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But the body being projected at an 2 of 45° degrees with the horizon or 90° — 45o = 45° with R, we have

R
P= R. sin. 45o =

2
R
x

6
12
R 62

(1)

or

2

a

or R=the latus rectum of the Ellipse.
Again, since the equation to an ellipse is generally

bap

pe =

2a - P

we have

62 R p? =

2a - R
R?

aR?
or
2

2
2a - R

.. 20 - R=a

.. a=R.

R? Hence 62 =

2 and b = R

(3) V2 Hence then the equation to the trajectory is

(2)

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2

2R and it may therefore be described. Again, the equation to an ellipse between pand e is

62

at ae cos. O ae being = va? 62, and 0 measured from the shortest distance. :. in this case we have

R?
P

R
2 (R +

cos. 6) 72

2 A 2

R

(5) 2 + y 2. cos. 8 Hence, to find the angular distance between the body's departure from and to the earth's surface, we have

R - 2

cos. 6 =

SV 2

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Let p = R; then

6 = cos.-- (F 12)

= 135° or 45°

Hence the distance required is

1350 45o = 90°. The same result may more easily be obtained geometrically from considering Fig. 95, and Problem (459).

515. The least tension will evidently be at the highest, and the greatest tension at the lowest point of the circle. It is also plain that the centrifugal force in the circle must be just equal at the highest point to counteract the gravitation, and it will therefore be equal to g. Hence, at the highest point if n be the number of particles in the revolving body, the tension will be t = ng = w, the weight

(1) Again, since in a circle the centrifugal force = centripetal, and velocity generally in any curve.= that acquired down chord of curvature, therefore the velocity in the circle when the force is g is

R v=N 2. g.

VOR

2

R being the length of the string.

In the descent, moreover, from the highest to the lowest point, the velocity acquired will be 25. (2R) = 2VgR. Hence the velocity of the body at the lowest point is

t' =3 Ngủ Now, generally the centrifugal force arising from the angular

1

velocity • (angular velocity) 2 x oC (linear velocity) 2 in

dist. the same circle. Hence if o be the centrifugal force at the lowest point, we have

g:0:: 09:0'? :: 9:9g ::1:9

..Q=9g and the Tension = np = 9. ng = 9W, and not 6 W as the enunciation has it.

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and u =

ľ t.

2 since u is the area described in the time t.

Hence when t is given u oc k the absolute force; and when also the absolute force is given

u is constant for all reciprocal spirals. Q. E. D.

517. The angular velocity about the first focus is (463)

с

and the angular vel. about the other focus :

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Hence the angular velocity about the second focus is

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V' =

27

b
V=
$(2ae)

s(24 - P Now, the mean angular velocity V' is that by which 360° would be uniformly described in the periodic time in the ellipse, or 27

NH
т
2621

aj

N" Hence when V = V', we have

6 va x p(2a -P) at and p.(2a – s) = ab

i. pe – 2ap - -

::$=a ai - ab which, by means of the equation of the ellipse determines two of the required points. The other two are immediately opposite to these.

If ß denote the given angular velocity, we have
Otxß

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:: B = 4/2

4 2

al
and we finally get for the whole area described

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