Then ca: CP ::1:f; and the three equations are rf = x; and ʊʊ= − 2gfż, and iv-. Henceƒ=—, and vʊ = vi

-2gri

T

T

; the correct fluent of which gives v = √(2g × 2)

2g

2g = PD2, the velocity at the point P; where PD

PD PD

Τ

CE

and CE are perpendicular to CA.

So that the velocity at any point P, is as the perpendicular PD at that point.

√2gr = √(2g.Ac)

When the body arrives at c, then v = √2gr= =25950 feet or 4.9148 miles per second, which is the great

est velocity, or that at the centre c.

Again, for the time, ¿ = = = √

r

2g

√(72-22)

and the

fluents give t = √ X arc to cosine = √ X arc AD.

So that the time of descent to any point P, is as the corresponding arc AD.

When p arrives at c, the above becomes t =

× quadrant AE = 2gr conds = 21′ 7′′, for the time of falling to the centre c.

2g

The time of falling to the centre is the same quantity 1-5708, from whatever point in the radius AC the body begins to move. For let n be any given distance from c at which the motion commences: then, by correction, v = 2g the fluents

T

~[~~ (n2 —,2^)]; and hence i = √2 × √

2g √(112-x2)

of which give t = √・ x arc to cosine; which, when

x = 0, gives t = √ quadrant 1·5708✓ for the time of descent to the centre c, the same as before.

As an equal force, acting in contrary directions, generates or destroys an equal quantity of motion, in the same time; it follows that, after passing the centre, the body will just ascend to the opposite surface at B, in the same time in which it fell to the centre from A; then from в it will return again in the same manner, through c to a ; and so vibrate continually between A and B, the velocity being always equal at equal distances from c on both sides; and the whole time

of a double oscillation, or of passing from ▲ and arriving at Again, will be quadruple the time of passing over the radius Ac, or = 2 x 3.1416 = 1h 24′ 29′′.

2g

PROBLEM XIV.

To find the Time of a Pendulum vibrating in the Arc of a

Let s be the point of suspension,

SA = the arc SB or SC the length of the pendulum,

CA = AB = SB or sc the semi-cycloid,

AD DS the diameter of its generating circle, to which

PKE, HIG are perpendiculars.

GP:

To any point & draw the tangent GP, also draw Ga parallel and PQ perpendicular to AD. Then PG is parallel to the chord AI by the nature of the curve. And, by the nature of forces, the force of gravity: force in direct. GP GQ :: AI : AH :: AD: AI; in like manner, the force of grav. : force in curve at E:: AD: AK; that is, the accelerative force in the curve, is as the corresponding chord AI or AK of the circle, or as the arc AG or AE of the cycloid, since ag is always 2A1. So that the process and conclusions for the velocity and time of describing any arc in this case, will be the same as in the last problem, the nature of the forces being the same, viz, as the distance to be passed over to the lowest point A.

From which it follows, that the time of a semi-vibration, in all arcs, AG, AE, &c, is the same constant quantity

28

whole vibration from в to C, or from c to B,

is 3·1416 where / AS AB is the length of the pendulum, g=16 feet, or 193 inches, and 3.1416 the circumference of a circle whose diameter is 1.

Since the time of a body's falling by gravity through 17, or half the length of the pendulum, is ✔ which being in 2 2g proportion to 3-1416, as 1 to 3:1416; therefore the di2g ameter of a circle is to its circumference, as the time of falling through half the length of a pendulum, to the time of one vibration.

If the time of the whole vibration be 1 second, this equa

2g

g 4.9348

2g

g

3.14162 4.9348

tion arises, 1′′ = S∙1416√, and hence 7 = and g 3.14162 × 1/ = 4.93481. So that if one of these, g or 7, be given by experiment, these equations will give the other. When g, for instance, is supposed to be 16 feet, or 193 inches, then is 7 = 39-11 the length of a pendulum to vibrate seconds. Or if l = 394, the length of the seconds pendulum for the latitude of London, then is g= 4·9348/193·07 inches = 16,2% feet, or nearly 161⁄2 feet, for the space descended by gravity in the first second of time in the latitude of London, also agreeing with experi

ment.

Hence the times of vibration of pendulums, are as the square roots of their lengths; and the number of vibrations made in a given time, reciprocally as the square roots of the lengths. And hence also, the length of a pendulum vibrating n times in a minute, or 60", is l = 394 ×

602

na

=

140850

nn

When a pendulum vibrates in a circular arc, as the length of the string is constantly the same, the time of vibration will be longer than in a cycloid; but the two times will approach nearer together as the circular arc is smaller; so that when it is very small, the times of vibration will be nearly equal. And hence 394 inches is the length of a pendulum vibrating seconds in the very small arc of a circle.

PROBLEM XV.

To find the Velocity and Time of a heavy Body descending down the Arc of a Circle, or vibrating in the Arc by a Line fired in the Centre.

A

Let D be the beginning of the descent, c the centre, and a the lowest point of the circle; draw DE and PQ perpendicular to AC. Then the velocity in P being the same as in a by falling through EQ, it will be v=2√(g XEQ) = 8√(ax), when a = AE, x = AQ.

But the flux. of the time i is = AP, and AP =

where r=the radius AC. Theref. t =

v

16 √(ax − x2) × √√(d=x)

X

√(2rx—x2) × √ (a − x)

(1 +

+

+

2d

2.4d2

2.4.

&c),

6d3

by developing √(1 − ) in a series.

is x arc to radius ža and

vers. x, or it is the arc whose rad. is 1 and vers.

2x which

a

call A. And let the fluents of the succeeding terms, without the coefficients, be B, C, D, E, &c. Then will the flux. of any one, as a, at n distance from A, be Q = x2À = xp, which dx" suppose also the flux. of bp

Hence, by equating the coefficients of the like terms,

1.3.5

2.4.6d3

5ac-2x2/(ax-x2)

6

found by art. 32, pa. 238, vol. 3 of the Course.

But when x = a, those terms become barely

22d 22.42d2 22.42.62d3

&c); which being sub

tracted, and r taken = 0, there arises for the whole time of descending down DA, or the corrected value of t

=

When the arc is small, as in the vibration of the pendulum of a clock, all the terms of the series may be omitted after the second, and then the time of a semi-vibration t is nearly × (1+ —). And theref. the times of vibration of a pendulum, in different arcs, are as 81 + a, or 8 times the radius added to the versed sine of the arc.

=

1.5708 g

4

2

If D be the degrees of the pendulum's vibration, on each side of the lowest point of the small arc, the radius being r, the diameter d, and 3·1416 = p; then is the length of that But the versed sine in terms of the arc

פייס

arc A

pdD

is

@=

+ &c =

d

360a 3.360+ &c, or only = 360o

the first term, by rejecting all the rest of the terms on ac

value of the time, it becomes t = nearly. And therefore the times of vibration in different small arcs, are as 52524 + D2, or as 52524 added to the square of the number of degrees in the arc.

Hence it follows that the time lost in each second, by vi brating in a circle, instead of the cycloid, is 52524; and consequently the time lost in a whole day of 24 hours, or 24 × 60 × 60 seconds, is D2 nearly. In like manner, the seconds lost per day by vibrating in the arc of A degrees, is {A2.