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L=

596.

By Venturoli, p. 140, we have

S S + Mk2
Mk

Mk where k is GS the distance of the centre of gravity from the axis of rotation, M the mass of the body, Sthe moment of inertia referred to an axis passing through the centre of gravity, and L the distance of the centre of oscillation from the axis.

Hence since S and M are the same for all values of k, L is constant for the same values of k, which proves the first part of the proposition.

Again, when the axis of rotation is any where in the circumference of the circle whose radius is GO, we have

S" = S - M (SG' - GO)

= $ - M x SO X SG + Mx SO. GO
S" being the momentum of inertia referred to this new axis.

But S' = M.SO.SG
i. S = M.SO . GO

S' M.SO.GO
and :. L' =

=SO = L; Mk M.GO or the pendulum will oscillate in the same time as before.

597. Letr be the radius of the circle, m the distance of either body from the axis of rotation ; then

S m?

= 2r. Mk m

27

Let a be the given length of the lever, M the weight

598. required,

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599.

Let a be the length of the lever, 2 the distance of the

fulcrum from the end to which P is attached; then the moving force of P is

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P + Q.(a –

and the mass moved reckoning the inertia is.

(a -.x)

x2
Hence the accelerating force is

P.x2–Q (a-x).
F=

P.x*+Q (a-x)
Again, the moving force which generates P's velocity is
Px'-Q (a-x).x

XP, PxP+Q (a-x) and that part of P which is sustained, or the tension of the string is

Px2 - Qa-X
P (1-

PQa (a-x)
Pro +Q(a-x)

Px + Q(a - x)
maximum by the question. Hence
Pro +Q(a - x)

= minimum,
Q-2
2.1P-2 (a-r) Q Pr' + (a-x)'

=0

(a-x) which gives

qoQ
ms
2ax =

P + Q

* )= a

+

a-X

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600. Generally let two bodies P, P' connected by a string passing over a fixed pulley move by the action of gravity along two given curves, as in 590. Then adopting the notation of that article, the moving force upon Palong the curve which causes it to

ds' move with the velocity is evidently

dt

dx das' Px (9

ds' dt2

-).

Hence the tension of the string, or the moving force along the string is

ds'
o = P x

gdx'
ds' dt da'

-(1)
d's
gd.t

ds
or = P x

ds
dt

da

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+

Hence also the pressure on the polley being the resultant of these equal tensions, is (see Fig. 100) gdx d's ds

PSP a = 2P x

+ ds

dt da

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2

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The equations (1) will give the general result (a) of art. 590.

Let us apply them in the investigation of the tension and pressure for different systems. As the simplest case, suppose both P and P' to descend or ascend vertically.

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Again, let P' descend vertically, draw P up an inclined plane whose equation is

Y = AX. In this case, which is that of the problem to be resolved, we have also

ds = da' dx',
and therefore
o= P (9 - 0)

PSP
and a = 2P. (g q) cos.

2
u'du'
But d'E and by (13) art. 590, we easily get

ds'

PAX-B)+P (1-1) -..
P+P.

(A' +1) ?

{(A + 1)X - Am + a} since a=vim-Y) +(a-X)”.

PSP
Also cos. PSP =

= 2 cos.

}

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(2)

m-AX

1;

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whence in making the substitutions and necessary reductions we shall obtain

and If the pulley be at the top of the inclined plane, the calculation becomes much less tedious. For in that case we have

da' = da = ds' = ds Also m = b = Aa, and ::

a = (a – X). N(A' + 1)
:: {(A®+1) X - a (A + 1)} = (a – X). (A+ 1)*

= (A' + 1) PAP {-(AX-B)P+P". (a-X. VA’+1-1)} which gives the velocity at any given point of the descent. Moreover v'dv'

-9_(PA + P' A+ 1) dX P + P

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