Let a be the centre of gyration, or the point into which all the particles a, B, C, &c. being collected, it shall receive the same angular motion from a force f acting at r, as the whole sys. tem receives. Now, by cor. 3, art. 228, the angular velocity generated in the system by f. sp the force f, is as A. SA2+ B SB2 &c.' and by the same, the angular velocity of the system placed in R, is f. sp : then, by making these two ex (A+B+C &c.). SR2 pressions equal to each other, the equation gives A. SA2+ B. SB2 + c. sc2 A+B+C for the distance of the centre of gyration below the axis of motion. 236. Corol. 1. Because A. SA2 + B. SB2 &c. sc. so. w, where a is the centre of gravity, o the centre of oscillation, and w the weight of body A + B + c &c. ; therefore SR2 SG. so; that is, the distance of the centre of gyration, from the point of suspension, is a mean proportional between those of gravity and oscillation. 237. Corol. 2. If p denote any particle of a body w, at d distance from the axis of motion; then SR sum of all the pd1 body w Or, if g be put for SR, the distance of the centre of gyra. tion from the point of suspension, wgA. SA2 + B. SB2 +c.sc2+ &c. sum of all the pd3. SCHOLIUM. 238. By means of the theory of the centre of gyration, and the values of g thence deduced in the note to prop. 2, under the heading "Maximum in Machines" in a subsequent part of this volume, the phænomena of rotatory motion become connected with those of accelerating forces: for then, if a weight or other moving power P act at a radius r to give rotation to a body, weight w, and dist. of centre of gyration from axis of motion = g, we shall have for the accelerating force, the expression weight or power P, in a given time t, we shall have the usual formula S == gt, introducing the above value of f. 239. For applications of these formulæ and their obvious modifications, as they are exceedingly useful in rotatory motions, the student may solve the following problems. Problems illustrative of the Principle of the Centre of 1. Suppose a cylinder that weighs 100lbs. to turn upon a horizontal axis, and imagine motion to be communicated by a weight of 10lbs. attached to a cord which coils upon the surface of the cylinder : how far will that weight descend in 10 seconds? Ans. 268 055 f. 2. Required the actuating weight such that when attached in the same way to the same cylinder, it shall descend 167', feet in 3 seconds. P = sw gr's = 61. 3. Another cylinder, which weighs 200lbs, is actuated in like manner by a weight of 30lbs. How far will the weight descend in 6 seconds? Ans. 133.6 feet. 4. Suppose the actuating weight to be 30 pounds; and that it descends through 48 feet in 2 seconds, what is the weight of the cylinder? Ans. 20 lbs. 5. Suppose a cylinder that weighs 20lbs. to have a weight of 30lbs. actuating it, by means of a cord coiled about the surface of the cylinder; what velocity will the descending weight have acquired at the end of the first second ? Ans. 241. 6. Of what weight will the axis be relieved in the case of the last example, when the system is completely in motion? Ans. 22 lbs. 7. A sphere, w, whose radius is three feet, and weight 500lbs. turns upon a horizontal axis, being put in motion by a weight of 20lbs. acting by means of a string that goes over a wheel whose radius is half a foot. How long will the weight, P, be in descending 50 feet? Ans. 33". 8. Of what weight will the axle be relieved as soon as motion is commenced? Ans. lbs. 9. If in example seventh the radius of the wheel be equal to that of the sphere, what ratio will the accelerating force bear to that of gravity? 10. A paraboloid, w, whose weight is 200lbs. and radius of base 20 inches, is put in motion upon a horizontal axis by a weight P of 15lbs. acting by a cord that passes over a wheel whose radius is 6 inches. After P has descended for 10 seconds, suppose it to reach a horizontal plane and cease to act, then how many revolutions. would the paraboloid: make in a minute? BALLISTIC PENDULUM. 240. PROP. To explain the construction of the Ballistic Pendulum, and show its use in determining the velocity with which a cannon or other ball strikes it. The ballistic pendulum is a heavy block of wood иN, suspended vertically by a strong. horizontal iron axis at s, to which it is connected by a firm iron stem. This problem. is the application of the preceding articles, and was invented by Mr. Robins, to determine the initial velocities of military projectiles; a circumstance very useful in that science; and it is the best method yet known for determining them with any degree of accuracy. M S N Let G, R, o, be the centres of gravity, gyration, and oscillation, as determined by the foregoing propositions; and let P be the point where the ball strikes the face of the pendulum; the momentum of which, or the product of its weight and velocity, is expressed by the forcef, acting at e, in the foregoing propositions. Now, Put p = the whole weight of the pendulum, b the weight of the ball, g SG the distance of the centre of gravity, so the distance of the centre of oscillation, = r = SR go the distance of centre of gyration, i = SP the distance of the point of impact, the velocity of the ball, u = that of the point of impact P, c = chord of the arc described by o. By art. 235, if the mass p be placed all at R, the pendulum will receive the same motion from the blow in the point the mass which being placed at P, the pendulum will still receive the same motion as before. Here then are two go quantities of matter, namely, 6 and p, the former moving with the velocity v, and striking the latter at rest; to determine their common velocity u, with which they will jointly proceed forward together after the stroke. In which case, by the law of the impact of non-elastic bodies, we have bii + gop go +bb:: v : u, and therefore v = -u the ve üp bii locity of the ball in terms of u, the velocity of the point P, and the known dimensions and weights of the bodies. But now to determine the value of u, we must have recourse to the angle through which the pendulum vibrates; for when the pendulum descends again to the vertical posi tion, it will have acquired the same velocity with which it began to ascend, and by the laws of falling bodies, the veloeity of the centre of oscillation is such as a heavy body would acquire by freely falling through the versed sine of the arc described by the same centre o. But the chord of that arc is c, and its radius is o; and, by the nature of the circle, the chord is a mean proportional between the versed cc sine and diameter, therefore 20: c:: e: the versed sine 20 of the arc described by o. Then, by the laws of falling bodies. point o in descending through the arc whose chord is c, where a = 16 feet and therefore o:i::C√ 2a ci 2a : 0 which is the velocity u, of the point P. Then, by substituting this value for u, the velocity of the that the velocity of the ball is directly as the chord of the arc described by the pendulum in its vibration. SCHOLIUM. 241. In the foregoing solution, the change in the centre of oscillation is omitted, which is caused by the ball lodging in the point P. But the allowance for that small change, and that of some other small quantities, may be seen in my Tracts, where all the circumstances of this method are treated at full length. For an example in numbers, suppose the weights and dimensions to be as follow: namely, P = 570lb. Then b 18oz. 14dr hii+gop = 1-131 X94.32+78×847 × 570 = 1.131lb. Xc= bio g= 78 inc. 1-131X94X847 18.73 0= 847 inc. = 656.56. =7.065 feet 12 193 And e 18.73 inc. =√. = 2.1337. 7.065 42.39 Therefore 656.56 × 2·1337, or 1401 feet, is the velocity, per second, with which the ball moved when it struck the pen. dulum. 242. When the impact is made upon the centre of oscil. lation, the computation becomes simplified. In that case, since the whole mass, p, of the pendulum, may be regarded as concen. tered at o, and the ball, b, strikes that point, we shall have bu = (b + p)v; v being the velocity of the ball before the impact, and v' that of the ball and pendulum together, after the impact. Now, if the centre of oscillation o, after the blow, describes the arc oo', before the motion is destroyed, the velocity v will be equal to that acquired by falling through the versed sine vo, of the S arc oo' or angle s to the radius so. But, if the time t of a very minute oscillation of the pendulum be known or inferred from that in an ascertained arc, we have (art. 233,, so = 391 inches 3 and (art. 154) v' = √(64 × 3·26042 versin s) |