where k is GS the distance of the centre of gravity from the axis of rotation, M the mass of the body, S the 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

SSM (SG' - GO3)

=SMX SO x SG + MX SO. GO

S" being the momentum of inertia referred to this new axis.
But S'M.SO.SG

or the pendulum will oscillate in the same time as before.

597.

Let r be the radius of the circle, m the distance of

either body from the axis of rotation; then

598. required,

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

599.

gt2

am2 m2 - gt2

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

x

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

Again, the moving force which generates P's velocity is

and that part of P which is sustained, or the tension of the string

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 P along the curve which causes it to

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

Hence also the pressure on the pulley being the resultant of these equal tensions, is (see Fig. 100)

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.

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

v'dv'
ds'

v'2 = 2g

But '=- and by (13) art. 590, we easily get

whence in making the substitutions and necessary reductions we shall obtain

and w.

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

.. { (A2+ 1) X − a (A + 1)}2 — (a − X)2. (A2 + 1)2

= (A3 + 1) x®

:: v22 = - {−(AX−ß)P+P'. (a−X . √ A2 + 1−1)}

2g P+P

which gives the velocity at any given point of the descent.