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a er du :. dt = - Х

24 for the Integration of which see Lacroix's Differential and Integral Calculus, 4to edit., or Whewell's Dynamics, p. 15.

F =

490. The tension is = the centripetal force necessary to retain the stone in the circle, or it is

472 g

T2 gr being the length of the string.

4.r
Also F : Weight ::

Ig :: 4:1
T

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and :: by substituting and performing the arithmetical operations we shall get the time required.

491. Generally let it be required to find the velocity acquired through (n + 1) XR to the earth's centre; R being its radius.

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and v = 2gR' X

1

(n + 1)R Let e = nR, and we have

1

29R
X

n+1 n.(n + 1)
and V =
29R

(6) n.(n + 1) the velocity acquiring in descending to the surface of the earth.

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Again, since within the Earth's surface Foc g, we have

F:g::p : R
:: F = ☆ *

хе
and v = - 1 l X (R? – 5).

(a')

Lets = 0. Then
V” = X.R.

and V' R = NgR .. (6)
the velocity acquired from the surface to the centre.

Hence

V + V' = VgR x (1+

✓nin+1 is the velocity required.

vnosti)

In the Problem

n = l.

:: V + V' =NgR x (1 +72)

492. Let a and 6 be the major and minor axes of the given ellipse. Then the greatest dist. = a + va'-b?=all+e) is known.

Again, the Periodic Time in the Ellipse is (484)

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T =

su and in the circle for the same force it is (considering it the limit of an ellipse)

2oat(1 + e)
TE

NE
::T: T::1: (1 + ey

493. and

By (443) if v and v' denote the angular velocities of

Po we have

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But p =

dp ..

podp 62

in the ellipse.
2a

e
dp 2ab2 2abop-
(22-3) 6*g*.

2a ps

62 p2
:: 6 = 0 px

62 PS
2
= . W αμ x

63
Vape x

p3 x(2a –p)

1 = 26 vaje x

(2ag – g) VOL. II.

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p3
p3

oglo

z

Hence u' is least when

2ag pa
is greatest, or when
d (2ap – p)

= 2a 2p=0
de
that is when

p= a; or the angular velocity of p in the ellipse, the force being in the focus, is least at the extremity of the minor-axis.

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Rp

1

p=

(aR - 1-1.p) and 0 =

1.

Si+vn N (nR? – n-1p)-RJ17 2n 11-vn ✓(nR' – ņ– 1.3") +R7nS Hence whenę = 00

1
X1.
1+n

1

1+Nn 2 dn 1-yn √n 1-1 and the number of revolutions is

1 It n

1.
2 n 1-n

0 =

2

n

N =

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.1.(1+2) *V 2

12.2.(72) +
But 1. (1+12)
1= 2

{**+ **)' + ** (in)* + &c.;

NE

Hence, if a few terms of this converging series be summed, and the sum be called S, we get

1
x1.2 +

N 2

XS 22 which may be arithmetically computed to any degree of accuracy, by means of Logarithmic Tables. The whole number in the result will answer the problem.

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Let a, b be the semiaxes of the Ellipse; then since

495. (by 436)

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and v = 2 X

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24 (r = p)

dt =

= V

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3
a + Na? 62

Те a + Vai-b?, which is the distance from the farther apside, being put = r. Again,

gde
24 vers 8?)

dr
x{N (rp-pa) X
2

v (re-
x) 2 유

P + C} 2u

2
see Hirsch's Integral Tables, pp. 141 and 140.

Let e = r; then
t = 0, and c = vers.-12 =

and t =

pe}

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2

vers.--1

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28
24
Let

?

= 0. Then the time to the focus from the farther

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