Chapter 4f

Proposition 96

Euler
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PROPOSITION 96. Problem

  1. According to the hypothesis of gravity g acting downwards uniformly, to determine the line that a body projected on some surface in vacuo describes.

Solution

Let APQ (Fig. 92) be a horizontal plane and the point M is given both on the surface, and on the line described by the body.

Hence MQ is vertical and therefore in the direction of the force of gravity g. On putting AP = x, PQ = y and QM = z and with the equation expressing the nature of the surface dz = Pdx + Qdy let the speed at M, in which the element Mm is traversed, correspond to the height v.

Therefore as this problem is a case of the preceding, for it becomes G = g, E = 0 and F = 0, these two equations are obtained : dv = − gdz (857) and (859). Again let a height equal to b correspond to the speed that the body is to have, if it arrives in the horizontal plane APQ; then v = b − gz . Now through the other equation :

Hence it becomes : Which equation with the help of the equation dz = Pdx + Qdy is changed into this : which integrated gives : Therefore in which a special case has to be investigated, or can be integrated. If that happens, then v is obtained by differentials of the first degree, and since it is v = b − gz , an equation of the differential of the first degree arises expressing the nature of the described curve. Now the pressing force on the surface along the normal is equal to :

which with the differentials of the second order removed, (861) changes into this : Q.E.I.

Corollary 1. [p. 483]

  1. Hence the speed of any motion the body according to the hypothesis of uniform gravity g directed downwards can be known on some surface can become known from the height only, as if everywhere the body is moving in the same plane. Corollary 2.

  2. If the minute time, in which the element Mm is completed, is put as dt, then Hence by the equation found, we have:

Corollary 3.

  1. From these equations uncovered, it is found that : and

Hence

Now the radius of osculation of this described curve is equal to : Scholium.

  1. We will explain more broadly in the following problems the use of these formulas in particular cases, in which a certain kind of surfaces is considered, to which we add examples of individual surfaces. [p. 484]

PROPOSITION 97 Problem

  1. According to the hypothesis of uniform gravity g acting downwards, to determine the motion of a body on the surface of cylinder of any kind, the axis of which is vertical.

Solution

Because the axis of the cylinder is put as vertical, all the sections are equal to each other; therefore let ABQC (Fig. 95) be the base of the cylinder, on the surface of which the body is moving. Putting AP = x, PQ = y and let z be the height of the body at the point Q on the surface of the cylinder. Hence the nature of this surface of the cylinder is expressed by this equation: 0 = Pdx + Qdy or Qdy = − Pdx . Moreover this equation arises from the general equation dz = Pdx + Qdy , if P and Q are made infinitely large quantities, or with the coefficient of dz vanishing, if we may assume this. On account of which in the equations found before, P and Q must be considered as infinitely large quantities, even if they are quantities of finite magnitude [i. e. relative to each other; recall that P = ∂∂xz and Q = ∂∂yz , here with a negative sign.] Moreover P and Q are functions of x et y only, and nor is z is present in these. Therefore from these there is obtained from the deductions in the calculation : v = b − gz and

on account of the ratio P : Q = dy : −dx. But the logarithm of the second equation is : [p. 485]

Thus the equation becomes :

Hence there arises : or and the integral of this is [corrected in the O. O edition]: where ∫ (dx + dy ) denotes the arc of the base BQC traversed in the horizontal motion. 2 2

If the increment of the time, in which the element Mm is completed, is put as dt, then [From the equation vc = dx 2 + dy 2 + dz 2 it follows that the constant α = dx 2 + dy 2 c .

Note by Paul St. in the O. O.] and again,

Whereby the times are in proportion to the corresponding arcs in the base

Moreover the equation gives the equation for the curve described on the surface of the cylinder, if this surface is considered to be set out in a plane; for then ∫ (dx + dy ) denotes the abscissa on the 2 2 horizontal axis and z the applied vertical line. Now we have the projection of the curve described in the vertical plane in which we have AC cutting the horizontal plane, if with

the help of the equation y is eliminated, in order that an equation is produced between x then and z, which are the coordinates of this projection. Clearly since dy = − Pdx Q and thus where this value of x must be substituted into P and Q in place of y. Now the pressing force that the surface sustains, arises only from the centrifugal force on account of the force g being placed in the surface itself, and is equal to : by which force the body is trying to recede from the axis of the cylinder, if this expression is positive. Q.E.I. [p. 486]

[There was at the time, and even now, unfortunately, the notion held by some people that bodies try to flee from the centre when involved in circular motion, and that seems to have originated with Huygens in his Horologium. The reader may wish to compare Euler’s early solutions with more modern solution to these and further problems, such as presented by Whittaker in his Analytical Dynamics, in Ch. 4, The Soluble Problems of Particle Dynamics. ]

Corollary 1

  1. Therefore the curve, that the body describes on the surface of the cylinder if the surface is set out as a plane, changes into a parabola, clearly that trajectory that a body describes in the vertical plane.

Corollary 2

  1. If the motion of the body on the surface of the cylinder is considered to be composed from a vertical motion, as it progresses either up or down, and from the horizontal, then the horizontal motion is uniform, since the times t are proportional to ∫ (dx + dy ) , i. e. 2 2 to the arcs traversed in the horizontal motion. Now the vertical motion is either uniformly accelerated or decelerated.

Corollary 3

  1. Hence if the horizontal motion vanishes, then the body either ascends or descends along a straight line, and generally if the body can freely ascend or descend. And this case is produced if c = 0, when ∫ (dx + dy ) vanishes. 2 2

Example. [p. 487]

  1. Let the base of the cylinder be a circle, the quadrant of which is BQC and the radius AB = a; then the equation is x 2 + y 2 = a 2 and xdx + ydy = 0. Hence the equations become P = x and Q = y = (a 2 − x 2 ) . Now the projection of the line described by this on the surface of the cylinder in the vertical plane erected from AC, is expressed by the equation : Therefore if we set c = 0, then dx = 0 and x is constant; whereby in this case the projection is a straight line. Now the curve, which is the projection for whatever value of c, is constructed with the help of the rectification of the circle. Moreover the pressing force, that the surface of the cylinder sustains, is equal to on account of the equation Whereby the pressing force is constant everywhere and proportional to c.

Corollary 4

  1. On account of the same equation the pressing force generally that any cylinder sustains,

Scholium

  1. But not only can the motion of a body on the surface of an erect cylinder be easily determined with the help of resolution, but also, if the axis of the cylinder should be horizontal, with the same ease the motion of the body becomes known.

If the body does not have a horizontal motion along the cylinder, then the body perpetually remains on the same section of the cylinder as that is moving on a given line. But if now the motion is agreed to be horizontal, this remains the same always and does not disturb the other motion; and from these motions taken together the motion of the body truly becomes known easily.

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