Since therefore the power of gravity in drawing down the body must also communicate to it the rolling motion just described; it seems evident, that the center of gravity of the body cannot be drawn down as swiftly, as when the power of gravity has no other effect to produce on the body, than merely to draw it downward. If therefore the whole matter of the body C B could be crouded into its center of gravity, so that being united into one point, this rolling motion here mentioned might give no hindrance to its descent; this center would descend faster, than it can now do. And the point, which now descends as fast, as if the whole matter or the body C B were crouded into it, will be farther removed from the point A, than the center of gravity of the body C B.
69. AGAIN, suppose the pendulum A B (in fig. 55.) to hang obliquely. Here the power of gravity will operate less upon the ball of the pendulum, than before: but the line D E being drawn so, as to stand perpendicular to the rod A C of the pendulum; the force of gravity upon the body C B, now it is in this situation, will produce the same effect, as if the body were to glide down an inclined plane in the position of D E. But here the motion of the body, when the rod is fixed to the point A, will not be equal to the uninterrupted descent of the body down this plane; for the body will here also receive the same kind of rotation in its motion, as before; so that the motion of the center of gravity will in like manner be retarded; and the point, which here descends with that degree of swiftness, which the body would have, if not hindered by being fixed to the point A; that is, the point, which descends as fast, as if the whole body were crouded into it, will be as far removed from the point A, as before.
70. THIS point, by which the length of the pendulum is to be estimated, is called the center of oscillation. And the mathematicians have laid down general directions, whereby to find this center in all bodies. If the globe A B (in fig. 56.) be hung by the string C D, whose weight need not be regarded, the center of oscillation is found thus. Let the straight line drawn from C to D be continued through the globe to F. That it will pass through the center of the globe is evident. Suppose E to be this center of the globe; and take the line G of such a length, that it shall bear the same proportion to E D, as E D bears to E C. Then E H being made equal to ⅖ of G, the point H shall be the center of oscillation[65]. If the weight of the rod C D is too considerable to be neglected, divide C D (fig. 57) in I, that D I be equal to ⅓, part of C D; and take K in the same proportion to C I, as the weight of the globe A B to the weight of the rod C D. Then having found H, the center of oscillation of the globe, as before, divide I K in I, so that I L shall bear the same proportion to L H, as the line C H bears to K; and L shall be the center of oscillation of the whole pendulum.