Page:Somerville Mechanism of the heavens.djvu/112

Hence the centrifugal force of a particle revolving about a centre, is equal to the square of its velocity divided by the radius.

89. The plane of the osculating circle, or the plane that passes through two consecutive and indefinitely small sides of the curve described by the particle, is perpendicular to the surface on which the particle moves. And the curve described by the particle is the shortest line that can be drawn between any two points of the surface, consequently this singular law in the motion of a particle on a surface depends on the principle of least action. With regard to the Earth, this curve drawn from point to point on its surface is called a perpendicular to the meridian; such are the lines which have been measured both in France and England, in order to ascertain the true figure of the globe.

90. It appears that when there are no constant or accelerating forces, the pressure of a particle on any point of a curved surface is equal to the square of the velocity divided by the radius of curvature at that point. If to this the pressure due to the accelerating forces be added, the whole pressure of the particle on the surface will be obtained, when the velocity is variable.

91. If the particle moves on a surface, the pressure due to the centrifugal force will be equal to what it would exert against the curve it describes resolved in the direction of the normal to the surface in that point; that is, it will be equal to the square of the velocity divided by the radius of the osculating circle, and multiplied by the sine of the angle that the plane of that circle makes with the tangent plane to the surface.

fig. 24.

Let $MN$, fig. 24, be the path of a particle on the surface; $mo$ the radius of the osculating circle at $m$, and $mD$ a tangent to the surface at $m$; then $om$ being radius, $oD$ is the sine of the inclination of the plane of the osculating circle on the plane that is tangent to the surface at $m$, the centrifugal force is equal to

${\frac {v^{2}\times o\mathrm {D} }{om}}.$

If to this, the part of the pressure which is due to the accelerat-