« PreviousContinue »
ing 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.
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
r being radius. But as the central force causes the particle to move through the space
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
t and t' being the times of revolution. If c and c' be the centrifugal forces of the two bodies, then
or, substituting for v2 and v2, we have
Thus the centrifugal forces are as the radii divided by the the times of revolution. 95. With regard to the Earth the times of rotation are everywhere the same; hence the centrifugal forces, in different latitudes, are as the radii of these parallels. These elegant theorems discovered by Huygens, led Newton to the general theory of motion in curves, and to the law of universal gravitation.
Motion of Projectiles.
96. From the general equation of motion is also derived the motion of projectiles.
Gravitation affords a perpetual example of a continued force; its influence on matter is the same whether at rest or in motion; it penetrates its most intimate recesses, and were it not for the resistance of the air, it would cause all bodies to fall with the same velocity: it is exerted at the greatest heights to which man has been able to ascend, and in the most profound depths to which he has penetrated. Its direction is perpendicular to the horizon, and therefore varies for every point on the earth's surface; but in the motion of projectiles it may be assumed to act in parallel straight lines; for, any curves that projectiles could describe on the earth may be esteemed as nothing in comparison of its circumference.
The mean radius of the earth is about 4000 miles, and MM. Biot and Gay Lussac ascended in a balloon to the height of about four miles, which is the greatest elevation that has been attained, but even that is only the 1000th part of the radius.
The power of gravitation at or near the earth's surface may, without sensible error, be considered as a uniform force; for the decrease of gravitation, inversely as the square of the distance, is hardly perceptible at any height within our reach.
97. Demonstration.-If a particle be projected in a straight line MT, fig. 26, forming any angle whatever with the horizon, it will constantly deviate from the direction MT by the action of the gravitating force, and will describe a curve MN, which is concave towards the horizon, and to which MT is tangent at M. On this particle there
cording to the perpendiculars Ed, Cf, &c. The resistance of the air may be resolved into three partial forces, in the direction of the three axes ox, oy, oz, but gravitation acts on the particle in the direction of oz alone. If A represents the resistance of the air, its
component force in the axis or is evidently A
ds be the space proportional to the resistance, then
Am: Ec :: A:A = A
but as this force acts in a direction contrary to the motion of the particle, it must be taken with a negative sign. The resistance in
gravitation, the forces acting on the particle are
X = − A dr; Y = − A dy; Z=g-Adz.
As the particle is free, each of the virtual velocities is zero; hence
for the determination of the motion of the projectile. If A be eli
minated between the two first, it appears that
dx = Cdy, and if we integrate a second time,
x = Cy + D,
in which C and D are the constant quantities introduced by double integration. As this is the equation to a straight line, it follows that the projection of the curve in which the body moves on the plane roy is a straight line, consequently the curve MN is in the plane zor, that is at right angles to roy; thus MN is a plane curve, and the motion of the projectile is in a plane at right angles to the horizon. Since the projection of MN on roy is the straight line ED, therefore y = 0, and the equation = А A dy is of no
use in the solution of the problem, there being no motion in the direction oy. Theoretical reasons, confirmed to a certain extent by experience, show that the resistance of the air supposed of uniform density is proportional to the square of the velocity;
h being a quantity that varies with the density, and is constant when it is uniform; thus the general equations become
C being an arbitrary constant quantity, and c the number whose hyperbolic logarithm is unity.
In order to integrate the second, let dz = uda, u being a function of z; then the differential according to t gives
If this be put in the second of equations (a), it becomes, in conse
or, eliminating de by means of the preceding integral, and making
The integral of this equation will give u in functions of x, and when substituted in
dz = udr,
it will furnish a new equation of the first order between z, x, and t, which will be the differential equation of the trajectory.
If the resistance of the medium be zero, h 0, and the preceding equation gives
u = 2ax + b,
for u, and integrating again
z = ax2 + bx + b'
b and b' being arbitrary constant quantities. This is the equation to a parabola whose axis is vertical, which is the curve a projectile would describe in vacuo. When
h = 0, diz gdť2;
and as the second differential of the preceding integral gives
If the particle begins to move from the origin of the co-ordinates, the time as well as x, y, z, are estimated from that point; hence b' and the two equations of motion become
and a' are zero,