Page:Somerville Mechanism of the heavens.djvu/117

Chap II.] forces be added, the sum will be the whole pressure on the surface.

92. It appears that the centrifugal force is that part of the pressure which depends on velocity alone ; and when there are no accelerating forces it is the pressure itself.

fig. 25

93. It is very easy to show that in a circle, the centrifugal force is equal and contrary to the central force. Demonstration.—By article 63 a central force F combined with an impulse, causes a particle to describe an indefinitely small arc $$mA$$, fig. 25, in the time $$dt$$. As the sine may be taken for the tangent, the space described from the impulse alone

$aA = vdt$;

$(aA)^2 = 2r. am$,

$$r$$ being radius. But as the central force causes the particle to move through the space

$am = \frac{1}{2}F. dt^2$,

in the same time,

$\frac{v^2}{r} = F$.

94. If $$v$$ and $$v'$$ be the velocities of two bodies, moving in circles whose radii are $$r$$ and $$r'$$, their velocities are as the circumferences divided by the times of their revolutions; that is, directly as the space, and inversely as the time, since circular motion is uniform. But the radii are as their circumferences, hence

$v^2 : v'^2 :: \frac{r^2}{t^2} : \frac{r'^2}{t'^2}$,

$$t$$ and $$t'$$ being the times of revolution. If $$c$$ and $$c'$$ be the centrifugal forces of the two bodies, then

$c : c' :: \frac{v^2}{r} : \frac{v'^2}{r'}$,}}

or, substituting for $$v$$ and $$v'$$, we have