Sz Sr Let the initial values of z,, be 0, 8; and those of r, ' δι be a, a; then, from the equations (2) and (4), after executing obvious operations, we shall obtain the first two of these equations give the position of the particle on the revolving plane, and therefore, by virtue of the equation (1), the absolute position of the particle at any time; while the third is the equation to the path which the particle describes on the plane. SECT. 2. Systems of Particles. (1) Two heavy particles P, P' (fig. 179), are attached to a rigid imponderable rod APP', which is oscillating in a vertical plane about a fixed point at its extremity A: to determine the motion. = Let m, m', be the masses of the two particles; let AP= a, AP a'. Draw AB vertically downwards and let ‹ PAB= 0. Let ds, ds', denote the elements of the circular paths described by P, P', in a small time dt, estimated in a direction corresponding to an increase of 0. Then the effective moving forces of the mag sin 0, - m'a'g sin 0. of the impressed forces will be Hence, for the equilibrium of the impressed forces, and the effective forces applied in directions opposite to their own, we have But ds = add, ds' = a'd0; hence 220 (ma + m'a'') + (ma + m'a') g sin 0 = 0; a result which shews that the rod will oscillate isochronously with a perfect pendulum of which the length is (2) Two particles, attached to the extremities of a fine inextensible thread, are placed upon two inclined planes with a common summit: to determine the motion of the particles and the tension of the thread at any time. Let m, m', be the masses of the particles; a, a', the inclinations of the planes to the horizon; x, x', the distances of the particles from the common summit of the planes at any time. Then the impressed accelerating forces on the particles m, m', estimated down the two planes, will be gsin a, y sin a', respectively, and the effective accelerating forces, estimated in the same direcd2x d'x' tions, will be Hence, for the equilibrium of the im , dt dta pressed moving forces, and the effective moving forces applied in directions opposite to their own, we have d2x ' da' dť dti ..(1). g (m sin a - m' sin a') = m m which determines the common acceleration of the two particles estimated in accordance with an increase of x: should the d'x expression for be a negative quantity, x will decrease and ' increase. If T denote the tension of the thread, we shall have, for the equilibrium of the impressed moving forces T, mg sin a, exerted d2x on the particle m, and the effective moving force m applied dta in a direction opposite to its own, which gives the value of T, which is therefore of invariable magnitude during the whole motion. Poisson; Traité de Mécanique, Tom. II. p. 12. (3) One body draws up another on the wheel and axle: to determine the motion of the weights and the tension of the strings. Let a, a', denote the radii of the wheel and axle; m, m', the masses of the bodies suspended from them; s the arc described, at the end of the time t, by a molecule μ of the mass of the wheel and axle, r the distance of the molecule from the axis of rotation; x, x', the vertical distances, below the horizontal plane through the axis, of the masses m, m'. Then the moment of the impressed forces about the axis of rotation will be mag -- m'a'g: and the moments of the effective forces, estimated in the same direction, will be ma d2x d's dt + Σμη dt2 Let 0 represent the whole angle through which the wheel and axle have rotated at the end of the time t; then, b, b', denoting the initial values of x, x', it is clear that Also, it is manifest that s ro, and therefore d's = where Mk2 denotes the moment of inertia of the wheel and axle together about the axis of rotation. and therefore, if the system be supposed to have no motion when t = 0, 0 = gt2 ma m'a' ma2 + m'a2 + Mk2 (4). Let T denote the tension of the string supporting m; then Similarly, the tension of the other string being denoted by T', ma (a' + a) + M2 T" = m'g m'a2 + ma2 + Mk2* (4) A thin uniform rod AB (fig. 180) slides down between the vertical and horizontal rods OBy, OAx, to which it is attached by small rings at A and B: to find the angular velocity of AB in any position. Let X the pressure of Oy on AB, = Let m denote the mass of an elemental length ds of the rod at P: let OM= x, PM=y, AB= a, ▲ OAB = 0, AP=s. the figure and let the same thing be done in regard to all the molecules of the descending rod. Then the system of forces will satisfy the conditions of equilibrium. Let X denote the mass of a unit of length of the rod: then |