ROTATION

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 perpendicular to the axis AB of rotation.

Rotation. This line will lie in the plane of the circle $P$ Rotation. $m$ $n$ 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.

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 $\backslash frac\{1\}\{4\}$, $\backslash frac\{1\}\{2\}$, 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.

How the rotation of different bodies 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.

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 $PCP$ may be expressed by $\backslash frac\{Pp\}\{PC\}$. 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 $\backslash frac\{1\}\{57.296\}$ of the radius, or that an arch of 57.296 degrees is equal to the radius.

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.

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 philosophers have little to do with this variety. The distinctions which are the immediate causes of fluidity, of hardness, 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), the 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.

Fig. 2. Let the forces $a\; A$, $b\; B$, $c\; C$ (fig. 2.), be applied to three particles of the solid body. Therefore $a\; A$ is immediately in equilibrium with an equal and opposite force $A\; a$, 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 $b\; B$ is immediately in equilibrium with $B\; \backslash beta$, the equivalent of the forces $BF$ and $BG$; and $c\; C$ is in immediate equilibrium with the equivalent $C\; \backslash times$ of the forces $CH$ and $CI$. We shall conceive it very clearly if we suppose the three forces $A\; a$, $B\; \backslash beta$, $C\; \backslash times$, 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 ($N^o$ 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 $A\; a$ were applied in place of $AD$, $AE$, the equilibrium would remain, because $A\; a$ is equivalent to $AD$ and $AE$. The same is true of $B\; \backslash beta$ and $C\; \backslash times$. Therefore if the three forces $A\; a$, $B\; \backslash beta$, $C\; \backslash times$, were applied to one point, they would be in equilibrium. Consequently if the three forces $a\; A$, $b\; B$, $c\; C$, which are respectively equal and opposite to $A\; a$, $B\; \backslash beta$, $C\; \backslash times$, 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, Poncelet, 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.

12 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.

13 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.

14 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 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 $m.GI$ 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.

properly expressed by $m.GI$. In like manner, $m.i$ 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 $C\backslash lambda$; and the body must, on this account, resist in the opposite direction $Ff$. 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 $Cc$, and $A$ in the direction $Aa$; and if we take $Cc$ to represent the action on $C$, $Aa$ must be taken equal to it. In like manner $AF$ is stretched and tends to contract, pulling $F$ in the direction $F\backslash phi$ and $A$ in the direction $Aa$ with equal forces. Thus the particle $A$ is pulled in the directions $Aa$ and $A\backslash phi$; the particle $F$ is pulled in the direction $F\backslash phi$, and pushed in the direction $Ff$; and $C$ is pulled in the direction $Cc$, and pushed in the direction $C\backslash lambda$. $Aa$ and $A\backslash phi$ have produced their equivalent $AL$, by which $A$ is dragged into motion; $Ff$ and $F\backslash phi$ produce their equivalent $Fg$, by which the external force is resisted, and $Fg$ is equal and opposite to $m.GI$; the forces $Cc$ and $C\backslash lambda$ produce their equivalent $Cd$ 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 $A.A.L$ are both external, viz. the impelling force $gF$, and the supporting force $dC$. $AL$ therefore is not only the immediate equivalent of $Aa$ and $A\backslash phi$, but also the remote equivalent of $gF$ and $dC$. We may therefore ascertain the proportion of $gF$ (that is, of $m.GI$) to $AL$ (that is, of $A.A.C$), independent of the property of the lever. $gF$ is to $AL$ in the ratio compounded of the ratios of $gF$ to $F\backslash phi$ or $A\backslash phi$, and of $A\backslash phi$ 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

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 $A$ express its quantity of matter, $A.CA$ will express its quantity of motion, and will represent the force which would produce it by acting uniformly during the moment of time.

We expressed the external moving force by $m.GI$. Part of it is employed in exciting the force $A.CA$, 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 $A.CA$ applied to $A$ in the direction $LA$. Therefore (by the property of the lever $ACP$, which is impelled at right angles at $A$ and $P$) we must have $CP$ to $CA$ as the force $A.CA$ to the balancing pressure, which must be exerted at $P$, or at any point in the line $FP$. This pressure is therefore $\backslash frac\{A.CA.CA\}\{CP\}$ or $\backslash frac\{A.CA^2\}\{CP\}$. As we took $m.GI$ for the mea-

sure of the whole external force, $GI$ being the velocity which it would communicate to the whole body moving in free space, we may take $GI$ for the velocity which would be communicated to the whole body by the pressure $\backslash frac\{A.CA^2\}\{CP\}$, and then this pressure will be

VOL. XVIII. Part I.

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.

27 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 $m.P\backslash rho$ may be $=m.GI$. That is, we must have $m.GI\; =\; \backslash int\; \backslash frac\{A.CA^2\}\{CP\}$; or, because $CP$ is given when the position of the line $FP$ is given, we must have $m.GI\; =\; \backslash int\; \backslash frac\{A.CA^2\}\{CP\}$, where both $A$ and $CA$ are variable quantities.

28 This equation gives us $m.GI.CP\; =\; \backslash int\; A.CA^2$. 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 $\backslash int\; A.CA^2$ is the sum of the momenta 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 $\backslash int\; A.CA^2$ may express it. Or, take $p$ to represent the quantity of matter in any particle, and $r$ to represent its radius vector, or distance from the axis of rotation, $\backslash int\; p.r^2$ 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 $m.GI.CP\; =\; \backslash int\; A.CA^2\; =\; \backslash int\; p.r^2$. 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 Rotation. 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 $GI\; =\; \backslash frac\{\backslash int\; p.r^2\}\{m.CP\}$.

Now $GI\; :\; CG\; =\; \backslash frac\{\backslash int\; p.r^2\}\{m.CP\}\; :\; CG\; =\; \backslash int\; p.r^2\; :\; m.CP.CG$; 29

but $CG$ represents the velocity of the centre. Hence we derive this fundamental proposition $\backslash int\; p.r^2\; :\; m.CP.CG\; =\; GI\; :\; CG$; or, that $\backslash int\; p.r^2$ 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 $=\; \backslash frac\{m.GI.CP.CG\}\{\backslash int\; p.r^2\}$. 20

The velocity of any point $B$ is $=\; \backslash frac\{m.GI.CP.CB\}\{\backslash int\; p.r^2\}$. 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 $=\; \backslash frac\{m.GI.CP\}\{\backslash int\; p.r^2\}$. 22

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 $\backslash frac\{1\}\{2\}$, $\backslash frac\{1\}\{4\}$, $\backslash frac\{1\}\{8\}$, &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 $Aa\text{'}$ perpendicular to $CG$, and join $AG$. Then $CA^2\; =\; GA^2\; +\; CG^2\; =\; 2CG\; \backslash times\; Ga$. Therefore $\backslash int\; CA^2\; =\; \backslash int\; GA^2\; +\; \backslash int\; CG^2\; =\; \backslash int\; 2CG\; \backslash times\; Ga\; =\; \backslash int\; GA^2\; +\; m.CG^2\; =\; \backslash int\; 2CG\; \backslash times\; Ga$. 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 $\backslash int\; 2CG\; \backslash times\; Ga$ is nothing; and therefore $\backslash int\; CA^2\; =\; \backslash int\; GA^2\; +\; m.CG^2$. 23

Therefore $\backslash int\; CA^2$ or $\backslash int\; p.r^2$ is smallest, and $\backslash frac\{m.GI.CP\}\{\backslash int\; p.r^2\}$ 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, 24

Rotation. dy, or the sum of the motions of all its particles, is
$\backslash frac\{m\; \backslash cdot\; GI\; \backslash cdot\; CP\; \backslash cdot\; f\; \backslash cdot\; p\; \backslash cdot\; r\}\{f\; \backslash cdot\; p\; \backslash cdot\; r^2\}$. For the motion of each particle is
$\backslash frac\{m\; \backslash cdot\; GI\; \backslash cdot\; CP\; \backslash cdot\; p\; \backslash cdot\; r\}\{f\; \backslash cdot\; p\; \backslash cdot\; r^2\}$.

25
Ratio of the resistance of a 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 $f\; \backslash cdot\; p\; \backslash cdot\; r^2$. 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 $m\; \backslash cdot\; GI$, and the other $m\; \backslash cdot\; \backslash gamma$.—Let both be applied at the distance $CP$. Let $r$ be the radius vector in the one body, and $p$ in the other; now the angular velocities $\backslash frac\{m\; \backslash cdot\; GI\; \backslash cdot\; CP\}\{f\; \backslash cdot\; p\; \backslash cdot\; r^2\}$ and $\backslash frac\{m\; \backslash cdot\; \backslash gamma\; \backslash cdot\; CP\}\{f\; \backslash cdot\; p\; \backslash cdot\; r^2\}$ are equal by supposition. Therefore $m\; \backslash cdot\; GI\; :\; m\; \backslash cdot\; \backslash gamma\; =\; f\; \backslash cdot\; p\; \backslash cdot\; r^2\; :\; f\; \backslash cdot\; p\; \backslash cdot\; r^2$.

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 lxxi. 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 supposed 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 $A\; \backslash times\; CA$, which is immediately urging the particle $A$, and the force $m\; \backslash cdot\; GI$ 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 centre of gravity, as is well known. The difference of a body between the motion of the centre of a free body and estimated by that of its centre of gravity, etc.
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.

29
Therefore join $I$ and $H$, and complete the parallelogram $GIHK$. It is plain that $m\; \backslash cdot\; GK$ must represent the forces exerted by the axis on the fixed points.

30
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) $\backslash frac\{m\; \backslash cdot\; GI\; \backslash cdot\; CP\; \backslash cdot\; CG\}\{f\; \backslash cdot\; p\; \backslash cdot\; r^2\}$ is equal to $GI$, or $\backslash frac\{m\; \backslash cdot\; CP\; \backslash cdot\; CG\}\{f\; \backslash cdot\; p\; \backslash cdot\; r^2\}\; =\; 1$, and $CP\; =\; \backslash frac\{f\; \backslash cdot\; p\; \backslash cdot\; r^2\}\{m\; \backslash cdot\; CG\}$.

The equation $CP\; =\; \backslash frac\{f\; \backslash cdot\; p\; \backslash cdot\; r^2\}\{m\; \backslash cdot\; CG\}$ gives us $m\; \backslash cdot\; CG\; \backslash cdot\; CP\; =\; f\; \backslash cdot\; p\; \backslash cdot\; r^2\; =\; f\; \backslash cdot\; A\; \backslash cdot\; CA^2$. But it was shown ($n^o$ 23), that $f\; \backslash cdot\; A\; \backslash cdot\; CA^2\; =\; f\; \backslash cdot\; A\; \backslash cdot\; GA^2\; +\; m\; \backslash cdot\; CG^2$. Therefore $f\; \backslash cdot\; A\; \backslash cdot\; GA^2\; =\; m\; \backslash cdot\; CG\; \backslash cdot\; CP\; -\; m\; \backslash cdot\; CG\; \backslash cdot\; CG\; =\; m\; \backslash cdot\; CG\; (CP\; -\; CG)\; =\; m\; \backslash cdot\; CG\; \backslash cdot\; GP$. Therefore we have (for another determination of the point of impulse $P$ so as to annihilate all pressure on the axis) $GP\; =\; \backslash frac\{f\; \backslash cdot\; A\; \backslash cdot\; GA^2\}\{m\; \backslash cdot\; CG\}$. This is generally the most easily obtained, the mathematical situation of the centre of gravity being well known.

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 $m.GI\; \backslash times\; 2\; \backslash sin\; GCP$.

Advantage of annulling or diminishing the pressure on the supports of the axis of motion.

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 $3\backslash frac\{1\}\{2\}$ lbs. For, if we call the radius of the pulley 1, the momentum of the moving force is $10\; \backslash times\; 1\; =\; 1\; \backslash times\; 1\; =\; 10$; and the momentum of inertia is $10\; \backslash times\; 1^2\; +\; 1\; \backslash times\; 1^2$. ($\backslash text\{No\; \}\; 18.$) = 11. Therefore the angular velocity is $\backslash frac\{1\}\{11\}$. But the distance CG of the centre of gravity from the axis of motion is also $\backslash frac\{1\}\{11\}$, because we may suppose the two weights in contact with the circumference of the pulley. Therefore the velocity of the centre of gravity is $\backslash frac\{1\}\{11\}\; \backslash times\; \backslash frac\{1\}\{11\}\; =\; \backslash frac\{1\}\{121\}$ of its natural velocity. It is therefore diminished $\backslash frac\{120\}\{121\}$ by the figure of the axis of the pulley, and the 11 pounds press it with $\backslash frac\{120\}\{121\}$ of their weight, that is, with $3\backslash frac\{1\}\{2\}$ lbs. pounds.

Of knowing the momentum of inertia;

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.

and consequently the force necessary to overcome it.

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 $\backslash frac\{m.GI.CP\}\{fpr^2\}$ of the angular velocity becomes $\backslash frac\{m.GI.CP\}\{m.CS^2\}$ ($\backslash text\{No\; \}\; 22.$); and these are equal by supposition. Therefore $fpr^2\; =\; m.CS^2$, and $CS\; =\; \backslash sqrt\{\backslash frac\{fpr^2\}\{m\}\}$.

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 $\backslash sqrt\{\backslash frac\{1\}\{3\}\}$ of its length.

In a circle or cylinder, such as the solid drum of a capstan, CS = $\backslash sqrt\{\backslash frac\{1\}\{2\}\}$ its radius, or nearly $\backslash frac\{1\}\{2\}$ radius. But if it turns round one of its diameters, CS = $\backslash frac\{1\}\{4\}$ radius.

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

If it turn round a diameter, CS = $\backslash sqrt\{\backslash frac\{1\}\{4\}\}$ radius. The surface of a sphere, or a thin spherical shell, turning round a diameter, has CS = $\backslash sqrt\{\backslash frac\{2\}\{3\}\}$ radius, or nearly $\backslash frac\{2\}\{3\}$ or $\backslash frac\{1\}\{2\}$.

A solid sphere turning round a diameter has CS = $\backslash sqrt\{\backslash frac\{2\}\{5\}\}$ radius, or nearly $\backslash frac\{2\}\{5\}$. 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 $\backslash frac\{fpr^2\}\{m\}$.

There is another way of making this correction of A simpler the motion of a machine, or allowing for the inertia of mode of allowing for the inertia of machine itself, which is rather simpler than the one now given. We can suppose a quantity of matter collected at the point to which the moving force is applied, such that its inertia will oppose the same resistance to 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 = $\backslash frac\{fpr^2\}\{CP^2\}$, 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 $\backslash frac\{m.GI.CP\}\{CP^2\}\; \backslash cdot\; \backslash frac\{fpr^2\}\{CP^2\}$ ($\backslash text\{No\; \}\; 22.$), which is = $\backslash frac\{m.GI.CP\}\{fpr^2\}$, the same as before.

A point O may be found, at such a distance from the 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 $\backslash frac\{m.GI.CO\}\{m.CO^2\}$ ($\backslash text\{No\; \}\; 22.$),

Rotation. (No 22.), which is $\backslash frac\{GI\}\{CO\}$. 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 $\backslash frac\{m.GI.CG\}\{f\backslash rho\; r^2\}$.
38 Therefore we have $\backslash frac\{GI\}\{CO\}\; =\; \backslash frac\{m.GI.CG\}\{f\backslash rho\; r^2\}$, and $\backslash frac\{CO\}\{GI\}\; =\; \backslash frac\{f\backslash rho\; r^2\}\{m.GI.CG\}$, and $CO\; =\; \backslash frac\{f\backslash rho\; r^2\}\{m.CG\}$. Also, as in No 31.
$GO\; =\; \backslash frac\{fA.GA^*\}\{m.CG\}$.

39 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 $\backslash frac\{fA.CA^*\}\{m.CG\}$, or $GO\; =\; \backslash frac\{fA.GA^*\}\{m.CG\}$, 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 $CO\; =\; \backslash frac\{fA.CA^*\}\{m.CG\}$ 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.

40 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 $CO\; =\; \backslash frac\{f\backslash rho\; r^2\}\{m.CG\}$.

a. If the body is a heavy straight line, suspended by one extremity, CO is $\backslash frac\{1\}\{2\}$ 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 $\backslash frac\{1\}\{2\}$ 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 $\backslash frac\{1\}\{2\}$ 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 $a\; +\; r\; +\; \backslash frac\{r^2\}\{a+r\}$; and the distance between its centres of bulk and oscillation is $\backslash frac\{r\}\{a+r\}$. Thus, in a common second's pendulum,

whose length at London is about $39\backslash frac\{1\}\{2\}$ inches, the centre of oscillation will be found about $\backslash frac\{1\}\{350\}$ of an inch below the centre of the ball, if it be two inches in diameter.