.CG^2 \pm \int 2CG \times Ga$ . But by the nature of the centre of gravity, the sum of all the $+Ga$ is equal to that of all the $-Ga$ ; and therefore $\pm \int 2CG \times Ga$ is nothing; and therefore $\int CA^2 = \int GA^2 + m.CG^2$ .

Therefore $\int CA^2$ or $\int \rho.r^2$ is smallest, and $\frac{m.GI.CP}{\int \rho.r^2}$ is greatest when $m.CG^2$ is nothing, or when $CG$ is nothing; that is, when $C$ and $G$ coincide.

The absolute quantity of motion in the whirling body,

(B) The word momentum is very carelessly used by our mechanical writers. It is frequently employed to express the product of the quantity of matter and velocity, that is, the quantity of motion; and it is also used (with strict propriety of language) to express the power, energy, or efficacy of a force to produce motion in the circumstances in which it acts. We wish to confine it to this use alone. Sir Isaac Newton adhered rigidly to this employment of the term (indeed no man exceeds him in precision of expression), even when he used it to express the quantity of motion: for in these instances the energy of this quantity of motion, as modified by the circumstances of its action, was always in the ratio of the quantity of motion.

Rotation. dy, or the sum of the motions of all its particles, is
$\frac{m \cdot GI \cdot CP \cdot f p \cdot r}{f p r^3}$ . For the motion of each particle is
$\frac{f p r^3}{m \cdot GI \cdot CP \cdot p r}$ .

25
ratio of the resistance of quantity of matter to a motion of rotation.
The resistance which a given quantity of matter makes to a motion of rotation is proportional to $f p r^3$ . For this must be measured by the forces which must be similarly applied in order to give it the same angular motion or angular velocity. Thus let one external force be $m \cdot GI$ , and the other $m \cdot \gamma i$ .—Let both be applied at the distance $CP$ . Let $r$ be the radius vector in the one body, and $\epsilon$ in the other; now the angular velocities $\frac{m \cdot GI \cdot CP}{f p r^3}$ and $\frac{m \cdot \gamma i \cdot CP}{f p \epsilon^3}$ are equal by supposition. Therefore $m \cdot GI : m \cdot \gamma i = f p r^3 : f p \epsilon^3$ .

As in the communication of motion to bodies in free space a given force always produces the same quantity of motion; so in the communication of motion to bodies obliged to turn round axes, a given force, applied at a given distance from the axes, always produces the same quantity of momentum. Whence it may easily be deduced (and we shall do it afterwards), that as in the communication of motion among free bodies the same quantity of motion is preserved, so in the communication of motion among whirling bodies the same quantity of whirling motion is preserved.

This is a proposition of the utmost importance in practical mechanics, and may indeed be considered as the fundamental proposition with respect to all machines of the rotatory kind when performing work; that is, of all machines which derive their efficacy from levers or wheels. There is a valuable set of experiments by Mr Smeaton in the Philosophical Transactions, Volume lxvi, which fully confirm it. We shall give an example by and bye of the utility of the proposition, showing how exceedingly imperfect the usual theories of mechanics are which do not proceed on this principle.

26
With respect to the general proposition from which all these deductions have been made, we must observe, that the demonstration is not restricted to the time necessary for causing each particle to describe an arch equal to the radius vector. We assumed the radius vector as the measure of the velocity merely to simplify the notation. Both the progressive motion of the free body and the rotation of the whirling body are uniformly accelerated, when we suppose the external force to act uniformly during any time whatever; and the spaces described by each motion in the same time are in a constant ratio. The formulae may therefore with equal propriety represent the momentary accelerations in the different cases.

27
All the particles of a body not necessarily applied in one plane.
It must also be observed, that it is not necessary to suppose that all the particles of the body are in one plane, and that the moving force acts in a line $FP$ lying also in this plane. This was tacitly allowed, merely to make the present investigation (which is addressed chiefly to the practical mechanic) more familiar and easy. The equilibrium between the force $A \times CA$ , which is immediately urging the particle $A$ , and the force $m \cdot G i$ employed at $P$ or $F$ , in order to excite that force at $A$ , would have been precisely the same although the lines $AC$ and $FP$ had been in different planes, pro-

vided only that these planes were parallel. This is known to every person in the least acquainted with the wheel and axle. But if the external moving force does not act in a plane parallel to the circles of rotation of the different particles, it must be resolved into two forces, one of which is perpendicular to these planes, or parallel to the axis of rotation, and the other lying in a plane of rotation. And it is this last only that we consider as the moving force; the other tends merely to push the body in the direction of its axis, but has no tendency to turn it round that axis. When we come to consider the rotation of a body perfectly free, it will be necessary to attend particularly to this circumstance. But there are several important mechanical propositions which do not require this.

28
The motion of any body is estimated by that of its motion centre of gravity, as is well known. The difference of a body between the motion of the centre of a free body and the motion of the centre of a body turning round an axis, is evidently owing to the connexion which the parts of the body have with this axis, and to the action of the points of support on this axis. This action must be considered as another external force, combined with that which acts on the particle $P$ , and therefore must be such as, if combined with it, would produce the very motion which we observe. That is, if we suppose the body unconnected with any fixed points, but as having its axis acted on by the same forces which these points exert, the body would turn as we observe it to do, the axis remaining at rest.

Therefore join $I$ and $H$ , and complete the parallelogram $GIHK$ . It is plain that $m \cdot GK$ must represent the forces exerted by the axis on the fixed points.

If therefore $GI$ should coincide with $GH$ , and the point $I$ with the point $H$ , the force $GK$ vanishes, and the body begins to turn round $C$ , without exerting any pressure on the points of support; and the initial motion is the same as if the body were free. Or, the axis at $C$ is then a spontaneous axis of conversion.

That this may be the case, it is necessary, in the first place, that the external force act in a direction perpendicular to $CG$ ; for $GI$ is always parallel to $FP$ ; it being a leading proposition in dynamics, that when a moving force acts on any part whatever of a solid body, unconnected with fixed points, the centre of gravity will proceed in a straight line parallel to the direction of that force. In the next place $GH$ must be equal to $GI$ ; that is, ( $N^o$ 21) $\frac{m \cdot GI \cdot CP \cdot CG}{f p r^3}$ is equal to $GI$ , or $\frac{m \cdot CP \cdot CG}{f p r^3} = 1$ , and $CP = \frac{f p r^3}{m \cdot CG}$ .

The equation $CP = \frac{f p r^3}{m \cdot CG}$ gives us $m \cdot CG \cdot CP = f p r^3 = f A \cdot CA^3$ . But it was shown ( $N^o$ 23), that $f A \cdot CA^3 = f A \cdot GA^3 + m \cdot CG^3$ . Therefore $f A \cdot GA^3 = m \cdot CG \cdot CP - m \cdot CG \cdot CG = m \cdot CG (CP - CG)$ , $= m \cdot CG \cdot GP$ . Therefore we have (for another determination of the point of impulse $P$ so as to annihilate all pressure on the axis) $GP = \frac{f A \cdot GA^3}{m \cdot CG}$ . This is generally the most easily obtained, the mathematical situation of the centre of gravity being well known.

NB. When $CP = \frac{\int p r^3}{m.GC}$ , we shall always have the

velocity of the centre the same as if the body were free, but there will always be a pressure on the points of support, unless $FP$ be also perpendicular to $CG$ . In other positions of $FP$ the pressure on the axis, or on its points of support, will be $m.GI \times 2 \sin. GCP$ .

It would be a desirable thing in our machines which derive their efficacy from a rotatory motion, to apply the pressures arising from the power and from the resistance opposed by the work in such a manner as to annihilate or diminish this pressure on the supports of the axis of motion. Attention to this theorem will point out what may be done; and it is at all times proper, nay necessary, to know what are the pressures in the points of support. If we are ignorant of this, we shall run the risk of our machine failing in those parts; and our anxiety to prevent this will make us load it with needless and ill disposed strength. In the ordinary theories of machines, deduced entirely from the principles of equilibrium, the pressure on the points of support (exclusive of what proceeds from the weight of the machine itself) is stated as the same as if the moving and resisting forces were applied immediately to these points in their own directions. But this is in all cases erroneous; and, in cases of swift motions, it is greatly so. We may be convinced of this by a very simple instance. Suppose a line laid over a pulley, and a pound weight at one end of it, and ten pounds at the other; the pressure of the axis on its support is eleven pounds, according to the usual rule; whereas we shall find it only $3\sqrt{2}$ . For, if we call the radius of the pulley $r$ , the momentum of the moving force is $10 \times r = 1 \times 10r$ , and the momentum of inertia is $10 \times r^2 + 1 \times r^2$ . ( $N^o 18.$ ) = $11r^2$ . Therefore the angular velocity is $\sqrt{2}r$ . But the distance $CG$ of the centre of gravity from the axis of motion is also $\sqrt{2}r$ , because we may suppose the two weights in contact with the circumference of the pulley. Therefore the velocity of the centre of gravity is $\sqrt{2}r \times \sqrt{2}r = 2r$ of its natural velocity. It is therefore diminished $\frac{4}{11}r$ by the figure of the axis of the pulley, and the 11 pounds press it with $\frac{4}{11}$ of their weight, that is, with $3\sqrt{2}$ pounds.

Since all our machines consist of inert matter, which requires force to put it in motion, or to stop it, or to change its motion, it is plain that some of our natural power is expended in producing this effect; and since the principles of equilibrium only state the proportion between the power and resistance which will preserve the machine at rest, our knowledge of the actual performance of a machine is imperfect, unless we know how much of our power is thus employed. It is only the remainder which can be stated in opposition to the resistance opposed by the work. This renders it proper to give some general propositions, which enable us to compute this with ease.

It would be very convenient, for instance, to know some point in which we might suppose the whole rotatory part of the machine concentrated; because then we could at once tell what the momentum of its inertia is, and what force we must apply to the impelled point of the machine, in order to move it with the desired velocity.

Let $S$ , fig. 3. be this point of a body turning round

the supported axis passing through $C$ ; that is, let $S$ be such a point, that if all the matter of the body were collected there, a force applied at $P$ will produce the same angular velocity as it would if applied at the same point of the body having its natural form.

The whole matter being collected at $S$ , the expression

$\frac{m.GI.CP}{\int p r^3}$ of the angular velocity becomes $\frac{m.GI.CP}{m.cS^3}$

( $N^o 22.$ ); and these are equal by supposition. There-

fore $\int p r^3 = m.cS^3$ , and $CS = \sqrt{\frac{\int p r^3}{m}}$ .

This point $S$ has been called the CENTRE of GYRATION.

In a line or slender rod, such as a working beam, or the spoke of a wheel in a machine, $CS$ is $\sqrt{\frac{r}{2}}$ of its length.

In a circle or cylinder, such as the solid drum of a capstan, $CS = \sqrt{\frac{r}{2}}$ its radius, or nearly $\frac{r}{\sqrt{2}}$ . But if it turns round one of its diameters, $CS = \frac{r}{2}$ radius.

In the periphery of a circle, or rim of a wheel, $CS =$ radius nearly.

If it turn round a diameter, $CS = \sqrt{\frac{r}{2}}$ radius. The surface of a sphere, or a thin spherical shell, turning round a diameter, has $CS = \sqrt{\frac{r}{2}}$ radius, or nearly $\frac{r}{\sqrt{2}}$ or $\frac{r}{2}$ .

A solid sphere turning round a diameter has $CS = \sqrt{\frac{r}{2}}$ radius, or nearly $\frac{r}{\sqrt{2}}$ . This is useful in the problem of the precession of the equinoxes. We may observe by the way, that if we consider the whirling body as a system of several bodies with rigid or inflexible connections, we may consider all the matter of each of these bodies as united in its centre of gyration, and the rotation of the whole will be the same; for this does not change the value of $\frac{\int p r^3}{m}$ .

There is another way of making this correction of A simpler the motion of a machine, or allowing for the inertia of mode of al- the machine itself, which is rather simpler than the one lowing for now given. We can suppose a quantity of matter col- the inertia lected at the point to which the moving force is applied, of ma- such that its inertia will oppose the same resistance to chines. rotation that the machine does in its natural form. Suppose the moving force applied at $P$ , as before, and that instead of the natural form of the body a quantity of matter = $\frac{\int p r^3}{CP}$ , collected at $P$ ; the moving force will produce the same angular velocity as on the body, in its natural form. For the angular velocity in this case

must be $\frac{m.GI.CP}{\int p r^3} \cdot CP^2$ ( $N^o 22.$ ) which is = $\frac{m.GI.CP}{\int p r^3}$ , the

same as before.

A point $O$ may be found, at such a distance from the centre of axis, that if all the matter of the body were collected there, and an external force $m.GI$ applied to it in a direction perpendicular or any how inclined to $CO$ , it will produce the same angular velocity as when applied to the centre of gravity $G$ , with the same inclination to the line $CG$ .

In this case, the angular velocity must be $\frac{m.GI.CO}{m.CO^2}$ , ( $N^o 22.$ ).

(No 22.), which is $\frac{GI}{CO}$ . This must be equal (by

supposition) to the angular velocity where the same force $m.GI$ is applied in the same inclination to $G$ .—
The angular velocity in this case must be $\frac{m.GI.CG}{fpr^2}$ .

Therefore we have $\frac{GI}{CO} = \frac{m.GI.CG}{fpr^2}$ , and $\frac{CO}{GI} = \frac{fpr^2}{m.GI.CG}$ , and $CO = \frac{fpr^2}{m.CG}$ . Also, as in No 31.
$GO = \frac{fA.GA^2}{m.CG}$ .

This point O has several remarkable properties.

In the first place, it is the point of a common heavy body swinging round C by its gravity, where, if all its weight be supposed to be concentrated, it will perform its oscillations in the same time. For while the body has its natural form, the whole force of gravity may be supposed to be exerted on its centre of gravity. When the matter of the body is collected at O, the force of gravity is concentrated there also; and if CG have the same inclination to the horizon in the first case that CO has in the second, the action of gravity will be applied in the same angle of inclination, and the two bodies will acquire the same angular velocity; that is, they will descend from this situation to the vertical situation (that is, through an equal angle) in the same time. These two bodies will therefore oscillate in equal times. For this reason, the point O so taken in the line CG, which is the radius vector of the centre of inertia, that CO is equal to $\frac{fA.CA^2}{m.CG}$ , or $GO = \frac{fA.GA^2}{m.CG}$ , is called the CENTRE OF OSCILLATION of the body; and a heavy point suspended by a thread of the length CO is called its equivalent or synchronous pendulum, or the simple pendulum, corresponding to the body itself, which is considered as a compound pendulum, or as consisting of a number of simple pendulums, which, by their rigid connection disturb each other's motions.

That CO may be the equivalent pendulum, and O the centre of oscillation, O must be in the line CG, otherwise it would not rest in the same position with the body, when no force was keeping it out of its vertical position. The equation $CO = \frac{fA.CA^2}{m.CG}$ only determines the distance of the centre of oscillation from the centre of suspension, or the length of the equivalent simple pendulum, but does not determine the precise point of the body occupied by the centre of oscillation; a circumstance also necessary in some cases.

Mathematicians have determined the situation of this point in many cases of frequent occurrence. Huyghens, in his Horologium Oscillatorum, and all the best writers of treatises of mechanics, have given the method of investigation at length. The general process is, to multiply every particle by the square of its distance from the axis of suspension, and to divide the sum of all these products by the product of the whole quantity of matter multiplied by the distance of its centre of gravity from the same axis. The quotient is the distance of the centre of oscillation, or the length of the equivalent simple pendulum: for $CO = \frac{fpr^2}{m.CG}$ .

a. If the body is a heavy straight line, suspended by one extremity, CO is $\frac{1}{3}$ of its length. Rotation.

b. This is nearly the case of a slender rod of a cylindrical or prismatic shape. It would be exactly so if all the points of a transverse section were equally distant from the axis of suspension.

c. If the pendulum is an isosceles triangle suspended by its apex, and vibrating perpendicularly to its own plane, CO is $\frac{1}{3}$ of its height.

d. This is nearly true of a very slender triangle (that is, whose height many times exceeds its base) swinging round its vertex in any direction.

e. In a very slender cone or pyramid swinging from its vertex, CO is $\frac{1}{3}$ of its height nearly.

f. If a sphere, of which $r$ is the radius, be suspended by a thread whose weight may be neglected, and whose length is $l$ , the distance between its centre of suspension and centres of oscillation is $a + r + \frac{r^2}{a+r}$ ; and the distance between its centres of bulk and oscillation is $\frac{r}{a+r}$ . Thus, in a common second's pendulum,

whose length at London is about 39 $\frac{1}{2}$ inches, the centre of oscillation will be found about $\frac{1}{50}$ of an inch below the centre of the ball, if it be two inches in diameter.

g. If the weight of the thread is to be taken into the account, we have the following distance between the centre of the ball and that of oscillation, where B is the weight of the ball, $a$ the distance of the point of suspension and its centre, $d$ the diameter of the ball, and $w$ the weight of the thread or rod, $GO = \frac{(\frac{1}{2}w + \frac{1}{2}B)d^2 - \frac{1}{2}w(ad + a^2)}{(\frac{1}{2}w + B)a - \frac{1}{2}dw}$ ; or, if we consider the weight of the thread as an unit, and the weight of the ball as its multiple (or as expressed by the number of times it contains the weight of the thread),

GO = \frac{\frac{1}{2}a}{B + \frac{1}{2}}

As the point O, determined as above, by making $CO = \frac{fpr^2}{m.CG}$ , is the centre of oscillation of the body turning round C, so C is the centre of oscillation of the same body turning round O: for resuming A.C.A in place of $pr$ , we have $fA.CA^2 = m.CO.CG$ . Now $fA.CA^2 = fA.OA^2 + fA.OC^2 - fA.OC.2O.d$ , (Euclid, II. 12. 13.), or $m.CO.CG = fA.OA^2 + fA.OC^2 - fA.OC.2O.d$ . But $fA.OC^2 = m.OC^2 = m.OC.OC$ ; and (by the nature of the centre of gravity) $fA.OC.2O.d = m.OC.2OG$ . Therefore we have $m.CO.CG = fA.OA^2 + m.OC.OC - m.OC.2OG$ ; and $fA.OA^2 = m.OC.CG + m.CO.2OG - m.CO.CO$ , $= m.CO(CG + 2OG - CO)$ . But $CG + 2OG$ is equal to $CO + OG$ , and $CG + 2OG - CO$ is equal to $OG$ . Therefore $fA.OA^2 = m.CO.OG$ , and

CO = \frac{fA.OA^2}{m.OG}

which is all that is wanted (according to No 39.) to make C the centre of oscillation when O is the centre of suspension.

If the point of suspension, or axis of rotation, be anywhere in the circumference of a circle of which G is the centre, the point O will be in the circumference of another circle of which G is the centre: for, by No 38.

43 Rotation. $GO = \frac{SA \cdot GA^2}{m \cdot CG}$ . Now $f A \cdot GA^2$ is a fixed quantity; and therefore while $CG$ is constant, $OG$ will also be constant.

We may also observe, that the distance of the axis from the centre $S$ of gyration is a mean proportional between its distance from the centre $G$ of gravity and the centre $O$ of oscillation: for we had ( $N^o$ .)

CS^2 = \frac{f p r^2}{m}, \text{ and } CO = \frac{f p - r^2}{m \cdot CG}, \text{ and therefore}

CO \cdot CG = \frac{f p r^2}{m} = CS^2 \text{ and } CO : CS = CS : CG.

44 We see also that the distance $CO$ is that at which an external force must be applied; so that there may not be any pressure excited in the axis upon its points of support, and the axis may be a spontaneous axis of conversion. This we learn, by comparing the value of $CO$ with that of $CP$ in art. 30. This being the case, it follows, that if an external force is applied in a direction passing through $O$ , perpendicularly to $CO$ , it will produce the same initial velocity of the centre as if the body were free: for as it exerts no pressure on the points of support, the initial motion must be the same as if they were not there.

45 If the external force be applied at a greater distance in the line $CG$ , the velocity of the centre will be greater than if the body were free. In this case the pressure excited in the axis will be backward, and consequently the points of support will react forward, and this reaction will be equivalent to another external force conspiring with the one applied at $O$ . Some curious consequences may be deduced from this.

46 Why this point is sometimes called the centre of percussion. If the external force be applied to a point in the line $GC$ lying beyond $C$ , the motion of the centre will be in the opposite direction to what it would have taken had the body been free, and so will be the pressures exerted by the points of support on the axis.

47 A force $m \cdot GI$ applied at $P$ produces the initial progressive motion $m \cdot GH$ ; and any force applied at $O$ , perpendicularly to $CG$ , produces the same motion of the centre as if the body were free. Therefore a force $m \cdot GH$ applied thus at $O$ will produce a motion $m \cdot GH$ in the centre, and therefore the same motion which $m \cdot GI$ applied at $P$ would produce; and it will produce the momentum $m \cdot GI$ at $P$ . Therefore if a force equal to the progressive motion of the body be applied at $O$ , perpendicularly to $CO$ , in the opposite direction, it will stop all this motion without exciting any strain on the axis or points of support. Therefore the equivalent of all the motions of each particle round $C$ is conceived as passing through $O$ in a direction perpendicular to $CO$ ; and the blow given by that point to any body opposed to its motion is considered as equal to the compounded effect of the rotatory motion, or to the progressive motion of the body combined with its rotation.

48 Improperity of the term. For such reasons $O$ has been called the CENTRE OF PERCUSSION of the body turning round $C$ . But the name of centre of momentum, or rotatory effort, would have been more proper.

We can feel this property of the point $O$ when we give a smart blow with a stick. If we give it a motion round the joint of the wrist only, and strike smartly

with a point considerably nearer or more remote than two-thirds of its length, we feel a painful shock or wrench in the hand; but if we strike with that point which is precisely at two-thirds of its length, we feel no such disagreeable strain.

Mechanical writers frequently say, that $O$ considered as the centre of percussion, is that with which the most violent blow is struck. But this is by no means true; $O$ is that point of a body turning round $C$ which gives a blow precisely equal to the progressive motion of the body, and in the same direction. As we have already said, it is the point where we may suppose the whole rotatory momentum of the body accumulated. Every particle of the body is moving in a particular direction, with a velocity proportional to its distance from the axis of rotation; and if the body were stopped in any point, each particle tending to continue its motion endeavours to drag the rest along with it. Whatever point we call the centre of percussion should have this property, that when it is stopped by a sufficient force, the whole motion and tendency to motion of every kind should be stopped; so that if at that instant the supports of the axis were annihilated, the body would remain in absolute rest.

The consideration of a very simple case will show that this point of stoppage cannot be taken indifferently. Suppose a square or rectangular board $CDD'C'$ , fig. 4, advancing in the direction $GH$ , perpendicular to its plane, without any rotation. Let $G$ be the centre of gravity, and the middle of the board. It is evident, that if a force be applied at $G$ , in the direction $HG$ , and equal to the quantity of motion of the board, all motion will be stopped: for when the point $G$ is stopped, no reason can be assigned why one part of the board shall advance more than another. The same thing must happen if the board be stopped by a straight edge put in its way, and passing through $G$ : for example, in the line $LGM$ , or $g G h$ . But if this edge be so placed that the board shall meet it with the line $IPK$ , then, because this line does not divide it equally, and because there is a greater quantity of motion in the part $CIKC'$ than in the part $IDD'K$ , though the progressive motion may be stopped, the upper part will advance, and a motion of rotation will commence, of which $IK$ will be the axis. Now suppose that the board, instead of having been moving along in the direction $GH$ , every part with the same velocity had been swinging round the axis $CC'$ like a pendulum, from the position $C d d' C'$ , and that it is stopped by a straight edge meeting it in the line $LGM$ parallel to $CO'$ , in the moment that it has attained the vertical position $CDD'C'$ ; all its motion will not be stopped: for, although $LGM$ divides the board equally, there is more motion in the lower part $LDD'M$ than in the upper part $CLMC'$ , because every particle of the lower part is describing larger circles and moving swifter. Therefore when the line $LGM$ is stopped, there will be a tendency of the lower part to advance, and the pivots $C$ and $C'$ of the axis will be pressed backwards on their holes; and if the holes were at that instant removed, a rotation would commence, of which $LM$ is the axis. The board must therefore be stopped in some line $IPK$ below $LGM$ , and so situated, that the sum of all the momenta on each side of it shall be equal. This alone

can hinder a rotation round the axis IPK. From what has been already demonstrated, it appears, that this will be prevented if the edge meets the board in a line IPK passing through O the centre of oscillation, which is situated in the line $g G h$ passing through the centre of gravity perpendicular to the axis $CC'$ . This line IOK may therefore be called the line or axis of percussion.

But any point of this line will not do. It is evident that if the board should meet the fixed edge in the line $g G O h$ , all motion will be stopped, for the motions on each side are equal, and neither can prevail. But if it be stopped in the line $p P q$ , there is more motion in the part $p q D C'$ than in the part $p q D C$ ; and if the supports at C and $C'$ were that instant taken away, there would commence a rotation round the axis $p q$ . Consequently, if the body were not stopped by an edge, but by a simple point at P, this rotation would take place. The motions above and below P would indeed balance each other, but the motions on the right and left sides of it would not. Therefore it is not enough for determining the centre of percussion that we have ascertained its distance $g O$ from the axis of rotation by the equation $g O = \frac{f p r^2}{m g G}$ . This equation only gives us the line IOK parallel to $CC'$ , but not the point of percussion. This point (suppose it P) must be such that if any line $p P q$ be drawn through it, and considered as an axis round which a rotation may commence, it shall not commence, because the sum of all the momenta round this axis on the right side is equal to the sum of the momenta on the left. Let us investigate in what manner this condition may be secured.

Let there be a body in a state of rotation round the axis $D d$ (fig. 5.), and let G be its centre of gravity; and CGO a line through the centre of gravity perpendicular to the axis $DC d$ . At the moment under consideration, the centre of gravity is moving in the direction GH, perpendicular to the radius vector GC, as also perpendicular to a plane passing through the lines $D d$ and CG. Let O be the centre of oscillation. Draw the line $n O$ parallel to $D d$ . The centre of percussion must be somewhere in this line. For the point of percussion, wherever it is, must be moving in the same direction with the progressive motion of the body, that is, in a direction parallel to GH, that is, perpendicular to the plane DCG. And its distance from the axis $D d$ must be the same with that of the centre of oscillation. These conditions require it therefore to be in some point of $n O$ . Suppose it at P. Draw $P p$ perpendicular to $D d$ . P must be so situated, that all the momenta tending to produce a rotation round the line $p P$ may balance each other, or their sum total be nothing.

Now let A be any particle of the body which is out of the plane DCG, in which lie all the lines CGO, $p P$ , $n O$ , &c. Draw its radius vector $A a$ perpendicular to $D a$ , and draw $a n$ parallel to CG, and therefore perpendicular to $D a$ . The plane $A a n$ is perpendicular to the plane $D a n$ (Euclid, XI. 4.). Draw $A L$ perpendicular to $A a$ , and $A l$ perpendicular to $a n$ . Then, while the body is beginning to turn round $D d$ , the incipient motion of the particle A is in the direction $A L$ , perpendicular to its radius vector $A a$ .

This motion $A L$ may be considered as compounded of the motion $A l$ , perpendicular to the plane DCG, and the motion $l L$ in this plane. It is evident that it is $A l$ only which is opposed by the external force stopping the body at P, because $A l$ alone makes any part of the progressive motion of the centre of gravity in the direction GH.

We have hitherto taken the radii vectores for the measures of the velocities or motions of the particles. Therefore the quantity of motion or the moving force of A is $A \cdot A a$ , and this is exerted in the direction $A L$ , and may be conceived as exerted on any point in this line, and therefore on the point L. That is, the point L might be considered as urged in this direction with the force $A \cdot A a$ , or with the two forces of which the force $A \cdot A a$ is compounded. The force in the direction $A L$ is to the force in the direction $A l$ as $A L$ to $A l$ , or as $a A$ to $a l$ , because the triangles $A l L$ and $a l A$ are similar. Therefore, instead of supposing the point L urged by the force $A \cdot A a$ , acting in the direction $A L$ , we may suppose it impelled by the force $A \cdot a l$ , acting perpendicularly to the line $A l$ , or to the plane DCG, and by the force $A \cdot A l$ acting in this plane, viz. in the direction $L n$ . This last force has nothing to do with the percussion at P. Therefore we need consider the point L as only impelled by the force $A \cdot A l$ . The momentum of this force, or its power to urge the plane DCG forward in the direction GH, by turning it round $D d$ , must be $A \cdot a l \cdot a L$ . (N. B. This is equal to $A \cdot A a^2$ , because $a l : a A = a L : a L$ , and $A \cdot A a^2$ , has been shown long ago to be the general expression of the rotatory momentum of a particle).

Draw $L m$ perpendicular to $P p$ . If we consider $P p$ as an axis about which a motion of rotation may be produced, it is plain that the momentum of the point L to produce such a rotation will be $A \cdot a l \cdot L m$ . In like manner, its momentum for producing a rotation round $n P$ would be $A \cdot a l \cdot L n$ . In general, its momentum for producing rotation round any axis is equal to the product of the perpendicular force at L (that is, $A \cdot a l$ ) and the distance of L from this axis.

In order therefore that P may be the centre of percussion, the sum of all the forces $A \cdot a l \cdot L m$ must be equal to nothing; that is, the sum of the forces $A \cdot a l \cdot L m$ on one side of this axis $P p$ must be balanced by the sum of forces $A \cdot a l \cdot L n$ on the other side. To express this in the usual manner, we must have $\int A \cdot a l \cdot L m = 0$ . But $n P = n O = O P$ . Therefore $\int A \cdot a l \cdot L m = \int A \cdot a l \cdot O P = 0$ , and $\int A \cdot a l \cdot L n = \int A \cdot a l \cdot O P$ . But $O P$ is the same wherever the particle A is situated; and because G is the centre of gravity, the sum of all the quantities is $A \cdot a l$ is $m \cdot G C$ , $m$ being the quantity of matter of the body; that is, $\int A \cdot a l = m \cdot G C$ , and $\int A \cdot a l \cdot O P = m \cdot G C \cdot O P = \int A \cdot a l \cdot n O$ . Hence we derive the final equation

O P = \frac{\int A \cdot a l \cdot n O}{m \cdot G C}.

Therefore the centre of percussion P of a body turning round the axis $D d$ is determined by these conditions: 1st, It is in the plane DCG passing through the axis and the centre of gravity; 2d, It is in a line $n O$ passing through the centre of oscillation, and parallel to the axis, and therefore its distance $P p$ from the axis of rotation.

rotation is $\frac{f A \cdot A a^2}{m \cdot CG}$ ; and, 3d, Its distance OP from the centre of oscillation is $\frac{f A \cdot a \cdot l \cdot n O}{m \cdot CG}$ .

In order therefore that the centres of oscillation and percussion may coincide, or be one and the same, OP must vanish, or $S A \cdot a \cdot l \cdot n O$ must be equal to nothing, that is, the sum of all the quantities $A \cdot a \cdot l \cdot n O$ on one side of the line CO must be equal to the sum of all the quantities $A \cdot a \cdot l \cdot n O$ on the other side.

Let $D d \delta \Delta$ be a plane passing through the axis $D d$ perpendicular to that other plane $DCG$ through it, in which the centre of gravity is situated, and let $C g \gamma \kappa$ be a third plane passing through the centre of gravity perpendicular to both the planes $D d \delta \Delta$ and $DCG$ . Draw $l r$ and $a a$ perpendicular to $a l$ , and $r a$ perpendicular to $c r$ , and then draw $A a$ , $A a$ perpendicular to $a a$ and $r a$ . It is evident that $A a$ and $A a$ are respectively equal to $a l$ and $l r$ , or to $a l$ and $n o$ ; so that the two factors or constituents of the momentum of a particle $A$ round the centre of percussion are the distances of the particles from the planes $D d \delta \Delta$ and $C g \gamma \kappa$ , both of which are perpendicular to that plane through the axis in which the centre of gravity is placed.

We may see, from these observations, that the centres of oscillation and percussion do not necessarily coincide, and the circumstance which is necessary for their coincidence, viz. that $f A \cdot A a \cdot A a$ is equal to $O$ . It is of importance to keep this in mind.

There occurs here another observation of great importance. Since every force is balanced by an equal force acting in the opposite direction, and since all motion progressive and rotatory is stopped by an external force applied at $P$ in the direction $q P$ , it follows that, if the body were at rest; and the same force be applied there, it will set the body in rotation round the axis $D d$ , in the opposite direction, with the same angular velocity, and without any pressure on the pivots $D$ and $d$ . For whatever motion of the particle $A$ , in the direction $A L$ , was stopped by a part of the external force applied at $P$ , the same motion will be produced by it in the quiescent particle $A$ in the opposite direction $L A$ . And as the pivots $D$ and $d$ had no motion in the case of the body turning round them, they will acquire no motion, or will have no tendency to motion, or no pressure will be exerted on them, in the last case. Therefore when an external force is applied at $P$ in a direction perpendicular to the line $P p$ , the line $D d$ will become a momentary spontaneous axis of conversion, and the incipient motion of the body will perfectly resemble the rotation of the same body round a fixed axis $D d$ .

There is another set of forces of which we have as yet taken no notice, viz. that part of each force $A L$ , which is directed along the plane $DCG$ , and is represented by $l L$ when the whole force is represented by $A L$ , or by $A l$ when the whole force is represented by $A a$ . These forces being all in the plane $DCG$ , and in the direction $C G$ or $G C$ , can have no effect on the rotation round any axis in that plane. But they tend, separately, to produce rotation round any axis passing through this plane perpendicularly. And the momentum of $A$ to produce a rotation round an axis

perpendicular to this plane, in $O$ for instance, must evidently be $A \cdot A \cdot l \cdot n O$ , and round $P$ it must be $A \cdot A \cdot l \cdot n P$ , &c. We shall have occasion to consider these afterwards.

It is usual in courses of experimental philosophy to illustrate the motions of bodies on inclined planes and curved surfaces by experiments with balls rolling down these surfaces. But the motions of such rolling balls are by no means just representations of the motions they represent. The ball not only goes down the inclined plane by the action of gravity, but it also turns round an axis. Force is unnecessary for producing this rotation; and as there is no other source but the weight of the ball, part of this weight is expended on the rotation, and the remainder only accelerates it down the plane. The point of the ball which rests on the plane is hindered from sliding down by friction; and therefore the ball tumbles, as it were, over this point of contact, and is instantly caught by another point of contact, over which it tumbles in the same manner. A cylinder rolls down in the very same way; and its motion is nearly the same as if a fine thread had been lapped round it, and one end of it made fast at the head of the inclined plane. The cylinder rolls down by unwinding this thread.

The mechanism of all such motions (and some of them are important) may be understood by considering them as follows: Let a body of any shape be connected with a cylinder $FCB$ (fig. 6.) whose axis passes through $G$ the centre of gravity of the body. Suppose that body suspended from a fixed point $A$ by a thread wound round the cylinder. This body will descend by the action of gravity, and it will also turn round, unwinding the thread. Draw the horizontal line $OGC$ . It will pass through the point of contact $C$ of the thread and cylinder, and $C$ is the point round which it begins to turn in descending. Let $O$ be its centre of oscillation corresponding to the momentary centre of rotation $C$ . It will begin to descend in the same manner as if all its matter were collected in $O$ : for it may be considered, in this instant, as a pendulum suspended at $C$ . But in this case $O$ will descend in the same manner as if the body were falling freely. Therefore the velocity of $G$ (that is, the velocity of descent) will be to the velocity with which a heavy body would fall as $CG$ to $CO$ . Now since the points $C$ , $G$ , $O$ , are always in a horizontal line, and the radius $CG$ is given; as also $CO$ (No 48.) the velocity of a body falling freely, and of the body unwinding from this thread, will always be in the same proportion of $CO$ to $CG$ , and so will the spaces described in any given time. And thus we can compare their motions in every case when we know the place of the centre of oscillation.

Cor. 1. The weight of the descending body will be to the tension of the thread as $CO$ to $GO$ : for the tension of the thread is the difference between the momentum of the rolling body and that of the body falling freely.

Observe, that this proportion between the weight of the body and the tension of the thread will be always the same: for it has been demonstrated already, No 42, that if $C$ be in the circumference of a circle whose centre is $G$ , $O$ will be in the circumference of another circle

Rotation. circle round the same centre, and therefore the ratio of CG to CO is constant.

61 Cor. 2. If a circular body FCB roll down an inclined plane by unfolding a thread, or by friction which prevents all sliding, the space described will be to that which the body would describe freely as CG to CO: for the tendency down the inclined plane is a determined proportion of the weight of the body. The motion of rotation in these cases, both progressive and whirling, is uniformly accelerated.

62 Case of pendulous bodies. Something of the same kind obtains in common pendulous bodies. A ball hung by a thread not only oscillates, but also makes part of a rotation; and for this reason its oscillations differ from those of a heavy point hanging by the same thread, and the centre of oscillation is a little below the centre of the ball. A ball hung by a thread, and oscillating between cycloidal cheeks, does not oscillate like a body in a cycloid, because its centre of oscillation is continually shifting its place. Huyghens avoided this by suspending his pendulous body from two points, so that it did not change its attitude during its oscillation. If our spring-carriages were hung in this manner, having the four lower staples to which the straps are fixed as far asunder as the four upper staples at the ends of the springs, the body of the carriage would perform its oscillations without kicking up and down in the disagreeable manner they now do, by which we are frequently in danger of striking the glasses with our heads. The swings would indeed be greater, but incomparably easier; and we could hold things almost as steadily in our hand as if the carriage were not swinging at all.

This will suffice for an account of the rotation round fixed axes, as the foundation for a theory of machines actually performing work. The limits of our undertaking will not allow us to do any more than just point out the method of applying it.

63 Method of applying the theory of rotation in practice. Let there be any machine of the rotatory kind, i. e. composed of levers or wheels, and let its construction be such, that the velocity of the point to which the power is applied (which we shall call the impelled point) is to the velocity of the working point in the ratio of $m$ to $n$ . It is well known that the energy of this machine will be the same with that of an axis in peritrochio, of which the radii are $m$ and $n$ .

Let $p$ express the actual pressure exerted on the impelled point by the moving power, and let $r$ be the actual pressure or resistance exerted on the working point by the work to be performed. Let $x$ be the inertia of the power, or the quantity of dead matter which must move with the velocity of the impelled point in order that the moving power may act. Thus the moving power may be the weight of a bucket of water in a water-wheel; then $x$ is the quantity of matter in this bucket of water. Let $y$ in like manner be the inertia of the work, or matter which must be moved with the velocity of the working-point, in order that the work may be performed. Thus $y$ may be a quantity of water which must be continually pushed along a pipe. This is quite different from the weight of the water, though it is proportional to it, and may be measured by it.

Let $f$ be a pressure giving the same resistance when applied at the working-point with the friction of the machine, and let $an^2$ be the momentum of the machine's

inertia, viz. the same as if a proper quantity of matter were attached to the working-point, or to any point at the same distance from the axis.

This state of things may be represented by the wheel and axle PQS (fig. 7.) where $x$ and $y$ and $a$ are represented by weights acting by lines. P is the impelled point, and R the working-point; CP is $m$ and CR is $n$ . The moving force is represented by PA, the resistance by RB, and the friction by BF.

It is evident that the momentum of the inertia of $x$ , $y$ , and $a$ are the same as if they were for a moment attached to the points P and R.

Hence we derive the following expressions:

work is proportional to the product of the resistance and the velocity with which it is overcome.

We shall give a very simple example of the utility of these formulae. Let us suppose that water is to be raised in a bucket by the descent of a weight, and that the machine is a simple pulley. Such a machine is described by Defagulier, who says he found it preferable to all other machines. The bucket dipped itself in the cistern. A chain from it went over a pulley, and at its extremity was a flag on which a man could step from the head of a stair. His preponderance brought down the flag and raised the bucket, which discharged its water into another cistern. The man quit the flag, and walked up stairs, and there he found it ready to receive him, because the empty bucket is made heavier than the empty flag.

Now, if there be no water in the bucket, it is evident, that although the motion of the machine will be the quickest possible, there will be no work performed. On the other hand, if the loaded flag and the full bucket are of equal weight, which is the usual statement of such a machine in elementary treatises of mechanics, the machine will stand still, and no work will be performed. In every intermediate state of things the machine will move, and work will be performed. Therefore the different values of the work performed must be a series of quantities which increase from nothing to a certain magnitude, and then diminish to nothing again. The maxim which is usually received as a fundamental proposition in mechanics, viz. that what is gained in force by the intervention of a machine is lost in time, is therefore false. There must be a particular proportion of the velocities of the impelled and working-points, which will give the greatest performance when the power and resistance are given; and there is a certain proportion of the power and resistance which will have the same effect when the structure of the machine has previously fixed the velocities of the impelled and working-points.

This proportion will be found by treating the formula which expresses the work as a fluxionary quantity, and finding its maximum. Thus, when the ratio of the power and resistance is given, and we wish to know what must be the proportion of the velocities.

Rotation. $m$ and $n$ , that we may construct the machine accordingly, we have only to consider $n$ as the variable quantity in the third formula. This gives us

n = m \times \frac{\sqrt{x^2 + r^2 + f^2 + p^2 x a + y^2} - x r + f}{p a + y}

This is a fundamental proposition in the theory of working machines: but the application requires much attention. Some natural powers are not accompanied by any inertia worth minding; in which case $x$ may be omitted. Some works, in like manner, are not accompanied by any inertia; and this is a very general case. In many cases the work exerts no contrary strain on the machine at rest, and $r$ is nothing. In most instances the intensity of the power varies with the velocity of the impelled point, and is diminished when this increases; the resistance or actual pressure at the working-point frequently increases with the velocity of the working-point. All these circumstances must be attended to; but still they only modify the general proposition. These are matters which do not come within the limits of the present article. We only took this opportunity of showing how imperfect is the theory of machines in equilibrio for giving us any knowledge of their performance or just principles of their construction.

One thing, however, must be particularly attended to in this theory. The forces which are applied to the body moveable round an axis are considered in the theory as pressures actually exerted on the impelled points of the body or machine, as when a weight is appended to a lever or wheel and axle, and, by descending uniformly, acts with its whole weight. In this case the weight multiplied by its distance from the axis will always express its momentum, and the rotation will (ceteris paribus) be proportional to this product. But in many important cases our machines are actuated by external impulses. A body in motion strikes on the impelled point of the machine, and causes it to turn round its axis. It is natural for us to consider the quantity of motion of this impelling body as the measure of our moving force. Supposing $n$ to be its quantity of matter, and $V$ its velocity, $nV$ appears a very proper measure of its intensity. And if it be applied at the distance $CP$ from the axis of rotation, $nV \cdot CP$ should express its energy, momentum, or power to turn the machine round $C$ ; and we should express the angular velocity by $\frac{nV \cdot CP}{f p r^2}$ . Accordingly, this is the

tion will therefore certainly be greater if $CP$ be shorter. Rotation. We need not examine the case more minutely.

We must therefore carefully distinguish between the quantity of motion of the impelling body and its moving power, as it is modified by its manner of acting. The moving power is the pressure actually exerted on the impelled point of the machine. Now the universal fact of the equality of action and reaction in the collision of bodies assures us, that their mutual pressure in their collision is measured by the change of motion which each sustains: for this change of motion is the only indication and measure of the pressure which we suppose to be its cause. A way therefore of ascertaining what is the real moving force on a machine actuated by the impulse of a moving body, is to discover what quantity of motion is lost by the body or gained by the machine; for these are equal. Having discovered this, we may proceed according to the propositions of rotatory motion.

Therefore let $AEF$ (fig. 8.) represent a body moveable round an axis passing through $C$ , perpendicular to the plane of the figure. Let this body be struck in the point $A$ by a body moving in the direction $FA$ , and let $BAD$ be a tangent to the two bodies in the point of collision. It is well known that the mutual actions of two solid bodies are always exerted in a direction perpendicular to the touching surfaces. Therefore the mutual pressure of the two bodies is in the direction $AP$ perpendicular to $AD$ . Therefore let the motion of the impelling body be resolved into the directions $AP$ and $AD$ . The force $AD$ has no share in the pressure. Therefore let $V$ be the velocity of the impelling body estimated in the direction $AP$ , and let $n$ be its quantity of matter. Its quantity of motion in the direction $AP$ will be $nV$ .

Did $AP$ pass through $C$ , it is evident that the only effect would be to press the axis on its supports. But $AP$ , the direction of the pressure, being inclined to $AC$ , the point $A$ is forced aside, and in some small moment of time describes the little arch $Aa$ round the centre $C$ . The point $P$ will therefore describe a small arch $Pp$ , subtending an angle $PCp = ACa$ . Draw $ao$ perpendicular to $AP$ , and $ad$ perpendicular to $AD$ . The triangles $dAo$ , $ACP$ are similar, and $Aa : Aa = AC : CP$ . But the angles $ACa$ , $PCp$ being equal, the arches are as their radii, and $Aa : Pp = AC : CP = Aa : Ao$ , therefore $Pp = Ao$ .

Now since, in consequence of the impulse, $A$ describes $Aa$ in the moment of time, it is plain that $Ao$ is the space through which the impelling body continues to advance in the direction of the pressure; and if $V$ be taken equal to the space which it described in an equal moment before the stroke, $v$ will express the remaining velocity, and $V-v$ is the velocity lost, and $n(V-v)$ is the quantity of motion lost by the impelling body, and is the true measure of the pressure exerted. This gives us the whole circumstances of the rotatory motion. The

angular velocity will be $\frac{n(V-v) \cdot CP}{f p r^2}$ , and the velocity

of the point $A$ will be $\frac{n(V-v) \cdot CP \cdot CA}{f p r^2}$ . Call this velocity $u$ . The similarity of triangles gives us $CA :$

$CP = Aa$ (or $u$ ) : $Ao$ (or $v$ ) and $u = \frac{v \cdot CA}{CP}$ . There-

Rotation. fore $\frac{V \cdot CA}{CP} = \frac{n(V-v)CP \cdot CA}{fp^2}$ . From this we deduce

$v = \frac{n \cdot V \cdot CP^2}{fp^2 + n \cdot CP^2}$ , and thus we have obtained the value of $v$ in known quantities; for $n$ was given, or supposed known; so also was $V$ : and since the direction $FA$ was given, its distance $CP$ from the axis is given; and the form of the body being known, we can find the value of $fp^2$ . Now we have seen that $v$ is also the velocity of the point $P$ ; therefore we know the absolute velocity of a given point of the body or machine, and consequently the whole rotatory motion.

73 We have the angular velocity $= \frac{n \cdot V \cdot CP}{fp^2 + n \cdot CP^2}$ : we shall find this a maximum when $fp^2 = n \cdot CP^2$ ; and in this case $CP = \sqrt{\frac{fp^2}{n}}$ , and $v = \frac{1}{2}V$ . So that the greatest velocity of rotation will be produced when the striking body loses $\frac{1}{2}$ of its velocity.

74 Authors treating of the application of that theory recommended. What we have now delivered is sufficient for explaining all the motions of bodies turning round fixed axes; and we presume it to be agreeable to our readers, that we have given the investigation of the centres of gyration, oscillation, and percussion. The curious reader will find the application of these theorems to the theory of machines in two very valuable dissertations by Mr Euler in the Memoirs of the Academy of Berlin, vol. viii. and x. and occasionally by other authors who have treated mechanics in a scientific and useful manner, going beyond the school-boy elements of equilibrium.

75 Of the rotation of free bodies. There remains a very important case of the rotation of bodies, without which the knowledge of the motion of solid bodies is incomplete; namely, the rotation of free bodies, that is, of bodies unconnected with any fixed points. We hardly see an instance of motion of a free body without some rotation. A stone thrown from the hand, a ball from a cannon, the planets themselves, are observed not only to advance, but also to whirl round. The famous problem of the precession of the equinoxes depends for its solution on this doctrine; and the theory of the working of ships has the same foundation. We can only touch on the leading propositions.

76 We need not begin by demonstrating, that when the direction of the external force passes through the centre of the body, the body will advance without any rotation. This we consider is familiarly known to every person versant in mechanics; nor is it necessary to demonstrate, that when the direction of the moving force does not pass through the centre of gravity, this centre will still advance in a direction parallel to that of the moving force, and with the same velocity as if the direction of the moving force had passed through it. This is the immediate consequence of the equality of action and reaction observed in all the mechanical phenomena of the universe.

But it is incumbent on us to demonstrate, that when the direction of the moving force does not pass through the centre of gravity, the body will not only advance in the direction of the moving force, but will also turn round an axis, and we must determine the position of this axis, and the relation subsisting between the progressive and rotatory motions.

Rotation. The celebrated John Bernoulli was the first who considered this subject; and in his Disquisitiones Mechanico-dynamicae, he has demonstrated several propositions concerning the spontaneous axis of conversion, and the motions arising from eccentric external forces: and although he assumed for the leading principle a proposition which is true only in a great number of cases, he has determined the rotation of spherical bodies with great accuracy.

This combination of bodies will be palpable in some simple cases, such as the following: Let two equal bodies $A$ and $B$ (fig. 9.) be connected by an inflexible rod (of which we may neglect the inertia for the present). Let $G$ be the middle point, and therefore the centre of gravity. Let an external force act on the point $P$ in the direction $FP$ perpendicular to $AB$ , and let $AP$ be double of $PB$ . Also let the force be such, that it would have caused the system to have moved from the situation $AB$ to the situation $ab$ , in an indefinitely small moment of time, had it acted immediately on the centre $G$ . $G$ would in this case have described $Gg$ . $A$ would have described $Aa$ , and $B$ would have described $Bb$ , and $ab$ would have been parallel to $AB$ : for the force impressed on $A$ would have been equal to the force impressed on $B$ ; but because the force acts on $P$ , the force impressed on $A$ is but one half of that impressed on $B$ by the property of the lever: therefore the initial motion or acceleration of $A$ will be only half of the initial motion of $B$ ; yet the centre $G$ must still be at $g$ . We shall therefore ascertain the initial motion of the system, by drawing through $g$ a line $\alpha\beta$ , so that $A\alpha$ shall be $\frac{1}{2}$ of $B\beta$ . This we shall do by making $AC = AB$ , and drawing $C\alpha\beta$ . Then $\alpha\beta$ will be the position of the system at the end of the moment of time. Thus we see that the body must have a motion of rotation combined with its progressive motion.

77 How motion is performed in these cases, And we deduce immediately from the premises that this rotation is performed round an axis passing through the centre of gravity $G$ : for since the centre describes a straight line, it is never either above or below the axis of rotation, and is therefore always in it. This is a fundamental theorem, and our subsequent investigation is by this means greatly simplified, being thus reduced to two problems: 1. To determine in what direction the axis passes through the centre of gravity. 2. To determine the angular velocity of the rotation, or how far the centre must advance while the body makes one turn round the axis. This establishes the relation between the progressive and rotatory motions. It will contribute to our better conception of both these problems to see the result in the present simple case.

78 It is evident, in the first place, that the impressions made on $A$ and $B$ are in lines $Aa$ , $Bb$ parallel to $FP$ and $Gg$ ; and therefore the motions of the points $A$ , $G$ , and $B$ , are made in one plane, viz. the plane $FPG$ . The axis of rotation therefore must be a line drawn through $G$ , perpendicular to this plane. If we give it any other position, one of the points $A$ , $B$ , or both of them, must quit this plane.

In the next place, in $ba$ produced take $bc = BC$ . Then supposing $AC$ to be a rigid line connected with the system, it is evident that if there had been no rotation, the line $BC$ would have kept parallel to its first position, and that at the end of the moment of time $C$ would

Rotation. would have been at $c$ . The point $C$ therefore has had, by the rotation, a backward motion $cC$ , relative to the centre $G$ or $g$ , and this motion is equal to the progressive motion $Gg$ of the centre; therefore if we make $Gg$ equal to the circumference of a circle whose radius is $CG$ , the body will make one rotation round the centre of gravity, while this centre moves along $Gg$ ; and thus the relation is established between the two motions.

Rotation. fion, and the motion $Ah$ of rotation round $Gg$ , are equivalent to, and really constitute, a momentary simple motion of rotation round the axis $Cc$ given in position, that is, determinable by the ratio of $Ae$ to $Ah$ .

On the other hand, the converse proposition is, that a simple motion of rotation round a fixed axis $Cc$ , such that the centre $G$ has the velocity and direction $GI$ perpendicular to $CG$ , is equivalent to, and produces a motion of rotation round an axis $Gg$ , along with the progressive motion $GI$ of this axis. This proposition is demonstrated in the very same way, from the consideration that, by the rotation round $Cc$ , we have $cA : cg = Af : gi$ . From this we deduce, that $Ah$ is perpendicular to $Ag$ , and that $fh : Ah = cg : gA$ ; and thus we resolve the motion $Af$ into a motion $Ah$ of rotation round $Gg$ , and a motion $Ae$ of progression common to the whole body.

But let us not confine the progressive motion to the direction perpendicular to the axis $Gg$ . Let us suppose that the whole body, while turning round $Gg$ , is carried forward in the direction and with the velocity $GK$ . We can always conceive a plane $LGC$ , which is perpendicular to the plane in which the axis $Gg$ and the direction $GK$ of the progressive motion are situated. And the motion $GK$ may be conceived as compounded of a motion $GI$ perpendicular to this plane and to the axis; and a motion of translation $GL$ , by which the axis slides along in its own direction. It is evident, that in consequence of the first motion $GI$ , there arises a motion of rotation round $Cc$ . It is also evident, that if, while the body is turning for a moment round $Cc$ , this line be slid along itself in the direction $cC$ , a motion equal to $GL$ will be induced on every particle $A$ , and compounded with its motion of rotation $AF$ , and that if $f\phi$ be drawn equal and parallel to $GL$ , $\phi$ will be the situation of the particle $A$ when $G$ is in $K$ .

And thus it appears, that when the progressive motion is perpendicular to the axis of rotation passing through the centre of gravity, the two motions progressive and rotatory are equivalent to a momentary simple motion of rotation round a spontaneous axis of conversion, which is at rest: but when the progressive motion is inclined to the axis passing through the centre, the spontaneous axis of conversion is sliding in its own direction.

We may conceive the whole of this very distinctly and accurately by attending to the motion of a garden roller. We may suppose it six feet in circumference, and that it is dragged along at the rate of three feet in a second from east to west, the axis of the roller lying north and south. Suppose a chalk line drawn on the surface of the roller parallel to its axis. The roller will turn once round in two seconds, and this line will be in contact with the ground at the intervals of every six feet. In that instant the line on the roller now spoken of is at rest, and the motion is the same as if it were fixed, and the roller really turning round it. In short, it is then a spontaneous axis of conversion.

Now, suppose the roller dragged in the same manner and in the same direction along a sheet of ice, while the ice is floating to the south at the rate of four feet in a second. It is now plain that the roller is turning round an axis through its centre of gravity, while the centre is carried in the direction $f36^{\circ} 52' W.$ at the rate of five

But farther, the point $C$ has, in fact, not moved out of its place. The incipient motion has therefore been such, that $C$ has become a spontaneous centre of conversion. It is easy to see that this must always be the case, whatever may be the form of the rigid body or system of particles connected by inflexible and inextensible lines. Since the system both advances and turns round an axis passing through its centre of gravity, there must be some point in the system, or which may be conceived as connected with it by an inflexible line, which moves backward, by the rotation, as fast as the centre advances forward. A line drawn through this point parallel to the axis must in this instant be at rest, and therefore must be a spontaneous axis of conversion. And, in this instant, the combined motions of rotation round an axis passing through the centre of gravity and the motion of progression, are equivalent to, and actually constitute, an incipient simple motion of rotation round another axis parallel to the former, whose position may be ascertained. But it is necessary to establish this proposition and its converse on clearer evidence.

Therefore let $G$ (fig. 10.) be the centre of gravity of a rigid system of particles of matter, such as we suppose a solid body to be. Let this system be supposed to turn round the axis $Gg$ , while the axis itself is moving forward in the direction and with the velocity $GI$ . Let the rotation be such, that a particle $A$ has the direction and velocity $Ah$ . Let us first suppose the progressive motion $GI$ to be perpendicular to the axis $Gg$ . It will therefore be parallel to the planes of the circles described round the axis by the different particles. Let $CGg$ be a plane perpendicular to $GI$ . It will cut the plane of the circle described by $A$ in a straight line $cg$ , and $g$ will be the centre round which $A$ is turning. Therefore $Ag$ will be the radius vector of $A$ , and $Ah$ is perpendicular to $Ag$ . Let $Ad$ be perpendicular to $cg$ , and in $Ad$ take $Ae$ equal to $GI$ or $gi$ . It is evident, that the absolute motion of $A$ is compounded of the motions $Ae$ and $Ah$ , and is the diagonal $Af$ of the parallelogram $Aefh$ . In the line $gc$ , which is perpendicular to $Gg$ , take $ge$ to $gA$ , as $Ae$ to $Ah$ , and draw $eC$ parallel to $gC$ , and produce $hA$ till it cut $eg$ in $n$ . We say that $Cc$ is in this moment a spontaneous axis of conversion; for, because $An$ is perpendicular to $Ag$ and $Ad$ to $Cg$ , the angle $cgA$ is equal to $dAn$ , or $fhA$ . Therefore, since $cg : gA = fh : hA$ , the triangles $cgA$ and $fhA$ are similar, and the angle $gAc$ is equal to $hAf$ . Take away the common angle $gAf$ , and the remaining angle $cAf$ is equal to the remaining angle $hAg$ , and $Af$ is perpendicular to $Ae$ , and the incipient motion of $A$ is the same in respect of direction as if it were turning round the axis $Cc$ . Moreover, $Af$ is to $fh$ or $gi$ as $Ae$ to $cg$ . Therefore, both the direction and velocity of the absolute motion of $A$ is the same as if the body were turning round the fixed axis $Cc$ ; and the combined motion $Ae$ of progres-

five feet per second. It is also plain, that when the line drawn on the surface of the stone is applied to the ice, its only motion is that which the ice itself has to the southward. The motion is now a motion of rotation round this spontaneous axis of conversion, compounded with the motion of four feet per second in the direction of this axis. And thus we see that any complication of motion of rotation round an axis passing through the centre of gravity, and a motion of progression of that centre, may always be reduced to a momentary or incipient motion of rotation round another axis parallel to the former, compounded with a motion of that axis in its own direction.

The demonstration which we have given of these two propositions points out the method of finding the axis $Cc$ , the incipient rotation round which is equivalent to the combined progressive motion of the body, and the rotation round the axis $Gg$ . We have only to note the rotatory velocity $Ah$ of some particle $A$ , and its distance $Ag$ from the axis, and the progressive velocity $GI$ of the whole body, and then to make $GC$ a fourth proportional to $Ah$ , $GI$ , and $gA$ , and to place $GC$ in a plane perpendicular to $GI$ , which is perpendicular to $Gg$ , and to place $C$ on that side of $Gg$ which is moving in the opposite direction to the axis.

In the simple case of this problem, which we exhibited in order to give us easy and familiar notions of the subject, it appeared that the retrograde velocity of rotation of the point $C$ was equal to the progressive velocity of the centre. This must be the case in every point of the circumference of the circle of which $CG$ , fig. 9, is the radius. Therefore, as the body advances, and turns round $G$ , this circle will apply itself in succession to the line $CK$ parallel to $Gg$ ; and any individual point of it, such as $C$ , will describe a cycloid of which this circle is the generating circle, $CK$ the base, and $CG$ half the altitude. The other points of the body will describe trochoids, elongated or contracted according as the describing points are nearer to or more remote from $G$ than the point $C$ is.

It is now evident that all this must obtain in every case, as well as in this simple one. And when we have ascertained the distance $GC$ between the axis of rotation passing through the centre, and the momentary spontaneous axis of conversion passing through $C$ , we can then ascertain the relation between the motions of rotation and progression. We then know that the body will make one rotation round its central axis, while its centre moves over a space equal to the circumference of a circle of a known diameter.

We must therefore proceed to the methods for determining the position of the point $C$ . This must depend on the proportion between the velocity of the general progressive motion, that is, the velocity of the centre, and the velocity of some point of the body. This must be ascertained by observation. In most cases which are interesting, we learn the position of the axis, the place of its poles, the comparative progressive velocity of the centre, and the velocity of rotation of the different points, in a variety of ways; and it would not much increase our knowledge to detail the rules which may be followed for this purpose. The circumstance which chiefly interests us at present is to know how these motions may be produced; what force is necessary, and how it must be applied, in order to produce a

given motion or rotation and progression; or what will be the motion which a given force, applied in a given manner, will produce. Rotation.

We have already given the principles on which we may proceed in this investigation. We have shown the circumstances which determine the place of the centre of percussion of a body turning round a given fixed axis. This centre of percussion is the point of the body where all the inherent forces of the whirling body precisely balance each other, or rather where they unite and compose one accumulated progressive force, which may then be opposed by an equal and opposite external force. If, therefore, the body is not whirling, but at rest on this fixed axis, and if this external force be applied at the centre of percussion, now become a point of impulsion, a rotation will commence round the fixed axis precisely equal to what had been stopped by this external force, but in the opposite direction; or, if the external force be applied in the direction in which the centre of percussion of the whirling body was moving at the instant of stoppage, the rotation produced by this impulse will be the same in every respect. And we found that in the instant of application of this external force, either to stop or to begin the motion, no pressure whatever was excited on the supports of the axis, and that the axis was, in this instant, a spontaneous axis of conversion.

Moreover, we have shown, art. 84, that a rotation round any axis, whether fixed or spontaneous, is equivalent to, or compounded of, a rotation round another axis parallel to it, and passing through the centre of gravity, and a progressive motion in the direction of the centre's motion at the instant of impulse.

Now, as the position of the fixed axis, and the known disposition of all the particles of the body with respect to this axis, determines the place of the centre of percussion, and furnishes all the mathematical conditions which must be implemented in its determination, and the direction and magnitude of the force which is produced and exerted at the centre of percussion; so, on the other hand, the knowledge of the magnitude and direction of an external force which is exerted on the point of impulsion of a body not connected with any fixed axis, and of the disposition of all the parts of this body with respect to this point of impulsion, will furnish us with the mathematical circumstances which determine the position of the spontaneous axis of conversion, and therefore determine the position of the axis through the centre (parallel to the spontaneous axis of conversion), round which the body will whirl, while its centre proceeds in the direction of the external force.

The process, therefore, for determining the axis of Mode of progressive rotation is just the converse of the process determining the axis of progressive rotation for determining the centre of percussion. 89

John Bernoulli was the first who considered the motion of free bodies impelled by forces whose line of direction did not pass through their centre of gravity; and converse of what he takes it for granted, that since the body both advances and turns round an axis passing through the centre of gravity, this axis is perpendicular to the plane passing through the direction of the force, and through the centre of impulsion and the centre of gravity. Other authors of the first name, such as Huyghens, Leibnitz, Roberval, &c. have thought themselves obliged to demonstrate this. Their demonstration is as follows:

Let a body whose centre of gravity is $G$ (fig. 11.) be impelled at the point $P$ by a force acting in the direction $PQ$ not passing through the centre. The inertia of the whole body will resist in the same manner as if the whole matter were collected in $G$ , and therefore the resistance will be propagated to the point $P$ in the direction $GP$ . The particle $P$ , therefore, is impelled in the direction $PQ$ , and resisted in the direction $PA$ , and must therefore begin to move in some direction $PB$ , which makes the diagonal of a parallelogram of which the sides have the directions $PQ$ and $PA$ . The diagonal and sides of a parallelogram are in one plane. $P$ is therefore moving in the plane $APQB$ or $GPQ$ , and it is turning round an axis which passes through $G$ .—Therefore this axis must be perpendicular to the plane $GPQ$ .

It would require a series of difficult propositions to show the fallacy of this reasoning in general terms, and to determine the position of the axis through $G$ . We shall content ourselves with a very simple case, where there can be no hesitation. Let $A$ and $B$ (fig. 12.) be two equal balls connected with the axis $ab$ by inflexible lines $Aa$ , $Bb$ , perpendicular to $ab$ . Let $Aa$ be 1, and $Bb$ 2. The centre of gravity $G$ will evidently be in the line $cG$ parallel to $Aa$ and $Bb$ , and in the middle of $ab$ , and $cG$ is $\frac{1}{2}$ . Let $O$ be the centre of oscillation. $cO$ is $= \frac{A \cdot Aa^2 + B \cdot Bb^2}{A + B \cdot cG} = \frac{1}{2}$ .

Draw $Am$ , $Bn$ perpendicular to $cG$ , and suppose the balls transferred to $m$ and $n$ . The centre of oscillation will be still at $O$ ; and we see that if the system in this form were stopped at $O$ , all would be in equilibrium. For the force with which the ball $A$ arrives (by swinging round the axis) at $m$ , is as its quantity of matter and velocity jointly, that is $A \cdot Aa$ , or 1. That of $B$ arriving at $n$ is $B \cdot Bb$ , or 2. The arm $mO$ of the lever turning round $O$ is $\frac{1}{2}$ , and the arm $nO$ is $\frac{1}{2}$ . The forces, therefore, are reciprocally as the arms of the lever on which they act, and their momenta, or powers to turn the line $mn$ round $O$ , are equal and opposite, and therefore balance each other; and therefore, at the instant of stopping, no pressure is exerted at $c$ . Therefore, if any impulse is made at $O$ , the balls at $m$ and $n$ will be put in motion with velocities 1 and 2, and $c$ will be a spontaneous centre of conversion. Let us see whether this will be the case when the balls are in their natural places $A$ and $B$ , or whether there will be any tendency to a rotation round the axis $cO$ . The momentum of $A$ , by which it tends to produce a rotation round $cO$ is $A \cdot Aa \cdot Am = 1 \times Am$ . That of $B$ is $B \cdot Bb \cdot Bn = 2 \times Bn$ . $Am$ and $Bn$ are equal, and therefore the momentum of $B$ is double that of $A$ , and there is a tendency of the system to turn round $cO$ ; and if, at the instant of stoppage, the supports of the axis $ab$ were removed, this rotation round $cO$ would take place, and the point $b$ would advance, and $a$ would recede, $c$ only remaining at rest. Therefore, if an impulse were made at $O$ , $ab$ would not become a spontaneous momentary axis of conversion, and $O$ is not the centre of percussion. This centre must be somewhere in the line $OP$ parallel to $ab$ , as at $P$ , and so situated that the momenta $A \cdot Aa \cdot Aa$ and $B \cdot Bb \cdot Bb$ may be equal, or that $Aa$ may be double of $Bb$ , or $ap$ double of $bp$ . If an impulse be now made at $P$ , the balls $AB$ will be urged by forces as 1 and 2, and

therefore will move as if round the axis $ab$ , and there will be no pressures produced at $a$ and $b$ , and $ab$ will really become a momentary spontaneous axis of conversion.

Now join $G$ and $P$ . Here then it is evident that a body or system $A, B$ , receiving an impulse at $P$ perpendicular to the plane $acG$ , acquires to itself a spontaneous axis of conversion which is not perpendicular to the line joining the point of impulsion and the centre of gravity. And we have shown, in art. 84. that this motion round $ab$ is compounded of a progressive motion of the whole body in the direction of the centre, and a rotation round an axis passing through the centre parallel to $ab$ . Therefore, in this system of free bodies, the axis of rotation is not perpendicular to the plane passing through the centre of gravity in the direction of the impelling force.

As we have already observed, it would be a laborious task to ascertain in general terms the position of the progressive axis of rotation. Although the process is the inverse of that for determining the centre of percussion when the axis of rotation is given, it is a most intricate business to convert the steps of this process. The general method is this: The momentum of a particle $A$ (fig. 5.) by which it tends to change the position of the axis $Dd$ , has for its factors $Aa \cdot A\Delta$ , and $A\Delta$ , which are its distances from three planes $Dd\Delta$ , $DCO$ , and $Cg\gamma\kappa$ , given in position. The sum of all these must be equal to nothing, by the compensation of positive and negative quantities. We must find three other planes (of which only one is in some measure determined in position, being perpendicular to $DCO$ ), so situated that the sums of similar products of the distances of the particles from them may in like manner be equal to nothing. This is a very intricate problem; so intricate, that mathematicians have long doubted and disputed about the certainty of the solutions. Euler, d'Alembert, Frisi, Landen, and others, have at last proved, that every body, however irregular its shape, has at least three axes passing through its centre of gravity, round which it will continue to revolve while proceeding forward, and that these are at right angles to each other; and they have given the conditions which must be implemented in the determination of these axes. But they still leave us exceedingly at a loss for means to discover the positions of the axes of a given body which have these conditions.

To solve this problem therefore in general terms, would lead to a disquisition altogether disproportionate to our work. We must restrict ourselves to those forms of body and situations of the point of impulsion which admit of the coincidence of the centres of oscillation and percussion; and we must leave out the cases where the axis has a motion in the direction of its length; that is, we shall always suppose the spontaneous axis of conversion to have no motion. Thus we shall comprehend the phenomena of the planetary motions, similar to the precession of our equinoctial points, and all the interesting cases of practical mechanics. The speculative mathematical reader will fill up the blanks of this investigation by consulting the writings of Euler and d'Alembert in the Berlin Memoirs, Frisi's Cosmographia, and the papers of Mr Landen, Mr Milner, and Mr Vince, in the Philosophical Transactions. But we hope, by means of a beautiful proposition on the composition

position of rotatory motions, to enable every reader to discover the position of the axis of progressive rotation in every case which may interest him, without the previous solution of the intricate problem mentioned above.

Let $ABPCpBA$ (fig. 13.) be a section of a body through its centre of gravity $G$ , so formed, that the part $ABPC$ is similar, and similarly placed with the part $AbpC$ , so that the plane $AC$ would divide it equally. Let this body be impelled at $P$ in the direction $HP$ , perpendicular to the plane $AC$ . The axis round which it will turn will be perpendicular to $G\pi$ . Suppose it at $A$ . Then drawing $AB$ and $Ab$ to similar points, it is plain that $B\beta$ , $b\beta$ are equal and opposite; these represent the forces which would raise or lower one end of the axis, as has been already observed. The axis therefore will remain perpendicular to $G\pi$ .

Let the body be so shaped, that if the parts to the right and left of the point of impulse $\pi$ (the impulse is here supposed not perpendicular to the plane $AC$ , but in this plane) are equal and similarly placed; then the momenta round $AC$ must balance each other, and the axis $EF$ will have no tendency to go out of the plane $ABCbA$ perpendicular to the impulse.

Any body whose shape has these two properties will turn round an axis perpendicular to the plane which passes through the centre of gravity in the direction of the impelling force. This condition is always found in the planets when disturbed by the gravitation to a distant planet: for they are all figures of revolution. The direction of the disturbing or impelling force is always in a plane passing through the axis and the disturbing body.

With such limitations therefore we propose the following problem:

Let $G$ (fig. 14.) be the centre of gravity of a body in free space, which is impelled by an external force $f$ , acting in the line $FP$ , which does not pass through the centre. Let $m$ be the number of equal particles in the body, or its quantity of matter. Let the force $f$ be such, that it would communicate to the body the velocity $v$ ; that is, would cause the centre to move with the velocity $v$ . It may be expressed by the quantity of motion which it produces, that is, by $mv$ , and it would produce the velocity $mv$ on one particle. It is required to determine the whole motion, progressive and rotatory, which it will produce, and the space which it will describe during one turn round its axis.

Draw $GI$ parallel and $PGC$ perpendicular to $FP$ , and let $GI$ be taken for the measure of the progressive velocity $v$ .

It has been demonstrated that the centre $G$ will proceed in the direction $GI$ with the velocity $v$ , and that the body will at the same time turn round an axis passing through $G$ , perpendicular to the plane of the figure, every particle describing circles in parallel planes round this axis, and with velocities of rotation proportional to their distances from it. There is therefore a certain distance $GB$ , such that the velocity with which a particle describes its circumference is equal to the progressive velocity $v$ . Let $BCD$ be this circumference. When the particle describing this circumference is in the line $CGP$ , and in that part of it which lies beyond $P$ from $G$ , its absolute velocity must be double that of

the centre $G$ ; but when it is in the opposite point $C$ , its retrograde velocity being equal to the progressive velocity of the centre, it must be at rest. In every position of the body, therefore, that point of the accompanying circumference which is at this extremity of the perpendicular drawn through the centre on the line of direction of the impelling force is at rest. It is at that instant a spontaneous centre of conversion, and the straight line drawn through it perpendicular to the plane of the figure is then a spontaneous axis of conversion, and every particle is in a momentary state of rotation round this axis, in directions perpendicular to the lines drawn to the axis at right angles, and with velocities proportional to these distances; and lastly, the body advances in the direction $GI$ through a space equal to the circumference $BCD$ , while it makes one turn round $G$ .

Let $A$ be one of the particles in the plane of the figure. Join $AC$ , $AG$ , $AP$ . Draw $Ab$ , $Ae$ , $Ad$ perpendicular to $CP$ , $CA$ , $GA$ . The absolute motion $Ae$ of $A$ is compounded of the progressive motion $Ab$ common to the whole body and equal to $GI$ , and the motion $Ad$ of rotation round the centre of gravity $G$ . Therefore since $Ab$ is equal to $v$ , and $Ae$ is the diagonal of a parallelogram given both in species and magnitude, it is also given, and (as appears also from the reasoning in art. 85.) it is to $GI$ as $CA$ to $CG$ .

By the application of the force $mv$ in the direction $FP$ , every particle of the body is dragged out of its place, and exerts a resistance equal to the motion which it acquires. A part of this force, which we may call $mv$ , is employed in communicating the motion $Ae$ to $A$ . And, from what has been lately shown, $CG : CA = GI : Ae$ , $= v : Ae$ , and therefore $Ae = \frac{v \cdot CA}{CG}$ .

But farther (agreeably to what was demonstrated in art. 16.) we have $CP : CA = Ae : mv$ , $= \frac{v \cdot CA}{CG} : mv$ ,

and therefore $mv = \frac{v \cdot CA^2}{CG \cdot CP}$ . Therefore the whole force employed in communicating to each particle the motion it really acquires, or $mv$ , is equal to the fluent of the quantity $\frac{v \cdot CA^2}{CP \cdot CG}$ or $mv = \frac{v \cdot f \cdot CA^2}{CP \cdot CG}$ , and $m \cdot CP \cdot CG = f \cdot CA^2$ , which by art. 23. is equal to $f \cdot GA^2 + m \cdot CG^2$ . Therefore we have $m \cdot CP \cdot CG = m \cdot CG = CG = f \cdot GA^2$ , or $m \cdot GP \cdot CG = f \cdot GA^2$ , and finally, $CG = \frac{f \cdot GA^2}{m \cdot GP}$ .

Now the form of the body gives us $f \cdot GA^2$ , and the position of the impelling force gives us $m \cdot GP$ . Therefore we can compute the value of $CG$ ; and if $\pi$ be the periphery of a circle whose radius is unity, we have $\pi \cdot CG$ equal to the space which the body must describe in the direction $GI$ , while it makes one rotation round its axis.

Cor. 1. The angular velocity, that is, the number of turns or the number of degrees which one of the radii will make in a given time, is proportional to the impelling force: for the length of $CG$ depends only on the form of the body and the situation of the point of impulse; while the time of describing $\pi$ times this length is inversely as the force.

2. The angular velocity with any given force is as $GP$ ;

Rotation. GP: for CG, and consequently the circumference $\pi \cdot CG$ , described during one turn, is inversely as GP.

99 3. PC is equal to $\frac{f \cdot PA^2}{m \cdot GP}$ : for we have $f \cdot PA^2 =$

f \cdot GA^2 + m \cdot GP^2. \text{ Therefore } \frac{f \cdot PA^2}{m \cdot GP} = \frac{f \cdot GA^2}{m \cdot GP} +

\frac{m \cdot GP^2}{m \cdot GP} = CG + GP, = CP.

100 4. If the point C is the centre of impulsion of the same body, P will be a spontaneous centre of conversion (see art. 41.).

101 5. A force equal and opposite to $m \cdot v$ , or to $f$ , applied at G, will stop the progressive motion, but will make no change in the rotation; but if it be applied at P, it will stop all motion both progressive and rotatory. If applied between P and G, it will stop the progressive motion, but will leave some motion of rotation. If applied beyond P it will leave a rotation in the opposite direction. If applied beyond G, or between G and C, it will increase the rotation. All this will be easily conceived by reflecting on its effect on the body at rest.

102 6. A whirling body which has no progressive motion cannot have been brought into this state by the action of a single force. It may have been put into this condition by the simultaneous operation of two equal and opposite forces. The equality and opposition of the forces is necessary for stopping all progressive motion. If one of them has acted at the centre, the rotatory motion has been the effect of the other only. If they have acted on opposite sides, they conspired with each other in producing the rotation; but have opposed each other if they acted on opposite sides.

In like manner, it is plain that a motion of rotation, together with a progressive motion of the centre in the direction of the axis, could not have been produced by the action of a single force.

103 7. When the space S which a body describes during one rotation has been observed, we can discover the point of impulse by which a single force may have acted in producing both the motions of progression and rotation: for $CG = \frac{S}{\pi}$ , and $GP = \frac{f \cdot GA^2}{m \cdot CG} = \frac{\pi \cdot f \cdot GA^2}{m \cdot S}$ .

104 Application of this doctrine to the heavenly motions. In this manner we can tell the distances from the centre at which the sun and planets may have received the single impulses which gave them both their motions of revolution in their orbits and rotation round their axes.

It was found (art. 40. f) that the distance OG of the centre of oscillation or percussion of a sphere swinging round the fixed point C from its centre G, is $\frac{1}{2}$ of the third proportional to CG, and the radius of the sphere, or that $OG = \frac{2}{3} \frac{RG^2}{CG}$ . Supposing the planets to be homogeneous and spherical, and calling the radius of the planet $r$ , and the radius of its orbit $R$ , the time of a rotation round its axis $t$ , and the time of a revolution in its orbit $T$ , and making $\pi : \pi$ the ratio of radius to the periphery of a circle, we shall have $\pi R$ for the circumference of the orbit, and $\pi R \frac{t}{T}$ for

the arch of this circumference described during one rotation round the axis. This is $S$ in the above-mentioned formula. Then, diminishing this in the ratio of the circumference to radius, we obtain $CG = R \frac{t}{T}$ , and $OG = \frac{2}{3} \frac{r^2}{CG} = \frac{2}{3} \frac{T r^2}{t R}$ . This is equivalent to $\frac{\pi \cdot f \cdot GA^2}{m \cdot S}$ , and easier obtained.

This gives us $G v$

For the Earth =	$\frac{r}{157}$	} nearly.
Moon	$\frac{r}{555}$
Mars	$\frac{r}{195}$
Jupiter	$\frac{r}{2.8125}$
Saturn	$\frac{r}{2.588}$

We have not data for determining this for the sun. But the very circumstance of his having a rotation in 27 d. 7 h. 47 m. makes it very probable that he, with all his attending planets, is also moved forward in the celestial spaces, perhaps round some centre of still more general and extensive gravitation: for the perfect opposition and equality of two forces, necessary for giving a rotation without a progressive motion, has the odds against it of infinity to unity. This corroborates the conjectures of philosophers, and the observations of Herschel and other astronomers, who think that the solar system is approaching to that quarter of the heavens in which the constellation Aquila is situated.

8. As in the communication of progressive motion among bodies, the same quantity of motion is preserved before and after collision, so in the communication of rotation among whirling bodies the quantity of rotatory momentum is preserved. This appears from the general tenor of our formulae: for if we suppose a body turning round an axis passing through its centre, without any progressive motion, we must suppose that the force $m \cdot v$ , which put it in motion, has been opposed by an equal and opposite force. Let this be supposed to have acted on the centre. Then the whole rotation has been the effect of the other acting at some distance GP from the centre. Its momentum is $m \cdot v \cdot GP$ . Had it acted alone, it would have produced a rotation compounded with a progressive motion of the centre with the velocity $v$ ; and the body acquires a momentary spontaneous axis of conversion at the distance GC from the centre of gravity. The absolute velocity AC of any particle is $\frac{v \cdot AC}{CG}$ ; its momentum is $\frac{v \cdot AC^2}{CG}$ , and the sum of all the momenta is $\frac{f \cdot v \cdot AC^2}{CG}$ , or $\frac{v \cdot f \cdot AC^2}{CG}$ , and this is equal to $m \cdot v \cdot GP$ . But when the progressive motion is stopped, Ab, which was a constituent of the absolute motion of A, is annihilated, and nothing remains but the motion Ad of rotation round G. But the triangles dac and GAC were demonstrated

strated (Nº 81.) to be similar; and therefore $AC:Ad = CA:GA$ . Therefore the absolute velocity of the particle, while turning round the quiescent centre of

gravity $G$ , is $\frac{v.GA}{GC}$ ; its momentum is $\frac{v.GA^2}{GC}$ ; the

sum of all the momenta is $\frac{v.GA^2}{GC}$ ; and this is still equal to $mv$ . Observe, that now $GC$ is not the distance of the centre of conversion from the centre of gravity, because there is now no such thing as the spontaneous axis of conversion, or rather it coincides with the axis of rotation. $GC$ is the distance from the centre of a particle whose velocity of rotation is equal to $v$ .

Now let the body be changed, either by a new distribution of its parts, or by an addition or abstraction of matter, or by both; and let the same force $mv$ act at the same distance $GP$ from the centre. We shall still have $mv.GP = \frac{v.GA^2}{GC}$ ; and therefore the sum

of the momenta of the particles of the whirling body is still the same, viz. equal to the momentum of the force $mv$ acting by the lever $GP$ . If therefore a free body has been turning round its centre of gravity, and has the distribution of its parts suddenly changed (the centre however remaining in the same place), or has a quantity of matter suddenly added or taken away, it will turn with such an angular velocity that the sum of the momenta is the same as before.

We have been so particular on this subject, because it affects the celebrated problem of the precession of the equinoxes; and Sir Isaac Newton's solution of it is erroneous on account of his mistake in this particular. He computes the velocity with which a quantity of matter equal to the excess of the terrestrial spheroid over the inscribed sphere would perform its librations, if detached from the spherical nucleus. He then supposes it suddenly to adhere to the sphere, and to drag it into the same librating motion; and he computes the libration of the whole mass, upon the supposition that the quantity of motion in the librating spheroid is the same with the previous quantity of motion of the librating redundant ring or shell; whereas he should have computed it on the supposition that it was the quantity of momenta that remained unchanged.

The same thing obtains in rotations round fixed axes, as appears by the perfect sameness of the formulae for both classes of motions.

This law, which, in imitation of the Leibnitzians, we might call the conservatio momentorum, makes it of importance to have expressions of the value of the accumulated momenta in such cases as most frequently occur. The most frequent is that of a sphere or spheroid in rotation round an axis or an equatorial diameter; and a knowledge of it is necessary for the solution of the problem of the precession of the equinoxes. See PRECESSION, Nº 33.

Let $APap$ (fig. 15.) be a sphere turning round the diameter $Pp$ , and let $DD', dd'$ be two circles parallel to the equator $Aa$ , very near each other, comprehending between them an elementary slice of the sphere. Let $CA = a$ , $CB = x$ , and $BD = y$ , and let $\pi$ be the circumference of a circle whose radius is $r$ . Lastly, let the velocity of the point $A$ be $v$ . Then

$\frac{v.y}{a}$ is the velocity at the distance $y$ from the axis, $\pi y$ is the quantity of matter in the circumference whose radius is $y$ ; for it is the length of that circumference when expanded.

$\frac{v\pi y^2}{a}$ , or $\frac{v.y}{a} \times \pi y$ , is the quantity of motion in this circumference turning round the axis $Pp$ .

$\frac{v\pi y^3}{a}$ is the momentum of the same circumference.

$\frac{v\pi y^3 y}{a}$ is the fluxion of the momentum of the circle whose radius is $y$ , turning in its own plane round the axis.

$\frac{v\pi y^4}{4a}$ is the fluent, or the momentum of the whole circle; and therefore it is the momentum of the circle $DD'$ .

$\frac{v\pi y^4 y}{4a}$ is the fluxion of the momentum of the hemisphere; for $Bb = x$ , and this fraction is the momentum of the slice $dDD'd'$ .

$y^4 = a^4 - x^4$ , and $y^4 = a^4 - 2a^2x^2 + x^4$ . Therefore $\frac{v\pi}{2a} \times (a^4 x - 2a^2 x^3 + x^5)$ is the fluxion of the momentum of the whole sphere. Of this the fluent for the segments whose heights are $CB$ , or $x$ , is $\frac{v\pi}{2a} (a^4 x - \frac{2a^2 x^3}{3} + \frac{x^5}{5})$ .

Let $x$ become $a$ , and we have for the momentum of the whole sphere $\frac{v\pi}{2a} (a^5 - \frac{2}{3}a^5 + \frac{1}{5}a^5) = v\pi (\frac{a^4}{2} - \frac{a^4}{3} + \frac{a^4}{10}) = v\pi \frac{7}{30} a^4$ .

Let us suppose that this rotation has been produced by the action of a force $mu$ ; that is, a force which would communicate the velocity $u$ to the whole matter of the sphere, had it acted in a direction passing through its centre; and let us suppose that this force acted on the equatorial point $A$ at right angles to $AC$ . Its momentum is $muA$ , and this is equal to $v\pi \frac{7}{30} a^4$ . Also, we know that $m = \frac{1}{3} \pi a^3$ . Therefore we have $u = \frac{7}{21} \pi a^4$ .

Let $EPQp$ be an oblate spheroid whose semi-axis $PC$ is $a$ , and equatorial radius $EC$ is $b$ , and let $v$ be the velocity on the equator of the inscribed sphere. Then since the momentum of the whirling circle $DD'$ is $\frac{v\pi y^4}{4a}$ , the momenta of the sphere and spheroid are in the quadruplicate ratio of their equatorial radii; and therefore that of the whole spheroid is $\frac{4}{3} \pi b^4 v$ . And if $w$ be the velocity at $E$ corresponding to the velocity $v$ at $A$ , so that $w = \frac{b}{a} v$ , we have the momentum of the spheroid, expressed in terms of the equatorial velocity at the surface, $\frac{4}{3} \pi b^4 a w$ .

If the same force $mu$ be made to act in the same manner

Rotation. manner at E, its momentum $mu b$ is $= \frac{4}{3} b^3 a w$ , and $w = \frac{15 mu}{4 \pi b^3 a}$ . Therefore the angular velocities $\frac{v}{a}$ , $\frac{w}{b}$ , which the same force $mu$ acting at A or E will produce in the sphere and the spheroid, are as $\frac{15 mu}{4 \pi a^4}$ and $\frac{15 mu}{4 b^3 a}$ , that is, in the triplicate ratio of the equatorial diameter $b$ to the polar axis $a$ .

112 Lastly, if the oblate spheroid is made to turn round an equatorial diameter passing through C perpendicular to the plane of the figure, it is plain that every section parallel to the meridian EPQ $p$ is an ellipse similar to this meridian. If this ellipse differs very little from the inscribed circle, as is the case of the earth in the problem of the precession of the equinoxes, the momentum of each ellipse may be considered as equal to that of a circle of the same area, or whose diameter is a mean proportional between the equatorial and polar diameters of the spheroid. This radius is to the radius of the circumscribed circle as $\sqrt{ba}$ to $b$ . Therefore the momenta of the section of the spheroid and of the circumscribed sphere are in the constant ratio of $b^3 a^2$ to $b^4$ , or of $a^2$ to $b^2$ . And if the velocity in the equator of this circumscribed sphere be called $w$ , the momentum of the sphere is $\frac{4}{3} \pi b^4 w$ ; and therefore that of the spheroid is $\frac{4}{3} \pi b^3 a w$ , agreeably to what was assumed in the article PRECESSION, No 33.

This value of the momentum of a spheroid round an equatorial diameter is only a very easy approximation; an exact value may be obtained by an infinite series. The whole matter of the spheroid may be considered as uniformly distributed on the surface of a similar spheroid whose diameter is $= \sqrt{\frac{2}{3}}$ of the diameter of the spheroid. It will have the same momentum, because a triangle in one of the ellipses, having an elementary arch of the circumference for its base, and the centre of the ellipse for its vertex, has its centre of gyration distant from the vertex $\sqrt{\frac{2}{3}}$ the length of the radius of the ellipse, and the problem is reduced to the finding the sum of all these lines. But even when the series for this sum involves the 3d power of the eccentricity, it is not more exact than the above approximation.

A similar proposition may be obtained for a prolate spheroid vibrating round an equatorial diameter, and applied to the conjectural shape of the moon, for explaining her oscillations.

113 All rotatory motions accompanied by centrifugal forces. The reader must have observed that the preceding disquisitions refer to those motions only which result from the action of external forces and to the state of incipient motion. All circular motions, such as those of rotation, are accompanied by centrifugal forces. A central force is necessary for retaining every particle in its circular path; such forces must therefore be excited in the body, and can arise only from the forces of cohesion by which its particles are held together. These forces are mutual, equal, and opposite; and as much as a particle A (fig. 5.) is retained by a force in the direction Aa of the line which connects it with the fixed axis Dd, or in the direction AG (fig. 10.), which connects it with the progressive axis; so much must the point a of the axis Dd be urged in the opposite direction aA, or so much must the whole body be urged in the direction GA. Every point therefore of the axis

Dd, or of the axis through G in fig. 10. is carried in a variety of directions perpendicular to itself. These forces may or may not balance each other. If this balance obtains with respect to the fixed axis, its supports will sustain no pressure but what arises from the external force; if not, one support will be more pressed than the other; and if both were removed, the axis would change its position. The same must be affirmed of the axis through G in fig. 10. This, having no support, must change its position.

And thus it may happen, that the axis of rotation passing through G which has been determined by the preceding disquisitions, is not permanent either in respect of the body, or in respect of absolute space. These two rotations are essentially different. The way to conceive both is this. Suppose a spherical surface described round the body, having its centre in the centre of gravity; and suppose this surface to revolve and to proceed forward along with the body: in short, let it be conceived as an immaterial surface attached to the body. The axis of rotation will pass through this surface in two points which we shall call its poles. Now, we say that the axis is permanent with respect to the body when it has always the same poles in this spherical surface. Suppose another spherical surface described round the same centre, and that this surface also accompanies the body in all its progressive motion, but does not turn with it. The axis is permanent with respect to absolute space when it has always the same poles in this surface: it is evident that these two facts are not inseparable. A boy's top spins on the same point and the same corporeal axis, while, towards the end of its motion, we observe it directing this round and round to different quarters of the room. And when we make an egg or a lemon spin with great rapidity on its side on a level table, we see it gradually rise up, till it stand quite on end, spinning all the while round an axis pointing to the zenith.

This change in the position of the axis is produced by the unbalanced actions of the centrifugal forces exerted by the particles. Suppose two equal balls A and B (fig. 16.) connected by an inflexible rod whose middle point is G, the centre of gravity of the balls. This system may be made to turn round the material axis Dd, A describing the circle AEF, and B describing the circle BHK. The rod AB may also be conceived as moveable round the point G by means of a pin at right angles to the axis. Suppose the balls passing through the situations A and B; their centrifugal forces urge them at the same time in the directions CA and OB, which impulses conspire to make the connecting rod recede from both ends of the axis Dd. And thus the balls, instead of describing parallel circles round this axis, will describe parallel spirals, gradually opening the angles DGA, dGB more and more, till the balls acquire the position $\alpha\beta$ at right angles to the axis. They will not stop there, for each came into that position with an oblique motion. They will pass it; and were it not for the resistance of the air and the friction of the joint at G, they would go on till the ball A came to describe the circle BHK, and the ball B to describe the circle AEF. The centrifugal forces will now have exhausted by opposition all the motion which they had acquired during their passage from the position AB to the position $\alpha\beta$ ; and now they will again describe spirals

Rotation. rals gradually opening, and then contracting, till the balls arrive at their original position AB, when the process will begin again. Thus they will continue a kind of oscillating rotation.

216 Thus the axis is continually changing with respect to the system of balls; but it is fixed in respect to absolute space, because the axis Dd is supported. It does not yet appear that it has any tendency to change its position, because the centrifugal tendency of the balls is completely yielded to by the joint at G. The material axis has indeed sustained no change; but the real axis, or mathematical line round which the rotation was going on every moment, has been continually shifting its place. This is not so obvious, and requires a more attentive consideration. To show accurately the gradual change of position of the real axis of rotation would require a long discussion. We shall content ourselves with exhibiting a case where the position of the momentary axis is unquestionably different from Dd, which we may suppose horizontal.

Take the balls in the position $\alpha\beta$ . They came into this position with a spiral motion, and therefore each of them was moving obliquely to the tangents $\alpha\phi$ , $\beta\gamma$ to the circle $\alpha\delta\beta\iota$ , suppose in the directions $\alpha\iota$ , $\beta\lambda$ . They are therefore moving round the centre G in a plane $\iota\alpha\beta\lambda$ , inclined to the plane $\phi\alpha\beta\gamma$ of the circle $\alpha\delta\beta\iota$ . The momentary axis of rotation is therefore perpendicular to this oblique plane, and therefore does not coincide with Dd.

217 Of the eva-
gation of
the axis.
We cannot enter upon the investigation of this evagination of the axis, although the subject is both curious and important to the speculative mathematicians. A knowledge of it is absolutely necessary to a complete solution of the great problem of the precession. But when treating that article, we contented ourselves with showing that the evagation which obtains in this natural phenomenon is so exceedingly minute, that although multiplied many thousands of times, it would escape the nicest observation of modern astronomers; and that it is a thing which does not accumulate beyond a certain limit, much too small for observation, and then diminishes again, and is periodical. Euler, D'Alembert, Frisi, and De la Grange, have shown the momentary position of the real variable axis corresponding to any given time; and Landen has with great ingenuity and elegance connected these momentary positions, and given the whole paths of evagation. Mr Segnor was, we believe, the first who showed (in a Dissertation De Motu Turbinum, Halle, 1755), that in every body there were at least three lines passing through the centre of gravity at right angles to each other, forming the solid angle of a cube, round which the centrifugal forces were accurately balanced, and therefore a rotation begun round either of these three lines would be continued, and they are permanent axes of rotation. Albert Euler gave the first demonstration in 1760, and since that time the investigation of these axes has been extended and improved by the different authors already named. It is an exceedingly difficult subject; and we recommend the synthetical investigation by Frisi in his Cosmographia as the fittest for instructing a curious reader to whom the subject is new. We shall conclude this dissertation with a beautiful theorem, the enunciation of which we owe to P. Frisi, which has amazingly improved the whole theory, and gives easy and elegant

solutions of the most difficult problems. It is analogous to the great theorem of the composition of motions and forces.

118 If a body turn round an axis AG a (fig. 17.) passing through its centre of gravity G with the angular velocity $a$ , while this axis is carried round another axis BG b with the angular velocity $b$ , and if GD be taken to GK as $a$ to $b$ (the points B and E being taken on that side of the centre where they are moving towards the same side of the plane of the figure), and the line DE be drawn, then the whole and every particle of the body will be in a state of rotation round a third axis CG c, lying in the plane of the other two, and parallel to DE, and the angular velocity $c$ round this axis will be to $a$ and to $b$ as DE is to GD and to GE.

For, let P be any particle of the body, and suppose a spherical surface to be described round G passing through P. Draw PR perpendicular to the plane of the figure. It is evident that PR is the common section of the circle of rotation IP i round the axis A a, and the circle KP k of rotation round the axis B b. Let I i, K k be the diameters of these circles of rotation, F and G their centres. Draw the radii PF and PG, and the tangents PM and PN. These tangents are in a plane MPN which touches the sphere in P, and cuts the plane of the axis in a line MN, to which a line drawn from the centre G of the sphere through the point R is perpendicular. Let PN represent the velocity of rotation of the point P round the axis B b, and Pf its velocity of rotation round A a. Complete the parallelogram PNfP. Then Pf is the direction and velocity of motion resulting from the composition of PN and Pf. Pf is in the plane MPN, because the diagonal of a parallelogram is in the plane of its sides PN and Pf.

Let perpendiculars fF, tT, be drawn to the plane of the axes, and the parallelogram PNfP will be orthographically projected on that plane, its projection being a parallelogram RNfF. (F here falls on the centre by accident). Draw the diagonal RT. It is evident that the plane PRtT is perpendicular to the plane of the two axes, because PR is so. Therefore the compound motion Pt is in the plane of a circle of revolution round some axis situated in the plane of the other two. Therefore produce TR, and draw GC cutting it at right angles in H, and let LP l be the circle, and PH a radius. Pt is therefore a tangent, and perpendicular to PH, and will meet RT in some point Q of the line MN. The particle P is in a state of rotation round the axis CG c, and its velocity is to the velocities round A a or B b as Pt to Pf or PN. The triangles PRN and OPN are similar. For PN the tangent is perpendicular to the radius OP, and PR is perpendicular to ON. Therefore $OP : PN = PR : RN$ , and $RN = \frac{PR \cdot PN}{OP}$ .

But the velocity of P round the axis B b is OP.b. Therefore $RN = \frac{PR \cdot OP \cdot b}{OP} = PR \cdot b$ . In like manner $RF = PR \cdot a$ . Therefore $RF : RN = a : b = GD : GE$ . But $NT : RN = \text{fine NRT} : \text{fine NTR}$ , and $GD : GE = \text{fine GED} : \text{fine GDE}$ . Therefore fine NRT : fine NTR = fine GED : fine GDE. But $RNT = EGD$ , for NR is perpendicular to EG and NT (being parallel

Rotation. to $IF$ ) is perpendicular to $DG$ . Therefore $TR$ is perpendicular to $ED$ , and $Cc$ is parallel to $ED$ , and the rotation of the particle $P$ is round an axis parallel to $ED$ .

And since $RN$ , $RF$ , $RI$ , are the velocities $b$ , $a$ , $c$ , round these different axes, and are proportional to $EG$ , $DG$ , $DE$ , we have $c$ to $a$ or to $b$ as $ED$ to $GD$ or $GE$ , and the proposition is demonstrated.

This theorem may be thus expressed in general terms.

If a body revolves round an axis passing through its centre of gravity with the angular velocity $a$ , while this axis is carried round another axis, also passing through its centre of gravity, with the angular velocity $b$ , these two motions compose a motion of every particle of the body round a third axis, lying in the plane of the other two, and inclined to each of the former axes in angles whose sines are inversely as the angular velocities round them; and the angular velocity round this new axis is to that round one of the primitive axes as the sine of inclination of the two primitive axes is to the sine of the inclination of the new axis to the other primitive axis.

When we say that we owe the enunciation of this theorem to P. Frisi, we grant at the same time that something like it has been supposed or assumed by other authors. Newton seems to have considered it as true, and even evident, in homogeneous spheres; and this has been tacitly acquiesced in by the authors who followed him in the problem of the precession. Inferior writers have carelessly assumed it as a truth. Thus Nolle, Gravefande, and others, in their contrivances for exhibiting experiments for illustrating the composition of vortices, proceeded on this assumption. Even authors of more scrupulous research have satisfied themselves with a very imperfect proof. Thus Mr Landen, in his excellent dissertation on rotatory motion, Philosophical Transactions, Vol. lxvii. contents himself with showing, that by the equality and opposite directions of the motions round the axes $Aa$ and $Bb$ , the point $C$ will be at rest, and from thence concludes that $CGc$ will be the new axis of rotation. But this is exceedingly hasty (note also, that this dissertation was many years posterior to that of P. Frisi): For although the separate motions of the point $C$ may be equal and opposite, it is by no means either a mathematical or a mechanical consequence that the body will turn round the axis $Cc$ . In order that the point $C$ may remain at rest, it is necessary that all tendencies to motion be annihilated: this is not even thought of in making the assumption. Frisi has shown, that in the motion of every particle round the axis $Cc$ , there is involved a motion round the two axes $Aa$ and $Bb$ , with the velocities $a$ and $b$ ; and it is a consequence of this, and of this only, that the impulses which would separately produce the rotations of every particle round $Aa$ and $Bb$ will, either in succession or in conjunction, produce a rotation round $Cc$ . Moreover, Mr Landen's not having attended to this, has led him, as we imagine, into a mistake respecting the velocity with which the axis changes its position; and though his process exhibits the path of evagination with accuracy, we apprehend that it does not assign the true times of the axes arriving at particular points of this path.

It follows from this proposition, that if every particle of a body, whether solid or fluid, receives in one instant a separate impulse, competent to the production of a motion of the particle round an axis with a cer-

tain angular velocity, and another impulse competent to the production of a motion round another axis with a certain velocity, the combined effect of all these impulses will be a motion of the whole system round a third axis given in position, with an angular velocity which is also given: and this motion will obtain without any separation or disunion of parts; for we see that a motion round two axes constitutes a motion round a third axis in every particle, and no separation would take place although the system were incoherent like a mass of sand, except by the action of the centrifugal forces arising from rotation. Mr Simpson therefore erred in his solution of the problem of the precession, by supposing another force necessary for enabling the particles of the fluid spheroid to accompany the equator when displaced from its former situation. The very force which makes the displacement produces the accompaniment, as far as it obtains, which we shall see presently is not to the extent that Mr Simpson and other authors who treat this problem have supposed.

For the same reason, if a body be turning round any axis, and every particle in one instant get an impulse precisely such as is competent to produce a given angular velocity round another axis, the body will turn round a third axis given in position, with a given angular velocity: for it is indifferent (as it is in the ordinary composition of motion) whether the forces act on a particle at once or in succession. The final motion is the same both in respect of direction and velocity.

Lastly, when a rigid body acquires a rotation round an axis by the action of an impulse on one part of it, and at the same time, or afterwards, gets an impulse on any part which, alone, would have produced a certain rotation round another axis, the effect of the combined actions will be a rotation round a third axis, in terms of this proposition; for when a rigid body acquires a motion round an axis, not by the simultaneous impulse of the precisely competent force on each particle, but by an impulse on one part, there has been propagated to every particle (by means of the connecting forces) an impulse precisely competent to produce the motion which the particle really acquires; and when a rigid body, already turning round an axis $Aa$ (fig. 17.), receives an impulse which makes it actually turn round another axis $Cc$ , there has been propagated to each particle a force precisely competent to produce, not the motion, but the change of motion which takes place in that particle, that is, a force which, when compounded with the inherent force of its primitive motion, produces the new motion: that is (by this theorem), a force which alone would have caused it to turn round a third axis $Bb$ , with a rotation making the other constituent of the actual rotation round $Cc$ .

This must be considered as one of the most important propositions in dynamics, and gives a great extension to the doctrine of the composition of motion. We see that rotations are compounded in the same manner as other motions, and it is extremely easy to discover the composition. We have only to suppose a sphere described round the centre of the body; and the equator of this sphere corresponding to any primitive position of the axis of rotation gives us the direction and velocity of the particles situated in it. Let another great circle cut this equator in any point; it will be the equator of another rotation. Set off an arch of each from the

point of intersection, proportional to the angular velocity of each rotation, and complete the spherical parallelogram. The great circle, which is the diagonal of this parallelogram, will be the equator of the rotation, which is actually compounded of the other two.

And thus may any two rotations be compounded. We have given an instance of this in the solution of the problem of the PRECESSION of the Equinoxes.

It appears plainly in the demonstration of this theorem that the axis $Cc$ is a new line in the body. The change of rotation is not accomplished by a transference of the poles and equator of the former rotation to a new situation, in which they are again the poles and equator of the rotation; for we see that in the rotation round the axis $Cc$ , the particle of the body which was formerly the pole $A$ is describing a circle round the axis $Cc$ . Not knowing this composition of rotations, Newton, Walmesly, Simpson, and other celebrated mathematicians, imagined, that the axis of the earth's rotation remained the same, but changed its position. In this they were confirmed by the constancy of the observed latitudes of places on the surface of the earth. But the axis of the earth's rotation really changes its place, and the poles shift through different points of its surface; but these different points are too near each other to make the change sensible to the nicest observation.

It would seem to result from these observations, that it is impossible that the axis of rotation can change its position in absolute space without changing its position in the body, contrary to what we experience in a thousand familiar instances; and indeed this is impossible by any one change. We cannot by the impulse of any one force make a body which is turning round the axis $Aa$ change its position and turn round the same material axis brought into the position $Cc$ . In the same way that a body must pass through a series of intermediate points, in going from one end of a line to the other, so it must acquire an infinite series of intermediate rotations (each of them momentary) before the same material axis passes into another position, so as to become an axis of rotation. A momentary impulse may make a great change of the position of the axis of rotation, as it may make in the velocity of a rectilinear motion. Thus although the rotation round $Aa$ be indefinitely small, if another equally small rotation be impressed round an axis $Bb$ perpendicular to $Aa$ , the axis will at once shift to $Cc$ half way between them; but a succession of rotations is necessary for carrying the primitive material axis into a new position, where it is again an axis. This transference, however, is possible, but gradual, and must be accomplished by a continuation of impulses totally different from what we would at first suppose. In order that $A$ may pass from $A$ to $C$ , it is not enough that it gets an impulse in the direction $AC$ . Such an impulse would carry it thither, if the body had not been whirling round $Aa$ by the mere perseverance of matter in its state of motion; but when the body is already whirling round $Aa$ , the particles in the circle $IP$ are moving in the circumference of that circle; and since that circle also partakes of the motion given to $A$ , every particle in it must be incessantly deflected from the path in which it is moving. The continual agency of a force is therefore necessary for this purpose; and if this force be discontinued, the point

$A$ will immediately quit the plane of the arch $AC$ , along which we are endeavouring to move it, and will start up.

This is the theorem which we formerly said would enable us to overcome the difficulties in the investigation of the axis of rotation.

Thus we can discover what Mr. Landen calls the evagations of the poles of rotation by the action of centrifugal forces: For in fig. 16. the known velocity of the ball $A$ and the radius $AC$ of its circle of rotation will give us the centrifugal force by which the ball will tend to turn in the plane $DA\delta BD$ . This gives the axis $Dd$ a tendency to move in a plane perpendicular to the plane of the figure; and its separation from the poles $D$ and $d$ does not depend on the separation of the connecting rod $AB$ from its present inclination to $Dd$ , but on the angle which the spiral path of the ball makes with the plane of a circle of rotation round $Dd$ . The distance of the new poles from $D$ and $d$ is an arch of a circle which measures the angle made by the spiral with the circle of rotation round the primitive axis. This will gradually increase, and the mathematical axis of rotation will be describing a spiral round $D$ and $d$ , gradually separating from these points, and again approaching them, and coinciding with them again, at the time that the balls themselves are most of all removed from their primitive situation, namely, when $A$ is in the place of $B$ .

The same theorem also enables us to find the incipient axis of rotation in the complicated cases which are almost inaccessible by means of the elementary principles of rotation.

Thus, when the centres of oscillation and percussion do not coincide, as we supposed in fig. 5. and 12. Suppose, first, that they do coincide, and find the position of the axis $ab$ , and the angular velocity of the rotation. Then find the centre of percussion, the axis $Pp$ , and the momentum round it, and the angular velocity which this momentum would produce. Thus we have obtained two rotations round given axes, and with given angular velocities. Compound these rotations by this theorem, and we obtain the required position of the true incipient axis of rotation, and the angular velocity, without the intricate process which would otherwise have been necessary.

If the body is of such a shape, that the forces in the plane $DCG$ do not balance each other, we shall then discover a momentum round an axis perpendicular to this plane. Compound this rotation in the same manner with the rotation round $Dd$ .

And from this simple view of the matter we learn Position of (what would be difficult to discover in the other way), the axis that when the centre of percussion does not coincide with that of rotation, the axis is in the plane $DGC$ , though not perpendicular to $PG$ . But when there is a momentum round an axis perpendicular to this plane, the incipient axis of rotation is neither perpendicular to $PC$ , nor in a plane perpendicular to that passing through the centre in the direction of the impelling force.

We must content ourselves with merely pointing out these tracks of investigation to the curious reader, and recommending the cultivation of this most fruitful theorem of Father Frisi.

These are by no means speculations of mere curiosity, interesting to none but mathematicians: the noblest art remarks on which

which is practised by man must receive great improvement from a complete knowledge of this subject. We mean the art of SEAMANSHIP. A ship, the most admirable of machines, must be considered as a body in free space, impelled by the winds and waters, and continually moved round spontaneous axes of conversion, and incessantly checked in these movements. The trimming of the sails, the action of the rudder, the very disposition of the loading, all affect her versatility. An experienced seaman knows by habit how to produce and facilitate these motions, and to check or stop such as are inconvenient. Experience, without any reflection or knowledge how and why, informs him what position of the rudder produces a deviation from the course. A sort of common sense tells him, that, in order to make the ship turn her head away from the wind, he must increase the surface or the obliquity of the head sails, and diminish the power of the sails near the stern. A few other operations are dictated to him by this kind of common sense; but few, even of old seamen, can tell why a ship has such a tendency to bring her head up in the wind, and why it is so necessary to crowd the fore part of the ship with sails; fewer still know that a certain shifting of the loading will facilitate some motions in different cases; that the crew of a great ship running suddenly to a particular place shall enable the ship to accomplish a movement in a stormy sea which could not be done otherwise; and perhaps not one in ten thousand can tell why this procedure will be successful. But the mathematical inquirer will see all this; and it would be a most valuable acquisition to the public, to have a manual of such propositions, deduced from a careful and judicious consideration of the circumstances, and freed from that great complication and intricacy which only the learned can unravel, and expressed in a familiar manner, clothed with such reasoning as will be intelligible to the unlearned; and though not accurate, yet persuasive. Mr Bouguer, in his Traité du Navire, and in his Manœuvre des Vaisseaux, has delivered a great deal of useful information on this subject; and Mr Bezout has made a very useful abstract of these works in his Cours de Mathématique. But the subject is left by them in a form far too abstruse to be of any general use: and it is unfortunately so combined with or founded on a false theory of the action and resistance of fluids, that many of the propositions are totally inconsistent with experience, and many maxims of seamanship are false. This has occasioned these doctrines to be neglected altogether. Few of our professional seamen have the preparatory knowledge necessary for improving the science; but it would be a work of immense utility, and would acquire great reputation to the person who successfully prosecuted it.

We shall mention under the article SEAMANSHIP the chief problems, and point out the mechanical principles by which they may be solved.