Chapter 2j

Proposition 22

Motion is the translation of a body from the place it occupies to another place. True rest is a body remaining at the same place.

Euler
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PROPOSITION 22. Problem.

  1. If a body is always drawn towards a centre of force by a centripetal force C (Fig. 27) and let the curve EAF be suited to oscillations, the determine the oscillatory motion of the body on this curve.

Solution

Let the centripetal force be proportional to some function of the distance from the centre C, and the speed of the body at equal distances from the centre C such as M and N is the same. Now at E and F the speed of the body is zero; [p. 88] and indeed it is a maximum at the point on the curve A nearest to the centre C; and the line CAO is drawn. Hence the body completes oscillations along the arc EAF, to which it is sufficient to investigate the motion to be defined on each curve AE and AF.

Let the maximum speed of the body which it has at A, correspond to the height b and the speed at some other point M correspond to the height v. The distance CM, which is equal to CP, is equal to y and the centripetal force at M is equal to P. Let CA = a and AP = x and AG = k taking CG = CE; then y = a + x and … CG = CE = a + k .

With the arc AM = s , let the tangent MT be equal to ds as determined

…

by the perpendicular sent from C to that and thus the tangential force is equal to ds ,
since with increasing y is opposite to the motion of the body; hence we have the equation
dv = − Pdy = − Pdx and v = b − Pdx with the integral Pdx thus taken, so that it vanishes at the position x = 0. If therefore on putting v = 0, the value of x is elicited from the equation b = Pdx and the interval AG or k is given. Therefore the time, in which the arc AM is traversed, is equal to ∫ ( b−ds∫ Pdx ) , from which the time for the whole arc AE is produced, if after the integration it is thus put in place, in order that the integral vanishes with x = 0, x = k or Pdx = b. In a similar manner the time to complete the arc AF can be found, and therefore from the sum of these times the time of one semi oscillation is given. Q.E.I.

Corollary 1

  1. If the curve AF is similar and equal to the curve AE, then the times to pass through each are equal, [p. 89] and thus the time of one semi-oscillation is equal to twice the time to pass through AE.

Example 1.

  1. If the arc EAF is indefinitely small, the force P acting on account of the invariable distance from the centre C is constant and equal to g. Let the radius of osculation of the curve at A or AO = h; then the arc of the circle AE is described by this radius. But from the nature of the circle it follows that CT = a 2 + 2ah − y 2 2h and …

But since y = a + x and x is indefinitely small with respect to a and h then MT = … and h

But since v = b − gx and thus b = gk, we have v = g( k − x ) and the element of time is equal to …

But ∫ ( kxdx− x ) on putting x = k is equal to π, for the periphery of the circle arising from .. the diameter 1.

Consequently the time to pass along the indefinitely small arc AE is equal to

Corollary 2

  1. If the centre of force is infinitely distant, in order that a =∝ , the direction of the force is parallel to a direction and thus the above time, in which the arc is completed, is equal to π 2h .

But if the arc of the circle EA is a straight line (Fig. 28) or h =∝ , then the time to traverse EA is equal to π 2a .

Corollary 3. [p. 90]

  1. Therefore, if the case is compared likewise with the oscillations of a pendulum acted on by some force g , but in a direction parallel to itself, then the length of the isochronous . For the time of one descent or ascent of the pendulum is equal pendulum is equal to aah ..

Example 2.

  1. Now let the centripetal force (Fig. 28) be proportional to some power of the distance or P = yn and the line EF is straight. Then AM = s = fn ( y 2 − a 2 ) and x = y − a.

Moreover again we have : and with v = 0 this becomes : Or with the said CE = c then

Consequently on account of ds = ..

, the time to traverse AM is equal to

…

Which integration is thus to be taken, in order that it is equal to zero on putting y = a. And then, on making y = c the time is had for the line EA. Now a semi-oscillation or the motion along EAF is equal to twice this time.

Corollary 4

  1. The centripetal force is put in proportion to the distance or n = 1 ;the time to pass along AM is equal to : or on putting AE = i as c 2 = a 2 + i 2 and y 2 = a 2 + s 2 , then the time to traverse AM is equal to hence the time to traverse AE is given by …

Therefore all the oscillations on this line are completed in the same time; clearly made in half the time of the oscillation π 2f .

Corollary 5

  1. If the oscillation is indefinitely small, the time of one semi-oscillation on the line is also π 2 f ; but since the centripetal force while it can be considered to be constant, let this be equal to g; then af = g and thus the time of one semi-oscillation is equal to π 2a …

as above (208).

Corollary 6

  1. Since the directions of gravity actually converge towards the centre of the earth, a body on the surface of the earth on a perfectly horizontal line is able to perform oscillations, unless resistance and friction act as impediments. Moreover the time of one such semi-oscillation shall be (on account of a = radius of the earth and g = 1) 2536 seconds (183).

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