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along p and S be the retarding force upon B arising from the inertia, we have

du SXBAX

dt

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A udu
:.SE

B de
Again, the centrifugal force is

a222
(499) =
83

p3
:: the centripetal force is

a*ß A udu F=0-S

g3

B de

a gede А :. udu = Fde =

udu p3

B and udu =

Ba'ßo de

A+BP3
Hence

a BB 1
F =

Х
A+B pu
с'dp

(see 436)
de
B de dp
A + Bg3

B
and C +
A + B e

pe
But when g = a, p = a.

B
1
:.C.=

X (1
a

A + B (A+B)a
Hence

a”.(A+B).
pe =

.(1) Aa' + A the equation to the curve, which is therefore one of Cotes' Spirals.

Again, in all curves

2

2

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Pe

Code + de)

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673. Let w be the given weight of the cylinder; then the accelerating force down the plane is

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R being the distance of the centre of gyration from the axis of rotation, and 0 the inclination of the plane.

22

2 :. F=

sin. A sin. 0.

3 78 +

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2

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574. Let W be the weight of the cylinder, " the radius of its base, then its inertia is w x re, and the accelerating

2

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:. 12. (2P + W) = Pg
:: W=

P
gP

x g - - 24). 12

12

575. Supposing m' not equal to m, and putting cm' = x, ac =cb = a, then since the efficacy of m in opposing the motion of m', is measured by

тх

m',

v (a? + x2) the whole moving force is

2mi

V (a? + x2) and the mass moved is

2m + m'
2mx

1
:. F = m'
va? + x2

2m + m But at the time the velocity is a maximum, the accelerating force = 0.

2mx

ma a? + 2?

am And .. x =

(4m", - m')' the distance required.

= {

}

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577. Let , be the radius of either globe, M its mass, k, k' the distances of their centres of gravity from the axis of rotation 1, r the distances of their centres of oscillation from the same, when the bodies are unconnected ; then since (Creswell's Translat. of Venturoli, p. 141), the length of the compound pendulum is

Mkl + Mk'l kl+k'T
LE

Mk + MK ktk and by the question

k = 3r, k' = 5r. Therefore

31 + 51 L =

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578.

Let a be the length of the rod; then supposing generally the distance of the point of suspension from its extre

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mity, the length of the pendulum is (see Venturoli)

a 3

(-9)

+
S
Mk

3n3

3

a

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579. Since the centre of Initial Rotation is distant from that point of impact by the same interval as the centre of oscillation, considering that point as the point of suspension, the distance required is (Venturoli) S Aa2 + B62

Aa! + Bbe
Mk

Aa - B6 Aa Bb
(A+B) X

A + B A, B being the masses, and a, b the lengths of the arms of the lever.

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580. Generally, if P denote the power moving the system, whose weight is w, acting at the distance r from the axis of rotation, then the force which accelerates P is

Pr2

Prs + W.R?
R being the distance of the centre of gynation from the axis.
But s denoting the space described in the time t, we have

s=gFt
and gFt: 0 :: 2 : 360°

gFt :.0 = 360° x

2πη or the body moves at the rate of

gFt

revolutions in a second.

271 Hence, by the question,

gFt

277
2071 2071 Pr! + WR2
9

Pro

= 10

gF

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32

= 6'. 32". 6173 nearly.

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