Proposition 21

Table of Contents

PROPOSITION 21. Problem

If a body is always drawn towards a fixed centre C by some force (Fig. 25) and it is moving on a given curve AM, to determine the motion of the body on this curve, and the force it exerts on individual points of the curve.

Solution

Let the initial speed of the body at A corresponds to the height b and the distance of the point A from the centre C be AC = a.

The speed of the body at any place on the curve M must correspond to the height v and the force, by which the body at M is attracted towards C, is equal to P with the force of gravity arising for the motion the body put equal to 1. The distance MC is called y and the arc AM s; the element Mm = ds and Mn = – dy.

With centre C the circular arcs MP and mp are described ; then AP = a − y , Pp = Mn = −dy . Now with the perpendicular CT drawn to that tangent MT, we have MC : MT = Mm : Mn and MC : CT = Mm : mn , [see vol. 1, (911) Cor. 3 for a similar argument] hence this becomes :

From which, if the centripetal force is resolved into the tangential component along MT Pdy and the normal component along MO, then the tangential force is equal to − ds and the normal force is equal to − P ( ds 2 − dy 2 ) .

Hence from the tangential force there is had : ds dv = − Pdy. [p. 84] Putting the interval AP = x, in which the body approaches closer to the centre; then a − y = x and dx = −dy . Whereby dv = Pdx , and if P depends on the ∫ Pdx can be found.

Therefore with the ∫ Pdx thus accepted, in order that it vanishes on putting x = 0, then v = b + ∫ Pdx . From which the time to traverse the distance MC, then arc AM is equal to

The total normal force P ( ds − dy ) is taken up in exerting a force on the curve along ..

Therefore since this force can be shown more conveniently, and since the centrifugal force can likewise be shown, I put the perpendicular CT = p; then the normal

…

force is equal to y . Then the radius of osculation MO is equal to dp , from which the centrifugal force is obtained : and the effect of this is contrary to the effect of the normal force. On account of which the curve at M is pressed towards MO by a force equal to : Q.E.I.

Corollary 1

Therefore if the force P depends only on the distance y, thus in order that the body is acted on equally at equal distances from the centre, then the speed of the body also depends only on the distance, and the body moving on the curve AM at equal distances from the centre has equal speeds.

Corollary 2

And at any point M the speed has such a size, as the same body acquires if it falls from A with the same initial speed b through the interval AP, [p. 85] clearly with CP = CM arising.

Corollary 3

Therefore even if the curve AM is itself unknown, yet it is possible to assign the speed of the motion at each point at a distance C from the centre. Clearly for the distance y, v = b + Pdx with x = a − y arising.

[The reader will no doubt have long since noted the implicit use of a type of potential energy function in Euler’s analysis, where unit mass is assumed; this invention thus relieving him of the task of finding the speed as a function of the time, while making calculations much easier as the speed is a function of a height. At the time there was no system of units to which all physical quantities could be referred; hence comparisons of the work done under uniform gravity and as in this case under a varying force, is found by integration. [Thus, we find that the only units are the second, and the acceleration of gravity, taken as 1.] These can then be compared as a ratio if needed, and the square root taken to give the speed. Thus, each speed corresponds to the body falling from rest from the height evaluated in the comparison. Only occasionally does Euler take the calculation to the extent of getting an actual speed in units such as Rhenish feet per second. The method has its origin in the work of Galileo rather than Newton, whose calculus involved extensive use of time derivatives.]

Corollary 4

If the curve AM is such that the compressive force exerted by the body on the curve is zero, then the curve is that described by the body itself beginning to move freely from A with speed b . Thus for the free motion, there is the equation :

Ppdy = 2bdp + 2dp Pdx , or as dx = − dy , it is found that Ppdy + 2dp Pdy = 2bdp. The integral of which is p 2 Pdy = bp 2 − bh 2 with the perpendicular arising h sent from C to the tangent at A. From these equations it is found that P = 2bh 2 dp , as we found in the preceding book for free motion (587).

Corollary 5

Therefore in the above motion, the compressive force for any curve AM, which the curve sustains at the point M along MO is equal to : [Note that in this differentiation, dy = – dx.]

Example 1.

Let the curve AM be a circle having centre C, the motion of the body is uniform on account of this always having the same distance from the centre of force C[p. 86]. Whereby we have v = b and Pdx = 0 and the time to traverse AM = s = AM . Then on

…

putting y = a, we have p = a and dp = dy. On account of which the compression, which the curve sustains along MO or towards the centre C, produced is equal to P − 2ab . From , that the body is free to move in this circle. which it is evident, if b = Pa

Example 2

Let the centripetal force P be proportional to some power of the distance y or the curve AM a logarithmic spiral around the centre C, thus in order that p = my and dp = mdy and ds =

…

Hence we have :

…

and the time to complete the arc AM is equal to

The compression, that the curve sustains along MO, is equal to :

…

Corollary 6

Therefore the body, when it arrives at the centre C, has a finite speed, if n + 1 is a positive number, for the height corresponding to this speed is

…

a negative number and also if it is equal to zero, the speed at C becomes infinitely great.

Corollary 7

Now the body is pressed upon by a force tending away from the centre, or the centrifugal force prevails, if n > – 3. But if n < – 3, then the normal force prevails, and the curve is pressed upon by an infinite force towards the centre.