ROTATION

ROTATION.

Rotation. ROTATION, in mechanical philosophy, denotes the motion of the different parts of a solid body about an axis, in contradistinction to the progressive motion of the body in a straight line, or in an orbit about a distant centre, of which the rotation is absolutely independent, although both motions may take place simultaneously. Thus the planets have a progressive motion, or a motion of translation, in the orbits which they describe about the sun, while at the same time each revolves about an axis which passes through its centre of gravity.

In consequence of this double motion of the planetary bodies, the two great problems in physical astronomy are, first, to determine the orbit described in space by each body under the influence of the sun's attraction, but disturbed by the attraction of every other body in the system; and, secondly, to determine the circumstances of the rotation of each body under the influence of the same disturbing forces. This last problem is not of less importance or difficulty than the first. In the case of the earth it includes the determination of the precession of the equinoxes and the nutation of the axis under the disturbing action of the sun and moon; and its analysis has led to the discovery of two important facts connected with the physical constitution of the universe, namely, the invariability of the length of the day, and the permanence of the poles of rotation on the earth's surface.

But the problem of determining the rotation of a solid body is not merely interesting in consequence of its forming a principal part of physical astronomy; it is also one of the highest importance in practical mechanics; for the principles to which we are led in considering the general theory of rotatory motion, enable us to determine the effect or performance of machines, and the proper relations and most advantageous disposition of their several parts, in order to accomplish the object in view with the smallest waste of force, and the least strain or injury to the machine itself.

In treating the subject, we shall first endeavour to demonstrate the fundamental propositions of rotatory motion in as elementary a manner as possible; secondly, we shall show the methods of computing the inertia of bodies of different forms, and the properties of principal axes; thirdly, we shall consider the motion of a body which revolves about a fixed axis, and the properties of the centres of gyration, percussion, and oscillation; and, lastly, we shall give the general equations which determine the rotation of a body acted upon by any number of forces, and which is at liberty to move in any direction.

1. When a solid body, or system of material particles, so connected that their mutual distances remain invariable, turns round an axis, every particle describes a circle having its centre in the axis, and its plane perpendicular to the axis; and at every instant of the motion the particle is in the direction of a tangent to the circle, or in a straight line perpendicular to its radius vector. In order, therefore, to ascertain the direction of the motion of any particle of the system, we have only to draw a straight line from the particle perpendicular to the axis of rotation. This line will lie in the plane of the circle, and be its radius vector; and the perpendicular to the radius vector at its extremity gives the direction of the motion of the particle.

2. Since the system is supposed to be invariable in form, every particle describes the circumference of a circle in precisely the same instant of time; and as this is also true of any given portion of the circumference, it follows that the absolute velocities of the different particles are proportional to their distances from the axis of rotation. Hence,

if $w$ denote the velocity of the particle whose radius vector or distance from the axis = $l$, then $rw$ is the velocity of the particle whose distance from the axis = $r$. The velocity denoted by $w$ is called the angular velocity of the body; and the rotations of different bodies are compared in respect of velocity by comparing their angular velocities.

3. The nature of the forces which connect the different particles or atoms of a solid body, and render it impossible for any single particle to change its position without a corresponding change in the positions of all the other particles, is very obscurely known; but whatever their nature may be, there is one general fact or law, established by unexceptional experience, which affords a sufficient foundation for a dynamical theory, namely, that the forces by which the particles of a body act reciprocally on each other are equal. An immediate consequence of this law is, that when a solid body is in equilibrium between two external forces $A$ and $B$, these forces are equal and opposite; for the force $A$ being in immediate equilibrium with the opposite forces exerted by that particle of the solid body to which it is applied, must therefore be 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. In like manner, the compound force by which this 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 so on until we come to the particle to which the external force $B$ is immediately applied. And this particle being in equilibrium, the external force applied to it must be equal and opposite to the compound force exerted on it by the series of contiguous particles on that side; therefore the two forces $A$ and $B$ are equal and opposite.

Hence it follows, that when any number of external forces are applied to a solid body, and it is in equilibrium between them, the forces are such as would be in equilibrium if they were all applied at one point.

4. If a body is maintained in equilibrium by the action of three forces, these forces lie in the same plane, and are either parallel, or meet in the same point. This follows immediately from the composition of forces; for the forces being in equilibrium, any one of the three must be equal and opposite to the equivalent, or resultant, as it is termed, of the other two; and this resultant is the diagonal of a parallelogram of which the other two component forces are the sides. But the diagonal and sides of a parallelogram are in the same plane, therefore the third force is in the plane of the other two; and, further, since it is equal and opposite to the resultant of the other two, it must be in the same straight line, and consequently pass through their point of concurrence. These simple propositions are the foundation of the theory of statics.

5. In applying these propositions to explain the motion of rotation, we must recollect a fundamental theorem in dynamics, that the force which produces any motion is equal and opposite to the force which would prevent it, if applied at the same point, and in the same straight line, or which would extinguish the motion in the same time in which we suppose it to be produced. The force therefore which, acting on any particle, causes the body to revolve about an axis with a given angular velocity, is equal to the force which, if applied at the same point in the opposite direction, would reduce the body to a state of rest in the same time as was necessary to produce the given velocity.

6. The only distinct notion we can form of the intensity of any moving force, is the quantity of motion which

Rotation. it can produce by acting uniformly during a given time. This will be ascertained when we know the velocity impressed upon a body containing a given quantity of matter. Thus we know that terrestrial gravity acting on a body for a second, will cause it to fall sixteen 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 thirty-two feet per second. The force of gravity is therefore such as to communicate a velocity of thirty-two feet per second; and in this case the mass of the body is not considered, for gravity acts equally on every particle of which it is composed. In the same manner, the best way of acquiring a distinct conception of the rotatory effect of a moving force, is to determine the quantity of rotatory motion which it can generate by acting uniformly during some known time.

7. If a solid body, or rigid system of invariable form, admit of a rotatory motion about an axis, and if at any point of the body a force $P$ be applied, which acts in a plane perpendicular to the axis of rotation, and tends to turn the body in one direction; and to any other point in the same plane a force $Q$ be applied, tending to turn the body in the opposite direction; then, if the two forces be to each other inversely as the distances of the lines of their directions respectively from the axis, the body will remain at rest, or the two forces will be in equilibrium.

For let $O$ (fig. 1) be the axis of rotation, and suppose the force $P$ to be applied at the point $P$ in the direction $PA$, and tending to turn the solid body about the axis $O$ perpendicular to the plane of the paper, and let the force $Q$ act in the direction $QB$, and consequently tend to turn the body in the opposite direction, and draw $OA$ perpendicular to $PA$, and $OB$ perpendicular to $QB$. Then, in order that there may be equilibrium, we must have, by the principle of the lever, $P\; :\; Q\; ::\; OB\; :\; OA$, whence $P\; \backslash cdot\; OA\; =\; Q\; \backslash cdot\; OB$.

8. It is obviously indifferent at what point of the plane the force is applied, provided its distance from the axis remains the same. Let $BO$ therefore be produced, and take $OC$ equal to $OA$, and draw $CR$ perpendicular to $BC$, or parallel to $BQ$. We may now suppose the force $P$ to be replaced by an equal force $R$, acting in the direction $RC$; then, since $R\; \backslash cdot\; OC\; =\; P\; \backslash cdot\; OA$, we have $R\; \backslash cdot\; OR\; =\; Q\; \backslash cdot\; OB$, and, consequently, the axis being supposed fixed, the body will remain in equilibrium between the two forces $Q$ and $R$. These forces are therefore equivalent to a single force applied at $O$, equal to their sum, and acting in a parallel direction. Hence the resultant of two parallel forces applied to a straight line is a force parallel to the component forces, equal to their sum, and whose line of direction divides the straight line into parts reciprocally proportional to the forces themselves.

9. If two forces, which are inversely as the distances of their respective directions from the axis of rotation, be applied to a body, the equilibrium will still be maintained, although the two forces are not in the same plane. Let $OO\text{'}$ be two points in the axis of rotation, through which let there be drawn two planes $Aa$ and $Bb$ at right angles to the axis. Suppose also these planes to be intersected by another plane (here supposed to be the plane of the figure) passing through the axis, and let $OA$, $O\text{'}B\text{'}$ be the intersections, and let the straight line which joins $A$ and $B$ intersect the axis in $C$. Let $P$ and $Q$ be two forces applied respectively at $A$ and $B$ in the planes $Aa$ and $Bb$, and at right angles to $OA$ and $O\text{'}B\text{'}$, or perpendicular to the plane of the paper, then these two forces are parallel; and because by hypothesis $P\; :\; Q\; ::\; O\text{'}B\text{'}\; :\; OA$,

we have $P\; :\; Q\; ::\; CB\; :\; CA$; therefore (8) the two forces $P$ and $Q$ are equivalent to a single parallel force passing through $C$. But a force passing through the axis has no tendency to turn the body round the axis; consequently the body remains at rest.

10. Let $m,\; m\text{'},\; m\text{'}\text{'},\; \backslash dots$ (fig. 3) be a system of bodies immovable in respect of each other, and suppose them to revolve about an axis passing through $O$; also let $m,\; m\text{'},\; m\text{'}\text{'},\; \backslash dots$ denote the quantities of matter in the bodies respectively, and let $r,\; r\text{'},\; r\text{'}\text{'},\; \backslash dots$ denote their distances from the axis of rotation. Let the angular velocity of the system be $\backslash omega$; then (2) the absolute velocities of the bodies are respectively $\backslash omega\; r,\; \backslash omega\; r\text{'},\; \backslash omega\; r\text{'}\text{'},\; \backslash dots$. Now let us suppose the system to be put in motion by an external force $F$ applied at the point $P$, in the direction $PA$, which is contained in a plane perpendicular to the axis of rotation; then the relation between the force $F$ and the velocity $\backslash omega$ may be found as follows. Considering, first, the body $m$; since its velocity $=\; \backslash omega\; r$, and its quantity of matter $=\; m$, its quantity of motion $=\; \backslash omega\; rm$. But it is a principle in dynamics, that the quantity of motion of a moving body is proportional to the impressed force, and may be taken as a measure of that force; whence $\backslash omega\; rm$ is the measure of a force which, if applied at $m$, in a direction perpendicular to $Om$, would cause $m$ to begin to move with the actual velocity $\backslash omega\; r$. Let $f$ be the force which, if applied at $P$ in the direction $PA$, would make equilibrium with the force $\backslash omega\; rm$ at $m$; draw $OA$ perpendicular to $PA$, and make $OA\; =\; h$. We have then (7)

whence $fh\; =\; \backslash omega\; r^2\; m$. In like manner, the quantities of motion in the bodies $m\text{'},\; m\text{'}\text{'},\; \backslash dots$ are $\backslash omega\; r\text{'}\; m\text{'},\; \backslash omega\; r\text{'}\text{'}\; m\text{'}\text{'},\; \backslash dots$; and if we denote by $f\text{'},\; f\text{'}\text{'},\; \backslash dots$ the forces which, if applied at $P$ in the direction $PA$, would respectively make equilibrium with the forces $\backslash omega\; r\text{'}\; m\text{'},\; \backslash omega\; r\text{'}\text{'}\; m\text{'}\text{'},\; \backslash dots$, we shall have $f\text{'}\; h\; =\; \backslash omega\; r\text{'}^2\; m\text{'},\; f\text{'}\text{'}\; h\; =\; \backslash omega\; r\text{'}\text{'}^2\; m\text{'}\text{'},\; \backslash dots$; therefore, if we assume

we shall have the equation

If we suppose $m\; =\; m\text{'}\; =\; m\text{'}\text{'}\; =\; \backslash dots$, and take $r$ to denote the distance of any particle from the axis of rotation, the equation becomes $F\; \backslash cdot\; h\; =\; \backslash omega\; \backslash sum\; (r^2\; m)$; the summation denoted by $\backslash Sigma$ extending to every particle in the system.

11. When the direction of the external force $F$ is not in a plane parallel to the circles of rotation, it must be resolved into two forces, one of which lies in that plane, and the other is parallel to the axis. It is only the first of these forces which in the present case is regarded as the moving force; the second merely tends to push the body in the direction of the axis, and does not influence the rotation. When we come to consider the rotation of a body perfectly free, it will be necessary to attend particularly to this circumstance.

12. A solid body may be regarded as composed of molecules of a very small but finite magnitude, connected so as to form an invariable system; and although a solid, according to his view of the constitution of body, is formed of discrete elements, the sums relative to the molecules may be changed into definite integrals without sensible error; that is to say, for the sums $\backslash sum\; r^2\; m$ we may substitute the definite integrals $\backslash int\; r^2\; dm$, where $dm$ is the element of the mass, and the integrals extend to the whole mass. We have then $F\; \backslash cdot\; h\; =\; \backslash omega\; \backslash int\; r^2\; dm$.

13. The energy of a force applied to a lever, or its power of producing motion round the fulcrum, is expressed by the product of the force into the perpendicular drawn from the and of the fulcrum upon the line of its direction. This product (denoted above by $F\; \backslash cdot\; h$) is called the moment, or momentum, or rotatory effort, of the force $F$. And because $\backslash int\; r^2\; dm$ is

Rotation. the sum of the moments of all the particles of the body in actual motion, this integral expresses the energy or effort of all the resistances made to the rotation by the inertia of the particles, and is therefore called the moment of inertia of the moving body with respect to the axis from which the distances $r$ are reckoned. We have therefore these definitions:

The moment of a force in respect of a given axis, is the product of the force into the perpendicular from the axis upon the line of its direction.

The moment of inertia of a body in respect of any axis is the sum of the products obtained by multiplying each particle of the body into the square of its distance from the axis.

14. From the equation $F\; \backslash cdot\; h\; =\; wfr^2\; dm$ (12), we have the following expression for the angular velocity, namely,

that is to say, the angular velocity about any axis is equal to the moment of the external or applied force divided by the moment of inertia. This fraction expresses a number; for the force $F$ may be represented by the product of a mass $M$ into a certain velocity $v$, so that $F\; =\; Mv$; and therefore, since $v$ represents a line, the numerator and denominator have both the same dimension. The fraction also expresses the part of the radius which is equal to the arc measuring the angle; and since the radius is equal to the arc of $57^\backslash circ\; 29.58$, if we make the radius $=\; 1$, we have only to multiply the fraction by $57.2958$ in order to obtain the value of $\backslash omega$ in degrees.

15. Since the relation between the angular velocity and the external impelling force is expressed in terms of the moment of inertia, it is evident that this moment is an important element in the theory of rotation. We shall therefore now proceed to consider the subject under a more general view, and point out the method by which it is computed for bodies of any form revolving about given axes.

Let the different points of a solid body be referred to three rectangular co-ordinates $x,\; y,\; z$, and let $dm$ be the differential element of the mass; then the distance of any particle from the axis $x$ is $\backslash sqrt\{(y^2\; +\; z^2)\}$, and, by the definition, the moment of inertia relative to the axis $x$, is $\backslash int\; (y^2\; +\; z^2)\; dm$, the integral extending to the whole mass of the body. In like manner, the moments of inertia relative to the axes $y$ and $z$ are respectively $\backslash int\; (x^2\; +\; z^2)\; dm$ and $\backslash int\; (y^2\; +\; x^2)\; dm$. It is thus evident that the determination of the moments of inertia, with respect to any body whatever, consists in forming the analytical expression of the product of the differential element of the body into the square of its distance from the axis, and integrating for the whole extent of the body. The integration may always be effected by the method of quadratures.

16. Since the same body may be referred to an infinite number of different axes, in respect of all of which the moments of inertia will have different values, it follows that the moment cannot be absolutely defined unless it be referred to a determinate axis. But when it is required to investigate the moments of inertia of a body in respect of several axes successively, it is not always necessary to compute the formula $\backslash int\; r^2\; dm$ for each case. For example, when the moment of inertia has been determined in respect of any one axis, we may find its value in respect of any other parallel axis without further calculation, if we know the mass of the body and the distances of both axes from the centre of gravity.

17. The moment of inertia of any body in respect of an axis $AO$ (fig. 4) being given, to find the moment of inertia in respect of another axis $oa$ parallel to $OA$.

Let the origin of the rectangular co-ordinates be at $O$, and let $x$ be taken on the axis $OA$, and $y$ in the plane of

Rotation. $OA$ and $oa$. Also let $Z$ be the place of an element $dm$, Rotation. and $Y$ the projection of $Z$ on the plane $xy$; and from $Y$ draw $YX$ perpendicular to $OA$, meeting $OA$ in $X$ and $oa$ in $x$; then $OX\; =\; x,\; XY\; =\; y,\; YZ\; =\; z$.

Fig. 4.

The moment of inertia is $\backslash int\; r^2\; dm\; =\; \backslash int\; (y^2\; +\; z^2)\; dm$. But by hypothesis this is given; suppose it $=\; Mh^2$ ($M$ being the mass of the body), and we have $\backslash int\; (y^2\; +\; z^2)\; dm\; =\; Mh^2$. In like manner, if we assume the distance of $oa$ from $OA$ to be $k$, we have $ox\; =\; x,\; xY\; =\; k\; +\; y,\; YZ\; =\; z$, and the moment of inertia in respect of $oa$ is $\backslash int\; \backslash \{(k\; +\; y)^2\; +\; z^2\backslash \}\; dm\; =\; k^2\; \backslash int\; dm\; +\; 2k\; \backslash int\; y\; dm\; +\; \backslash int\; (y^2\; +\; z^2)\; dm$. But $\backslash int\; k^2\; dm\; =\; Mk^2$, and $\backslash int\; (y^2\; +\; z^2)\; dm\; =\; Mh^2$; therefore the moment in respect of the new axis is $Mh^2\; +\; Mk^2\; +\; 2k\; \backslash int\; y\; dm$. Suppose the centre of gravity of the body to be at $G$; let $GF$ be perpendicular to the plane $xy$, and draw $FE$ perpendicular to the axes, meeting $OA$ in $E$ and $oa$ in $e$. From the nature of the centre of gravity $\backslash int\; y\; dm\; =\; M\; \backslash cdot\; EF$; therefore the moment in respect of $oa$ becomes $Mh^2\; +\; Mk^2\; +\; 2k\; \backslash cdot\; M\; \backslash cdot\; EF$. But $k\; =\; eE$, and $2M\; \backslash cdot\; k\; \backslash cdot\; EF\; =\; 2M\; \backslash cdot\; eE\; \backslash cdot\; EF\; =\; M(eE^2\; -\; k^2\; -\; EF^2)$; whence we have finally for the moment of inertia in respect of the new axis $oa$,

Corollary 1. If the original axis $OA$ passes through the centre of gravity $G$, then $EF$ vanishes, and the moment in respect of the new axis $oa$ becomes $Mh^2\; +\; M\; \backslash cdot\; eE^2$, or $Mh^2\; +\; Mk^2$. And if $oa$ passes through the centre of gravity, then $eE$ vanishes, and the moment in respect of $oa$ becomes $Mh^2\; -\; M\; \backslash cdot\; EF^2$. Hence we infer that the moment of inertia in respect of the axis which passes through the centre of gravity is less than the moment in respect of any other parallel axis.

Corol. 2. The moment of inertia is the same in respect of every axis parallel to that which passes through the centre of gravity, and is at the same distance from it; or the same in respect of every straight line which lies in the surface of a cylinder having the centre of gravity in its axis.

18. We shall now give a few examples of the application of the preceding formula.

I. Let it be proposed to find the moment of inertia of a very slender rod (the breadth and thickness of which may be neglected), in respect of an axis passing through its extremity and perpendicular to its length.

Let $a$ be the length of the rod, and $x$ the distance of any point from the axis, then the element of the body is $dx$, and the moment of inertia $=\; \backslash int\; x^2\; dx$, the limits of the integral being $x\; =\; 0$ and $x\; =\; a$. For those limits we have $\backslash int\; x^2\; dx\; =\; \backslash frac\{1\}\{3\}\; a^3$, which therefore is the moment of inertia.

II. To find the moment of inertia of a very thin rectangle, the thickness of which may be neglected, in respect of an axis coinciding with one of its sides.

Let the side which coincides with the axis $=\; a$, and the other side $=\; b$, and let $x$ be taken in the axis. The co-ordinates of any point in the rectangle being $x,\; y$, the element of the body is $dx\; dy$, and the distance from the axis $=\; y$, therefore the moment of inertia is $\backslash int\; y^2\; dx\; dy$. The integral of this in respect of $x$ is $\backslash int\; y^2\; dy$, which from $x\; =\; 0$ to $x\; =\; a$ gives $a\; \backslash int\; y^2\; dy$. A second integration gives $\backslash frac\{1\}\{3\}\; ay^3$, which from $y\; =\; 0$ to $y\; =\; b$, becomes $\backslash frac\{1\}\{3\}\; ab^3$; therefore, since $M\; =\; ab$, $\backslash int\; y^2\; dx\; dy\; =\; \backslash frac\{1\}\{3\}\; ab^2\; =\; \backslash frac\{1\}\{3\}\; M\; \backslash cdot\; b^2$.

III. To find the moment of inertia of a circle in respect of one of its diameters.

Let $a$ be the radius of the given circle, $r$ the radius of a concentric circle passing through a point $P$ whose co-ordinates are $x$ and $y$, $x$ being taken on the axis of rotation, and $r\; +\; dr$ the radius of a second concentric circle, very near the other. If we now assume $dz\; =\; \backslash sqrt\{(dx^2\; +\; dy^2)\}$, then $dz\; \backslash cdot\; dr$ is the area of the elementary surface included between

Rotation. the circumferences of the two circles whose radii are respectively $r$ and $r\; +\; dr$, and two straight lines drawn from the centre to the extremities of a very small arc at $P$; that is to say, we have $dm\; =\; dzdr$. But the distance of this element from the axis is $y$; therefore the moment of inertia is $ffy^2dzdr$. Now, by the nature of the circle, $dz\; :\; dx\; ::\; r\; :\; y$, whence $ydz\; =\; rdx$, and the double integral becomes $ffydx\; \backslash times\; fdr$. Now, on the supposition that $r$ is constant, the integral $ffydx\; =\; \backslash pi\; r^2$ = area of the circle whose radius is $r$; therefore the integration in respect of $dx$ gives $\backslash pi\; r^2\; dr$; and on integrating this from $r\; =\; 0$ to $r\; =\; a$, we get ultimately $\backslash frac\{1\}\{2\}\; \backslash pi\; a^4$ for the moment of the circle whose radius is $a$.

Since in this case the mass $M\; =\; \backslash pi\; a^2$ = the area, we have $\backslash frac\{1\}\{2\}\; \backslash pi\; a^4\; =\; \backslash frac\{1\}\{2\}\; Ma^2$, which is another expression for the moment of inertia of a circle in respect of one of its diameters.

IV. To find the moment of inertia of a circle in respect of an axis perpendicular to its plane, and passing through its centre.

Supposing, as in the last proposition, the surface to be composed of elementary rings, we have $dm\; =\; dzdr$, whence $ff^2dm\; =\; ff^2dr\; \backslash times\; fdx\; =\; 2\backslash pi\; ff^2dr$ (since $fdx\; =\; 2\backslash pi\; r$), the integral of which, from $r\; =\; 0$ to $r\; =\; a$, gives $ff^2dm\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; a^4\; =\; \backslash frac\{1\}\{2\}\; Ma^2$.

V. To find the moment of inertia of a solid body generated by the revolution of a curve line in respect of the axis of rotation.

Let $AB$ (fig. 5) be a section perpendicular to the axis of rotation $OX$, and meeting it in $C$, and make $OC\; =\; x$. Suppose the solid to be divided into thin slices by planes parallel to $AB$, and let $ab$ be the section next to $AB$, so that $Cc\; =\; dx$. Let $P$ be a point in the plane $AB$, and make $CP\; =\; r$, $Cp\; =\; r\; +\; dr$. The length of the circumference described by $P$ in one revolution is $2\backslash pi\; r$, therefore the mass of the ring comprised between $AB$ and $ab$, and traced by the revolution of the points $P$ and $p$, is $2\backslash pi\; r\; dr\; dx$; and its distance from the axis is $r$, therefore the moment of inertia of the ring is $2\backslash pi\; r^2\; dr\; dx$. Integrating this in respect of $r$ from $r\; =\; 0$ to $r\; =\; CA\; =\; y$, we get $\backslash frac\{1\}\{2\}\; \backslash pi\; y^2\; dx$ for the moment of the slice comprised between the two sections of the solid, whose distances from $O$ are $x$ and $x\; +\; dx$. Hence the moment of inertia of all the slices, or of the solid body, is $\backslash frac\{1\}\{2\}\; \backslash pi\; \backslash int\; y^2\; dx$.

Applying this formula to a sphere whose radius = $a$, we get from the equation $y^2\; =\; 2ax\; -\; x^2$, $y^2\; =\; 4a^2x^2\; -\; 4ax^3\; +\; x^4$, on substituting which, and integrating, we have $\backslash frac\{1\}\{2\}\; \backslash pi\; \backslash int\; y^2\; dx\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; a^2\; x^2\; -\; \backslash frac\{1\}\{2\}\; \backslash pi\; ax^3\; +\; \backslash frac\{1\}\{10\}\; \backslash pi\; x^5$, which from $x\; =\; 0$ to $x\; =\; a$ gives $\backslash frac\{1\}\{2\}\; \backslash pi\; \backslash int\; y^2\; dx\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; a^5$ for the hemisphere, and consequently the moment of inertia of the whole sphere = $\backslash frac\{1\}\{2\}\; \backslash pi\; a^5\; =\; \backslash frac\{1\}\{2\}\; M\; \backslash cdot\; a^2$, since $M\; =\; \backslash frac\{4\}\{3\}\; \backslash pi\; a^3$.

The moment of inertia of a spheroid revolving about its shorter axis is $\backslash frac\{1\}\{2\}\; M(a^2\; +\; b^2)$. See the calculation of $fx^2dm$ and $fy^2dm$ in this case in the article PRECESSION, vol. xviii. p. 508.

In all the preceding examples, the density of the body has been supposed uniform, and = 1. For bodies of different densities the formulae must be multiplied in each case by the density. Thus, in the last example, the moment of inertia of a sphere whose radius = $a$, and density = $\backslash rho$, the moment of inertia in respect of one of the diameters is $\backslash frac\{1\}\{2\}\; \backslash rho\; \backslash pi\; a^5$.

19. The examples which have now been given will suffice to illustrate the method of computing the moments of inertia of bodies of any form in respect of particular axes; and they render it obvious that the computation will in general be greatly facilitated when the mass is symmetrically disposed about the axis for which the inertia is to be computed. It is likewise evident that the amount of rotatory inertia de-

pends on the position of the axis of rotation. Hence, for Rotation. every point of a mass there must exist at least one axis in respect of which the inertia is less, or greater, than in respect of any other axis passing through that point. The axes for which the moments of inertia are the greatest or least possible are called principal axes; and they possess some very remarkable properties of great importance in the theory of rotation. We now proceed to investigate the position and properties of the principal axes.

20. The moments of inertia of a mass $M$ being given in respect of three rectangular axes $OX$, $OY$, $OZ$ (fig. 6) passing through $O$, to find the moment in respect of another axis $OG$ given in position, and also passing through the origin $O$.

Let $P$ be the position of a particle $dm$; through $P$ draw $PQ$ parallel to the axis $OZ$, meeting the plane $XOY$ in $Q$, and draw $QR$ parallel to $OY$, meeting $OX$ in $R$; then $OR\; =\; x$, $RQ\; =\; y$, $QP\; =\; z$. Also suppose the plane which passes through $OG$, and is perpendicular to $XOY$, to meet the plane $XOY$ in $OF$, and let the angle $XOF\; =\; \backslash phi$, and the angle $FOG\; =\; \backslash psi$. The axis $OG$ is entirely determined by these two angles, and consequently, when $OG$ is given in position, $\backslash phi$ and $\backslash psi$ are also given. In order to compute the moment of inertia in respect of the axis $OG$, it is convenient to refer the different points of the mass, as $P$, to a new system of rectangular axes, of which one is $OG$. To effect this, in the first place let $OY\text{'}$ be drawn perpendicular to $OF$ in the plane $XOY$, and let $P$ be referred to the axes $OF$, $OY\text{'}$, $OZ$, and draw $QR\text{'}$ perpendicular to $OF$. We have then $OR\text{'}\; =\; x\text{'}$, $R\text{'}Q\; =\; y\text{'}$, $QP\; =\; z$, and the new co-ordinates expressed in terms of the former, and the angles $\backslash phi$ and $\backslash psi$ are

In the second place, we may pass in like manner from these to a third system of co-ordinates $x\text{'}\text{'}$, $y\text{'}\text{'}$, $z\text{'}\text{'}$, of which $x\text{'}\text{'}$ is in the line $OG$, $y\text{'}\text{'}$ remains parallel to $OY\text{'}$, and $z\text{'}\text{'}$ is perpendicular to $OG$ in the plane $FOG$. We have, then, similarly,

Substituting in these last equations the values of $x\text{'}$, $y\text{'}$, and $z\text{'}$, given by the three former, we have

If we now assume the distance of $P$ from the axis $OG$ to be $r$, we shall have $r^2\; =\; y\text{'}\text{'}^2\; +\; z\text{'}\text{'}^2$; whence, on forming from the two last equations the expression for $ff^2dm\; =\; f(y\text{'}\text{'}^2\; +\; z\text{'}\text{'}^2)dm$, and, for the sake of abridging, putting

the expression for the moment of inertia in respect of the axis $OG$ becomes

This expression for the moment of inertia in respect of the axis $OG$ is given by Euler (Theoria Motus Corporum Solidorum, p. 169).

21. The values of the angles $\backslash phi$ and $\backslash psi$, which give the position of the axis for which the moment $ff^2dm$ is a maximum or minimum, may be deduced from the above equation in the usual manner, that is to say, by differentiating successively for each of the variables, and making the differential co-efficient = 0. The differentiation in respect of $\backslash phi$ (supposing $\backslash psi$ constant) gives the equation

Rotation. and the differentiation in respect of $\backslash psi$ (supposing $\backslash phi$ constant) gives

On eliminating $\backslash psi$ from these two equations, there results an equation of the following form, in which $u\; =\; \backslash tan.\; \backslash phi$, and $e,\; f,\; g,\; h$, are co-efficients depending upon the quantities $A,\; B,\; C,\; D,\; E,\; F$, viz.

This being a cubic equation, must have at least one real root. But it is easy to see, from the nature of the thing, that all the roots are real; for if there be one axis for which the moment of inertia is a maximum, there must evidently be a second, in respect of which it is a minimum, and vice versa. Hence the equation must have two real roots, and therefore, by the theory of cubic equations, all the roots are real. We are therefore led to this very important conclusion, that through any point of every solid body, however irregular in its form, there may be drawn a system of three axes, such that the differential co-efficient of the moment of inertia in respect of each of them is nothing. In respect of one of these, the moment is a maximum, in respect of another, it is a minimum, and though, in respect of the third it is neither a maximum nor a minimum, it is still such that its value is not altered by a small change in the position of the axis. According to the definition in (19), these are the principal axes of the body.

22. It would be difficult, in general, to determine the position of the principal axes from the cubic equation found above; but they possess certain properties, by means of which they may be readily found in almost every case of practical application. On comparing the two equations for $\backslash tan.\; \backslash psi$ and $\backslash tan.\; 2\backslash psi$ (21), with the expressions for $x\text{'},\; y\text{'},\; z\text{'}$ (20), it will be seen that the first is identical with $2\; \backslash cos.\; \backslash psi\; \backslash frac\{dx\text{'}\; y\text{'}\; dm\}\{dx\text{'}\; z\text{'}\; dm\}$, and the second with $-2\; \backslash frac\{dx\text{'}\; z\text{'}\; dm\}\{dx\text{'}\; y\text{'}\; dm\}$; whence $dx\text{'}\; y\text{'}\; dm\; =\; 0$, and $dx\text{'}\; z\text{'}\; dm\; =\; 0$; and it follows, that if $x$ be taken in one of the principal axes, we have, in respect of that axis, $fxz\; dm\; =\; 0,\; fzx\; dm\; =\; 0$, or the quantities denoted by $D$ and $E$ respectively vanish.

23. From this it follows, that if all the matter in a body of uniform density be symmetrically disposed about any line, that line is one of the principal axes of rotation (for if $x$ be taken in the axis of symmetry, then to every element whose co-ordinates are $x,\; y$, there corresponds another element whose co-ordinates are $x,\; -y$, whence $fxz\; dm\; =\; 0$); therefore, in the case of symmetrical figures, the position of one of the principal axes is known, and the positions of the other two are thence readily determined by means of the two equations (21). On making $D\; =\; 0,\; E\; =\; 0$, the first of these equations gives

whence $\backslash cos.\; \backslash phi\; =\; 0$, and $\backslash sin.\; \backslash phi\; =\; 1$; or $\backslash phi\; =\; 90^\backslash circ$. Substituting this value of $\backslash phi$ in the second of the equations (21), and making at the same time $D\; =\; 0,\; E\; =\; 0$, we get $\backslash tan.\; 2\backslash psi\; =\; 2F\; \backslash div\; (B\; -\; C)$; or $2\backslash psi$ is the angle which has $2F\; \backslash div\; (B\; -\; C)$ for its tangent. Hence the angle $\backslash psi$ has a double value, one corresponding to FOG, and the other to FOG $\backslash pm\; 90^\backslash circ$.

24. Hence also it appears, that when one principal axis OX is determined, two others may be found having the same property of maximum or minimum, and, by reason of $\backslash phi\; =\; 90^\backslash circ$, these axes are both in a plane perpendicular to OX. As this conclusion holds good with respect to each of the three axes, it follows that the three principal axes are at right angles to each other, and also, since $fxz\; dm\; =\; 0$, and $fzx\; dm\; =\; 0$, that $fyx\; dm\; =\; 0$. We have therefore, in respect of the principal axes, the three equations

If the origin of the co-ordinates is placed at the centre of gravity, we have likewise

by the well-known property of the centre of gravity.

25. By means of these equations, the proposition in (20) is considerably simplified when the three axes for which the moments are respectively given are principal axes. Let OX, OY, OZ (fig. 6) be the three principal axes passing through O, then $D\; =\; 0,\; E\; =\; 0,\; F\; =\; 0$, and the equation (20) for the moment, in respect of the axis OG, becomes

Suppose the given moments in respect of the principal axes to be respectively $M\text{'}a^4,\; M\text{'}b^4,\; M\text{'}c^4$; then, from the values of $A,\; B,\; C$ (20) we have $M\text{'}a^2\; =\; B\; +\; C,\; M\text{'}b^2\; =\; A\; +\; C,\; M\text{'}c^2\; =\; A\; +\; B$, whence

on substituting which in the above equation, we get $fr^2\; dm\; =\; M\; (a^2\; \backslash cos^2\; \backslash phi\; \backslash cos^2\; \backslash psi\; +\; b^2\; \backslash sin^2\; \backslash phi\; \backslash cos^2\; \backslash psi\; +\; c^2\; \backslash sin^2\; \backslash psi)$. If we now refer the new axis OG to the co-ordinates OX, OY, OZ, and make the angle XOG = $\backslash alpha,\; YOG\; =\; \backslash beta,\; ZOG\; =\; \backslash gamma$, and observe that because the two planes XOF and FOG are at right angles, $\backslash cos.\; \backslash alpha\; =\; \backslash cos.\; \backslash phi\; \backslash cos.\; \backslash psi,\; \backslash cos.\; \backslash beta\; =\; \backslash sin.\; \backslash phi\; \backslash cos.\; \backslash psi,\; \backslash cos.\; \backslash gamma\; =\; \backslash sin.\; \backslash psi$, the equation becomes

It is to be remarked, that as the co-ordinates are rectangular, $\backslash cos^2\; \backslash alpha\; +\; \backslash cos^2\; \backslash beta\; +\; \backslash cos^2\; \backslash gamma\; =\; 1$, so that the position of OG is determined by any two of these angles. If we therefore suppose the moments in respect of two of the principal axes to be equal, for example, if $a^2\; =\; b^2$, the formula for the moment in respect of OG becomes

Now when $\backslash gamma\; =\; 90^\backslash circ$, that is, when OG lies in the plane XOY, then $fr^2\; dm\; =\; M\; a^2$. Hence, if the moments in respect of the axes OX and OY be equal, the moment will be the same for every axis passing through O, contained in the plane XOY.

When the moments in respect of the three principal axes are equal, that is, when $a^2\; =\; b^2\; =\; c^2$, then

so that the moment is the same in respect of any axis whatever which passes through O. Thus it appears, that for every point of a solid body there are either three principal axes of rotation only, or else an infinite number.

26. These conclusions are obviously true in the case of some simple figures. Thus, in the spheroid of revolution, the moments are equal in respect of every diameter which lies in the plane of the equator; and in the sphere they are the same in respect of any diameter whatever, the bodies being always supposed homogeneous. The existence of three principal axes in every body was first noticed by Segner in 1755, but demonstrated for the first time by Albert Euler, in a memoir which received the prize of the Academy of Sciences of Paris in 1760.

27. Having established these general propositions respecting the moments of inertia and principal axes of bodies endowed with a rotatory motion, we now proceed to consider more particularly the rotation of a solid body about a fixed axis.

When a force is impressed on a body which can only move round a fixed axis, the difference between the motion which actually takes place, and that which would take place if the body were free, is evidently owing to the connexion which its parts have with the axis, and to the action of the points of support upon the axis. This action must be viewed in the light of another external force, such that, if combined with the force impressed on the body, it would produce in the body unconnected with any fixed points the motion which actually takes place.

Rotation. 28. In order to determine the strain on the axis, let $OY$ (fig. 7) be the axis of rotation, retained in a fixed position by the two pivots $S,\; T$, and suppose a force $F$ applied to the body at $P$, in the direction $PQ$, which is supposed to lie in a plane perpendicular to the axis, and containing $PQ$, let the rectangular axes $OX,\; OY$ be drawn, and let $x,\; y,\; z$ be the co-ordinates of a particle $dm$, the projection of which on the plane $XY$ is at $A$, and let $r$ denote its distance from the axis of rotation, so that $r\; =\; OA\; =\; \backslash sqrt\{(x^2\; +\; y^2)\}$; also let $\backslash phi$ be the angle which $OA$ makes with $OX$. If we now suppose $\backslash omega$ to be the angular velocity communicated to the body, then $r\backslash omega$ is the velocity with which the point $A$ revolves, and the direction of this velocity is in the perpendicular to $OA$, and from $OX$ towards $OY$, whence the velocity of $A$ in the direction $OX$ is $-r\backslash omega\; \backslash sin\; \backslash phi$, and $r\backslash omega\; \backslash cos\; \backslash phi$ in the direction $OY$. But $r\; \backslash sin\; \backslash phi\; =\; y$, and $r\; \backslash cos\; \backslash phi\; =\; x$; therefore the velocity of $A$ in the direction $OX$ is $-wy$, and in the direction $OY$ it is $+wx$. The quantities of motion in these directions are therefore respectively $-wrydm$ and $+wrxdm$; and the integrals of those expressions extended to the whole mass of the moving body give the whole quantities of motion in those directions, namely, $-wrydm$, and $+wrxdm$. On denoting the mass by $M$, and the co-ordinates of its centre of gravity by $x\_1$ and $y\_1$, we have (from the well-known expressions for these co-ordinates) $frydm\; =\; My\_1$, and $fxzdm\; =\; Mx\_1$, therefore $-wrydm\; =\; -wMy\_1$, and $+wrxdm\; =\; +wMx\_1$.

29. Now, since the body is supposed to be put in motion by a force $F$ acting in a plane parallel to $OX$, let $\backslash alpha$ and $\backslash beta$ be the angles which its direction makes with parallels to $OX$ and $OY$ respectively; then the components of $F$ in those directions are $F\; \backslash cos\; \backslash alpha$ and $F\; \backslash cos\; \backslash beta$. But these two forces are evidently equal respectively to the two former, together with the pressures on the pivots estimated in the same directions; therefore, calling $p$ and $q$ the pressures sustained by the point $S$ in the directions $OX$ and $OY$, and $p\text{'},\; q\text{'}$ the pressures sustained by $T$ in the same directions, we shall have

In order to determine $p,\; p\text{'},\; q,\; q\text{'}$, it is still necessary to find two other equations. These will be obtained by considering, that in respect of the plane $XOY$, the moments of the forces acting on the two pivots, resolved in the directions $x$ and $y$, must be equal to the moments of the resultants of all the forces or quantities of motion due to the rotation estimated in the same directions. Let $k\; =\; OS$, and $k\text{'}\; =\; OT$, the moments of the resolved forces which act on the pivots in the directions parallel to $OX$ and $OY$ are $pk\; +\; p\text{'}k\text{'}$ and $qk\; +\; q\text{'}k\text{'}$; and the resultants of the quantities of motion in the same two directions being $-wrydm$ and $+wrxdm$, their moments in respect of the plane $XOY$ are respectively $-wryzdm$ and $+wrxzdm$; therefore

Combining these with the preceding two equations, $p,\; p\text{'},\; q,\; q\text{'}$ are obtained in terms of known quantities.

30. If the axis of rotation $OZ$ be one of the principal axes of the system which intersect at $O$, then, by (24), $fryzdm\; =\; 0$ and $fxzdm\; =\; 0$, whence $pk\; +\; p\text{'}k\text{'}\; =\; 0$, and $qk\; +\; q\text{'}k\text{'}\; =\; 0$, and consequently $k$ and $k\text{'}$ both vanish, and the pressures sustained by the axis pass through the origin $O$. Let $R$ represent the single force or shock which the axis supports in this case, and $\backslash phi$ the angle which its direction makes with the axis $OX$; then, by (29), $R\; \backslash cos\; \backslash phi\; =\; F\; \backslash cos\; \backslash alpha\; +\; wMy\_1$, $R\; \backslash sin\; \backslash phi\; =\; F\; \backslash cos\; \backslash beta\; -\; wMx\_1$, from which two equations $R$ and $\backslash phi$ are both determined.

31. It would be a desirable thing in machines which

derive their efficacy from a rotatory motion, to apply the Rotation. pressures arising from the power and from the resistance Applica- opposed by the work in such a manner as to annihilate or tion to ma- diminish this pressure on the supports of the axis of motion. chines. 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 to be 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 $3\backslash frac\{1\}\{2\}$ pounds. For the direction of the pressure $R$ being in this case parallel to the axis $y$, and the force $F$ also acting in the same direction, we have $\backslash cos\; \backslash phi\; =\; 0$, $\backslash cos\; \backslash alpha\; =\; 0$, $\backslash sin\; \backslash phi\; =\; 1$, $\backslash cos\; \backslash beta\; =\; 1$, $y\_1\; =\; 0$, and the two equations of (30) give the single equation

If we now call the radius of the pulley 1, the moment of the moving force is $10\; \backslash times\; 1\; =\; 1\; \backslash times\; 1\; =\; 9$; and the moment of inertia (18) is $10\; \backslash times\; 1^2\; +\; 1\; \backslash times\; 1^2\; =\; 11$; therefore (14) the angular velocity $\backslash omega\; =\; \backslash frac\{9\}\{11\}$. But the distance of the centre of gravity from the axis of motion is also $\backslash frac\{1\}\{11\}$, because the two weights may be supposed in contact with the circumference of the pulley; therefore, since $M\; =\; 11$, $wMx\_1\; =\; \backslash frac\{9\}\{11\}\; \backslash times\; \backslash frac\{1\}\{11\}\; \backslash times\; 11\; =\; \backslash frac\{9\}\{11\}$. Now $F\; =\; 10\; +\; 1\; =\; 11$, therefore $R\; =\; 11\; -\; \backslash frac\{9\}\{11\}\; =\; \backslash frac\{100\}\{11\}\; =\; 9\backslash frac\{1\}\{11\}$ pounds, which is the pressure sustained by the axis.

32. If we suppose $R\; =\; 0$, the two equations (30) will express the conditions that must be fulfilled in order that the axis may receive no shock from the action of the impressed force $F$. This supposition gives the equations

from which, since $\backslash cos\; \backslash beta\; =\; \backslash sin\; \backslash alpha$, we deduce

Now, if a plane be conceived to pass through the axis $OZ$ and the centre of gravity of the body, and if we denote by $\backslash psi$ the angle which it makes with the plane $XOZ$, we shall have $x\_1\; =\; y\_1\; \backslash tan\; \backslash psi$; and therefore, since $x\_1\; =\; -y\_1\; \backslash tan\; \backslash alpha$, $\backslash tan\; \backslash psi\; =\; -\backslash tan\; \backslash alpha$. From this equation it follows that the direction $PQ$ of the impressed force is perpendicular to the plane which passes through the axis and the centre of gravity. It now remains only to find the distance of the point of impulse $P$ from the axis. Let $h$ be this distance, and $r\_1$ = distance of centre of gravity from the axis; the above equations give $F^2\; \backslash cos^2\; \backslash alpha\; =\; w^2\; M^2\; y\_1^2$, $F^2\; \backslash sin^2\; \backslash alpha\; =\; w^2\; M^2\; x\_1^2$, whence $F^2\; =\; w^2\; M^2\; (x\_1^2\; +\; y\_1^2)\; =\; w^2\; M^2\; r\_1^2$, and $F\; =\; wMr\_1$. But (14) $w\; =\; Fh\; \backslash div\; fr^2\; dm$; therefore, by substitution,

33. Hence it appears that the conditions which must be fulfilled in order that the axis may receive no shock when the body is struck, are, 1st, that the direction of the force be in the plane of the two axes which make with the axis of rotation a system of principal axes; 2d, that its direction be perpendicular to the plane which passes through the axis and the centre of gravity; and, 3d, that the distance of the point of its application from the axis be $fr^2\; dm\; \backslash div\; Mr\_1$.

34. The point determined by these conditions is called Centre of the centre of percussion of the system. The centre of per-

Rotation. cussion is therefore the point at which, if an obstacle were opposed sufficient to resist the rotation of the system, no motion would be communicated to the axis; or, which is the same thing, if the axis were not fixed, the system would acquire from the shock no tendency to turn, the effect being the same as if the whole rotatory effort had been accumulated at that point.

Let HKL (fig. 8) be a section of the body perpendicular to the axis, and containing the centre of gravity G. Let O be the axis, and in the line OG take $OP\; =\; h\; =\; \backslash frac\{fr^2\; dm\}\{M\; \backslash cdot\; OG\}$, then P is the centre of percussion, and a force applied at P, in the plane HKL, and in the direction perpendicular to OP, will produce the same initial velocity of the centre of gravity G as if the body were free; for as the force exerts no pressure on the axis or points of support, the initial velocity will be the same as if they were not there.

Fig. 8.

35. The distance of the centre of percussion from the axis is always greater than the distance of the centre of gravity from the axis; for the moment of inertia in respect of the axis passing through O being $fr^2\; dm$, by (17) the moment of inertia in respect of the parallel axis passing through G is $fr^2\; dm\; -\; M\; \backslash cdot\; OG^2$; therefore, since the moment of inertia is necessarily positive, $fr^2\; dm\; >\; M\; \backslash cdot\; OG^2$. But by the definition $fr^2\; dm\; =\; M\; \backslash cdot\; OG\; \backslash cdot\; OP$, therefore $M\; \backslash cdot\; OG\; \backslash cdot\; GP\; >\; M\; \backslash cdot\; OG^2$, or $OP\; >\; OG$.

36. The property of the centre of percussion, that a force applied there produces no strain on the axis, may be rendered sensible by a simple experiment. If a straight line, or rod of uniform thickness, be suspended from one extremity, the centre of percussion is at the distance of two thirds of its length from the axis; for the centre of gravity being at the middle of the line, $OG\; =\; \backslash frac\{1\}\{2\}l$, so that if we call its mass $M\; =\; 1$, the formula $OP\; =\; \backslash frac\{fr^2\; dm\}\{M\; \backslash cdot\; OG\}$ becomes $OP\; =\; \backslash frac\{2\}\{3\}l$, which, on taking the integral from $x\; =\; 0$ to $x\; =\; l$, gives $GP\; =\; \backslash frac\{1\}\{3\}l$. Now if the rod be held in the hand, and we give it a motion round the joint of the wrist only, and strike smartly against an obstacle 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 exactly at two thirds of its length, we feel no such disagreeable strain.

37. It is sometimes said that the point P, considered as the centre of percussion, is that with which the most violent blow is struck. But this is by no means true. P is that point of a body turning round O 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.

38. The expressions in (29) for the stress on the axis have reference only to the initial movement of the body, and the forces which they represent are soon destroyed by the resistance of the axis. But when a body is in permanent rotation about a fixed axis, there is a strain exerted on the axis which depends only upon the centrifugal force of rotation, and which therefore continues as long as the rotation continues. This constant pressure is determined as follows: Since $\backslash omega$ is the angular velocity, and $r$ the distance

of a particle $dm$ from the axis, the velocity of the particle $=\; r\backslash omega$; and, by mechanics, the centrifugal force of a body moving in a circle is proportional to the square of the velocity divided by the radius; therefore the centrifugal force of the particle $dm$, in the direction of the radius $r$, is $\backslash omega^2\; r\; dm$. The resolved parts of this force in the directions $OX$ and $OY$ (fig. 7) are $\backslash omega^2\; x\; dm$ and $\backslash omega^2\; y\; dm$; therefore, denoting the pressures in those directions on the pivot S by $p$ and $q$, and those on the pivot T by $p\text{'}$ and $q\text{'}$, we shall have, as in (29), the following equations (in which $x\_1$ and $y\_1$ are the co-ordinates of the centre of gravity, and $k,\; k\text{'}$ the distances of the pivots S and T respectively from O) to determine $p,\; p\text{'},\; q,\; q\text{'}$, viz.

If we suppose $OZ$ to be one of the three principal axes passing through O, then $(22)\; f\; x\; x\; dm\; =\; 0$ and $f\; y\; y\; dm\; =\; 0$; whence $pk\; +\; p\text{'}k\text{'}\; =\; 0$, $qk\; +\; q\text{'}k\text{'}\; =\; 0$. If, therefore, we further suppose $k\; =\; 0$, or that the origin of the co-ordinates is placed at S, we shall also have $p\text{'}\; =\; 0,\; q\text{'}\; =\; 0$, whence the pivot T sustains no pressure, the whole being exerted on S.

If $OZ$, besides being a principal axis, also passes through the centre of gravity, then $x\_1\; =\; 0,\; y\_1\; =\; 0$, whence $p\; +\; p\text{'}\; =\; 0,\; q\; +\; q\text{'}\; =\; 0$; and since $p\text{'}$ and $q\text{'}$ vanish when $OZ$ is a principal axis, therefore $p\; =\; 0,\; q\; =\; 0$, so that no pressure is exerted on either of the pivots; and if they were removed, the body would still continue to revolve about the same axis.

The forces thus determined are in some sort the inverse of those arising from the percussions found in (29), the values of $p\; +\; p\text{'}$ being in this case expressed in terms of $x\_1$ and in the former in terms of $y\_1$; and the value of $q\; +\; q\text{'}$, which in the case of the percussions was expressed in terms of $x\_1$, is here given in terms of $y\_1$.

39. In computing the effect of machines, it is often convenient to determine the situation of a point in which the whole mass of the machine might be concentrated without altering the rotatory effort. In the straight line $OG$ (fig. 8) suppose a point Q to be taken, such that $OQ^2\; =\; fr^2\; dm\; =$ moment of inertia in respect of the axis O; then, if the whole matter in the body were collected at Q, a force applied at any point P in the line $OG$, would produce exactly the same angular velocity as if applied to the same point of the body having its natural form. For if the whole mass M were concentrated at Q, the moment of inertia would become $M\; \backslash cdot\; OQ^2$ (13), and the moment of the force F applied at P is $F\; \backslash cdot\; OP$; therefore the angular velocity (14) would be $\backslash frac\{F\; \backslash cdot\; OP\}\{M\; \backslash cdot\; OQ^2\}$. But the angular velocity produced by the force

F applied at P, the body having its natural form, is $\backslash frac\{F\; \backslash cdot\; OP\}\{fr^2\; dm\}$; therefore, since $M\; \backslash cdot\; OQ^2\; =\; fr^2\; dm$, the two expressions are equal, and the angular velocity is not altered by supposing the whole mass concentrated at Q.

40. The point Q is called the centre of gyration of the revolving mass; and since $OQ\; =\; \backslash sqrt\{fr^2\; dm\}\; \backslash div\; M$, its distance from the axis O is found by extracting the square root of the quotient which is obtained by dividing the moment of inertia by the mass of the body. It is also connected with the centres of gravity and percussion by a simple relation. Suppose P to be the centre of percussion, then $(34)\; OP\; =\; \backslash frac\{fr^2\; dm\}\{M\; \backslash cdot\; OG\}$, therefore $OG\; \backslash cdot\; OP\; =\; \backslash frac\{fr^2\; dm\}\{M\}\; =\; OQ^2$; whence the distance of the centre of gyration from the axis of rotation is a mean proportional between the distances of the centre of gravity and percussion from the axis.

In a straight line or slender rod, as a working beam, or the spoke of a wheel in a machine, $OQ$ is $\backslash sqrt\{\backslash frac\{1\}\{2\}\}$ of its length; for in this case $fr^2\; dm\; =\; \backslash frac\{1\}\{2\}\; r^2$; and $r$ being the whole length, $M\; =\; r$; therefore $\backslash sqrt\{fr^2\; dm\}\; \backslash div\; M\; =\; r\; \backslash sqrt\{\backslash frac\{1\}\{2\}\}$.

Rotation. In a circle or cylinder turning about its axis, $OQ\; =\; \backslash sqrt\{\backslash frac\{1\}\{2\}Mr^2\}$; for by (18) we have in this case $\backslash int\; r^2\; dm\; =\; \backslash frac\{1\}\{2\}Mr^2$; therefore $\backslash sqrt\{\backslash int\; r^2\; dm\}\; \backslash div\; M\; =\; r\backslash sqrt\{\backslash frac\{1\}\{2\}\}$. But if the circle turns round one of its diameters, then $OQ\; =\; \backslash frac\{1\}\{2\}r$, the radius; for in this case $\backslash int\; r^2\; dm\; =\; \backslash frac\{1\}\{4\}Mr^2$; therefore $\backslash sqrt\{\backslash int\; r^2\; dm\}\; \backslash div\; M\; =\; \backslash frac\{1\}\{2\}r$.

In the periphery of a circle, or rim of a wheel, turning about a diameter perpendicular to its plane, $OQ\; =\; \backslash text\{radius\}$; for $\backslash int\; r^2\; dm\; =\; r^2\; \backslash times\; 2\backslash pi\; r\; =\; 2\backslash pi\; r^3$, and $M\; =\; 2\backslash pi\; r$; therefore $\backslash sqrt\{\backslash int\; r^2\; dm\}\; \backslash div\; M\; =\; r$.

A solid sphere turning round a diameter has $OQ\; =\; r\backslash sqrt\{\backslash frac\{2\}\{5\}\}$; for (18) the moment of inertia of the sphere is $\backslash frac\{2\}\{5\}\backslash pi\; r^5$; and the mass $M\; =\; \backslash frac\{4\}\{3\}\backslash pi\; r^3$; therefore $OQ\; =\; \backslash sqrt\{\backslash frac\{2\}\{5\}\backslash pi\; r^5\}\; \backslash div\; M\; =\; r\backslash sqrt\{\backslash frac\{2\}\{5\}\}$.

41. In what precedes, we have supposed that the force applied to a rigid system produces its effect by instantaneous impact, and then ceases to act, so that the motion communicated to the system is uniform. But if the action of the external force continues after the first impact, like the force of gravity on a falling body, the motion of the system will be accelerated. In the case of uniform velocity, it has been shown (14) that the angular velocity is equal to the moment of the applied force divided by the moment of inertia; that is, assuming $F$ to denote the force, and $P$ the perpendicular from the axis on the line of its direction, $\backslash omega\; =\; P\; \backslash div\; \backslash int\; r^2\; dm$. Suppose now the body to be subjected to the action of an accelerating force, and let $dw$ be the increment of the angular velocity in the very small time $dt$. The increment of the actual velocity of the particle $dm$, the distance of which from the axis is $r$, is then $rdw$, and its accelerating force (the ratio of the increment of the velocity to the increment of the time) is $r\; \backslash frac\{dw\}\{dt\}$. Its moving

force is consequently $\backslash frac\{dw\}\{dt\}\; \backslash cdot\; rdm$; and since the direction of this force is perpendicular to $r$, its moment = $\backslash frac\{dw\}\{dt\}\; \backslash cdot\; r^2\; dm$.

But the sum of all these moments, in respect of every particle $dm$, must be equal to the moment of the impressed force, continuing to act during the same instant of time, therefore $\backslash frac\{dw\}\{dt\}\; \backslash int\; r^2\; dm\; =\; P\; \backslash cdot\; F$, whence $\backslash frac\{dw\}\{dt\}\; =\; \backslash frac\{P\; \backslash cdot\; F\}\{\backslash int\; r^2\; dm\}$.

42. In the case of a heavy body oscillating about an axis by the force of gravity, let $HKL$ be a section of the body passing through the centre of gravity $G$, and let the axis of suspension be at $O$, and perpendicular to the plane $HKL$. Also, let $OX$ be horizontal, and $OY$ vertical. Then the accelerating force being denoted by $g$ ($g\; =\; 32$ feet in a second of time), the moving force in respect of a particle $dm$ at $A$ is $gdm$; and the direction of this force being parallel to $OY$, the distance of its direction from the axis is $x$; therefore its moment = $gxdm$. The sum of all these moments is the moment of the whole gravitating force of the body, that is, $g\; \backslash int\; x\; dm\; =\; P\; \backslash cdot\; F$, and we have consequently, since $\backslash int\; x\; dm\; =\; Mx\_1$,

When the centre of gravity is in the straight line $OY$, or in the vertical passing through the axis, then $x\_1\; =\; 0$, and the angular velocity is uniform, and the body, if it rest in that position, would have no tendency to turn. But suppose the centre of gravity to be drawn aside from the vertical, and let $OG\; =\; k$, and the angle $GOY\; =\; \backslash theta$, then $x\_1\; =\; k\; \backslash sin\; \backslash theta$, and we have

whence the acceleration of the angular velocity is proportional to the sine of the angle which the line $OG$ makes with the vertical.

43. Suppose the whole mass of the body to be concentrated at a point $S$ in the straight line $OG$; we shall then have $r\; =\; OS$, $\backslash int\; r^2\; dm\; =\; M\; \backslash cdot\; OS^2$, and the formula will become $\backslash frac\{dw\}\{dt\}\; =\; \backslash frac\{g\; \backslash sin\; \backslash theta\}\{OS\}$. Now, if the point $S$ be chosen such that $OS\; =\; \backslash int\; r^2\; dm\; \backslash div\; M\; \backslash cdot\; k$, then the two formulæ will be identical, and the solid body will oscillate exactly in the same time as if all its matter were united at the point $S$, and connected with the axis by an inflexible rod without weight, that is, in the same time as a simple pendulum, of which the length is $OS$.

Since $r^2\; =\; x^2\; +\; y^2$, and $M\; \backslash cdot\; k\; =\; \backslash int\; x\; dm$, this expression for $OS$ may be put under the following form, in which it is usually exhibited, namely,

44. The point $S$ is called the centre of oscillation. On comparing the expression for $OS$ with that for $OP$ in (34), it will be seen that it is at the same distance from the axis as the centre of percussion. But the centre of oscillation is not, strictly speaking, limited to a single point; for it through $S$ a line be drawn perpendicular to the plane $HKL$, or parallel to the axis of rotation, every point in this line will move with the same angular velocity as if it were entirely unconnected with the other points of the oscillating body, and the line itself is denominated the axis of oscillation. It is usual, however, to understand by the term centre of oscillation, the point $S$ which is situated in the straight line, perpendicular to the axis, which contains the centre of gravity.

45. If we denote the moment of inertia of the body in respect of the axis which passes through $G$ the centre of gravity, perpendicular to the plane $HKL$, by $M\; \backslash cdot\; k^2$, then $k$ is a given line; and if we also make $OG\; =\; k$, then (17) the moment of inertia in respect of the axis passing through $O$ is $M(h^2\; +\; k^2)$. Hence we find $OS\; =\; \backslash frac\{M(h^2\; +\; k^2)\}\{M\; \backslash cdot\; k\}\; =\; k\; +\; \backslash frac\{k^2\}\{k\}$; and consequently $GS\; =\; \backslash frac\{k^2\}\{k\}$. This formula gives the distance of the centre of oscillation below the centre of gravity.

46. The point $S$ possesses several very remarkable properties, one of which is, that as $S$ is the centre of oscillation of the body turning about $O$, so $O$ is the centre of oscillation of the body turning about the parallel axis passing through $S$; in other words, the centres of oscillation and suspension are convertible. For it has been shown (45) that $GS\; =\; \backslash frac\{k^2\}\{k\}$

= $\backslash frac\{k^2\}\{OG\}$; therefore $OG\; =\; \backslash frac\{k^2\}\{GS\}$, that is to say, if the body oscillates about the axis passing through $S$, then $O$ is the centre of oscillation. This curious property, which was first demonstrated by Huygens in his Horologium Oscillatorum, was ingeniously applied by Captain Kater as a practical means of determining the length of the seconds pendulum. Two axes being assumed in a bar or rod of metal, the bar is adjusted by filing away one of its ends until the oscillations about each axis are performed exactly in the same time; and when this equality has been established, the distance between the axes is the length of the synchronous simple pendulum. See PENDULUM.

47. We shall now determine the centre of oscillation in a few particular cases.

Rotation. If the body is a heavy straight line suspended by one extremity, the OS = $\backslash frac\{2\}\{3\}$ of its length; for making $a$ = the length of the rod, we have (18) $\backslash int\; r^2\; dm\; =\; \backslash frac\{1\}\{2\}\; a^3$, and in this case the mass $M\; =\; a$, and the centre of gravity is at the middle of the line, or $k\; =\; \backslash frac\{1\}\{2\}\; a$, whence $Mk\; =\; \backslash frac\{1\}\{2\}\; a^2$, and $OS\; =\; \backslash int\; r^2\; dm\; \backslash div\; Mk\; =\; \backslash frac\{1\}\{2\}\; a^3\; \backslash div\; \backslash frac\{1\}\{2\}\; a^2\; =\; \backslash frac\{2\}\{3\}\; a$.

Computation of centres of oscillation. This is nearly the case of a slender rod of a cylindrical or prismatic shape, and it would be accurately so if all the points of a transverse section were equally distant from the axis of suspension.

If the oscillating body is an isosceles triangle suspended by its apex, and vibrating perpendicularly to its plane, $O\; =\; \backslash frac\{2\}\{3\}$ of its height. For let $a$ = the height, $na$ = the base, $x$ = a section parallel to the base, then $y\; =\; nx$, $r\; =\; x$, and $dm\; =\; ydx$; whence $\backslash int\; r^2\; dm\; =\; \backslash int\; nx^2\; dx\; =\; \backslash frac\{1\}\{3\}\; na^4$ (when $x\; =\; a$). But in this case $M\; =\; \backslash text\{area\}\; =\; \backslash frac\{1\}\{2\}\; a\; \backslash times\; na\; =\; \backslash frac\{1\}\{2\}\; na^2$; and the distance of the centre of gravity from the vertex being two thirds of the height, we have $k\; =\; \backslash frac\{2\}\{3\}\; a$, whence $Mk\; =\; \backslash frac\{1\}\{2\}\; na^3$, and consequently $OS\; =\; \backslash int\; r^2\; dm\; \backslash div\; Mk\; =\; \backslash frac\{2\}\{3\}\; a$.

48. The problem to find the centre of oscillation of a sphere suspended by a thread or an inflexible rod, is important on account of its application to the measurement of the seconds pendulum. Let us first suppose that the weight of the thread is insensible in comparison of that of the ball. Let $r$ = radius of ball, $l$ = length of thread from the point of suspension to the ball, and $a\; =\; l\; +\; r$ = distance from the point of suspension to the centre of the ball. The moment of inertia of the sphere in respect of an axis passing through its centre (18) is $\backslash frac\{2\}\{5\}\; Mr^2$; therefore (17) the moment in respect of a parallel axis at the distance $a$ is $M(a^2\; +\; \backslash frac\{2\}\{5\}\; r^2)$; and on dividing this by $Mk\; (=\; Ma)$, we get the distance from the point of suspension to the centre

of oscillation = $a\; +\; \backslash frac\{2r^2\}\{5a\}$, or the centre of oscillation is below the centre of gravity by the distance $\backslash frac\{2r^2\}\{5a\}\; \backslash div\; a$.

49. Let us now suppose the weight of the thread or rod to be taken into account. Let $C$ (fig. 10) be the centre of the ball, $G$ the centre of gravity of the ball and rod $OA$, $S$ the centre of oscillation; also let $b$ = weight of ball, and $u$ = weight of rod. The moment of the sphere, oscillating about $O$, supposing the weight of the rod insensible, is found above = $M(a^2\; +\; \backslash frac\{2\}\{5\}\; r^2)$; and (18) the moment of the rod alone is $\backslash frac\{1\}\{3\}\; OA^3\; =\; \backslash frac\{1\}\{3\}\; b^3$ (putting $M$ = mass of the rod); therefore, substituting the weights for the masses, that is, making $M\; =\; b$, and $M\text{'}\; =\; u$, the moment of the compound body is $b(a^2\; +\; \backslash frac\{2\}\{5\}\; r^2)\; +\; \backslash frac\{1\}\{3\}\; bu^3$. Now, to find the position of $G$, the centre of gravity, we have this theorem in statics, that the product of the whole mass multiplied into the distance of its centre of gravity from the axis, is equal to the sum of the products of the masses of the several parts into the respective distances of their centres of gravity; that is, putting $OG\; =\; k$, we have $(M\; +\; M\text{'})k\; =\; ba\; +\; \backslash frac\{1\}\{3\}\; bu^3$, and consequently (45),

Rotation on inclined planes. 50. Hitherto we have supposed the axes of rotation to be absolutely fixed; but the rotation may be performed about axes which are themselves in motion, as in the case of a ball or cylinder rolling down an inclined plane. When a ball rolls down an inclined 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 $HOK$ (fig. 11), whose axis passes through $G$, the centre of gravity of the body. Suppose the body suspended from a fixed point $A$ by a thread wound round the cylinder. The body will descend by the action of gravity, and it will also turn round, unwinding the thread. Draw the horizontal line $SGO$; this will pass through the point of contact $O$ of the thread and cylinder, and $O$ is the point round which the cylinder begins to turn in descending. Let $S$ be its centre of oscillation corresponding to the momentary centre of rotation $O$.

The body will begin to descend in the same manner as if all its matter were collected in $S$; for it may be considered, in this instant, as a pendulum suspended at $O$. But in this case $S$ 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 $OG$ to $OS$. Now since the points $O$, $G$, $S$, are always in a horizontal line, and the radius $OG$ is given, as also $OS$ (43), the velocity of a body falling freely, and of the body unwinding from this thread, will always be in the same proportion of $OS$ to $OG$, 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.

It follows from this that the weight of the descending body will be to the tension of the thread as $OS$ to $GS$; for the tension of the thread is the difference between the moment of the rolling body and that of the body falling freely. It is to be remarked, 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 (46), that if $O$ be in the circumference of a circle whose centre is $G$, $S$ will be in the circumference of another circle round the same centre, and therefore the ratio of $OG$ to $OS$ is constant.

If a circular body $HOK$ 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 $OG$ to $OS$; 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.

51. In order to give an example of the application of the Inertia of preceding formulae for the rotation of bodies about fixed machines, axes to the theory of machines actually performing work, we shall suppose the machine to be the wheel and axle, and that a weight $W$, attached to a chain passing over the cylinder, is to be raised by means of a power acting on another chain which passes over the wheel. Let the radius of the wheel = $a$, the radius of the cylinder = $b$, and suppose the moving power to be another weight $P$, or a mass of matter descending by the accelerating force of gravity.

Since the machine is in this case impelled by an accelerating force, the angular velocity $\backslash omega$ is obtained from the equation (41), namely,

Now, because the weight $P$ acts in a straight line whose distance from the axis = $a$, the moment of its force is $Pa$; and in like manner the moment of the force exerted by $W$ is $Wb$. But this last moment retards the motion, and must be taken as negative; therefore the moment of the impelling force is $Pa\; -\; Wb$. Again, the moment of inertia is made up of three parts: first, the moment of inertia of the

Fig. 10.

Fig. 11.

Rotation. machine, which depends on the quantity of dead matter contained in the wheel and cylinder, and which we may denote by $K$; second, the moment of inertia of the weight $P$, which may be considered as attached to the extremity of the radius of the wheel; third, the moment of inertia of the weight $W$, which may be considered as attached to the extremity of the radius of the cylinder. Let $M$ and $M\text{'}$ be the masses of the bodies whose weights are $P$ and $W$, then the moments of inertia are respectively $Ma^2$ and $M\text{'}b^2$. Let also $g$ = the acceleration of gravity, then the weight of a body being the product of its mass into the force of gravity, we have $P\; =\; gM$, $W\; =\; gM\text{'}$, whence the whole moment of inertia is

and we have consequently

For the sake of abridging, let us assume

given by the equation $\backslash omega\; =\; g\; f\; U\; dt$. Also let $v$ = the velocity of the impelled point of the machine, or the velocity with which $P$ descends, and $u$ = the velocity of the working point, or velocity with which $W$ is raised; then $v\; =\; aw$, and $u\; =\; bw$, at the end of the time $t$; and if we denote the respective accelerating forces by $\backslash phi$ and $\backslash psi$, we shall have

52. To find the proportion of the velocities of the impelled and working points which give the greatest performance when the weight and power are both given, we may treat the formula which expresses the work, that is, the accelerating force, or $W$, as a fluxionary quantity, and find its maximum. Thus, to find the radius of the wheel by which the weight will be raised with the greatest velocity, we have to differentiate the quantity denoted by $U$, on the supposition that $a$ is variable, and make the result equal to zero. This will give an equation from which $a$ will be obtained in terms of $P$, $W$, and $b$; and it will be observed, that as $K$ varies with $a$, the value of $K$ in terms of $a$ must be substituted in the value of $U$ before the differentiation.

53. The friction may be taken account of by considering it as a quantity to be added to the weight $W$. Let $f$ = the co-efficient of friction, and $e$ = radius of the pivots supporting the machine, then the friction will be expressed by $fe(P\; +\; W)$, and we have only to substitute $Wb\; +\; fe(P\; +\; W)$ for $Wb$ in the above expression for $U$.

54. In the case of the fixed pulley, the velocities of $P$ and $W$ are equal, and we have also $a\; =\; b$. If we denote the weight of the pulley by $Q$, then its mass will be $Q\; \backslash div\; g$, and (18) its moment of inertia $=\; \backslash frac\{1\}\{2\}\; Qa^2\; \backslash div\; g$; therefore the formula for the accelerating force with which $P$ descends or $W$ rises will be

Motion of a solid body which is entirely free. 55. We now proceed to consider the motion of a solid body which is entirely free.

When a solid body or rigid system is impressed at a single point by a force of which the direction does not pass through the centre of gravity, the body, if free to move in every direction, acquires two motions, one progressive, and equal to that which would have resulted if the direction of the force had passed through the centre of gravity, and the other rotatory, and equal to that which would have resulted if the body had been only at liberty to move round a fixed axis passing through the centre of gravity, and perpendicular to the plane which contains that centre, and the line of the direction of the impelling force.

56. In order to demonstrate this proposition, let $HKL$ (fig. Rotation. 12) be a section of the body passing through $G$ the centre of gravity, and containing the line in which the force is directed; let $P$ be the point at which the given force $F$ is applied, and $PA$ its direction; through $G$ draw $GA$ perpendicular to $PA$, and let $PA$ represent the intensity of the force $F$. Bisect $AP$ in $B$, produce $AG$ until $GC\; =\; AG$, and through $C$ draw $CD$, $CE$ perpendicular to $AC$, and make each equal to $AB$. If we now conceive a force represented by $DC$ to be applied at $D$ in the direction $DC$, and another force represented by $EC$ to be applied at $E$ in the direction $EC$, these two forces being equal, and acting in opposite directions, will destroy each other, or produce no movement in the body, and the motion which actually takes place in consequence of the application of the force $PA$ at $P$, will be the same as would result from the application of the four forces $PB$, $BA$, $DC$, $EC$. Now it is evident, that if the two forces $BA$ and $DC$ acted alone, they would produce no rotatory motion about the axis passing through $G$, for their moments $BA\; \backslash cdot\; AG$ and $DC\; \backslash cdot\; CG$ are equal, and they tend to turn the body in opposite directions. But the same two forces both tend to give the body a progressive motion; and as the resultant of two equal and parallel forces is a force equal to their sum, and acting at the point which bisects the distance between their directions, the initial progressive motion of the body in consequence of the forces $BA$ and $DC$ will be the same as if the force $PA\; =\; F$ were applied at the point $G$ in the direction parallel to $PA$. We have now to consider the two remaining forces $PB$ and $EC$. As these two forces act on the line $AC$ in opposite directions, their resultant is nothing, or their separate tendencies to produce a progressive motion in the centre of gravity of the body, are counteracted by their joint effort. But they conspire in tending to produce a rotatory motion in the same direction about the axis passing through $G$; and the moment of the first in respect of that axis being $PB\; \backslash cdot\; AG$, and that of the second $EC\; \backslash cdot\; CG$, the sum of their moments is $(PB\; +\; EC)\; AG$ or $PA\; \backslash cdot\; AG$. The whole effect of these two forces is therefore to produce the same initial angular velocity about the axis through $G$, perpendicular to $HKL$, as would be produced by the single force $PA$ applied at the point $A$; and it has been shown, that the whole effect of the forces $BA$ and $DC$ is to produce the same progressive movement of the centre of gravity as if the single force $PA$ were applied at that point, consequently the proposition is demonstrated, and it appears that the two motions produced by the application of a force which is not directed towards the centre of gravity may be considered independently of each other.

Fig. 12.

57. Let $V$ = the progressive velocity of the centre of gravity in the direction parallel to $PA$; then the mass of the body being $M$, we have the force $F\; =\; MV$, and consequently $V\; =\; F\; \backslash div\; M$. Again, if we make $AG\; =\; l$, the angular velocity produced by the force $F$ applied at $A$ is $\backslash omega\; =\; F\; \backslash cdot\; l\; \backslash div\; f^2\; dm$ (14), therefore $V\; :\; \backslash omega\; ::\; f^2\; dm\; :\; Ml$. Now, since $f^2\; dm$ and $M$ are both known from the nature of the body, it follows, that when the distance $l$ is given, the ratio of the progressive motion of the centre of gravity to the velocity of angular rotation can be found; and conversely, when the ratio of the two velocities is given, the distance $l$ can be found. For example, let $s$ be the space described by the centre of gravity while the body makes one revolution about its axis, then $2\backslash pi\; s$ is the space described, in consequence of the rotatory motion, by a point of the body whose distance from the axis = $l$, and the above proportion gives $l\; =\; \backslash frac\{2\backslash pi\; \backslash cdot\; f^2\; dm\}\{s\}\; \backslash cdot\; M$.

Rotation. 58. As it is not necessary that the progressive motion of the centre of gravity be in a straight line, this formula enables us to determine the distances from the centre at which the planets may have received the single impulses which gave them at the same time their motions of revolution in their orbits, and of rotation round their axes. Thus, taking the case of the earth, and making $R$ the radius of its orbit, the circumference of the orbit is $2\backslash pi\; R$, and the part of this which is described by the centre of gravity, while the earth revolves once about its axis, is $\backslash frac\{2\backslash pi\; R\}\{365\}\; =\; s$; and by (18)

If we assume $r\; =\; 4000$ miles, and $R\; =\; 95,000,000$ miles, this formula gives $l\; =\; 24\backslash frac\{1\}\{2\}$ miles nearly.

Centre of spontaneous rotation. 59. The direction of the force being $PA$ (fig. 12), if a point be taken in the line $AG$ between $A$ and $G$, the motion of the point arising from the rotation of the body about the axis is in the first instant of time parallel to the direction of the progressive motion, and if $r$ be its distance from $G$, the velocity due to both motions will be $V\; +\; rw$. But if the point be taken on the other side of $G$, the direction of the rotatory motion will be opposite to that of the progressive motion, and the velocity of the point will be $V\; -\; rw$. If therefore the point $T$ be taken such that making $GT\; =\; r$, we have the equation $V\; =\; rw$; then, for a single instant, every point in the straight line passing through $T$ perpendicular to $HKL$ is at rest, or is carried as far back by the rotatory motion in the first instant of time, as it is carried forward by the progressive motion; and, for a single instant, the double motion of the body may be regarded as a simple rotatory motion about that straight line. This point $T$ is called the centre of spontaneous rotation, and the straight line passing through $T$, perpendicular to the plane $HKL$, is called the axis of spontaneous rotation.

Progressive motion of centre of gravity. 60. The preceding formulae enable us to determine completely the two motions of translation and rotation when a force is applied to a single point of the free body; we have now to determine what the effect will be when any number of forces given in magnitude and direction are applied to different points of the body; and as it appears from what precedes, that during the first instant the two motions take place independently of each other, we shall first give the equations which fix the position of the centre of gravity of the moving body, and then those which define the rotatory motion, or determine the position of any point of the body with reference to its centre of gravity.

Let $G$ be the centre of gravity of the body, $M$ its mass, and let the motion be referred to three rectangular axes $OX,\; OY,\; OZ$. Let us now suppose forces given in magnitude and direction to be applied at the points $A,\; A\text{'},\; A\text{'}\text{'},$ &c. of the body, and that each of them is resolved into three others respectively parallel to the directions of the axes, and let $p,\; q,\; r$, be the resolved forces applied at $A$; $p\text{'},\; q\text{'},\; r\text{'}$, the resolved forces applied at $A\text{'}$, and so on; also let , , ; then, since the moving force is equal to the product of the mass into the velocity, and since by (56) the progressive motion resulting from the force applied at $A$ is the same as if it had been applied to the centre of gravity, the velocity of the centre of gravity from the joint action of all the forces, in the directions $OX,\; OY,\; OZ$, will be respectively $\backslash frac\{P\}\{M\},\; \backslash frac\{Q\}\{M\},\; \backslash frac\{R\}\{M\}$.

But the resulting velocity is the diagonal of the parallelopiped of which these component velocities are the sides; therefore, denoting it by $V$, we have

To determine the direction of this velocity, let $\backslash alpha,\; \backslash beta,\; \backslash gamma$, be

the angles which its direction makes with the axis $x,\; y,\; z$; Rotation. then, since the forces $P,\; Q,\; R$ meet in a point (the centre of gravity), we have

These equations give the direction of the velocity; hence, since $V$ has already been found, the initial velocity of the centre of gravity and its direction are both completely determined.

61. When the forces applied to the body are simply percussions, which cease to act after the first impact, and there is no resistance, no change will take place in the velocity or its direction during the second and succeeding instants, and the centre of gravity will describe a straight line with a uniform motion; but if the forces continue to act after the motion has commenced, we must have recourse to the formulae given by the principles of dynamics, which express the varied motion of a body subjected to the action of accelerating forces. At the end of the time $t$, reckoned from the commencement of the motion, let $x,\; y,\; z$ be the co-ordinates of the element of the body $dm$, referred to the rectangular axes $OX,\; OY,\; OZ$; then the velocity at that instant being expressed by the differential of the space described divided by the element of the time, the components

of the velocity at the same instant will be $\backslash frac\{dx\}\{dt\},\; \backslash frac\{dy\}\{dt\},\; \backslash frac\{dz\}\{dt\}$. Now

the forces acting on the body, whatever be their nature, may always be decomposed into others parallel to the three axes; suppose therefore $P,\; Q,\; R$ to be the components of the accelerating forces in those directions acting on the point $dm$, then the increment of velocity being equal to the force into the element of the time, if the particle $dm$

were entirely free, the velocities $\backslash frac\{dx\}\{dt\},\; \backslash frac\{dy\}\{dt\},\; \backslash frac\{dz\}\{dt\}$ would acquire the increments $Pdt,\; Qdt,\; Rdt$ in the instant $dt$. But the increments of velocity actually acquired in the instant $dt$ are the differentials of the velocities, or $d\; \backslash cdot\; \backslash frac\{dx\}\{dt\},\; d\; \backslash cdot\; \backslash frac\{dy\}\{dt\},\; d\; \backslash cdot\; \backslash frac\{dz\}\{dt\}$, and therefore the velocities lost in consequence of the mutual connection of the particles are respectively

Multiplying each of these expressions by $dm$, we get the quantities of motion due to the respective velocities. Now, according to a well-known principle of dynamics, the discovery of which is due to D'Alembert, the infinitely small quantities of motion lost in every instant by all the particles in consequence of their mutual connection must make equilibrium; the integrals of the preceding expressions multiplied by $dm$ must therefore be severally equal to zero, and we have accordingly (on effecting the operations indicated, and dividing by $dt$) the three equations

62. In order to find the place of the centre of gravity at the end of the time $t$, let its co-ordinates at that instant be $x\_1,\; y\_1,\; z\_1$; then from the property of the centre of gravity we have

Regarding the variables as functions of the time $t$, differentiating twice, and dividing each time by $dt$, we obtain

Rotation. Comparing these with the above equations, we have

Now, the first members of these equations denote respectively the quantities of motion in the direction of the axes $x,\; y,\; z$, of a material point whose mass = $M$, and the integrals $fPdm,\; fQdm,\; fRdm$ denote the quantities of motion of the body or rigid system under consideration, subjected to the accelerating forces $P,\; Q,\; R$; it follows, therefore, that during the whole continuance of the motion the centre of gravity moves as if the whole of the matter of the body were concentrated there, and the forces which act on the body were all applied at that point.

63. Having thus given the equations which enable us to determine the motion of the centre of gravity of a free system under the action of given forces, we now proceed to consider the rotatory movement of the system, and to investigate the position of any point of it with respect to the centre of gravity at the end of any given interval of time. The investigation of this problem is greatly facilitated by means of the following elegant theorem respecting the composition of rotatory motion, discovered by Frisi.

PROPOSITION. If a solid body revolve round an axis $OA$ (fig. 13) passing through its centre of gravity with an angular velocity = $\backslash omega$, and if a force be impressed upon it which alone would cause it to revolve about an axis $OB$ also passing through its centre of gravity with an angular velocity = $\backslash omega\text{'}$, then the body will not revolve about either $OA$ or $OB$, but about a third axis $OC$ 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 those axes, or so that $\backslash sin.\; AOC\; :\; \backslash sin.\; BOC\; ::\; \backslash omega\text{'}\; :\; \backslash omega$; and the angular velocity $u$ round the 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, that is, $u\; :\; \backslash omega\; ::\; \backslash sin.\; AOB\; :\; \backslash sin.\; COB$, or $u\; :\; \backslash omega\text{'}\; ::\; \backslash sin.\; AOB\; :\; \backslash sin.\; AOC$.

64. In order to prove this proposition, it is necessary to show, 1st, that any point in the line $OC$ is at rest while the two motions are taking place; 2d, that every particle of the body not situated in $AC$ revolves about the axis $AC$; and 3d, to assign the velocity of rotation.

(1.) Let the angular motion about the axis $OA$ be supposed to be in the direction by which every point in the angle $AOB$ would be raised above the plane of the paper, and that about $OB$ to be in the direction by which every point within the same angle would be depressed below the plane of the paper, and take any point $E$ in $OC$, and draw $EH$ and $EK$ respectively perpendicular to $OA$ and $OB$. The absolute velocity with which the point $E$ moves in consequence of the rotation about $OA$ is $HE\; \backslash cdot\; \backslash omega$, and the absolute velocity with which it moves about $OB$ is $KE\; \backslash cdot\; \backslash omega\text{'}$. But $HF\; =\; OE\; \backslash sin.\; AOC$, and $KE\; =\; OE\; \backslash sin.\; BOC$; and by hypothesis $\backslash omega\text{'}\; :\; \backslash omega\; ::\; \backslash sin.\; AOC\; :\; \backslash sin.\; BOC$, whence $\backslash omega\text{'}\; \backslash sin.\; BOC\; =\; \backslash omega\; \backslash sin.\; AOC$, therefore the two velocities $HE\; \backslash cdot\; \backslash omega$ and $KE\; \backslash cdot\; \backslash omega\text{'}$ are equal; and they are in opposite directions, consequently the point $E$ remains at rest. In the same manner, it is shown that every other point in $OC$ remains at rest.

(2.) Let $P$ be the place of any particle of the body, and conceive the surface of a sphere to pass through $P$, intersecting the axes in $A$ and $B$, and join $PA$ and $PB$ by arcs of great circles. Draw $PM$ perpendicular to $AP$, and make $PM\; =\; \backslash omega\; \backslash cdot\; \backslash sin.\; AP$ = absolute velocity of the particle $P$ about the axis $OA$ in 1^o of time; also draw $PN$ perpendicular to $BP$, and make $PN\; =\; \backslash omega\text{'}\; \backslash cdot\; \backslash sin.\; BP$ = absolute velocity of $P$ about $OB$ in 1^o (the radius being unity), and complete the parallelogram $PMRN$.

Since the particle $dm$, in virtue of the rotation about $OA$, has a motion which would cause it to describe the line $PM$ in a second, and in virtue of the rotation about $OB$ has also a motion which would cause it to describe $PN$ in a second, the particle will describe $PR$ the diagonal of the parallelogram $PMRN$. Through $P$ let an arc of a great circle be described perpendicular to $PR$, and produced to meet the plane $AOB$ in the point $C$, which at present we suppose indeterminate, and join $OC$; then $OC$ is the straight line in which a plane passing through $P$, perpendicular to $PR$, intersects the plane $AOB$.

Because $APM$ and $CPR$ are equal, being right angles, take from each the common angle $APR$, and there remains $RPM\; =\; APC$; and because $BNP$ and $CPR$ are right angles, add to each $NPC$, and we have $RPN\; =\; BPC$. Now $PM\; =\; \backslash omega\; \backslash sin.\; AP$, and $PN\; =\; RM\; =\; \backslash omega\text{'}\; \backslash sin.\; BP$, therefore $PM\; :\; RM\; ::\; \backslash omega\; \backslash sin.\; AP\; :\; \backslash omega\text{'}\; \backslash sin.\; BP$; but $PM\; :\; RM\; ::\; \backslash sin.\; MRP\; (=\; \backslash sin.\; RPN)\; :\; \backslash sin.\; RPM$, or $PM\; :\; RM\; ::\; \backslash sin.\; BPC\; :\; \backslash sin.\; APC$; therefore $\backslash sin.\; BPC\; :\; \backslash sin.\; APC\; ::\; \backslash omega\; \backslash sin.\; AP\; :\; \backslash omega\text{'}\; \backslash sin.\; BP$, whence $\backslash omega\text{'}\; \backslash sin.\; BP\; \backslash cdot\; \backslash sin.\; BPC\; =\; \backslash omega\; \backslash sin.\; AP\; \backslash cdot\; \backslash sin.\; APC$.

Now in the spherical triangles $BPC$ and $APC$ we have $\backslash sin.\; BP\; \backslash cdot\; \backslash sin.\; BPC\; =\; \backslash sin.\; CB\; \backslash cdot\; \backslash sin.\; PCB$, and $\backslash sin.\; AP\; \backslash cdot\; \backslash sin.\; APC\; =\; \backslash sin.\; AC\; \backslash cdot\; \backslash sin.\; ACP$; therefore $\backslash omega\text{'}\; \backslash sin.\; CB\; \backslash cdot\; \backslash sin.\; PCB\; =\; \backslash omega\; \backslash sin.\; AC\; \backslash cdot\; \backslash sin.\; ACP$. But $\backslash sin.\; PCB\; =\; \backslash sin.\; ACP$ (the one angle being the supplement of the other); also $\backslash sin.\; CB\; =\; \backslash sin.\; BOC$, $\backslash sin.\; AC\; =\; \backslash sin.\; AOC$, therefore $\backslash omega\text{'}\; \backslash sin.\; BOC\; =\; \backslash omega\; \backslash sin.\; AOC$, and consequently $\backslash omega\; :\; \backslash omega\text{'}\; ::\; \backslash sin.\; BOC\; :\; \backslash sin.\; AOC$.

From this it follows that $OC$ divides the angle $AOB$ into parts having a given ratio, and is therefore given in position; and as it may be shown in like manner that the plane passing through every other particle of the body perpendicular to the direction of the motion of that point intersects the plane $AOB$ in $OC$, $OC$ is therefore the axis of rotation.

(3.) To find $u$, the angular velocity about $OC$. In the triangle $RPM$ we have $RP\; :\; PM\; ::\; \backslash sin.\; RMP\; :\; \backslash sin.\; MRP$; but $\backslash sin.\; RMP\; =\; \backslash sin.\; MPN\; =\; \backslash sin.\; APB$, and $\backslash sin.\; MRP\; =\; \backslash sin.\; RPN\; =\; \backslash sin.\; CPB$, therefore $RP\; :\; PM\; ::\; \backslash sin.\; APB\; :\; \backslash sin.\; CPD$. Now $RP\; =\; u\; \backslash sin.\; CP$ = absolute velocity about $OC$ in 1^o, and $PM\; =\; \backslash omega\; \backslash sin.\; AP$ = absolute velocity about $OA$ in 1^o. Substituting these expressions therefore in the last proportion, and multiplying extremes and means, we get $u\; \backslash sin.\; CP\; \backslash cdot\; \backslash sin.\; CPB\; =\; \backslash omega\; \backslash sin.\; AP\; \backslash cdot\; \backslash sin.\; APB$.

But the spherical triangles $CPB$ and $APB$ give $\backslash sin.\; CP\; \backslash sin.\; CPB\; =\; \backslash sin.\; CB\; \backslash sin.\; CBP$, and $\backslash sin.\; AP\; \backslash sin.\; APB\; =\; \backslash sin.\; AB\; \backslash sin.\; ABP$. Therefore, substituting, and leaving out the common term, we have $u\; \backslash sin.\; CB\; =\; \backslash omega\; \backslash sin.\; AB$, or $u\; \backslash sin.\; COB\; =\; \backslash omega\; \backslash sin.\; AOB$, that is, $u\; :\; \backslash omega\; ::\; \backslash sin.\; AOB\; :\; \backslash sin.\; COB$.

In like manner, it may be shown that $u\; :\; \backslash omega\text{'}\; ::\; \backslash sin.\; AOB\; :\; \backslash sin.\; AOC$, whence the proposition is demonstrated.

65. When the two primitive axes $OA$ and $OB$ are at right angles, $\backslash sin.\; AOB\; =\; 1$, $\backslash sin.\; COB\; =\; \backslash cos.\; AOC$, and the above two proportions give $u\; :\; \backslash omega\; ::\; 1\; :\; \backslash cos.\; AOC$, $u\; :\; \backslash omega\text{'}\; ::\; 1\; :\; \backslash sin.\; AOC$, whence $u\; \backslash cos.\; AOC\; =\; \backslash omega$, and $u\; \backslash sin.\; AOC\; =\; \backslash omega\text{'}$, and consequently $u^2\; =\; \backslash omega^2\; +\; \backslash omega\text{'}^2$, or $u\; =\; \backslash sqrt\{\backslash omega^2\; +\; \backslash omega\text{'}^2\}$.

66. Suppose a third axis $OD$ at right angles to $OA$ and $OB$, and that a force is impressed on the body, which alone would cause it to revolve about $OD$ with an angular velocity = $\backslash omega\text{'}\text{'}$; then, as the body is revolving about $OC$ with the velocity $u$, the effect of this new force will be to cause it to revolve about an axis in the plane $DOC$; and if we make $\backslash psi$ = the angle which this new axis makes with the line $OC$, and $W$ = the velocity about the new axis, we

Rotation. shall have from the preceding formulæ $W\; \backslash cos.\; \backslash psi\; =\; u$, and $W\; \backslash sin.\; \backslash psi\; =\; w\text{'}$, whence $W\; =\; \backslash sqrt\{(u^2\; +\; w\text{'}^2)\}$, or, substituting for $u$ its value found above, $W\; =\; \backslash sqrt\{(w^2\; +\; w\text{'}^2\; +\; w\text{'}\text{'}^2)\}$. The position of the new axis is, however, more conveniently defined by the angles which it makes with the three primitive axes. Let $\backslash alpha,\; \backslash beta,\; \backslash gamma$ be the angles which it makes respectively with the primitive axes $OA,\; OB,\; OC$, then the three spherical triangles which it forms with the three axes give $\backslash cos.\; \backslash alpha\; =\; \backslash cos.\; AOC\; \backslash cos.\; \backslash psi\; +\; \backslash sin.\; AOC\; \backslash sin.\; \backslash psi\; \backslash cos.\; \backslash beta$, $\backslash cos.\; \backslash beta\; =\; \backslash cos.\; BOC\; \backslash cos.\; \backslash psi\; +\; \backslash sin.\; BOC\; \backslash sin.\; \backslash psi\; \backslash cos.\; \backslash alpha$, $\backslash cos.\; \backslash gamma\; =\; \backslash cos.\; AOC\; \backslash cos.\; \backslash psi\; +\; \backslash sin.\; AOC\; \backslash sin.\; \backslash psi\; \backslash cos.\; \backslash beta$. Now, from the preceding equations (65), we have $\backslash cos.\; AOC\; =\; \backslash frac\{w\}\{W\}$, $\backslash cos.\; BOC\; =\; \backslash frac\{w\text{'}\}\{W\}$, $\backslash sin.\; AOC\; =\; \backslash frac\{w\text{'}\text{'}\}\{W\}$, $\backslash sin.\; BOC\; =\; \backslash frac\{w\text{'}\text{'}\}\{W\}$. Therefore, substituting and writing for $W$ its value $\backslash sqrt\{(w^2\; +\; w\text{'}^2\; +\; w\text{'}\text{'}^2)\}$, we obtain

Hence it follows that the change of position of any point in consequence of the rotation $Wdt$, in the element of time $dt$, is equivalent to the change that would be effected by three simultaneous and independent rotations $Wdt\; \backslash cos.\; \backslash alpha$, $Wdt\; \backslash cos.\; \backslash beta$, $Wdt\; \backslash cos.\; \backslash gamma$, about three rectangular axes, with which the axis corresponding to the velocity $W$ makes respectively the angles $\backslash alpha,\; \backslash beta,\; \backslash gamma$.

67. In determining the rotation of a body about a fixed centre, it is most convenient to refer every point of the body to a system of three principal axes passing through the centre of gravity; there are therefore two distinct steps in the investigation, the first being to find the situation of the principal axes in respect of absolute space at the end of any given time, and the second to find the position of any point of the body referred to these axes, whose positions have been determined. To accomplish the first of these, the following proposition is required.

68. If at the end of any time $t$ the position of the axis about which the body is revolving be known in respect of the three principal axes supposed to be fixed in the system, together with the angular velocity of rotation; to determine the change of situation of the principal axes in respect of absolute space, in the element of time $dt$.

Imagine a sphere, whose radius = 1, to be described about $O$ the centre of gravity of the body; let $KTL$ (fig. 14) be a great circle, and in $KTL$ assume a point $T$ as that to which the position of the principal axes is to be referred. At the end of the time $t$, let $X,\; Y,\; Z$ be the points in which the three principal axes passing through $O$ meet the surface of the sphere; then, on joining these points by arcs of great circles, the three arcs $XY,\; YZ,\; ZX$ will be quadrantal arcs, since the principal axes are at right angles to each other. Join also $X,\; Y,\; Z$ with $T$ by arcs of great circles, and assume

Fig. 14.

Let $U$ be the point in which the axis about which the body is revolving at the end of the time $t$ meets the surface of the sphere, and let $U$ be joined by arcs of great circles with $X,\; Y,\; Z$. Put $W$ = angular velocity of rotation about $OU$ (which we suppose to be in the direction $XYZ$), and let $\backslash alpha,\; \backslash beta,\; \backslash gamma$ be the angles which $OU$ makes with the principal axes, namely, $\backslash alpha\; =\; UX,\; \backslash beta\; =\; UY,\; \backslash gamma\; =\; UZ$.

To investigate the motion of the point $X$, let $Xe$ be the small arc described by $X$ in the instant $dt$; then this arc

being perpendicular to $UX$, we have the angle $YXe\; =\; UXZ$, and $\backslash sin.\; YXe\; =\; \backslash cos.\; YXU$; also, since $YXZ$ is a right angle; $\backslash sin.\; TXY\; =\; \backslash cos.\; TXZ$. Hence the formulæ of spherical trigonometry give the following equations,

from which, since $TXe\; =\; YXe\; -\; TXY$, we deduce

Draw $ef$ perpendicular to $TX$; then, because $Xe\; =\; Wdt\; \backslash sin.\; \backslash alpha$, we have $Xf\; =\; Wdt\; \backslash sin.\; \backslash alpha\; \backslash cos.\; TXe$, $ef\; =\; Wdt\; \backslash sin.\; \backslash alpha\; \backslash sin.\; TXe$. But $Xf\; =\; -dl,\; ef\; =\; -\backslash sin.\; ld\; \backslash lambda$; therefore, by reason of the above values of $\backslash sin.\; TXe$ and $\backslash cos.\; TXe$, and because $W\; \backslash cos.\; \backslash beta\; =\; w\text{'},\; W\; \backslash cos.\; \backslash gamma\; =\; w\text{'}\text{'}$, we have $\backslash sin.\; ld\; \backslash lambda\; =\; dt(w\text{'}\; \backslash cos.\; n\; -\; w\text{'}\text{'}\; \backslash cos.\; m)$, $\backslash sin.\; ld\; \backslash lambda\; =\; -dt(w\text{'}\; \backslash cos.\; m\; +\; w\text{'}\text{'}\; \backslash cos.\; n)$.

It is evident that the consideration of the motion of each of the other two points $Y$ and $Z$ will lead to expressions entirely similar, and which may be concluded from analogy: we have therefore the following six equations to determine $l,\; m,\; n,\; \backslash lambda,\; \backslash mu,\; \backslash nu$, namely,

Because $\backslash cos^2\; l\; +\; \backslash cos^2\; m\; +\; \backslash cos^2\; n\; =\; 1$, it is only necessary to solve two of the first three equations, in order to find the arcs $l,\; m,\; n$; and when the values of $l,\; m,\; n$ have been obtained, it is only necessary to compute one of the three angles $\backslash lambda,\; \backslash mu,\; \backslash nu$, in order to have the other two; for when one is known, the others are obtained from the equations $\backslash cos.\; (\backslash mu\; -\; \backslash lambda)\; =\; -\backslash cot.\; l\; \backslash cot.\; m,\; \backslash cos.\; (\backslash nu\; -\; \backslash mu)\; =\; -\backslash cot.\; m\; \backslash cot.\; n$, given by the triangles $TYX$ and $TYZ$, of which the sides $XY$ and $YZ$ are arcs of $90^\backslash circ$.

69. By means of the preceding propositions, we are enabled to determine the motion of a body which is at liberty about a fixed point only to turn about a fixed point. The statical principle on which the determination must be made is, that a body so circumstanced will be in a state of equilibrium, when the sums of the moments of the forces with reference to three rectangular axes drawn through the fixed point are nothing relatively to each of the three axes. It is therefore necessary in the first place to find expressions for these moments.

70. Suppose a material point whose co-ordinates are $x$ and $y$ to be urged by two forces, namely, a force $X$, in the direction $OX$ (fig. 15), and another force $Y$ in the direction $OY$, perpendicular to $OX$; then the efficiency of the two forces to turn the particle about the axis $OZ$ perpendicular to the plane $XOY$, in the direction from $X$ towards $Y$, is equal to $Yx\; -\; Xy$. For let $M$ be the situation of the particle, and take $MN$ to represent the intensity and direction of the single force into which $X$ and $Y$ may be compounded; through $M$ and $N$ draw $MA,\; NB$ perpendicular to $OX$ and $MC,\; ND$ perpendicular to $OY$; let $E$ be the point in which $MC$ and $NB$ intersect; join $OM,\; ON,\; OE$, and complete the parallelogram $OF$. Now the force represented by $MN$ is compounded of the two forces represented by $ME$ and $EN$; we have therefore, by construction,

Fig. 15.

ME = -X = force in direction -OX,
EN = +Y = force in direction OY.

But the moment of the force MN in respect of the axis OZ is equal to MN multiplied into the line drawn from O perpendicular to MN, that is, equal to twice the triangle OMN, or to twice the sum of the triangles MOE, EON, MEN, or to the parallelograms MB, NC, EF, or to MB and CF. Now MB = AM · ME = -Xy, and CF = CD · CM = Yz; therefore, the moment of the force MN, or of the two forces X and Y, is expressed by Yz - Xy.

71. If we put Z = a force acting in the direction OZ perpendicular to the plane XOY, we shall evidently obtain similar expressions for the moments in respect of the other axes OX and OY; so that if we suppose the motion about OX to be in the direction from Z towards Y, and that about OY to be from X towards Z, and that about OZ to be from Y towards X, we shall have the three expressions,

moment about OX = Yz - Xy,
about OY = Zx - Xz,
about OZ = Xy - Yz.

72. If we now suppose P, Q, R to be the accelerating forces acting on a particle dm, in the directions of the three axes respectively, then (61) the quantities of motion lost by the particle at each instant, in consequence of the connexion of the parts of the system, or the differences between the impressed and effective forces, are

Substituting these differences for X, Y, Z respectively in the above expressions (71) for the moments of rotation, and forming the integrals in respect of dm, we shall get the sums of the moments in respect of all the particles about each of the three axes. But by the principle of D'Alembert (61), these sums must be severally = 0; therefore, if for brevity we put

there will result the three equations

73. We have now to express these formulae in terms of the angular velocities about the axes. Referring to fig. 14, let $x\; =\; \backslash cos.\; l$, $y\; =\; \backslash cos.\; m$, $z\; =\; \backslash cos.\; n$, then $x,\; y,\; z$ will be the co-ordinates of the point T referred to the rectangular axes OX, OY, OZ; and we shall have $dx\; =\; -\backslash sin.\; l\; dl$, $dy\; =\; -\backslash sin.\; m\; dm$, $dz\; =\; -\backslash sin.\; n\; dn$. Substituting these values of $x,\; y,\; z$, and their differentials, in the three first of the system of equations in (68), and supposing the direction of the motion to be changed (namely, the rotation about OX to be in the direction from Z towards Y, and so with the other as in (71)), those three equations will become

If we now observe that OX, OY, OZ are principal axes, and consequently that $fxy\; dm\; =\; 0$, $fzx\; dm\; =\; 0$, $fyz\; dm\; =\; 0$, we shall find, on forming the expressions for $ddx,\; ddy,\; ddz$, and substituting always for $dx,\; dy,\; dz$ their values as given by these equations,

and consequently

Similar expressions are obtained in the same manner for G dt and H dt. Now it is to be observed that $f(y^2\; +\; z^2)\; dm$ is the moment of inertia with respect to the axis OX; if therefore we assume A, B, C to denote the moments of inertia in respect of the three axes respectively, that is, if we put

$A\; =\; f(y^2\; +\; z^2)\; dm$, $B\; =\; f(x^2\; +\; z^2)\; dm$, $C\; =\; f(x^2\; +\; y^2)\; dm$, the three equations of the last number will be transformed into the following:

74. These three equations enable us to determine the angular velocities of rotation about each of the principal axes when the forces F, G, H are known, and consequently suffice, when joined to the equations of (68) and the equations (A), for the determination of any point of the revolving body at any given instant of time reckoned from the commencement of the motion; for the values of $w,\; w\text{'},\; w\text{'}\text{'}$ being supposed to be found in terms of $t$, and substituted in the two sets of equations of (68), the integrals of these equations will give the situation of the three principal axes at that instant, and the situation of the instantaneous axis of rotation, in respect of the principal axes, will be found by integrating the equations (A) after substituting in them the same values of $w,\; w\text{'},\; w\text{'}\text{'}$.

75. When the body is subject to no accelerating forces, or the origin of the co-ordinates is placed at the centre of gravity, the integrations indicated in the last paragraph are obtained without difficulty; for in this case the quantities denoted by F, G, H, respectively vanish, and the equations (B) become

Now if we multiply the first of these by $w$, the second by $w\text{'}$, and the third by $w\text{'}\text{'}$, and add the products, we shall have

the integral of which is

K being an arbitrary constant, and always positive. Again, let the same equations (a) be multiplied respectively by $Aw,\; Bw\text{'},\; Cw\text{'}\text{'}$, and the products added and integrated; we shall then have, in like manner,

$L^2$ being another arbitrary constant, and always positive. From these two last equations we deduce

and on substituting the values of $w\text{'}$ and $w\text{'}\text{'}$ thus found in the first of the equations (a), and resolving the equations in respect of $dt$, we get

an equation of which the integral will be obtained in a finite form when any two of the three quantities A, B, C are equal (as in the case of a solid of revolution), or when $L^2$ is equal to any one of the three quantities AK, BK, CK. In any case, the relation between $t$ and $w$ is expressed by a single differential equation of the first order, the integral of which gives the value of $w$ in terms of $t$, whence the values of $w\text{'}$ and $w\text{'}\text{'}$ are also given by the preceding formulae.

76. From the two equations (b) and (c) of last para-Mechanical graph, a conclusion is deduced of considerable interest in properties of the physical theory of the universe respecting the stability of the rotation about the axes of greatest and least of rotation.

Rotation. moment. Let the first of those equations be multiplied by $C$, and subtracted from the second; we shall have, on making $L^2\; =\; CK\; =\; D$,

Now, if we suppose the axis of instantaneous rotation to be situated at the commencement of the motion, very near the axis $OZ$, the angular velocity about which $=\; w\text{'}$, then the angle $\backslash gamma$ which it makes with that axis being very small,

and since (66) $\backslash sin.\; \backslash gamma\; =\; \backslash frac\{\backslash sqrt\{(w^2\; +\; w\text{'}^2)\}\}\{\backslash sqrt\{(w^2\; +\; w\text{'}^2\; +\; w\text{'}^2)\}\}$, it follows that

the angular velocities $w$ and $w\text{'}$ will be very small in comparison of $w\text{'}\text{'}$, and consequently $D$ must be a very small quantity. Now $D$ is constant; and therefore if the two differences $A\; -\; C$ and $B\; -\; C$ have the same sign, which will be the case when the value of $B$ is intermediate between $A$ and $C$, the two velocities $w$ and $w\text{'}$ must always be very small; and hence (64) the instantaneous axis will never deviate much from $OZ$ when this is the axis of greatest or of least moment. But if $A\; -\; C$ and $B\; -\; C$ have opposite signs, as will be the case when the value of $C$ lies between the values of $A$ and $B$, then $D$ may be a very small quantity, and the equation still be satisfied, although the values of $w$ and $w\text{'}$ be large; and hence, although the axis of instantaneous rotation may have been very near the axis $C$ at the commencement of the motion, it may deviate to any extent from that axis during the motion. Hence we infer that the rotation about the axes, both of greatest and least moment, is permanent, but that the rotation about the intermediate axis is not permanent.

77. The same conclusions may also be obtained by the direct integration of the equations (a) of (75), if we neglect, as insensible, the squares and products of the small velocities $w$ and $w\text{'}$. Restricting the calculus to this approximation, the third equation (a) becomes $dw\text{'}\text{'}\; =\; 0$, whence $w\text{'}\text{'}\; =\; k$, $k$ being an assured arbitrary constant. Substituting this value of $w\text{'}\text{'}$ in the first two equations (a) we get

In order to integrate these, assume

the quantities $\backslash beta,\; \backslash gamma,\; q,\; i$ being indeterminate constants. We have then

On substituting these values of $w,\; w\text{'},\; dw,\; dw\text{'}$, and rejecting the common factors, the equations (d) become

from which we deduce $q\; =\; k\backslash sqrt\{(A\; -\; C)(B\; -\; C)\}\; \backslash div\; \backslash sqrt\{AB\}$; and also, on assuming another indeterminate constant $\backslash alpha$, $\backslash beta\; =\; \backslash alpha\backslash sqrt\{B(B\; -\; C)\}$, $\backslash gamma\; =\; \backslash alpha\backslash sqrt\{A(A\; -\; C)\}$. Put $\backslash lambda\; =\; \backslash sqrt\{(A\; -\; C)(B\; -\; C)\}\; \backslash div\; \backslash sqrt\{AB\}$, then $q\; =\; \backslash lambda\; k$; and the substitution of these values of $q,\; \backslash beta,\; \backslash gamma$ in the equations (c) gives

which are the complete integrals of the equations (d).

If in these expressions the quantity denoted by $\backslash lambda$ is real, which will be the case when $C$ is the least or the greatest of the three moments, the values of $w$ and $w\text{'}$ must always continue small, because they are proportional to the sines and cosines of a real angle; but if $C$ be intermediate between $A$ and $B$, the $\backslash lambda$ will be an impossible quantity, and the sine and cosine of the arc $\backslash lambda\; kt\; +\; i$ will be changed into exponentials which increase continually with the time $t$, whence the velocities $w$ and $w\text{'}$ will also increase indefinitely; and consequently, by Frisi's theorem, the deviation of the instantaneous axis of rotation from the original axis will increase indefinitely.

If we suppose that at the commencement of the motion

the axis of instantaneous rotation coincided accurately with the principal axis $OZ$, we should have $w\; =\; 0,\; w\text{'}\; =\; 0$, when $t\; =\; 0$; and as in this case we have also $\backslash alpha\; =\; 0,\; w$ and $w\text{'}$ will always be zero, or the body will continue for ever to revolve uniformly about $OZ$. We have therefore, from what precedes, these mechanical properties relative to the rotation of solid bodies: 1. If a body begins to revolve about any one of the three principal axes, and is affected by no external force, it will continue always to revolve about that axis; 2. If it begins to revolve about an axis which is very near the axis of the greatest or least moment, then, although the motion is disturbed a little by the action of an extraneous force, the instantaneous axis of rotation will continue very near the original axis; 3. If the rotation commences about the axis of the mean moment of inertia, then, if the smallest disturbance takes place, the effect will increase indefinitely with the time, and the poles of rotation will entirely change their position on the surface of the body. These properties establish an essential difference between the three principal axes.

78. Our limits will not permit us to enter further into Conclusion. the discussion of this subject, nor is it within the scope of the present article to apply the formulae to the rotation of the planets, an example of which application has already been given in the article PRECESSION OF THE EQUINOXES. It has now been shown that every circumstance relative to the motion of a solid body, acted upon by any number of forces, is expressed by means of two systems of equations (each containing three differential equations of the first order), the first system (62) defining the motion in space of the centre of gravity, and the second (73) the rotation about that centre. At the commencement of the motion these two systems are independent of each other; but it is important to remark, that this independence does not necessarily continue during the motion. If we suppose the moving forces to act upon different points of the body, and to be functions of the co-ordinates of those points, that is, to vary with their distance from the origin, then the co-ordinates enter at the same time into both systems of equations, which therefore cannot be integrated separately, the two motions affecting each other. The integrals can then only be obtained by approximation. There are, however, two important cases in which the two motions continue to be permanently independent of each other. The first is that of a projectile which, during its motion, is acted upon by no other force than gravity; for in this case the equations (62) will be those of an isolated material point, and the weight of the body, acting as a single force applied to the centre of gravity, will have no influence on the rotation, which will be the same as if the centre of gravity were fixed. The second is the case of a sphere, either homogeneous or composed of concentric strata of different densities, the density of each stratum being constant; for in this case the resultant of all the attractive forces, as was shown by Newton, passes through the centre of gravity, which will therefore move as an isolated point, and the motion of rotation will be the same as if the centre of gravity were fixed.

The theory of the rotation of solid bodies forming an important branch of mechanics, is treated more or less in every systematic work on that science. As the more important works in which it is especially considered, we may refer the reader to Euler's Theoria Motus Corporum Solidorum, 1765; D'Alembert, Opuscula, tom. v. and vii.; Vince On the Principles of Progressive and Rotatory Motion, Phil. Trans. 1780; Frisi Opera, tom. ii. 1783; Atwood's Treatise on the Rectilinear Motion and Rotation of Bodies, 1784; Landen's Mathematical Memoirs, vol. ii. 1789; Laplace, Mécanique Céleste, tom. i.; Lagrange, Mécanique Analytique; Poisson, Traité de Mécanique; Venturoli's Mécanique, translated by Cresswell; and the article MECHANICS. (r. 6.)