William Kingdon Clifford. # Elements of dynamic; an introduction to the study of motion and rest in solid and fluid bodies (Volume 2) online

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we tilt the plane a little round q in any direction, so that

it cuts the ellipsoid E, the surface of the system which

touches the tilted plane will be an ellipsoid wholly inside

of E. Hence k will be less for the tilted plane, because

the axes of the touching ellipsoid are diminished. There-

fore P is the plane through q which has the greatest

swing-radius. Now the plane through any point which

has the greatest swing-radius, is normal to the longest

axis of the swing-ellipsoid at the point. Therefore the

longest axis of the swing-ellipsoid (and consequently the

shortest axis of the ellipsoid of gyration) at the point q,

is normal to the ellipsoid E, confocal to the null-quadric,

which passes through q.

Let R be the tangent plane at q to the two-sheeted

hyperboloid, H, of the system, which passes through q.

This surface, like the ellipsoid, lies entirely on one side of

the tangent plane in the neighbourhood of the point of

contact. Hence if we tilt the plane a little it will cut the

surface in a small oval curve; and therefore the surface

of the system which touches the new position of the plane

will be a two-sheeted hyperboloid lying further away from

the centre than H, and therefore having a larger trans-

verse axis. Consequently k is increased by the tilting;

whence it follows that R is the plane of least swing-

radius that can be drawn through q. This plane is normal

to the shortest axis of the swing-ellipsoid, or longest axis

of the ellipsoid of gyration.

We may put these results together by saying that at

any point the axis of least moment is normal to the ellip-

soid, and the axis of greatest moment to the two-sheeted

hyperboloid, which can be drawn through the point con-

focal to the null-quadric.

It follows that the ellipsoids and the two-sheeted hyper-

boloids of the system cut each other everywhere at right

angles. We shall now prove that the third principal axis

is normal to the one-sheeted hyperboloid through q, so

that the three systems of surfaces cut each other every-

where at right angles. [Cetera desunt.]

DYNAMIC.

CORE OF A SOLID.

The poles of all those planes which do not cut the

boundary of a solid fill up a certain space which is called

its core. If the solid has indentations we must suppose

a membrane to be tightly wrapped round it; then the

boundary of the core will be traced out by the pole of a

plane which moves about so as always to touch this

membrane.

The main use of knowing the core of the simpler class

of solids is that it is more easily remembered than a

formula, and supplies a ready means of finding the swing-

radius about any axis.

We have seen that the core of a finite line, which is

of one dimension, is the middle third ; and that the core

of a triangle or ellipse, which may be regarded as typical

figures of two dimensions, is the middle fourth. We shall

shew that the core of a tetrahedron or ellipsoid, which are

figures in three dimensions, is the middle fifth; that is, it

is a figure similar to the original, having the same mass-

centre, and of one-fifth of the linear dimensions.

A parallelepiped is to be taken as a figure of one

dimension in regard to each of the lines joining the centres

of opposite faces. Its core is accordingly

an octahedron having the middle thirds

of those lines for its three diagonals.

This figure is the reciprocal of the

parallelepiped, in the sense that the

vertices, edges and faces of one of

them are polar to, the faces, edges, and

vertices of the other. Thus the paral-

lelepiped has six four-sided faces and

eight three-legged vertices, while the

octahedron has six four-legged vertices and eight three-

sided faces.

For a triangular prism, or elliptic cylinder, we must as

before take the middle third of the line joining the mass-

centres of the opposite plane ends, and then join the ends

CORE OF A SOLID. 45

of this middle third to the middle fourth of a section of

the solid made by a plane parallel to the two ends and

midway between them. We shall thus form a core which

is in one case a pentacron, like that considered on p. 8,

and in the other case is made of two cones with a common

base and vertices on opposite sides of it. The pentacron,

having two three-legged and three four-legged vertices,

with six three-sided faces, is the reciprocal of the triangular

prism, which has two three-sided and three four-sided

faces, with six three-legged vertices.

In general, we may obtain the core of any prism or

cylinder, bounded partly by two parallel plane faces or

ends, and partly by a surface made up of parallel straight

lines, by taking the ends of the middle third of the line

which joins the mass-centres of the plane ends, and joining

them to the core of the middle section, which is parallel

to the plane ends and midway between them. The figure

so formed consists always of two pyramids or cones, having

a common base, and vertices on either side of it.

To determine the core of a tetrahedron, it is sufficient

to find the pole of a plane through one vertex a parallel

to the opposite face, bed. If the tetrahedron when loaded

from this plane be divided into slices of equal thickness

by planes parallel to it, the areas of the slices will vary as

the squares of their distances from a, and their densities

directly as those distances ; therefore their masses will be

proportional to the cubes of the distances from a. If g be

the mass-centre of the tetrahedron, a of the face bed, the

mass-centres of all these slices will be in aa, and conse-

quently the tetrahedron may be replaced by a rod aa.

whose density varies as the cube of the distance from a.

Thus the mass-centre of the loaded tetrahedron (which is

the pole of the plane through a parallel to bed) is | of

the way from a towards a, and its distance from a is

therefore | . ag = || ag. Thus the squared swing-radius,

measured parallel to ag, of a plane through g parallel to

bed is j-^ag 2 . Therefore the pole of bed, whose distance

from g measured along ag is $ag, must be at a point

whose distance is ^ga. Now the core of the tetrahedron

is clearly a tetrahedron whose vertices are the poles of

46 DYNAMIC.

the faces ; and we see that these poles are on the lines

ga, gb, gc, gd, at distances from g equal to one-fifth of

those lines respectively. Thus the core is the middle

fifth.

For a sphere we must proceed in the same way as for

a circle. The squared swing-radius of the centre in regard

to the sphere is the sum of the squared swing-radii of

three mutually perpendicular planes through it, and is

therefore three times that of either of them. If we

suppose the sphere made of concentric shells of equal

thickness, the masses of these shells are as the squares of

their distances from the centre ; the problem is therefore

the same as that of finding the squared swing-radius of a

rod, whose density varies as the squared distance from one

end, in regard to that end. If a is the length of the rod,

the second moment is ^a 5 , and the mass is ^a 3 ; thus the

squared swing-radius is fa 2 . This is therefore the squared

swing-radius of a sphere of radius a in regard to its

centre. In regard to a diametral plane it is one-third of

this, or ^a 2 ; from which it follows at once that the core

is a concentric sphere of the radius \a.

If any body be transformed by a homogeneous strain,

the core of the strained body is the strained position of the

core of the unstrained body. For let x be the distance of

a particle Sm of the body from a given plane, measured

in a given direction, and k the swing-radius in regard to

that plane, measured in the same direction. Then

mk 2 =fx 2 8m.

If the body receive a homogeneous strain, so that the

masses of corresponding volumes are unaltered, the lines

x and k, which are all parallel, will remain parallel and

be altered in the same ratio; so that if every x becomes

\x, k will become \k. Therefore m. (\kf =/(X#) 2 Sm, or

\k is the swing-radius measured parallel to the \x in the

strained body. Hence it follows that the pole of the given

plane, the swing-ellipsoid at every point, the null-quadric,

and the core of the body, all undergo the same homo-

geneous strain as the body does. It should be noticed

that this is not true of the ellipsoid of gyration.

CORE OF A SOLID. 47

Hence it follows, by homogeneous strain of a sphere,

that the core of an ellipsoid is a similar, similarly situated,

and concentric ellipsoid, of one-fifth the linear dimensions;

or, as we may say, it is the middle fifth of the body.

We may exemplify the use of the core by finding the

swing- radius of a sphere about certain axes. For an axis

through the centre the squared swing-radius is twice that

of a diametral plane, and therefore fa 2 , since the core is

the middle fifth. Hence about an axis touching the. sphere

the squared swing-radius is |a 2 .

Similarly in the case of a right circular cylinder,

height 2h, radius a, the squared swing-radius of its middle

section is |7i 2 , and of a plane through its axis a 2 . Hence

about the axis it is ^-a 2 , about a line through the centre

perpendicular to it a 2 + ^A 2 , and about a diameter of

either plane face Ja 2 + |& 2 .

CHAPTER III. MOMENTUM.

MOMENTUM OF TRANSLATION-VELOCITY.

THE product of the mass of a particle by its velocity is

called the momentum of the particle. Like the velocity, it

is a directed quantity ; but whereas the velocity is a vector,

and may be specified by a line of proper length and direc-

tion drawn anywhere in space, the momentum is what we

have called a rotor, and is situated in the straight line

through the particle along which it is moving. In fact,

the velocity is a magnitude associated with a direction,

and the mass is a magnitude associated with a definite

point ; thus the momentum, which is the product of these

two, is the product of their magnitudes associated with a

line drawn through the point in the given direction.

To compound together the momenta of two or more

particles, therefore, we must treat them as if they were

rotations about axes through the particles [p. 124]. If, for

example, particles of masses I, m situated at a, b, have the

same velocity, their momenta are proportional to the

masses I, m, and may be represented by lengths I, m

measured on lines through a, b in the direction of the

common velocity. If the lengths represented spins about

these lines as axes, the resultant spin would have a mag-

nitude equal to their sum, and would be about a parallel

line through the point c, which divides ab in the ratio

MOMENT OF MOMENTUM. 49

of m : I. Now c is the mass-centre

of the particles at a. b, and its velo-

city is the same as that of a and b.

Hence the resultant of the momenta

of two particles which have the same

velocity is the momentum of their re-

sultant mass. c 6

If we add to these a third particle, with the same

velocity, we may compound its momentum with that of the

resultant mass of the first two, and so find that the result-

ant of all the momenta is the momentum of the resultant

mass of the three ; and so on. Hence if a body have a

translation-velocity, the resultant of the momenta of its par-

ticles is equal to the momentum of its resultant mass ; that

is to say, its magnitude is the mass of the whole body

multiplied by the magnitude of the given velocity, and acts

in a straight line through the mass-centre in the direc-

tion of the given velocity.

MOMENT OF MOMENTUM.

The moment of the momentum of a particle in

regard to any point is defined in the same way as the

moment of any similar magnitude [p. 92]. Namely, if p

be the position of the particle, pt a

line representing its momentum, the

moment of the momentum about o is

twice the area of the triangle opt.

This is to be regarded as a vector

perpendicular to the plane opt, but not

localized. If m is the mass of a particle whose position-

vector is p, and velocity p, the moment of its momentum

in regard to the origin is mVpp.

Since momenta are to be compounded like spins, it

follows that the moment of the resultant of any number of

momenta in regard to any point is equal to the sum of the

moments of the components in regard to the same point.

We know that two spins about parallel axes, of the

same magnitude but of opposite senses, compound into a

c. 4

50 DYNAMIC.

translation-velocity perpendicular to the plane containing

them. This translation- velocity is equal to the sum of

their moments in regard to any point in space, and its

magnitude is the product of the magnitude of either spin

into the distance between their axes. Two such spins are

therefore equivalent to any other two equal and opposite

spins about parallel axes in any plane parallel to that of

the first pair, provided that this product is the same for

the two pairs.

In the same way, if the momenta of two particles are

on two parallel lines, equal in magnitude, but opposite in

direction, the sum of their moments is the same in regard

to every point of space, and the two momenta are equiva-

lent to any other such pair the sum of whose moments in

regard to all points is the same as that of the first pair.

They must therefore be regarded as together constituting

a vector quantity, of the nature of a moment of momentum ;

namely, it is precisely this constant sum of their moments.

If we start with two opposite momenta along parallel

lines, which are only nearly equal in magnitude, the effect

of makingithem more nearly equal is" to send their result-

ant farthest off and to diminish its magnitude. Hence a

moment of momentum may be regarded as the limit of a

very small momentum along a line which is very far off;

just as a translation-velocity may be regarded as an in-

finitely small spin about an infinitely distant axis.

Every momentum may be resolved into an equal and

parallel momentum through the origin, together with its

moment in regard to the origin. Thus if the particle m at

the end of the vector p have the velocity p, its momentum

is equivalent to mp through the origin together with the

moment of momentum m Vpp.

It is sometimes convenient to indicate a rotor through

the origin by prefixing the letter ft to the symbol of a

vector having the same magnitude and direction. Thus if

a is any vector, ft a means the rotor through the origin

which is one way of representing that vector. In this

notation, the momentum of the particle whose mass-vector

is mp will be denoted by ft mp + mVpp. And similarly the

velocity of a rigid body which has a spin 6 about an axis

S

ROTOR PART OF MOMENTUM. 51

passing through the end of the vector p will be denoted by

16+ VpO. In general, since any system of rotors and

vectors is equivalent to a rotor through the origin and a

vector, the expression for the resultant motor of such a

system is Ha -f- ft.

The moment of a vector in regard to any point is

simply the vector itself. Hence the moment of the system

Ha + fi, in regard to the point whose position-vector is 7,

is Fa7 + .

ROTOR PART OF MOMENTUM.

When we resolve the momentum of every particle of a

body into a parallel momentum through the origin, together

with a moment of momentum, and then add the results for

the whole body ; the rotor part of this is independent of

the particular origin we take. For it is simply the sum

of all the momenta regarded as vectors.

We shall prove that this rotor part is equal (in magni-

tude and direction) to the momentum of the resultant mass ;

that is to say, it is the same as if the whole mass were

collected at the mass-centre, and moving with its velocity.

To prove this, it is only necessary to remember that the

mass-vector of the resultant mass is the sum of the mass-

vectors of all the particles, or it is fpdm. The momentum

of the resultant mass is the rate of change of this, or

fpdm; but this is the vector-sum of the momenta of all the

particles.

Now every motion of a body can be resolved into two ;

a translation equal to the velocity of the mass-centre, and

the motion relative to the mass-centre. The momentum

of the whole body is compounded of the momenta due to

these motions. The momentum of the translation is a

pure rotor, passing through the mass-centre ; the proposi-

tion just proved shews us that the momentum of the motion

relative to the mass-centre is a pure vector. For the rotor

part of the whole momentum has been shewn to be equal

to the momentum of the translation.

In calculating the momentum of any motion, therefore,

it is sufficient to find the moment of momentum about the

42

52 DYNAMIC.

mass-centre ; for since the momentum of motion relative

to the mass-centre is a pure vector, its moment is the

same in regard to every point of space.

MOMENTUM OF SPINS ABOUT FIXED POINT.

We now consider a rigid body rotating about a fixed

point, and propose to calculate its moment of momentum

in regard to the point.

Every spin about an axis through the point may be

resolved into three component spins about the principal

axes ; and its moment of momentum will be the vector-

sum of the moments of momenta due to these. The first

thing to be done, therefore, is to find the moment of

momentum of a spin about a principal axis.

In the first place we may remark that the component

along the axis of the moment of momentum of a spin about

any straight line, is equal to the angular velocity multi-

plied by the second moment of the body about that line.

For let a be the origin, ab the spin, p a

rotating particle. The moment of its mo-

mentum about a is i . ap multiplied by the

magnitude of the momentum. Hence the

component of this along ab is i . mp multi-

plied by the same magnitude, which is

ab . mp into the mass of the particle. Thus

the component is ab . mp 2 into the mass of

the particle, or the angular velocity multi-

plied by the second moment. Adding this result for all

the particles, we have the proposition as stated for the

whole body.

The component in any direction perpendicular to the

axis is the angular velocity multiplied by the rnixed mo-

ment of two planes, one of which is the plane through a

perpendicular to the axis, and the other is the plane

through the axis perpendicular to the given direction.

For the whole component perpendicular to the axis is

i . am x magnitude of momentum ; that is, it is parallel to

pm, and of magnitude ab . am .pm. Now am is the

distance of p from the normal plane through a, and the

MOMENTUM OF SPINS ABOUT FIXED POINT. 53

component of pm in any direction perpendicular to ab is

the distance of p from a plane through ab perpendicular

to that direction ; whence the proposition follows.

Now if the axis is a principal axis, the mixed moment

of a plane through it and a plane perpendicular to it is

zero. Hence the moment about any point of momentum

due to a spin about a principal axis through the point is

directed along the axis, and is equal in magnitude to the

angular velocity multiplied by the second moment.

We may put these propositions together as follows.

Let oX, o Y, oZ be three rectangular lines through o, and

let us consider a unit spin about oZ, which is denoted by

k. The velocity of the end of p, = xi + yj + zTc, due to this

spin, is Vk (xi + yj + zk), = xj yi. Let Sm be the mass of

the particle there situated, then the moment of its mo-

mentum is

V (xi + yj + zk) (xj yi) 8m ={ xz.i yz.j+(x* + y z ) k} 8m.

Hence the moment of momentum of the whole body is

Jxzdm . i fyzdm . j + /(a; 2 + y 2 ) dm . k,

which gives the components as above stated.

Now let the axes be principal axes, and let the swing-

radii about them be a, b, c. Then if the body have a spin

0,=pi+qj + rk, its moment of momentum about the

origin will be m (a 2 pi + b 2 qj + c 2 rk) ; since it is the result-

ant of the moments of momentum due to the three spins

pi, qj, rk. The components of the spin must therefore be

respectively multiplied by ma 2 , mb 2 , me 2 , to give the com-

ponents of the moment of momentum.

Thus the moment of momentum is a pure linear function

of the spin. The ellipsoid representing this function must

have its semi-axes inversely proportional to the square

roots of the second moments ma 2 , mb 2 , me 2 , that is, in-

versely proportional to a, b, c. Their actual values, in ac-

cordance with the formula [p. 176], are mbc, mca, mab

each divided by TT. Thus the relation between x, y, z for

any point on this ellipsoid is

7T 2 (aV + b*y 2 + cV) = (mabcf.

54 DYNAMIC.

The ellipsoid thus determined is called the momental

ellipsoid,

There is no great convenience in thus fixing the size of

the momental ellipsoid, on account of the two scales of

representation involved ; angular velocity is represented by

a line, and mass by the inverse of a line (since mbc is one

semi-axis). If we take any ellipsoid similar to this one,

we may so choose the scale on which angular velocities are

represented that the momentum due to a spin represented

by any radius vector of the ellipsoid shall be equal to the

mass of the body multiplied by the area of the conjugate

section. We shall therefore give the name " momental

ellipsoid" to any ellipsoid having axes along oX, oY, oZ

inversely proportional to a, b, c ; so that we may determine

its size in particular cases by other considerations of con-

venience.

The momental ellipsoid is the reciprocal of the ellipsoid

of gyration. For the ellipsoid of gyration has semi-axes

a, 6, c, and therefore if p be a perpendicular from the

origin on a tangent plane to it, and make angles a/3y with

the axes, we must have p 2 = a* cos 2 a + 6 2 cos' 2 /3 + c 2 cos 2 7.

Now let us measure on the line p a distance r from the

origin, such that pr = C, and let s be the point at the end

of this distance, so that os = r. Then the locus of s will be

the reciprocal of the ellipsoid of gyration [p. 101]. The

coordinates of s are

x = r cos ot, y = r cos /3, z = r cos 7,

or px = G cos ai,py=G cos /3, pz = C cos 7.

Therefore

p 2 (aV + by + cV) = <7 2 (a 2 cos 2 a + 6 2 cos 2 /9 + c 2 cos 2 7 ) = C*p*

and consequently aV + 6 2 y 2 + cV = C 2

so that the locus of s is an ellipsoid whose axes are in-

versely proportional to a, b, c, that is, an ellipsoid of

momentum.

MOMENTUM OF TWIST. 55

MOMENTUM OF TWIST.

Let us now take the mass-centre of a rigid body for

origin. The instantaneous state of motion of the body is

made up of a translation-velocity equal to the velocity of

the mass-centre, and a spin about some axis through it.

Let these, regarded as vectors, be denoted by a and /3 re-

spectively ; then to indicate that the latter is really a rotor

passing through the origin, we may denote the whole

motion of the body by ftyS -,'- a. This is the general symbol

for a twist-velocity.

It is important to observe that the vector part of this

velocity gives rise to the rotor part of the momentum, and

vice versa. The translation is not localized, but its mo-

mentum has to pass through the mass-centre. The rota-

tion is about a definite axis ; but its momentum is a mere

moment of momentum, which has only magnitude and

direction, no definite position.

Thus the momentum due to the translation a is ftraa,

where m is the mass of the body. The momentum due

to the spin ft/3 is 0/3, where is the pure function which

converts any spin into its moment of momentum. Hence

the whole momentum due to the twist ft/3 +a is

ftma + 0/3.

Given any screw, the momentum due to a twist about

it may be associated with another screw. Namely the

momentum is equivalent to a rotor part along a certain

axis, together with a vector part parallel to the axis ; the

proof of this is precisely like that for twists [p. 126].

We now propose to determine in what cases these two

screws are identical. When this is so, the momentum

ftraa + 0/3 is a numerical multiple of the velocity ft/3 +a ;

say ftraa + 0/3 = x (ft/3 + a),

whence ma. = #/3, 0/3 = #a,

and therefore ra/3 = x 2 ft.

56 DYNAMIC.

Hence the spin /3 must be about one of the principal axes

at the mass-centre. Suppose it to be oX, then </3 = ma 2 /3,

and x = ma. Consequently a = + a/3, and the pitch is

+ a. There are therefore six screws having this property ;

they are called the principal screws of the body. These

axes are the principal axes at the mass-centre, and their

pitches are the swing-radii about them.

57

APPENDIX I. (A.)

ACCELERATION DEPENDING ON STRAIN.

Imagine a spiral spring, like that of a spring-balance,

with circular disks fastened to its ends and a small hole

through one of them. Let it be partially compressed, and

held in that position by a string attaching one of the two

it cuts the ellipsoid E, the surface of the system which

touches the tilted plane will be an ellipsoid wholly inside

of E. Hence k will be less for the tilted plane, because

the axes of the touching ellipsoid are diminished. There-

fore P is the plane through q which has the greatest

swing-radius. Now the plane through any point which

has the greatest swing-radius, is normal to the longest

axis of the swing-ellipsoid at the point. Therefore the

longest axis of the swing-ellipsoid (and consequently the

shortest axis of the ellipsoid of gyration) at the point q,

is normal to the ellipsoid E, confocal to the null-quadric,

which passes through q.

Let R be the tangent plane at q to the two-sheeted

hyperboloid, H, of the system, which passes through q.

This surface, like the ellipsoid, lies entirely on one side of

the tangent plane in the neighbourhood of the point of

contact. Hence if we tilt the plane a little it will cut the

surface in a small oval curve; and therefore the surface

of the system which touches the new position of the plane

will be a two-sheeted hyperboloid lying further away from

the centre than H, and therefore having a larger trans-

verse axis. Consequently k is increased by the tilting;

whence it follows that R is the plane of least swing-

radius that can be drawn through q. This plane is normal

to the shortest axis of the swing-ellipsoid, or longest axis

of the ellipsoid of gyration.

We may put these results together by saying that at

any point the axis of least moment is normal to the ellip-

soid, and the axis of greatest moment to the two-sheeted

hyperboloid, which can be drawn through the point con-

focal to the null-quadric.

It follows that the ellipsoids and the two-sheeted hyper-

boloids of the system cut each other everywhere at right

angles. We shall now prove that the third principal axis

is normal to the one-sheeted hyperboloid through q, so

that the three systems of surfaces cut each other every-

where at right angles. [Cetera desunt.]

DYNAMIC.

CORE OF A SOLID.

The poles of all those planes which do not cut the

boundary of a solid fill up a certain space which is called

its core. If the solid has indentations we must suppose

a membrane to be tightly wrapped round it; then the

boundary of the core will be traced out by the pole of a

plane which moves about so as always to touch this

membrane.

The main use of knowing the core of the simpler class

of solids is that it is more easily remembered than a

formula, and supplies a ready means of finding the swing-

radius about any axis.

We have seen that the core of a finite line, which is

of one dimension, is the middle third ; and that the core

of a triangle or ellipse, which may be regarded as typical

figures of two dimensions, is the middle fourth. We shall

shew that the core of a tetrahedron or ellipsoid, which are

figures in three dimensions, is the middle fifth; that is, it

is a figure similar to the original, having the same mass-

centre, and of one-fifth of the linear dimensions.

A parallelepiped is to be taken as a figure of one

dimension in regard to each of the lines joining the centres

of opposite faces. Its core is accordingly

an octahedron having the middle thirds

of those lines for its three diagonals.

This figure is the reciprocal of the

parallelepiped, in the sense that the

vertices, edges and faces of one of

them are polar to, the faces, edges, and

vertices of the other. Thus the paral-

lelepiped has six four-sided faces and

eight three-legged vertices, while the

octahedron has six four-legged vertices and eight three-

sided faces.

For a triangular prism, or elliptic cylinder, we must as

before take the middle third of the line joining the mass-

centres of the opposite plane ends, and then join the ends

CORE OF A SOLID. 45

of this middle third to the middle fourth of a section of

the solid made by a plane parallel to the two ends and

midway between them. We shall thus form a core which

is in one case a pentacron, like that considered on p. 8,

and in the other case is made of two cones with a common

base and vertices on opposite sides of it. The pentacron,

having two three-legged and three four-legged vertices,

with six three-sided faces, is the reciprocal of the triangular

prism, which has two three-sided and three four-sided

faces, with six three-legged vertices.

In general, we may obtain the core of any prism or

cylinder, bounded partly by two parallel plane faces or

ends, and partly by a surface made up of parallel straight

lines, by taking the ends of the middle third of the line

which joins the mass-centres of the plane ends, and joining

them to the core of the middle section, which is parallel

to the plane ends and midway between them. The figure

so formed consists always of two pyramids or cones, having

a common base, and vertices on either side of it.

To determine the core of a tetrahedron, it is sufficient

to find the pole of a plane through one vertex a parallel

to the opposite face, bed. If the tetrahedron when loaded

from this plane be divided into slices of equal thickness

by planes parallel to it, the areas of the slices will vary as

the squares of their distances from a, and their densities

directly as those distances ; therefore their masses will be

proportional to the cubes of the distances from a. If g be

the mass-centre of the tetrahedron, a of the face bed, the

mass-centres of all these slices will be in aa, and conse-

quently the tetrahedron may be replaced by a rod aa.

whose density varies as the cube of the distance from a.

Thus the mass-centre of the loaded tetrahedron (which is

the pole of the plane through a parallel to bed) is | of

the way from a towards a, and its distance from a is

therefore | . ag = || ag. Thus the squared swing-radius,

measured parallel to ag, of a plane through g parallel to

bed is j-^ag 2 . Therefore the pole of bed, whose distance

from g measured along ag is $ag, must be at a point

whose distance is ^ga. Now the core of the tetrahedron

is clearly a tetrahedron whose vertices are the poles of

46 DYNAMIC.

the faces ; and we see that these poles are on the lines

ga, gb, gc, gd, at distances from g equal to one-fifth of

those lines respectively. Thus the core is the middle

fifth.

For a sphere we must proceed in the same way as for

a circle. The squared swing-radius of the centre in regard

to the sphere is the sum of the squared swing-radii of

three mutually perpendicular planes through it, and is

therefore three times that of either of them. If we

suppose the sphere made of concentric shells of equal

thickness, the masses of these shells are as the squares of

their distances from the centre ; the problem is therefore

the same as that of finding the squared swing-radius of a

rod, whose density varies as the squared distance from one

end, in regard to that end. If a is the length of the rod,

the second moment is ^a 5 , and the mass is ^a 3 ; thus the

squared swing-radius is fa 2 . This is therefore the squared

swing-radius of a sphere of radius a in regard to its

centre. In regard to a diametral plane it is one-third of

this, or ^a 2 ; from which it follows at once that the core

is a concentric sphere of the radius \a.

If any body be transformed by a homogeneous strain,

the core of the strained body is the strained position of the

core of the unstrained body. For let x be the distance of

a particle Sm of the body from a given plane, measured

in a given direction, and k the swing-radius in regard to

that plane, measured in the same direction. Then

mk 2 =fx 2 8m.

If the body receive a homogeneous strain, so that the

masses of corresponding volumes are unaltered, the lines

x and k, which are all parallel, will remain parallel and

be altered in the same ratio; so that if every x becomes

\x, k will become \k. Therefore m. (\kf =/(X#) 2 Sm, or

\k is the swing-radius measured parallel to the \x in the

strained body. Hence it follows that the pole of the given

plane, the swing-ellipsoid at every point, the null-quadric,

and the core of the body, all undergo the same homo-

geneous strain as the body does. It should be noticed

that this is not true of the ellipsoid of gyration.

CORE OF A SOLID. 47

Hence it follows, by homogeneous strain of a sphere,

that the core of an ellipsoid is a similar, similarly situated,

and concentric ellipsoid, of one-fifth the linear dimensions;

or, as we may say, it is the middle fifth of the body.

We may exemplify the use of the core by finding the

swing- radius of a sphere about certain axes. For an axis

through the centre the squared swing-radius is twice that

of a diametral plane, and therefore fa 2 , since the core is

the middle fifth. Hence about an axis touching the. sphere

the squared swing-radius is |a 2 .

Similarly in the case of a right circular cylinder,

height 2h, radius a, the squared swing-radius of its middle

section is |7i 2 , and of a plane through its axis a 2 . Hence

about the axis it is ^-a 2 , about a line through the centre

perpendicular to it a 2 + ^A 2 , and about a diameter of

either plane face Ja 2 + |& 2 .

CHAPTER III. MOMENTUM.

MOMENTUM OF TRANSLATION-VELOCITY.

THE product of the mass of a particle by its velocity is

called the momentum of the particle. Like the velocity, it

is a directed quantity ; but whereas the velocity is a vector,

and may be specified by a line of proper length and direc-

tion drawn anywhere in space, the momentum is what we

have called a rotor, and is situated in the straight line

through the particle along which it is moving. In fact,

the velocity is a magnitude associated with a direction,

and the mass is a magnitude associated with a definite

point ; thus the momentum, which is the product of these

two, is the product of their magnitudes associated with a

line drawn through the point in the given direction.

To compound together the momenta of two or more

particles, therefore, we must treat them as if they were

rotations about axes through the particles [p. 124]. If, for

example, particles of masses I, m situated at a, b, have the

same velocity, their momenta are proportional to the

masses I, m, and may be represented by lengths I, m

measured on lines through a, b in the direction of the

common velocity. If the lengths represented spins about

these lines as axes, the resultant spin would have a mag-

nitude equal to their sum, and would be about a parallel

line through the point c, which divides ab in the ratio

MOMENT OF MOMENTUM. 49

of m : I. Now c is the mass-centre

of the particles at a. b, and its velo-

city is the same as that of a and b.

Hence the resultant of the momenta

of two particles which have the same

velocity is the momentum of their re-

sultant mass. c 6

If we add to these a third particle, with the same

velocity, we may compound its momentum with that of the

resultant mass of the first two, and so find that the result-

ant of all the momenta is the momentum of the resultant

mass of the three ; and so on. Hence if a body have a

translation-velocity, the resultant of the momenta of its par-

ticles is equal to the momentum of its resultant mass ; that

is to say, its magnitude is the mass of the whole body

multiplied by the magnitude of the given velocity, and acts

in a straight line through the mass-centre in the direc-

tion of the given velocity.

MOMENT OF MOMENTUM.

The moment of the momentum of a particle in

regard to any point is defined in the same way as the

moment of any similar magnitude [p. 92]. Namely, if p

be the position of the particle, pt a

line representing its momentum, the

moment of the momentum about o is

twice the area of the triangle opt.

This is to be regarded as a vector

perpendicular to the plane opt, but not

localized. If m is the mass of a particle whose position-

vector is p, and velocity p, the moment of its momentum

in regard to the origin is mVpp.

Since momenta are to be compounded like spins, it

follows that the moment of the resultant of any number of

momenta in regard to any point is equal to the sum of the

moments of the components in regard to the same point.

We know that two spins about parallel axes, of the

same magnitude but of opposite senses, compound into a

c. 4

50 DYNAMIC.

translation-velocity perpendicular to the plane containing

them. This translation- velocity is equal to the sum of

their moments in regard to any point in space, and its

magnitude is the product of the magnitude of either spin

into the distance between their axes. Two such spins are

therefore equivalent to any other two equal and opposite

spins about parallel axes in any plane parallel to that of

the first pair, provided that this product is the same for

the two pairs.

In the same way, if the momenta of two particles are

on two parallel lines, equal in magnitude, but opposite in

direction, the sum of their moments is the same in regard

to every point of space, and the two momenta are equiva-

lent to any other such pair the sum of whose moments in

regard to all points is the same as that of the first pair.

They must therefore be regarded as together constituting

a vector quantity, of the nature of a moment of momentum ;

namely, it is precisely this constant sum of their moments.

If we start with two opposite momenta along parallel

lines, which are only nearly equal in magnitude, the effect

of makingithem more nearly equal is" to send their result-

ant farthest off and to diminish its magnitude. Hence a

moment of momentum may be regarded as the limit of a

very small momentum along a line which is very far off;

just as a translation-velocity may be regarded as an in-

finitely small spin about an infinitely distant axis.

Every momentum may be resolved into an equal and

parallel momentum through the origin, together with its

moment in regard to the origin. Thus if the particle m at

the end of the vector p have the velocity p, its momentum

is equivalent to mp through the origin together with the

moment of momentum m Vpp.

It is sometimes convenient to indicate a rotor through

the origin by prefixing the letter ft to the symbol of a

vector having the same magnitude and direction. Thus if

a is any vector, ft a means the rotor through the origin

which is one way of representing that vector. In this

notation, the momentum of the particle whose mass-vector

is mp will be denoted by ft mp + mVpp. And similarly the

velocity of a rigid body which has a spin 6 about an axis

S

ROTOR PART OF MOMENTUM. 51

passing through the end of the vector p will be denoted by

16+ VpO. In general, since any system of rotors and

vectors is equivalent to a rotor through the origin and a

vector, the expression for the resultant motor of such a

system is Ha -f- ft.

The moment of a vector in regard to any point is

simply the vector itself. Hence the moment of the system

Ha + fi, in regard to the point whose position-vector is 7,

is Fa7 + .

ROTOR PART OF MOMENTUM.

When we resolve the momentum of every particle of a

body into a parallel momentum through the origin, together

with a moment of momentum, and then add the results for

the whole body ; the rotor part of this is independent of

the particular origin we take. For it is simply the sum

of all the momenta regarded as vectors.

We shall prove that this rotor part is equal (in magni-

tude and direction) to the momentum of the resultant mass ;

that is to say, it is the same as if the whole mass were

collected at the mass-centre, and moving with its velocity.

To prove this, it is only necessary to remember that the

mass-vector of the resultant mass is the sum of the mass-

vectors of all the particles, or it is fpdm. The momentum

of the resultant mass is the rate of change of this, or

fpdm; but this is the vector-sum of the momenta of all the

particles.

Now every motion of a body can be resolved into two ;

a translation equal to the velocity of the mass-centre, and

the motion relative to the mass-centre. The momentum

of the whole body is compounded of the momenta due to

these motions. The momentum of the translation is a

pure rotor, passing through the mass-centre ; the proposi-

tion just proved shews us that the momentum of the motion

relative to the mass-centre is a pure vector. For the rotor

part of the whole momentum has been shewn to be equal

to the momentum of the translation.

In calculating the momentum of any motion, therefore,

it is sufficient to find the moment of momentum about the

42

52 DYNAMIC.

mass-centre ; for since the momentum of motion relative

to the mass-centre is a pure vector, its moment is the

same in regard to every point of space.

MOMENTUM OF SPINS ABOUT FIXED POINT.

We now consider a rigid body rotating about a fixed

point, and propose to calculate its moment of momentum

in regard to the point.

Every spin about an axis through the point may be

resolved into three component spins about the principal

axes ; and its moment of momentum will be the vector-

sum of the moments of momenta due to these. The first

thing to be done, therefore, is to find the moment of

momentum of a spin about a principal axis.

In the first place we may remark that the component

along the axis of the moment of momentum of a spin about

any straight line, is equal to the angular velocity multi-

plied by the second moment of the body about that line.

For let a be the origin, ab the spin, p a

rotating particle. The moment of its mo-

mentum about a is i . ap multiplied by the

magnitude of the momentum. Hence the

component of this along ab is i . mp multi-

plied by the same magnitude, which is

ab . mp into the mass of the particle. Thus

the component is ab . mp 2 into the mass of

the particle, or the angular velocity multi-

plied by the second moment. Adding this result for all

the particles, we have the proposition as stated for the

whole body.

The component in any direction perpendicular to the

axis is the angular velocity multiplied by the rnixed mo-

ment of two planes, one of which is the plane through a

perpendicular to the axis, and the other is the plane

through the axis perpendicular to the given direction.

For the whole component perpendicular to the axis is

i . am x magnitude of momentum ; that is, it is parallel to

pm, and of magnitude ab . am .pm. Now am is the

distance of p from the normal plane through a, and the

MOMENTUM OF SPINS ABOUT FIXED POINT. 53

component of pm in any direction perpendicular to ab is

the distance of p from a plane through ab perpendicular

to that direction ; whence the proposition follows.

Now if the axis is a principal axis, the mixed moment

of a plane through it and a plane perpendicular to it is

zero. Hence the moment about any point of momentum

due to a spin about a principal axis through the point is

directed along the axis, and is equal in magnitude to the

angular velocity multiplied by the second moment.

We may put these propositions together as follows.

Let oX, o Y, oZ be three rectangular lines through o, and

let us consider a unit spin about oZ, which is denoted by

k. The velocity of the end of p, = xi + yj + zTc, due to this

spin, is Vk (xi + yj + zk), = xj yi. Let Sm be the mass of

the particle there situated, then the moment of its mo-

mentum is

V (xi + yj + zk) (xj yi) 8m ={ xz.i yz.j+(x* + y z ) k} 8m.

Hence the moment of momentum of the whole body is

Jxzdm . i fyzdm . j + /(a; 2 + y 2 ) dm . k,

which gives the components as above stated.

Now let the axes be principal axes, and let the swing-

radii about them be a, b, c. Then if the body have a spin

0,=pi+qj + rk, its moment of momentum about the

origin will be m (a 2 pi + b 2 qj + c 2 rk) ; since it is the result-

ant of the moments of momentum due to the three spins

pi, qj, rk. The components of the spin must therefore be

respectively multiplied by ma 2 , mb 2 , me 2 , to give the com-

ponents of the moment of momentum.

Thus the moment of momentum is a pure linear function

of the spin. The ellipsoid representing this function must

have its semi-axes inversely proportional to the square

roots of the second moments ma 2 , mb 2 , me 2 , that is, in-

versely proportional to a, b, c. Their actual values, in ac-

cordance with the formula [p. 176], are mbc, mca, mab

each divided by TT. Thus the relation between x, y, z for

any point on this ellipsoid is

7T 2 (aV + b*y 2 + cV) = (mabcf.

54 DYNAMIC.

The ellipsoid thus determined is called the momental

ellipsoid,

There is no great convenience in thus fixing the size of

the momental ellipsoid, on account of the two scales of

representation involved ; angular velocity is represented by

a line, and mass by the inverse of a line (since mbc is one

semi-axis). If we take any ellipsoid similar to this one,

we may so choose the scale on which angular velocities are

represented that the momentum due to a spin represented

by any radius vector of the ellipsoid shall be equal to the

mass of the body multiplied by the area of the conjugate

section. We shall therefore give the name " momental

ellipsoid" to any ellipsoid having axes along oX, oY, oZ

inversely proportional to a, b, c ; so that we may determine

its size in particular cases by other considerations of con-

venience.

The momental ellipsoid is the reciprocal of the ellipsoid

of gyration. For the ellipsoid of gyration has semi-axes

a, 6, c, and therefore if p be a perpendicular from the

origin on a tangent plane to it, and make angles a/3y with

the axes, we must have p 2 = a* cos 2 a + 6 2 cos' 2 /3 + c 2 cos 2 7.

Now let us measure on the line p a distance r from the

origin, such that pr = C, and let s be the point at the end

of this distance, so that os = r. Then the locus of s will be

the reciprocal of the ellipsoid of gyration [p. 101]. The

coordinates of s are

x = r cos ot, y = r cos /3, z = r cos 7,

or px = G cos ai,py=G cos /3, pz = C cos 7.

Therefore

p 2 (aV + by + cV) = <7 2 (a 2 cos 2 a + 6 2 cos 2 /9 + c 2 cos 2 7 ) = C*p*

and consequently aV + 6 2 y 2 + cV = C 2

so that the locus of s is an ellipsoid whose axes are in-

versely proportional to a, b, c, that is, an ellipsoid of

momentum.

MOMENTUM OF TWIST. 55

MOMENTUM OF TWIST.

Let us now take the mass-centre of a rigid body for

origin. The instantaneous state of motion of the body is

made up of a translation-velocity equal to the velocity of

the mass-centre, and a spin about some axis through it.

Let these, regarded as vectors, be denoted by a and /3 re-

spectively ; then to indicate that the latter is really a rotor

passing through the origin, we may denote the whole

motion of the body by ftyS -,'- a. This is the general symbol

for a twist-velocity.

It is important to observe that the vector part of this

velocity gives rise to the rotor part of the momentum, and

vice versa. The translation is not localized, but its mo-

mentum has to pass through the mass-centre. The rota-

tion is about a definite axis ; but its momentum is a mere

moment of momentum, which has only magnitude and

direction, no definite position.

Thus the momentum due to the translation a is ftraa,

where m is the mass of the body. The momentum due

to the spin ft/3 is 0/3, where is the pure function which

converts any spin into its moment of momentum. Hence

the whole momentum due to the twist ft/3 +a is

ftma + 0/3.

Given any screw, the momentum due to a twist about

it may be associated with another screw. Namely the

momentum is equivalent to a rotor part along a certain

axis, together with a vector part parallel to the axis ; the

proof of this is precisely like that for twists [p. 126].

We now propose to determine in what cases these two

screws are identical. When this is so, the momentum

ftraa + 0/3 is a numerical multiple of the velocity ft/3 +a ;

say ftraa + 0/3 = x (ft/3 + a),

whence ma. = #/3, 0/3 = #a,

and therefore ra/3 = x 2 ft.

56 DYNAMIC.

Hence the spin /3 must be about one of the principal axes

at the mass-centre. Suppose it to be oX, then </3 = ma 2 /3,

and x = ma. Consequently a = + a/3, and the pitch is

+ a. There are therefore six screws having this property ;

they are called the principal screws of the body. These

axes are the principal axes at the mass-centre, and their

pitches are the swing-radii about them.

57

APPENDIX I. (A.)

ACCELERATION DEPENDING ON STRAIN.

Imagine a spiral spring, like that of a spring-balance,

with circular disks fastened to its ends and a small hole

through one of them. Let it be partially compressed, and

held in that position by a string attaching one of the two