Page:Great Neapolitan Earthquake of 1857.djvu/174

where $\omega$ is the angular velocity of the body at starting, $k$ the radius of gyration, with respect to the point or arris on which it turns, and $g$ the velocity acquired by a falling body in one second of time.

Equating these two values of the work done we find

$\omega ^{2}k^{2}=2ga(1-\cos {\phi })$

(I.)

but $\omega$, the angular velocity, is equal to the statical couple applied, divided by the moment of inertia, or

$\omega ={\frac {Va\cos {\phi }}{k^{2}}}$

squaring and substituting

$V^{2}=2g\times {\frac {k^{2}}{a}}\times {\frac {1-\cos {\phi }}{\cos ^{2}{\phi }}}$

and since the length of the corresponding simple pendulum is

$l={\frac {k^{2}}{a}}$

(II.)

$V^{2}=2gl\times {\frac {1-\cos {\phi }}{\cos ^{2}{\phi }}}$

In order to apply this formula to any given case we must determine the corresponding value of $l$, the simple pendulum applying to that case.

1st. In the case of a solid cube overturned (Fig. 94) whose side is $\alpha$;

$k^{2}={\frac {2\alpha ^{2}}{3}}$

$a={\frac {\alpha }{\sqrt {2}}}$