Sidebilder
PDF
ePub
[merged small][merged small][ocr errors]

be the axis of the cycloid, then the velocity

[merged small][merged small][ocr errors][merged small]

g.x*

y = x tan. 450

2lg cos. 450

1 the horizontal range is x = l. Hence the time of flight is

21 lg cos. 45°

g Again, the time of an oscillation is

[ocr errors]

t,

[ocr errors]

425.

Since, generally, the time of an oscillation is

[ocr errors][merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small]

426. Let l be the length of the pendulum ; then 21 = length of the whole cycloid, and, since the time of an oscillation is

1

[ocr errors]

and that down 2l is

[ocr errors]

we have

[merged small][ocr errors]

427.

Let the body descend from rest through the arc FA (Fig. 90); then the greatest velocity will be at the lowest point A;

and, since s oc = v*, if BR = BA, the velocity of a body de

a scending from rest through BA at R will = velocity at A. Hence, through R drawing RM 1 AB, and meeting the curve in M, the point M is determined in which the velocity of the body descending down the cycloid = the greatest velocity.

2

428. Since generally, vo = 2gs, the velocity due to 2r (r being the radius of the generating of the cycloid), is

v 4gr = Vgr. Now the velocity V of the moving point in the cycloid : velocity of the moving point in the circle :: ds : dz, s and z denoting the cycloidal and circular arcs respectively. But

2 chord of

supp 12

S4r

of z

z

4 cos.

[merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors][ocr errors][ocr errors][merged small][ocr errors][ocr errors][merged small][ocr errors][merged small][merged small][ocr errors][ocr errors]

x being the abscissa corresponding to s, measured from the vertex along the axis of the cycloid

Hence V = 29 (2r velocity due to 2r - x = that of a body oscillating in the cycloid. (Venturoli, p. 102.)

[ocr errors]

429,

In the generating O AB (Fig. 90,) inflect AP= its radius, and draw R'PM' I AB; then since AM' = 2AP= AB

AF, AF is bisected in M'.

Now (Translat. of Venturoli, p. 103.) the time through any arc of the cycloid, whose abscissa, measured from the lowest point is x, is

d

- d d

[ocr errors]
[ocr errors]

cos. -1

[blocks in formation]

But when z= 0, the time (t)) down FA is

d
cos.-- (- 1) =

X 180°
29

29

[ocr errors]
[merged small][ocr errors]

430. In Newton's Construction (Prop. 50, Princep.), we have

CA: CO :: CO : CR
or AO + CO: CO :: CO: CO - OR

.. AO : CO :: OR: CO - OR. But in the common cycloid, SOQ is a straight line, and :: CO

.. CO= CO – OR, and AO = OR; Also BW is parallel to RA, and BV = AO = OR = VW

;. PT = 2PV = PS, &c. &c.

431. Let R be the radius of the base, r that of the wheel; then (by Prop. 49, Princip. Newt.) it easily appears that AP = 4r (R—-). vers.

AB

R

2

vers.

AB 4r (R – 9)

to rad. 2

R Also by Prop. 52 of that work, we learn that the time through

AL AP o arc vers. AP to rad.

AB 4r (R

QAB, 2

(

[ocr errors]
[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]

Hence s = 9T= 1000 , the space required.

433. Let l be the length of either pendulum ; then the distance descended vertically from the highest to the lowest points of the O and cycloid, will be

[merged small][ocr errors]

and the velocities in the curves (v, d) will be those due to such distances. :: v::: 2gl :

2g

:: N 2:1.

Now the chord (C) of the quadrant = 1 2 the arc (A) of

1 the cycloid = 2.

2

:: C:A :: N2 : 1

viu' :: C:A

434. The semi-cycloidal arc FA (Fig. 90) is bisected by

AB inflecting AP = &c. &c. (4.29.)

[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][subsumed][merged small][merged small]
« ForrigeFortsett »