THE MOTION OF A POINT ON A GIVEN LINE IN A VACUUM

by Euler

March 20, 2025 37 minutes • 7702 words

PROPOSITION 18. Problem.

With a uniform force present acting in the downwards direction, to determine the time of the ascent or the descent through any arc of a circle EA (Fig.23), ending at the lowest point A.

Solution. [p. 70]

Let C be the centre of the circle, CA is the radius of the vertical or the line parallel to the direction of the force g. Putting AC = a and the arc AE equal to the height AG = b, the speed at the lowest point A corresponds to the height gb, since the body descending from E has such a speed when it arrives at A . And the body must have such a speed at A, in order that it can rise as far as E. Some element Mm of the arc AE is considered and calling AP = x ; then PM = ( 2ax − x 2 ) and Mm = adx.

Now the speed at M
( 2 ax − x 2 )
corresponds to the height g .GP = gb − gx (93). Therefore the time in which the element
Mm is traversed either in the ascent or in the descent is equal to
adx
.
g ( b − x )( 2 ax − x 2 )
Which, since it cannot be integrated, we express the integral by a series. Moreover, with
putting 2a = c :EULER’S MECHANICA VOL. 2.
Chapter 2b.
Translated and annotated by Ian Bruce.
page 104
Hence this is multiplied by adx and the integration gives the time, in which the arc AM is
g
completed, to equal :
Now the time in which the whole arc EA is traversed can be produced, if we put x = b
and the ratio of the periphery to the diameter = π : 1 , with which in place there is
obtained : [p. 71]
9 , etc are the squares of the coefficients 1, 1 , 3 , which is
Where the coefficients 1, 14 , 64
2 8
produced if ( 1 − z )− 2 in resolved into a series. Now the time can therefore be found
approximately from this series. Q.E.I.
1
Corollary 1.
162. Therefore where the arc EA is made larger, then the time too is greater, in which it is
traversed. Indeed on putting b = 2a = c , the time is infinite, since the body in descending
is by no means able to complete the semicircle.
Corollary 2.
163. Therefore if the body in an oscillatory motion is moving in the arc EAF of the circle,
then the time of one to or fro motion is twice as great as the time of one ascent or one
descent, since the time to pass through ANF is equal to the time to pass through AME.
Whereby the time of one to or fro motion, or the time for half an oscillation, is equal to
Truly the time for one oscillation to be completed is twice as great.
Scholium 1.
164. The series expressing this time can at once be found in this way. An element of time
can be resolved into these factors :
and of these only the latter should be converted into a series, clearly this :
with 2a = c. Moreover, because after the integration, on placing x = b, then [p. 72]EULER’S MECHANICA VOL. 2.
Chapter 2b.
page 105
Translated and annotated by Ian Bruce.
From which the whole descent time can be gathered together to be equal to :
Scholium 2.
165. From which it is apparent that the summation of the series depends on the
construction of the equation
I put
b = tt
c 1+ tt
qdt
and the sum of the series is equal to e ∫ t with e denoting the number of which the log is
equal to1. With these in place, the series is to be summed by my method explained in
Comment. Acad. Petrop. Tom. VII [1740, p. 123; Opera Omnia series I, vol. 14; E41 is
translated in this series; however, this appears to be a misquote by Paul Stackel, as this
paper does not present the method used to sum the present series. One should look
instead in E025 perhaps], and the following equation is found from the exposition:
q 2 dt
dq + t = tdt 2 .
( 1+ tt )
From which equation, if it can be solved, q is found in terms of t and hence the sum itself
is found in terms of t or bc . Moreover since the construction of the equation does not
follow from inspection, it is yet apparent that it can be done, since the sum of the series
for the time can be assigned with the help of quadrature. Indeed the given sum of the
series is found to follow from the construction of that equation.
Corollary 3.
166. If the arc AE, in which the descent or the ascent is completed, is put infinitely small,
yet the time for that motion is not infinitely small. For in the expression for the time, only
b vanishes, and the time in which a vanishing arc AE is completed is equal to π 2a .
2 g
[i. e. the radius of the circle a remains unchanged, while the distance fallen b tends
towards zero.]EULER’S MECHANICA VOL. 2.
Chapter 2b.
page 106
Translated and annotated by Ian Bruce.
Corollary 4. [p. 73]
167. With the other part AF of the circle joined with AE the oscillations through the arc
EAF can be made indefinitely small; still with a finite completion time. Clearly the time
for one ’to’ or ‘fro’ motion, or the time for half an oscillation, is equal to π 2a .
g
Corollary 5.
168. Therefore the times of this kind of infinitely small oscillations are in the square root
ratio directly as the radius and inversely as the force [of gravity] acting.
Corollary 6.
169. These same formulae prevail, if the force acting should not be uniform. For
whatever variable force is put in place, yet while the body driven along an infinitely small
arc, it has the same constant value.
Corollary 7.
170. It is to be understood that even if the curve EAF is not a circle, but any curve, then
also these results reported here pertain to infinitely small oscillations on this curve. Then
indeed in place of the radius the radius of osculation of this curve is to be taken at the
lowest point A.
Corollary 8.
171. Oscillations upon an infinitely small arc of the curve EAF are effected with the aid
of a pendulum, the length of which is the radius AC. [p. 74]Therefore the times of
indefinitely small oscillations of the pendulum vary directly as the square root of the
length of the pendulum and inversely as the square root of the force acting.
Corollary 9.
172. If the curve ANF is not equal to the curve AME, [i. e. no longer circular arcs, and
each with its own radius of curvature] it is still sufficient to consider the radius of
osculation at the point A for infinitely small oscillations. Let this length be equal to α,
then the ascent time through the indefinitely small arc AF is equal to π 2α , and since
2 g
the descent time through the vanishing arc EMA is π 2a , then the time for one journey
2 g
or half an oscillation on the composite curve EAF
π( a+ α )
2 g
.EULER’S MECHANICA VOL. 2.
Chapter 2b.
page 107
Translated and annotated by Ian Bruce.
Corollary 10.
173. If the oscillations are not indefinitely small on the circle BAD, the oscillation times
are greater, as the arcs of the oscillations are greater. And if the oscillations are yet
definitely small, the time of such an oscillation to the time of an indefinitely small
oscillation to is as the square of the diameter of the circle increased by the versed sine of
the arc traversed to the square of the diameter itself.
Corollary 11.
174. The height, from which a body descends in the same time by the same force g
acting, as it descends along an indefinitely small arc EMA , is equal to π8a , or is to the
2
eighth part of the radius as the square of the circumference to the square of the diameter ;
[p. 75] this height is hence approximately equal to 54 a .
Corollary 12.
175. Moreover the body descends along the chord of the arc EMA in the same time that it
descends along the diameter of the circle (102). Whereby the descent time along an
indefinitely small arc is to the descent time along the corresponding arc is as 2 2a to
g
π 2a , i. e. as the diameter to the fourth part of the circumference. And the descent time
2 g
from the diameter or from twice the length of the pendulum is to the time of one whole
indefinitely small oscillation composed from a to and fro motion is as the diameter to the
circumference.
Scholium 3.
176. If two circular arcs AE and FA (Fig. 24), upon
which connected oscillations are carried out, are not
equal, these oscillations can be made with the aid of
a pendulum, if in the centre of K of the arc AF a nail
is driven in, in order that the thread CA, after it has
described the arc EA about the centre, is retained at
K, and describes the arc AF about the centre K.EULER’S MECHANICA VOL. 2.
Chapter 2b.
Translated and annotated by Ian Bruce.
page 108
PROPOSITION 19.
Problem.
177. For a given force acting, to find the length of the pendulum making indefinitely
small oscillations, which completes a to and fro motion in a time of one second.
Solution. [p. 76]
With the length of the pendulum a sought and the force acting g, with the force of
gravity denoted by one, the time of one indefinitely small oscillation is equal to π 2a .
g
Now this has to be expressed in seconds of minutes, with the length a expressed in
thousandth parts of Rhenish feet and the formula π 2a is to be divided by 250, as is
g
apparent from the first book (221). On account of which the time of one half oscillation is
obtained π 2a seconds. Whereby, since the time has to be one second, that is
250 g
π 2a = 250 g and a =
31250 g
π2
= 3166 14 g thousandth parts of Rhenish feet.
Therefore, this is the length of the pendulum completing a semi-oscillation in a time of
one second. Q.E.I.
Corollary 1.
178. Hence the lengths of the pendulums executing oscillations in the same time, but with
different forces acting, are in the ratio of the forces.
Corollary 2.
179. If the force acting g is equal to the force of gravity 1, which case agrees with
oscillations on the surface of the earth, the length of the pendulum which makes a single
to and fro journey [i. e. half an oscillation in one second] is equal to 3.16625 Rhenish
feet, or three and one sixth feet. [p. 77]
Scholium 1.
180. This length agrees extremely well with that found by Huygens from experiment;
from which it is apparent that we have assumed correctly the number in the preceding
book (220) of 15625 scruples of Rhenish feet that a body falls, acted on by the force of
gravity, for a time of one second from rest; for indeed this number departs from the
number 250, by which the expressions for the time must be divided, in order that a time
of one second is presented. [Recall that the number 250 was just a useful number
introduced by Euler as a memory aid, and so was only approximately correct.] Therefore
since it may be generally wished to have the sixth part of a foot for the Huygens length of
the pendulum of 3.166, clearly the length of this must be determined by observationsEULER’S MECHANICA VOL. 2.
Chapter 2b.
page 109
Translated and annotated by Ian Bruce.
everywhere on the surface of the earth, from which it generally consists of 1055
thousandth parts of a Rhenish foot.
Scholion 2.
181. Now from observations the universal foot can be determined in the following
manner. A pendulum of length f is taken, which is set in motion to make the smallest
oscillations, and let the number of these counted in a time of one hour be n, thus in order
that a single semi-oscillation is completed in a time of 3600
seconds. Now let the length
n
of the pendulum completing semi-oscillations in one second be z. Whereby, since the
times of oscillations of different pendulums acted on by the same force are as the square
root ratio of the lengths of the pendulums (171), then the ratio is 3600: 1 = f : z and n n2 f n2 f thus z = 12960000 , [p. 78] and consequently the universal foot is equal to 38880000 . Corollary 3.

Therefore a pendulum four times longer than 3166 14 scruples of Rhenish feet completes semi-oscillations in two seconds, since the times of the oscillations are in the square root ratio of the lengths of the pendulums. Corollary 4.
Since the radius of the earth is 20382230 Rhenish feet., if a pendulum of such a length is conceived, a single semi-oscillation of this will last for 2536 sec. Whereby in 24 hours almost 17 whole oscillations are completed. Corollary 5.
[There is no section 184.] Since the time of half an oscillation is π 2a , the time of g whole oscillations is 2π 2a . But this time is equal to the time of the revolution g performed on the periphery of a circle of radius a by a body in free motion, which is drawn towards the centre by a force equal to g, as from the preceding book is apparent (612). On this account the time of a whole oscillation of the pendulum equal to the radius of the earth is equal to the time that a body projected on the surface carries out a complete revolution. Now Huygens also showed that a body completes almost 17 revolutions in a time of 24 hours in performing this motion.[p. 79]EULER’S MECHANICA VOL. 2. Chapter 2b. page 110 Translated and annotated by Ian Bruce. Corollary 6.
Since the force of gravity shall be to the force that a body on the surface of the sun is urged towards the centre of the sun, as 41 to 1000, the length of the pendulum which on the surface of the sun performs a semi-oscillation in a time of one second is equal to 77.226 Rhenish feet. In a similar manner on account of gravity on the surface of Jupiter , for such a pendulum the length is 6.448 feet. And on the surface of being equal to 167 82 Saturn on account of gravity equal to 105 , the length of such a pendulum is 4.054 feet. 82 PROPOSITION 20. Problem.
If the curve BAD (Fig. 25), upon which oscillations are made, is a cycloid described by the circle with diameter AC on the horizontal base BD, to determine the time of the oscillation through each arc EAF, with a uniform force acting downwards. Solution. Let the radius of osculation at A, truly AO, = a, which is twice the diameter of the generating circle AC; hence AC = 12 a and with the abscissa AP = x and with the corresponding arc AM = s, from the nature of the cycloid, we have s 2 = 2ax . Now let the abscissa for the arc EAF, which is traversed in the oscillatory motion correspond to AG = b; the speed at the lowest point A corresponds to the height gb and the speed at M corresponds to the height g( b − x ) . [In the sense that the ratio of the speeds is as the square root of the ratio of the heights.] Whereby, since ds = adx , [p. 80] the time in 2ax which the arc AM is traversed, is equal to :EULER’S MECHANICA VOL. 2. Chapter 2b. page 111 Translated and annotated by Ian Bruce. Now, if after integration on putting x = b, then the time in which the whole arc AE is travelled through, is produced : or the circumference of the circle divided by the diameter. Whereby the time of a single ascent or descent is equal to π 2a and the time of one journey along the arc EAF is equal 2 g to π 2a . And the time for a complete oscillation is equal to 2π 2a . Q.E.I. g g Corollary 1.
Since in this expression of the time the letter b which determines the magnitude of the arc EAF is not present, all the times of the oscillations which are performed on the same cycloid are equal to each other. Corollary 2.
Therefore the time of any one oscillation is equal to the time of the oscillation through an indefinitely small arc. But the indefinitely small arclet agrees with the arc of the circle with radius OA to be described. Whereby the time of any oscillations on the cycloid BAD is equal to the time in which a pendulum of length a completes the smallest oscillation. It has also been made evident in the previous proposition that the time of one of the smallest oscillations of the pendulum a is equal to 2π 2a (167), in which we have g found the time of a single whole oscillation by the same formula. [p. 81] Corollary 3.
Therefore if the pendulum is thus adjusted, in order that the oscillating body is moving on the cycloid, all the oscillations of this, whether they are large or small, are completed in equal intervals of time. [One may recall that Huygens had to resort to reductio ad absurdum arguments to prove this in the Horologium.] Whereby if AO is 3166 14 g scruples of Rhenish feet, individual semi-oscillations are completed in times of one second. Corollary 4.
Therefore all the descents to the lowest point A on the cycloid are of equal times or isochronous, and likewise all the ascents from the lowest point A, until the speed is spent. Truly the time of one ascent or descent is 2π 2a . 2 gEULER’S MECHANICA VOL. 2. Chapter 2b. page 112 Translated and annotated by Ian Bruce. Scholium 1.
On account of this property the cycloid is usually given the name tautochrone, since all the oscillations are completed on these in the same time. Huygens first uncovered this extraordinary property of the cycloid and understood at once that the cycloid could be substituted in place of the circle, that he effected in clocks. Yet now the clockmakers have abandoned this way of making oscillations, as they have learned almost nothing of this use. And surely in a vacuum with any curve, isochronous oscillations are produced, since they are always present with the same magnitude. Now in a resisting medium, in which the oscillations decrease, the cycloid loses this property and thus there is no advantage in the use. Scholium 2.
Also it is understood, if two dissimilar cycloids AE and AF (Fig. 24) are joined at the lower points, the oscillations upon the composite curve EAF are completed in equal times. For since both times of ascent or descent are constant quantities, also the sum of these, clearly the times of half an oscillation and the whole oscillation are equal to each other. Let twice the diameter of the circle generating the cycloid be AF = α, then the time of one ascent or descent on AF = π 2α . Whereby the to and fro journey on the 2 g composite curve EAF is completed in a time equal to time for the whole oscillation equal to π ( 2a + 2α ) 2 g , and now with the 2π ( 2a + 2α ) . g Scholium 3.
Order requires that before we can progress to forces acting in different directions, we should explain the effect of forces acting parallel in the same direction, but [in which the magnitudes] are variable, and we should investigate the motion of bodies acted on by forces of this kind upon given curves. But since the contents worthy of note in the examples of motion hitherto set out may lie hidden from us, the principles shall now be explained with the help of which the motion on any curve can be understood, while we defer these other considerations to a fuller treatment, and here we take curves to be investigated, [p. 83] upon which a body is acted on by some kind of force, and advances according to a given law.EULER’S MECHANICA VOL. 2. Chapter 2b. page 113 Translated and annotated by Ian Bruce. 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. Now 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 toEULER’S MECHANICA VOL. 2. Chapter 2b. page 114 Translated and annotated by Ian Bruce. The total normal force P ( ds − dy ) is taken up in exerting a force on the curve along ds 2 2 MO. 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 Pp ydy 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.]EULER’S MECHANICA VOL. 2. Chapter 2b. page 115 Translated and annotated by Ian Bruce. 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 p 3 dy 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 ∫ b b 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 2 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 = dy ( 1− m 2 ) . Hence we have : and the time to complete the arc AM is equal to :EULER’S MECHANICA VOL. 2. Chapter 2b. page 116 Translated and annotated by Ian Bruce. Now the compression, that the curve sustains along MO, is equal to : Corollary 6. [p. 87]
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 n+1 + b . But if n + 1 is ( n +1 ) f n 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. PROPOSITION 22. Problem.
If a body is always drawn towards a centre of force by a centripetal force C (Fig.

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 ydy CG = CE = a + k . With the arc AM = s , let the tangent MT be equal to ds as determinedEULER’S MECHANICA VOL. 2. Chapter 2b. Translated and annotated by Ian Bruce. page 117 Pdy 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.

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.
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 = 2ahx( a + h ) 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 2 the diameter 1.EULER’S MECHANICA VOL. 2. Chapter 2b. Translated and annotated by Ian Bruce. page 118 Consequently the time to pass along the indefinitely small arc AE is equal to Corollary 2.
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 . 2 g 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 . 2 g Corollary 3. [p. 90]
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 +h to π 2ah 2 g( a+ h ) . (166) Example 2.
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 = ydy ( y 2 −a 2 ) , the time to traverse AM is equal toEULER’S MECHANICA VOL. 2. Chapter 2b. Translated and annotated by Ian Bruce. page 119 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. [p. 91]
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 π 2f 2 . 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.
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 g as above (208). Corollary 6.
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).EULER’S MECHANICA VOL. 2. Chapter 2b. Translated and annotated by Ian Bruce. page 120 PROPOSITION 23. [p. 92] Problem.
If a body is acted on by any two forces, of which the direction of one is along the vertical MQ (Fig. 29), and the other MP is horizontal ,to define the motion of the body from these forces acting on a given curve AMB. Solution. Let the speed at B be zero, and at M it corresponds to the height v. The force acting along MQ is equal to P and that along MP is equal to Q. Put BR = t, RM = z, the arc BM = w, which letters we use for the descent of the body from rest at B. But for the ascent from A with any initial speed, which motion is referring to oscillations, let AP = x = QM, PM = AQ =y and the arc AM = s; now the speed of the body at A corresponds to the height b; hence t + x = const., likewise z + y = const. and w + s = const., thus dt + dx = 0 [, dz + dy = 0] and dw + ds = 0. With the forces P and Q resolved into normal and the and tangential components, the tangential force that arises from P is equal to Pdt dw normal force that arises from P is equal to Pdz pulling along MN. Then the tangential dw Qdz , which is force arising from Q is equal to dw and the normal force from Q is equal to Pdt dw contrary to that normal force. And beyond that the tangential force along BM accelerates the motion and thus dv = Pdt + Qdz and v = Pdt + Qdz ∫ ∫ [p. 93], with these integrations thus made, in order that they vanish with t and z = 0. And for the ascent from A there is : v = b − Pdx − Qdy , with these integrations thus made, in order that they vanish with x ∫ ∫ and y put equal to 0. Therefore with t = BD and z = AD put in this equation : v = Pdt + Qdz , we find that v = b. Whereby the time to traverse BM is equal to : ∫ ∫ and the time for AM is equal to :EULER’S MECHANICA VOL. 2. Chapter 2b. page 121 Translated and annotated by Ian Bruce. 3 dw and thus With the element dt or dx taken as constant, the radius of osculation at M = dtddz the centrifugal force, the direction of which is along MN , is equal to : Hence the total force, by which the curve is pressed upon at M along MN, is equal to : Q.E.I. Corollary 1.
But if P is some function of x or t and Q some function of y or z, so that Pdx as well as Qdy can be integrate; and thus the speed v can be exhibited and with the help of the equation for the curve the time too. Corollary 2.
Because whatever and however many forces are acting, but if the directions of these are in that plane as the curve AMB, these forces can be resolved into two forces of this kind, and this proposition extends widely and embraces all the cases in which the directions of the forces and the curve are in the same plane. Scholium. [p. 94]
Also it is apparent that this proposition is of wider applicability if a few cases are added on and examined, in which not all the directions of the forces are in the plane of the curve. For then these forces are to be resolved into two components, of which the one are in the plane of the curve itself, and the other normal to this plane. Therefore these situated in the plane of the curve, so that in the proposition we have used, the analysis gives the acceleration of the body and the compression force along MN ; the other forces, because they are normal to the curve, are only devoted to pressing on the curve. Whereby hence a twofold compression arises, which the curve sustains, the one directed along MN , and the other normal to the plane of the curve. Therefore of these two compressions, if the direction of the mean is taken, there is produced, and there is produced the direction of the equivalent force by which the curve is pressed. On this account there is no need for us to explain cases of this kind, but we will mention briefly a few in which the motion of bodies are on a curve which is not itself placed in the plane in which the forces act, which we take as constant and in the downwards direction.EULER’S MECHANICA VOL. 2. Chapter 2b. Translated and annotated by Ian Bruce. page 122 PROPOSITION 24. Problem.
With a uniform force acting in a downwards direction, to determine the motion of a body on some curve AM (Fig. 30)not set up in the same plane. Solution. [p. 95] Let the projection of the curve AM be the curve AQ in the horizontal plane, and with the perpendiculars MQ and mq sent from some nearby points M and m to this plane, and there are drawn to the axis AP taken as you please, the normals QP and qp and put AP = x, PQ = y and QM = z. Let the speed of the body at A correspond to the height b, and the speed at M correspond to the height v. Now the force is equal to g, by which the body at M is acted on along MQ. With the tangent MT drawn, and in that from Q to the perpendicular QT the force g in is resolved into tangential and normal [components]. Since the tangential force is equal to : And since the normal force is equal to : Moreover since the tangential force slows the motion, then we have dv = − gdz and v = b − gz , hence the time in which the arc AM is completed, is made equal to : Now the normal force brings about a compression of the curve by the body at M with so much force along a direction normal to Mm and situated in the plane QMmq. Now the curve is acted on in addition by the centrifugal force along the opposite direction of the position of the radius of osculation by a force equal to 2( b − gz ) , with r designating the r radius of osculation at M. Moreover we found above (71) the position of the radius of osculation, from which the direction of the centrifugal force can hence become known.EULER’S MECHANICA VOL. 2. Chapter 2b. Translated and annotated by Ian Bruce. page 123 Now the magnitude of the centrifugal force is given from the radius of osculation, which has been found (72) ; clearly it is given by : Q.E.I. Corollary 1. [p. 96]
Therefore the speed of the body in this case also depends only on the height. And the speed at M is of such as magnitude as the body has ascending through QM , when it has a speed corresponding to the height b at Q. Corollary 2.
Hence the body is unable to ascend to a greater height than to bg . For if we set b − gz = 0 , the body has lost all its speed at that height and begins to descend again. Corollary 3.
Also it is understood, if the force cannot be taken as constant, but is the variable P, then the speed at M can be found corresponding to the height b − Pdz . ∫ Scholium 1.
If the curve AM is considered to be in the vertical plane (Fig. 31) and with the curve related to the horizontal axis AQ, and AQ is equal to the curve AQ in the previous figure, and QM = QM in the preced. fig., then the curve AM is also equal to the preceding curve AM. If now the body on the curve AM ascends with an initial speed at A corresponding to the height b and acted on by the same force g, then it also has the speed at M corresponding to the height b − gz . And thus the time of ascent along AM also agrees with the times of ascent along AM in the preced. fig. Therefore by this reason the motion of the body not in the same plane can be reduced to motion on a curve placed in the same plane. [p. 97] For it is not possible to distinguish between the motions ; but the forces acting on these two curves are different. On account of which this compression can be varies as it pleases, with the motion on the curve remaining the same. Scholium 2.
Up to the present we have put the curve in place upon which the body is moving, and the force acting in one given direction, and from these we have deduced the motion of the body and the compression of the curve. Therefore now, since these should suffice, we progress to other questions, in which other quantities are taken as given, and the remaining quantities are to be found. First indeed the compression is given at individual points on the curve and the force acting; from which the curve itself and the motion on the curve must be found. Then from other combinations made from these things, which are come upon in the computation, we will form other questions.