From the equation to the path of the projectile, (the top of the plane being origin,) 22 - 2(h cota cos2a+c sin 2a) z + h2 cot'a = 0... (1): and, differentiating with respect to h, in order to find the maximum value of z, therefore, substituting this value of z in (1), and reducing, efficient of continues nearly equal to c sin 2a, and therefore dz to +, for z is sta continues positive, while the remainder of equation (2), viz. · cota cos 2a.z + h cot' a, increases from tionary and h is increasing; 17. A slender ring, moveable in a vertical plane, has a fixed rough cylinder passing through it, the axis of the cylinder being perpendicular to the plane of the ring; the ring whirls round in its own plane so as always to be in contact with the cylinder, and to roll on it without sliding: if VV2 be the velocities of the centre of the ring when in its highest and lowest positions respectively, and if P be the point of contact, O the centre of the ring, when the tendency to slide is greatest, and OA a vertical drawn downwards through 0, shew that 1 2 V2 + V 2 3V2. 1 1 Let a, b, be the radii of the ring and of the cylinder, the angle POA (fig. 33), the angle which a particular radius fixed in the ring makes with a fixed line in the plane of the ring, F, R, the friction, and normal action at P, estimated as accelerating forces. F The tendency to slide will be greatest when is a maxi mum, provided it never become infinite; we must therefore F find an expression for and make it a maximum. Applying D'Alembert's principle, and resolving forces parallel and perpendicular to OP, and taking moments about 0, we obtain 2 (a−b) (dd) * = g cos$ + R = R...... .(1), It remains to introduce V, and V, instead of C'; 2 1 2 1 2 2 2 pression for cos is less than 1 only when V2 <3 V2; hence arises the question, What happens when V2 is greater than 1 3V2? Now, by substituting for V2, after eliminating C from equations (7), the expression for cos 2 becomes g(a-b) therefore, as V decreases from ∞ to g(a - b), & decreases from 2 1 to 0; and if V2 be less than g(a - b), & becomes impossible, 1 and at the same time V becomes greater than 3V. Hence, if the ring revolve very rapidly, the risk of sliding is greatest when the centre of the ring is only just above a horizontal line through the centre of the cylinder; if the speed be diminished, the position in which sliding is more probable than in any other becomes nearer to the highest position of the ring; if the speed be such that V,* = g(a−b), the risk of sliding is greater when the ring is highest, than at any other part of the revolution. In this case (i.e. when V = g(a - b),) C = 2g by (7), and therefore when the ring is highest, 1 A smaller value of V, besides making V2>3V,, would make the expression for R, given by (6), negative; hence we see that the contact between the ring and the cylinder would be broken before the ring completed a revolution, and that the risk of sliding would never be a maximum, in the proper sense of the word, but would increase without limit as the ring approached the critical position at which it would fall. 18. A cylindrical vessel is moveable about a horizontal axis passing through its centre of gravity, and is placed so as to have its axis vertical; if water be poured in, shew that the equilibrium is at first unstable; and find the condition which must be satisfied, in order that it may be possible to make the equilibrium stable by pouring in enough water. Let CFD (fig. 34) be the base of the vessel, G its centre of gravity, AEB the surface of the water, H its centre of gravity, The equilibrium will be stable if, on the vessel being turned round through a very small angle, the resultant of the fluid pressures tends to bring the vessel back to its former position, the weight of the vessel producing no effect, because the centre of gravity lies on the axis of motion. Now the line of action of the resultant pressure is the same as if a solid cylinder, that would just fit into the given cylinder, were floating on a fluid, in such a manner that the volume of fluid displaced were equal to the volume contained in the given cylindrical vessel; for the pressures would be the same in the two cases, except that in the one they would act downwards and outwards, in the other they would act upwards and inwards; therefore in the existing case the downward resultant acts through M, the metacentre of the space AD; and by the usual formula we have Now by the Theory of Equations this is always positive unless h be between Now if it be possible to make the equilibrium stable, these two quantities must be real; hence the required condition is |