Precisely similar reasoning may be used in the case of a sliding sphere, and of the general motion of a rigid body.

There are some difficulties in this proof, which the following exact investigation may clear up. The question is, what velocity-systems are consistent with rigidity? We shall secure that the body does not change in size or shape, if we make sure that no straight line in the body is altered in length. Let a and b be two points in the body, then the motion of b relative to a must be at right angles to ab; for its component along ab is the flux of the length ab, which has to be zero. We shall find it convenient to denote the velocity of the point a by ȧ. This being so, it is necessary and sufficient for rigidity that b-à should be either zero, or perpendicular to ab, where a, b are any two points in the moving body. It follows at once that if two velocity-systems are consistent with rigidity, their resultant is consistent with rigidity.

Now suppose a plane to be sliding on a plane, and combine with its velocity-system a translation equal and opposite to the velocity of any point a. Then the new motion is consistent with rigidity, and the point a is at rest. Consequently the new motion is a spin about the point a. The original motion, therefore, is the resultant of this spin and of a translation equal to the velocity of a; it is therefore a spin of the same magnitude w, about a point o situate on a line through a perpendicular to its direction of motion, at a distance such that à iw.oa.

=

To determine the motion of the instantaneous centre, we must find the acceleration of any point in the plane. The instantaneous centre shall be called c in the fixed plane, and c, in the moving plane; and at a certain instant of time it shall be supposed to be at a point o in the moving plane. Then at that instant c, c,, o are the same point; but c means the velocity of the instantaneous centre in the fixed plane, c, its velocity in the moving plane, and o the velocity of o in the moving plane, which we know to be zero.

ROLLING OF CENTRODES.

139

Now if p be any point in the moving plane, we know that at every instant piw.cp. To find the acceleration of p we must remember that the flux of cp is p-ċ. Therefore

p = iw.cp+ iw (p − ċ) = (iw — w2) cp — iw .ċ.

Now let p coincide with o, that is (for the instant) with c. Then ö=-iw.c, or the acceleration of o is at right angles to the velocity of c, and equal to the product of it by the angular velocity.

If we suppose the moving plane to be fixed, and the fixed plane to slide upon it so that the relative motion is the same, then if p, is the point of the fixed plane which at a given instant coincides with p in the moving plane, the velocity and accleration of p, on one supposition are equal and opposite to the velocity and acceleration of p on the other supposition; also w becomes w. Hence we shall have ö+iw.c1, but öö. Therefore ċ1 = ċ, or the velocity of the instantaneous centre in the moving plane is the same in magnitude and direction as its velocity in the fixed plane.

=

Because these velocities are the same in direction, the two centrodes touch one another; and because they are the same in magnitude, the moving centrode rolls on the fixed one without sliding. For let s, s, be the arcs ac, bc

measured from points a, b which have been in contact; then s=s, and therefore (since they vanish together) s=s1·

The angular velocity w is equal to s multiplied by the difference of the curvatures of the two centrodes. For suppose them to roll simultaneously on the tangent ct; then their angular velocities & and will be respectively equal to their curvatures multiplied by s, and the relative angular velocity will be the difference of these. When the curvatures are in opposite directions one of them must be con

sidered negative. The same result may be obtained by calculating the flux of the acceleration of o.

1

Thus if r, r, are radii of curvature of the fixed and rolling centrodes, we have

We may derive some important consequences from the expression just obtained for the the acceleration of a point in the moving plane, namely

This consists of three parts; w. pc is the acceleration towards c due to rotation about it as a fixed point; iw.cp is in the direction np perpendicular to cp, due to the change in the angular velocity; and iw.ċ is in the

direction cn, due to the change in position of c as the centrode rolls. Hence the normal acceleration of p, that is, the component along pc, is in magnitude w2. pc - w.ċ cose. It vanishes for those points p for which w.pc =ċ cose, or for which cn=ċ : w. These points lie on a circle having cn for diameter; the curvature

CHANGE OF INSTANTANEOUS SPIN.

141

of this circle is 1: cn or 2w: ċ, that is, it is twice the difference between the curvatures of the centrodes. All the points of this circle, therefore, are at the given instant passing through points of inflexion on their paths.

The path of any point p is called a roulette, as being traced by rolling motion. We can now determine the curvature of a roulette at any point. For since the normal acceleration is the squared velocity multiplied by the curvature, we have

where r, r, are the radii of curvature of the fixed and rolling centrodes.

The tangential acceleration of p is w.cp-wċ sine. If therefore we make ct= wc: w, the locus of points whose tangential acceleration is zero is a circle on ct as diameter. The point at c belongs to both circles; it is a cusp on its path, being a point where there is no normal acceleration, but also no velocity. It has, however, as we know, a tangential acceleration - iwc. The other intersection of the two circles has no acceleration at all.

INSTANTANEOUS AXIS.

In the case of a body moving with one point fixed, we may combine with its velocity-system a spin about any axis through the point, such that the velocity of a certain point a due to the spin is equal and opposite to its actual velocity in the motion of the body. The resultant velocity-system is consistent with rigidity, and the point a is at rest; it is therefore a spin about the axis oa. Consequently the actual motion of the body is a spin about some axis in the plane of these two.

=

Let oc, =w, be the instantaneous spin in magnitude and direction, op, p, the position vector of any point p. Then we know that the velocity of p is the moment of oc about p, that is, twice the area of the triangle ocp. This quantity, which is in magnitude oc. op sin cop, and in direction perpendicular to oc and op, is what we have called the vector product of oc and op, and denoted by Vop. We have therefore

p = Vwp.

To find the acceleration of p, therefore, is to find the flux of the triangle ocp, due to the motion of p and c. Now suppose that c moves to c, in a certain interval; then oc pocp+coc, +c,pc, all the areas being of course regarded as vectors. But if we draw pd equal and parallel to cc, we shall have coc, + c1pc=pod, for the three triangles stand on the same base cc, or pd, and the height op is the sum of oc and cp. It follows that the flux of ocp, due to the motion of c, is equal to the moment about o of the velocity of c supposed to be transferred to p. That is, the flux of Vop, due to the change of w, is Vop. In a similar way it may be shewn that the flux due to change of p is Vap. Hence1 altogether, since p = Vwp, we have

μ= Vip+ Væp = Vip + V.wVwp.

The expression V. wVwp means the vector product of the two vectors, o and Vwp. Thus it appears that the acceleration of p consists of two parts; V. Vop along the perpendicular from p to the axis oc, due to the rotation w; and Vop, perpendicular to op and to the velocity of c, due to the change of w.

Let a be the point of the moving body at which c is instantaneously situated, then ä = Vow, or the acceleration of a is equal to the moment about o of the velocity of c.

1 The flux of a vector product has been already found by a different method on p. 97.