Fig. 3
Let the distances A B, B C, C D, and D E in this figure represent four successive seconds of time. They may just as well be conceived to represent any other equal units, however small. Seconds are taken only for convenience. At the commencement of the first second, let a body start from a state of rest at A, under the action of a constant force, sufficient to move it in one second through a distance of one foot. This distance also is taken only for convenience. At the end of this second, the body will have acquired a velocity of two feet per second. This is obvious because, in order to move through one foot in this second, the body must have had during the second an average velocity of one foot per second. But at the commencement of the second it had no velocity. Its motion increased uniformly. Therefore, at the termination of the second its velocity must have reached two feet per second. Let the triangle A B F represent this accelerated motion, and the distance, of one foot, moved through during the first second, and let the line B F represent the velocity of two feet per second, acquired by the body at the end of it. Now let us imagine the action of the accelerating force suddenly to cease, and the body to move on merely with the velocity it has acquired. During the next second it will move through two feet, as represented by the square B F C I. But in fact, the action of the accelerating force does not cease. This force continues to be exerted, and produces on the body during the next second the same effect that it did during the first second, causing it to move through an additional foot of distance, represented by the triangle F I G, and to have its velocity accelerated two additional feet per second, as represented by the line I G. So in two seconds the body has moved through four feet. We may follow the operation of this law as far as we choose. The figure shows it during four seconds, or any other unit, of time, and also for any unit of distance. Thus:
Time 1 Distance 1
" 2 " 4
" 3 " 9
" 4 " 16
So it is obvious that the distance moved through by a body whose motion is uniformly accelerated increases as the square of the time.
But, you are asking, what has all this to do with a revolving body? As soon as your minds can be started from a state of rest, you will perceive that it has everything to do with a revolving body. The centripetal force, which acts upon a revolving body to draw it to the center, is a constant force, and under it the revolving body must move or be deflected through distances which increase as the squares of the times, just as any body must do when acted on by a constant force. To prove that a revolving body obeys this law, I have only to draw your attention to Fig. 2. Let the equal arcs, A B and B C, in this figure represent now equal times, as they will do in case of a body revolving in this circle with a uniform velocity. The versed sines of the angles, A O B and A O C, show that in the time, A C, the revolving body was deflected four times as far from the tangent to the circle at A as it was in the time, A B. So the deflection increased as the square of the time. If on the table already given, we take the seconds of arc to represent equal times, we see the versed sine, or the amount of deflection of a revolving body, to increase, in these minute angles, absolutely so far as appears up to the fifteenth place of decimals, as the square of the time.
The standard from which all computations are made of the distances passed through in given times by bodies whose motion is uniformly accelerated, and from which the velocity acquired is computed when the accelerating force is known, and the force is found when the velocity acquired or the rate of acceleration is known, is the velocity of a body falling to the earth. It has been established by experiment, that in this latitude near the level of the sea, a falling body in one second falls through a distance of 16.083 feet, and acquires a velocity of 32.166 feet per second; or, rather, that it would do so if it did not meet the resistance of the atmosphere. In the case of a falling body, its weight furnishes, first, the inertia, or the resistance to motion, that has to be overcome, and affords the measure of this resistance, and, second, it furnishes the measure of the attraction of the earth, or the force exerted to overcome its resistance. Here, as in all possible cases, the force and the resistance are identical with each other. The above is, therefore, found in this way to be the rate at which the motion of any body will be accelerated when it is acted on by a constant force equal to its weight, and encounters no resistance.
It follows that a revolving body, when moving uniformly in any circle at a speed at which its deflection from a straight line of motion is such that in one second this would amount to 16.083 feet, requires the exertion of a centripetal force equal to its weight to produce such deflection. The deflection varying as the square of the time, in 0.01 of a second this deflection will be through a distance of 0.0016083 of a foot.
Now, at what speed must a body revolve, in a circle of one foot radius, in order that in 0.01 of one second of time its deflection from a tangential direction shall be 0.0016083 of a foot? This decimal is the versed sine of the arc of 3°15', or of 3.25°. This angle is so small that the departure from the law that the deflection is equal to the versed sine of the angle is too slight to appear in our computation. Therefore, the arc of 3.25° is the arc of a circle of one foot radius through which a body must revolve in 0.01 of a second of time, in order that the centripetal force, and so the centrifugal force, shall be equal to its weight. At this rate of revolution, in one second the body will revolve through 325°, which is at the rate of 54.166 revolutions per minute.
Now there remains only one question more to be answered. If at 54.166 revolutions per minute the centrifugal force of a body is equal to its weight, what will its centrifugal force be at one revolution per minute in the same circle?