Page:EB1911 - Volume 17.djvu/1006

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KINETICS]
MECHANICS
 987


the mass-centre. Take, for example, the case of a sphere rolling on a plane; and let the axes Ox, Oy be drawn through the centre parallel to the plane, so that the equation of the latter is z = −a. We will suppose that the extraneous forces consist of a known force (X, Y, Z) at the centre, and of the reactions (F1, F2, R) at the point of contact. Hence

M0 = X + F1,   M0 = Y + F2,   0 = Z + R,
C = F2a,   C = −F1a,   C = 0.
(19)  

The last equation shows that the angular velocity about the normal to the plane is constant. Again, since the point of the sphere which is in contact with the plane is instantaneously at rest, we have the geometrical relations

u + qa = 0,   v + pa = 0,   w = 0,
(20)

by (12). Eliminating p, q, we get

(M0 + Ca−2) = X,   (M0 + Ca−2) = Y.
(21)

The acceleration of the centre is therefore the same as if the plane were smooth and the mass of the sphere were increased by C/α2. Thus the centre of a sphere rolling under gravity on a plane of inclination a describes a parabola with an acceleration

g sin α/(1 + C/Ma2)

parallel to the lines of greatest slope.

Take next the case of a sphere rolling on a fixed spherical surface. Let a be the radius of the rolling sphere, c that of the spherical surface which is the locus of its centre, and let x, y, z be the co-ordinates of this centre relative to axes through O, the centre of the fixed sphere. If the only extraneous forces are the reactions (P, Q, R) at the point of contact, we have

M0 = P,   M0ÿ = Q,   M0 = R,
C = − a (yR − zQ),   C = − a (zP − xR),   C = − a (xQ − yP),
c c c
(22)

the standard case being that where the rolling sphere is outside the fixed surface. The opposite case is obtained by reversing the sign of a. We have also the geometrical relations

= (a/c) (qzry),   = (a/c) (rxpz),   ż = (a/c) (pygx),
(23)

If we eliminate P, Q, R from (22), the resulting equations are integrable with respect to t; thus

p = − M0a (zẏ) + α,   q = − M0a (zẋ) + β,   r = − M0a (xẏyẋ) + γ,
Cc Cc Cc
(24)

where α, β, γ are arbitrary constants. Substituting in (23) we find

( 1 + M0a2 ) = a (βzγy),   ( 1 + M0a2 ) = a (γxαz),   ( 1 + M0a2 ) ż = a (αyβx).
C c C c C c
(25)

Hence α + β + γż = 0, or

αx + βy + γz = const.;
(26)

which shows that the centre of the rolling sphere describes a circle. If the axis of z be taken normal to the plane of this circle we have α = 0, β = 0, and

( 1 + M0a2 ) = −γ a y,   ( 1 + M0a2 ) = γ a x.
C c C c
(27)

The solution of these equations is of the type

x = b cos (στ + ε),   y = b sin (σt + ε),
(28)

where b, ε are arbitrary, and

σ = γa/c .
1 + M0a2/C
(29)


The circle is described with the constant angular velocity σ.

When the gravity of the rolling sphere is to be taken into account the preceding method is not in general convenient, unless the whole motion of G is small. As an example of this latter type, suppose that a sphere is placed on the highest point of a fixed sphere and set spinning about the vertical diameter with the angular velocity n; it will appear that under a certain condition the motion of G consequent on a slight disturbance will be oscillatory. If Oz be drawn vertically upwards, then in the beginning of the disturbed motion the quantities x, y, p, q, P, Q will all be small. Hence, omitting terms of the second order, we find

M0 = P,   M0 = Q,   R = M0g,
C = −(M0ga/c) y + aQ,   C = (M0ga/c) xaP,   C = 0.
(30)  

The last equation shows that the component r of the angular velocity retains (to the first order) the constant value n. The geometrical relations reduce to

= aq − (na/c) y,   = −ap + (na/c) x.
(31)

Eliminating p, g, P, Q, we obtain the equations

(C + M0a2) + (Cna/c) y − (M0ga2/c) x = 0,
(C + M0a2) ÿ − (Cna/c) x − (M0ga2/c) y = 0,
(32)


which are both contained in

{ (C + M0a2) d2 i Cna   d M0ga2 } (x + iy) = 0.
dt2 c dt c
(33)


This has two solutions of the type x + iy = αei(σt + ε), where α, ε are arbitrary, and σ is a root of the quadratic

(C + M0a2) σ2 − (Cna/c) σ + M0ga2/c = 0.
(34)

If

n2 > (4Mgc/C) (1 + M0a2/C),
(35)

both roots are real, and have the same sign as n. The motion of G then consists of two superposed circular vibrations of the type

x = α cos (σt + ε),   y = α sin (σt + ε),
(36)

in each of which the direction of revolution is the same as that of the initial spin of the sphere. It follows therefore that the original position is stable provided the spin n exceed the limit defined by (35). The case of a sphere spinning about a vertical axis at the lowest point of a spherical bowl is obtained by reversing the signs of α and c. It appears that this position is always stable.

It is to be remarked, however, that in the first form of the problem the stability above investigated is practically of a limited or temporary kind. The slightest frictional forces—such as the resistance of the air—even if they act in lines through the centre of the rolling sphere, and so do not directly affect its angular momentum, will cause the centre gradually to descend in an ever-widening spiral path.

§ 19. Free Motion of a Solid.—Before proceeding to further problems of motion under extraneous forces it is convenient to investigate the free motion of a solid relative to its mass-centre O, in the most general case. This is the same as the motion about a fixed point under the action of extraneous forces which have zero moment about that point. The question was first discussed by Euler (1750); the geometrical representation to be given is due to Poinsot (1851).

The kinetic energy T of the motion relative to O will be constant. Now T = 1/2Iω2, where ω is the angular velocity and I is the moment of inertia about the instantaneous axis. If ρ be the radius-vector OJ of the momental ellipsoid

Ax2 + By2 + Cz2 = Mε4
(1)

drawn in the direction of the instantaneous axis, we have I = Mε4/ρ2 (§ 11); hence ω varies as ρ. The locus of J may therefore be taken as the “polhode” (§ 18). Again, the vector which represents the angular momentum with respect to O will be constant in every respect. We have seen (§ 18) that this vector coincides in direction with the perpendicular OH to the tangent plane of the momental ellipsoid at J; also that

OH = 2T · ρ ,
Γ ω
(2)


where Γ is the resultant angular momentum about O. Since ω varies as ρ, it follows that OH is constant, and the tangent plane at J is therefore fixed in space. The motion of the body relative to O is therefore completely represented if we imagine the momental ellipsoid at O to roll without sliding on a plane fixed in space, with an angular velocity proportional at each instant to the radius-vector of the point of contact. The fixed plane is parallel to the invariable plane at O, and the line OH is called the invariable line. The trace of the point of contact J on the fixed plane is the “herpolhode.”

If p, q, r be the component angular velocities about the principal axes at O, we have

(A2p2 + B2q2 + C2r2) / Γ2 = (Ap2 + Bq2 + Cr2) / 2T,
(3)

each side being in fact equal to unity. At a point on the polhode cone x : y : z = p : q : r, and the equation of this cone is therefore

A2 ( 1 − Γ2 ) x2 + B2 ( 1 − Γ2 ) y2 + C2 ( 1 − Γ2 ) z2 = 0.
2AT 2BT 2CT
(4)


Since 2AT − Γ2 = B (A − B)q2 + C(A − C)r2, it appears that if A > B > C the coefficient of x2 in (4) is positive, that of z2 is negative, whilst that of y2 is positive or negative according as 2BT ≷ Γ2. Hence the polhode cone surrounds the axis of greatest or least moment according as 2BT ≷ Γ2. In the critical case of 2BT = Γ2 it breaks up into two planes through the axis of mean moment (Oy). The herpolhode curve in the fixed plane is obviously confined between two concentric circles which it alternately touches; it is not in general a re-entrant curve. It has been shown by De Sparre that, owing to the limitation imposed on the possible forms of the momental ellipsoid by the relation B + C > A, the curve has no points of inflexion. The invariable line OH describes another cone in the