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Volume 14, 1881
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Art. XV.—Fallacies in the Theory of Circular Motion.

[Read before the Wellington Philosophical Society, 21st January, 1882.]

If a body be set in motion in empty space it would move on in a straight line for ever, if not subject to any action outside itself. If the body be deflected out of a straight course some force outside itself must have acted upon it. When a body is deflected from a straight course continuously, so

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as to move in a circular orbit, of course round a central point, a certain force must have acted on this body. Mr. Todhunter says: (1.) “If a body of mass m describes a circle of radius r with uniform velocity v, then whatever be the forces acting on the body their resultant tends to the centre of the circle and is equal to m v2/r. No single fact in the whole range of dynamics is of greater importance than this.” It will be advisable to illustrate this measure of circular force by an example. A slinger whirls round a stone in a sling. “The stone pulls at the string one way, the controlling hand at the centre of its circle, the other (2). Were the string too weak it would break, and the stone prematurely released would fly off in a tangential direction. If a mechanist were told the weight of the stone (say a pound), the length of the string (say a yard, including the motion of the hand), and the number of turns made by the stone in a certain time (say sixty in a minute or one in a second), he would be able to tell precisely what ought to be the strength of the string so as just not to break: that is to say, what weight it ought at least to be able to lift without breaking. In the case I have mentioned it ought to be capable of sustaining 3 lbs. 10 oz. 386 grs. If it be weaker it will break. And this is the force or effort which the hand must steadily exert to draw the stone in towards itself, out of the direction in which it would naturally proceed if let go, and to keep it revolving in a circle at that distance.” The result of the foregoing example is obtained from the preceding formula in this way. The formula stated that the acceleration, that is the pull on the hand of the slinger, is equal to the mass (one pound) multiplied by the square of the velocity divided by the radius. As the radius was three feet, the circumference of the circle would be equal to 18.8496 feet nearly, and as the mass made one revolution in a second, this is the velocity. The square of this velocity is equal to 355.30742016 feet, and on dividing this by the radius, three feet, the acceleration is found to be equal to 118.43580672 feet. The acceleration of the force of gravity at the surface of the earth is very nearly equal to 32.2 feet, and it is thus seen that the force with which the stone has to be pulled into the hand greatly exceeds the force with which a pound weight is pulled to the earth by gravity. On dividing 118. etc., by 32.2 the answer is 3.6781306, which is the ratio of the greater acceleration to the less—that is, the force said to be pulling the stone (a pound mass) into the hand is 3.67, etc., times the force with which gravity pulls a pound mass towards the earth. Now the force of gravity pulling at the stone gave it a weight equal to one pound, and consequently the central force pulling at the stone would give it a weight of 3.6781306 pounds, meaning that the hand has to bear this weight to keep the stone from breaking away. This reduced is equal to 3 lbs. 10 oz. 371.9142 grains, and more exact figures would have given a closer result.

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Reference to the formula will show that the centripetal force varies as the square of the velocity. It will be well to see very clearly what this means. Taking the particular example already given, if the velocity be doubled the acceleration is quadrupled; that is, the acceleration would be 474 feet, nearly. This means that the force necessary to retain the stone in its revolution round the hand, if it constantly acted on a mass in the same way that the force of gravity does, would add to the velocity of the mass moving freely, 474 feet per second, which is nearly fifteen times the effect produced by gravity, and the hand would experience a pull equal to that which it would feel in supporting fifteen pounds weight (nearly). If the velocity had been trebled, the acceleration would be nine times 118 feet, and the weight nine times 3lbs. 11 oz. (nearly). If the velocity were increased nearly twenty-five times,—that is, if the pound mass made twenty-five revolutions per second (471 feet per second),—the pull exerted by the stone would be equal to the pull which would be exerted by a ton weight suspended from, say, a stout beam. If the pound mass revolved one hundred times per second, it would exert a pull equal to that which would be exerted by a weight of 16 tons. This velocity, 1885 feet per second, is about 300 feet faster than a fast cannon ball, and the weight, 16 tons, is almost sufficient to break, by stretching, ordinary bar iron one inch square. These figures are remarkable, but science teaches remarkable things, and we are not much surprised at them. The principle itself will have to be looked at closely.

Momentum is the measure of force. It is the measure of the force of gravity. The acceleration produced by gravity, namely, 32 feet per second, really means the momentum generated in a mass by gravity. Let us use the word momentum instead of acceleration, meaning, when applied to the effect produced by the force of gravity, the velocity with which the mass was moving at the end of the time. This can easily be done by taking a one pound mass. If the revolving pound mass have its velocity successively increased by blows, its momentum may be represented by v, av, bv, cv, nv, and in this case the corresponding accelerations may be represented by v1, a2 v1, b2 v1, c2 v1, n2 v1 when v1 stands for acceleration, ordinarily represented by f, and in the special case of gravity by g. Both these sets of measures are momentum measures of force. Does it not seem strange, that if the momentum of the revolving pound mass be increased n times, the momentum to be generated by the centripetal force will be increased the square of n times. It may be conceived possible to make n indefinitely great, and still it is asserted that the momentum generated by the centripetal force will be as the square of n. Does not this seem very much like creating force out of nothing. It may be urged against this, that

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a blow is of the nature of an infinite force, while it might be said that the centripetal force is infinitesimal, it being the force of a pressure. In such statements as these, however, the phrase “of the nature of an infinite force,” is itself a vague and indefinite expression.

The clearest objection that can be raised against the common measure of centripetal force, is that which can be urged against the assertion that a mass revolving in a circle with uniform velocity, is every instant trying to fly off at a tangent to its orbit. The words “instant” and “tangent” here have their mathematical meanings. If the string holding the revolving mass in its circular orbit is inextensible in a mathematical sense, then it is difficult to conceive how the stone can possibly be doing otherwise than trying to go off at a tangent to its orbit. But this will only make it less necessary for the centripetal force to be great in amount, even when the velocity is considerable. If the stone is every “instant” trying to fly off at a “tangent” in a strictly mathematical sense, then every “instant” the stone is going in a direction at right-angles to the string; how then can it possibly exert a pull along the string? Of course practically no string is inextensible, but theoretically, the more the string is made inextensible, the less should be the force necessary to retain the stone in its orbit. And yet for all that, some force is necessary, for how could a stone be deflected from a straight line unless a force acted upon it. Let a ball strike a smooth surface very obliquely, in a direction almost parallel with the surface, and it will be deflected from its straight course; but how small relatively to that of the striking ball would be the force that deflected the ball in the least degree only out of its course. Is not this effect similar to the effect of a centripetal force? If so, can the force possibly be so great as shown in a previous paragraph it would be if the formula is correct?

Perhaps the foregoing reasons may be considered a sufficient cause for a reconsideration of the formula giving the measure of centripetal force. Let us examine with great care, and step by step, the process by which this formula has been obtained. A large number of treatises on astronomy and mechanics, including the best and most commonly used, deduce the formula from one of the two following propositions:—

Theorem.—If two straight lines cut one another within a circle, the rectangle contained by the segments of one of them, shall be equal to the rectangle contained by the segments of the other. (Euclid, Third Book, Prop. 35.) The case is taken where the diameter bisects a chord.

Theorem.—If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it, the rectangle contained by the whole line which cuts the circle and the part of it without the circle, shall be equal to the square on the line which touches it.

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(Euclid, Third Book, Prop. 36.) The particular case taken is where the point outside the circle is the extremity of the tangent, and a diameter produced to meet the point. Mr. Todhunter deduces the formula apparently from the parallelogram of velocities.

The method of deriving the formula for the measure of centripetal force is very clearly and precisely given by Mr. (I think now Professor) Goodeve in his “Principles of Mechanics.” He takes the thirty-sixth Proposition of the Third Book of Euclid, where it is proved that the square of the tangent is equal to the rectangle contained by the diameter produced to meet the tangent and the part produced. Accordingly, in the annexed figure, the square on the line AP will be equal to the rectangle DP, PB.

In the work referred to, the explanation given is nearly as follows:—The rectangle DP, PB is equal to DB, PB together with the square on PB. When the angle is made very small, the square on PB may be neglected, and then the square on the tangent AP is equal to the rectangle DB, BP. Now DB is the diameter of the circle; then (AP)2 = 2r. PB. (1.) In the limit this is mathematically exact. Let the body revolving with uniform motion be supposed passing through the point B by the end of the time t. If no force had deflected the body it would have pursued a straight course along the tangent, and would have reached the point P at the end of the time t, or to speak more exactly it would very nearly have reached that point, because AP the tangent is greater than AB the are. When the are, however, is extremely small, the difference between the arc of the angle and its tangent is inappreciable—in the limit they coincide. The body was deflected from its course the length PB. It is pulled through the distance PB, that is it falls through that distance. From this geometrical construction we can now derive an algebraical equation. The line AP is equal to tv, and the distance fallen through, namely PB, is equal to ½ ft2. Here v of course is the velocity of the body, and f stands as usual for the acceleration. The time that would have been taken by the body to move from A to P is, of course, the same that it took to fall from P to B, that is, if the angle represented by the arc, or tangent, be very small. In the figure the angle is very much exaggerated for the sake of clearness, but the arc taken should not be greater than a degree when the error will be very small. Bearing these considerations in mind, we can proceed to evolve from the equation we have obtained the measure of centripetal force. The equation before given may be conveniently put distinctly. Thus:—

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AP = tv (2)

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PB = ½ ft2 (3)

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These two equations may be put in the form of a ratio, as follows:— As AP: PB: tv: ½ ft2

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or in a better form: AP/PB = tvft2 (4)

Both sides of this equation are really identical, the capital letters forming one side of the equation being lines, and the small letters forming the other side of the equation being the algebraical,—that is the numerical,—value of those lines. The square of the tangent AP is equal to the square of tv. See equation (2). Equation (4) can now be put as follows:—

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(AP)2/PB=(tv)2ft2 (5)

On referring back to equation (1), it will be seen that (AP)2 is equal to 2r.PB. Substituting this value of (AP)2 in equation (5), we have now:

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2r.PB/PB=t2v2ft2 (6)

or, as PB cancels out, the simple form will be:

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2r=t2v2ft2 (7)

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It will be necessary to pause here. A careful study of these three last equations, namely (5) (6) and (7), shows us that t2v2 is the square of the tangent, and that ½ ft2 is the distance fallen through. The last equation then reads thus, if the numerical value of the square of the tangent be divided by the numerical value of the distance fallen through, the quotient will be equal to 2r. Here 2r is, of course, the diameter of the circle. In the last three equations the quantity t2 could have been cancelled out, but, by retaining this quantity, the whole algebraical expression on the right-hand side of the equation can be directly transformed into its geometrical equivalent. The term ½ ft2 is certainly the distance fallen through represented by the line PP, and is it not equally true that tv is the length of the tangent represented by the line AP, t2v2 being the value of the square on the tangent, which is equal to (AP)2. The fraction in the denominator of the fraction on the right-hand side of equation (7) is got rid of, and the equation will then stand thus: r = t2v2/ft2 (8)

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It may be read thus: If the value of the square of the tangent be divided by the value of twice the distance fallen through, the quotient will give the value of the radius of the circle. Cancelling out the time equation (8) becomes r = v2/f (9) and from this equation the formula for the measure of centripetal force is obtained, that is the value of f is found to be as follows: f = v2/r (10)

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Equation (7) will be the best to take for a special consideration. The equation is 2r=t2v2ft2

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This may be read,—The diameter is equal to the square of the tangent divided by the distance fallen through. Let us double the velocity and the body would traverse the tangent AP in half the time, and it would have to fall the distance PB also in half the time. The velocity would be represented by 2v. The equation would then stand 2r=(2v)2t)2ft)2

Here the two numeral factors, 2 and ½, cancel one another, and the tangent is unaltered. It will be seen, also, that to give the same quotient, 2 r,f must be increased four-fold; that is, the velocity being doubled, the acceleration had to be increased in the ratio of the square of the velocity. Other cases can be made up in the same way. In all of them the acceleration would have to increase as the square of the velocity.

Mr. Proctor observes, in effect, of the radius vector sweeping over equal areas in equal times, as follows: “Because a line, which is attached to a fixed point at one end and at the other to a body in motion, sweeps over equal areas in equal times, it does not therefore follow that the body is going in any orbit. For if the body moved in a straight line, the line joining the body to the fixed point would still move over equal areas in equal times. Let there be any fixed point A, and another B at some distance from it, and join AB. Let now a body be projected from B along a straight line BZ, at right angles to the line AB. It will move with uniform velocity in this direction, and of course will move over equal distances in equal times. Let it move from from B to C in one second. Mark off on the line AZ a number of spaces CD, DE, EF, etc., each equal to BC. Join AC, AD, AE, AF, etc. The body passes through the points C, D, E, F, etc., in successive seconds. The triangles constructed on these bases, BC, etc., are all equal to one another, because they have equal bases and are between the same parallels. The line joining the fixed point with the body moving in the straight line AZ, will therefore sweep over equal areas in equal times. It is not necessary, therefore, for the body to move in any orbit, because the line joining it to a fixed point passes over equal areas in equal times.

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Let a body be held at rest at the point A and let a uniform constant force act in the direction AZ. This line AZ may be conceived as being vertical to the surface of the earth, or any other planet, or the sun. Let the body at the point A be set free to the action of the accelerating force, and let it be drawn, or fall through, the points B, C, D, at the end of the first, second, and third seconds respectively. The formula s=½ft2 is perfectly general. If the accelerating force be equal to the force of gravity

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at the surface of the earth, the point B will be 16 feet from A, the point C 64 feet, and the point D 144 feet. Let it now be required to make the body pass through these several points in half the time. What must be the acceleration? At B, s = 16 = ½ f (½)2 = frac18; f and f = 128 " C, s = 64 = ½ f 12 = ½ f and f = 128 " D, s = 144 = ½ f (3/2)2 = 9/8 f and f = 128 From this we gather that when the body falls through the same spaces in half the times the acceleration must be four-fold, for 128 is four times 32. Let the body be now drawn by an accelerating force through the points B, C, D, in one third of the times it was drawn through those points in the first case. What must now be the acceleration? " B, s = 16 = ½ f (frac18;)2 = 1/18 f and f = 288 " C, s = 64 = ½ f (⅔)2 = 4/18 f and f = 288 " D, s = 144 = ½ f 12 = ½ f and f = 288 We see from this that when the body is drawn through the same spaces in one-third the times that the acceleration must be increased ninefold. Reverting now to the geometrical figure already given (see page 138), if the velocity of the revolving body be increased to twice or thrice the velocity it had at first, it will have to be drawn from P to B in half or one-third the time. But if the body had not been revolving at all, but had been at rest at P, the acceleration would have had to be increased fourfold or ninefold. It is not necessary, therefore, for a body to be revolving in any orbit to satisfy the condition, that if it be required to draw the body through the same space in one-half or one-third the time, the acceleration must be increased fourfold or ninefold. That the acceleration should increase directly as the square of the velocity, or inversely as the square of the times, it is not necessary, therefore, that the body acted upon by an accelerating force should be moving in any orbit.

Let a circle be drawn, and let a polygon of n sides be inscribed in the circle. Produce each of the sides to a distance equal to itself. This lengthened side is divided equally by the circumference—the side of the polygon is one-half, and the part produced outside the circle is the other half. From the end of the produced side draw a line to meet the angular point of the polygon opposite to it. This line will not coincide with the radius—it will not form part of the radius produced through the angular point of the polygon. Let a particle B be moving with any velocity along one of the sides of the polygon, and when it comes to the angular point let it be struck by another particle H so as to cause it to move along the next side of the polygon. When the particle B comes to the next angular point of the polygon, let it be struck by another particle (of course equal to H), so as to cause it to move along the next side of the polygon. And so on in

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the same direction along the other sides without loss of motion. Let the length of each side be represented by b, and let the length of line passed over by particle H, during the time particle B traverses a side, be represented by h. Then h will be the length of the line drawn from the end of a produced side to the angular point opposite. When B has traversed all the sides of the polygon, it has passed over a distance equal to n × b and the sum of the distances passed over by the other particles (which are the same as H, and have the same velocity) is equal to n × h. The ratio of the distance passed over by the striking or deflecting particles to the distance passed over by the revolving (struck or deflected) particle is constant. Whether the velocity be augmented or diminished the ratio is the same. The force necessary to deflect by successive impulses a particle along the sides of a polygon does not therefore have to vary in strength as the square of the velocity of the deflected particle. It is only reasonable, therefore, to suppose that the force necessary to deflect a particle so as to cause it to move in a circle does not vary as the square of the velocity.