Page images
[ocr errors][ocr errors][ocr errors][merged small]

232. By the method of Fluxions, the centre of oscillation, for a regular body, will be found from cor. 1. But for an irregular one ; suspend it at the given point; and hang up also a simple pendulum of such a length, that making them both vibrate, they may keep time together. Then the length of the simple pendulum is equal to the distance of the centre of oscillation of the body, below the point of suspension.

233. Or it will be still better sound thus : Suspend the body very freely by the given point, and make it vibrate in very small arcs, counting the number of vibrations it makes in any time, as a minute, by a good stop watch ; and let that number of vibrations - made in a minute be called n : Then

shall the distance of the centre of oscillation,be so = o

inches. For, the length of the pendulum vibrating seconds, or 60 times in a minute, being 394 inches; and the lengths of pendulums being reciprocally as the square of the number of vibrations made in the same time ; therefore - - -

[ocr errors]

pendulum which vibrates n times in a minute, or the distance of the centre of oscillation below the axis of motion. Or, so = 39; to, in inches, t being the time of one oscillation in a very small arc. 234. The foregoing determination of the point into which all the matter of a body being collected, it shall oscillate in the same manner as before, only respects the case in which the body is put in motion by the gravity of its own particles, and the point is the centre of oscillation : but when the body is put in motion by some other extraneous force, instead of its gravity, and made to rotate instead of oscillate, then the point is different from the former, and is called the Centre of Gyration ; which is determined in the following manner :

235. Prop. To determine the centre of gyration of a compound body or of a system of bodies.

Vol. II. 32

Let R 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 facting at r, as the whole sys. tem receives. . Now, by cor. 3, art. 228, the angular velocity generated in the system by

- • sp

the forcef is as ATsA* + F se’ &c.” and by the same, the angular velocity of the system placed

[ocr errors]
[ocr errors]

sR = V/ A + b + c centre of gyration below the axis of motion. 236. Corol. 1. Because A. sa” + B. se” &c. =sg. so. w, where G is the centre of gravity, o the centre of oscillation, and w the weight of body A + b + c &c.; therefore sR* = 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 sa” _ sum of all the pd”

[ocr errors]
[ocr errors]

in R, is : then, by making these two ex


238. By means of the theory of the centre of gyration, and the values of f thence deduced in the note to prop. 2, under the heading “Maximum in Machines” in a subsequent part of this volume, the phaenomena 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 = ?, we shall have for the accelerating force, the expression f= Pro .

pr” + wi''

and consequently for the space descended by the actuating weight or power P, in a given time t, we shall have the usual

[ocr errors]

239. For applications of these formulae and their obvious modifications, as they are exceedingly useful in rotatory mo. tions, 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 10|bs. 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 attacned in the same way to the same cylinder, it shall descend 16+.

[ocr errors]

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 te be 30 pounds; and that it descends through 48 feet in 2 seconds, what is the weight of the cylinder 1 Ans. 201’, lbs. 5. Suppose a cylinder that weighs 201bs. 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 1 - Ans. 24}.

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. #'rlbs.

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 7


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 MN, 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.

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 force f acting at P, in the foregoing propositions. Now, Put p = the whole weight of the pendulum,

b = the weight of the ball,

g = so the distance of the centre of gravity,
o = so the distance of the centre of oscillation,
r = sR = x/ go the distance of centre of gyration,
i = sp the distance of the point of impact,
v = the velocity of the ball,
w = 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 pendu. lum will receive the same motion from the blow in the point 2 2 o r ; and as sp": sr” : ; p : #. . p or #p or #p. (art. 236), the mass which being placed at P, the pendulum will still receive the same motion as before. Here then are two


- - - o quantities of matter, namely, b 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

£p + b : b : : v : u, and therefore v _bit + gop, the ve. 2? 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 position, it will have acquired the same velocity with which it began to ascend, and by the laws of falling bodies, the velocity 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 of and, by the nature of the circle, the chord is a mean proportional between the versed

cC 20 of the arc described by o. Then, by the laws of falling bodies

sine and diameter, therefore 20 : c : : c : -, the versed sine

2 - v 16,'; : V i. :: 32} : c V * the velocity acquired by the point oin descending through the arc whose chord is c, 2

where a = 16*, feet: and therefore of i :: c vo : # vo, which is the velocity u, of the point P.

Then, by substituting this value for u, the velocity of the

- - 2

ball, before found, becomes v = ot X c V o So that the velocity of the ball is directly as the chord of the arc described by the pendulum in its vibration.

« PreviousContinue »