and reducing, we have,

49724453

tana =

6495

Hence, a = 4° 34' 10", and 85° 25′ 50′′. Ans.

Centripetal and Centrifugal Forces.

134. Curvilinear motion can only result from the action of an incessant force, whose direction differs from that of the original impulse. This force is called the deflecting force, and may arise from one or more active forces, or it may result from the resistance offered by a rigid body, as when a ball is compelled to run in a curved groove. Whatever may be the nature of the deflecting forces, we can always conceive them to be replaced by a single incessant force acting transversely to the path of the body. Let the deflecting force be resolved into two components, one normal to the path of the body, and the other tangential to it. The latter force will act to accelerate or retard the motion of the body, according to the direction of the deflecting force; the former alone is effective in changing the direction of the motion. The normal component is always directed towards the concave side of the curve, and is called the centripetal force. The body resists this force, by virtue of its inertia, and, from the law of inertia, the resistance must be equal and directly opposed to the centripetal force. This force of resistance is called the centrifugal force. Hence, we may define the centrifugal force to be the resistance which a body offers to a force which tends to deflect it from a rectilineal path. The centripetal and centrifugal forces taken together, are called central forces.

Measure of the Centrifugal Force.

135. To deduce an expression for the measure of the centrifugal force, let us first consider the case of a single material point, which is constrained to move in a circular

A

E

-T

path by a force constantly directed towards the centre, as when a solid body is confined by a string and whirled around a fixed point. In this case, the tangential component of the deflecting force is always 0. There will be no loss of velocity in consequence of a change of direction in the motion. (Art. 120). Hence, the motion of the point will be uniform. Let ABD represent the path of the body, and V its centre. Suppose the circumference of the circle to be a regular polygon, having an infinite number of sides, of which AB is one; and denote each of these sides by ds. When the body reaches A, it tends, by virtue of its inertia, to move in the direction of the tangent AT; but, in consequence of the action of the centripetal force directed towards V, it is constrained to describe the side ds in the time dt. If we draw BC parallel to AT, it will be perpendicular to the diameter AD, and AC will represent the space through which the body has been drawn from the tangent, in the time dt. If we denote the acceleration due to the centripetal force by f, and suppose it to be constant during the time đt, we shall have, from Art. 114,

AC=fdt

D

Fig. 119.

From a property of right-angled triangles, we have, since AB = ds,

Substituting this value of AC in (122), and solving with

ds2

But

dt2

v (Art. 113), in which v denotes the velocity

of the moving point. Substituting in the preceding equation, we have,

Here is the acceleration due to the deflecting force; and, since this is exactly equal to the centrifugal force, we have the acceleration due to the centrifugal force equal to the square of the velocity, divided by the radius of the circle.

If the mass of the body be denoted by M, and the entire centrifugal force by F, we shall have (Art. 24),

If we suppose the body to be moving on any curve whatever, we may, whilst it is passing over any two consecutive elements, regard it as moving on the arc of the osculatory circle to the curve which contains these elements; and, further, we may regard the velocity as. uniform during the infinitely small time required to describe these elements. The direction of the centrifugal force being normal to the curve, must pass through the centre of the osculatory circle. Hence, all the circumstances of motion are the same as before, and Equations (123) and (124) will be applicable, provided r be taken as the radius of the curvature. Hence, we may enunciate the law of the centrifugal force as follows:

The acceleration due to the centrifugal force is equal to the square of the velocity of the body divided by the radius of curvature.

The entire centrifugal force is equal to the acceleration, multiplied by the mass of the body.

In the case of a body whirled around a centre, and restrained by a string, the tension of the string, or the force

exerted to break it, will be measured by the centrifugal force. The radius remaining constant, the tension will increase as the square of the velocity.

Centrifugal Force at points of the Earth's Surface.

136. Let it be required to determine the centrifugal force at different points of the earth's surface, due to its rotation on its axis.

Suppose the earth spherical. Let A be any point on the surface, PQP' a meridian

F

P

r'

B

E

section through A, PP' the axis, FQ the equator, and AB perpendicular to PP', the radius of the parallel of latitude through A. Denote the radius of the earth by r, the radius of the parallel through by r', and the latitude of A, or the angle ACQ, by . The time of revolution being the same for surface, the velocities of Q and their distances from the axis. Denoting these velocities by v and v', we have,

P'
Fig. 120.

every point on the earth's A will be to each other as

But, from the right-angled triangle CAB, since the angle at A is equal to l, we have,

r' = r cosl.

Substituting this value of ' in the value of v', and reducing, we have,

v' = v cosl.

If we denote the acceleration due to the centrifugal force at the equator by f, we shall have, Equation (123),

In like manner, if we denote the acceleration due to the centrifugal force at A, by f', we shall have,

Substituting for v' and r' their values, previously deduced, we get,

Comparing Equations (125) and (126), we find,

ff' 1 cosl,

(126.)

That is, the centrifugal force at any point on the earth's. surface is equal to the centrifugal force at the equator, multiplied by the cosine of the latitude of the place.

Let AE, perpendicular to PP', represent the value of f', and resolve it into two components, one tangential, and the other normal to the meridian section. Prolong CA, and draw AD perpendicular to it at A. Complete the rectangle FD on AE as a diagonal. Then will AD represent the tangential, and AF the normal component of f'. In the right-angled triangle AFE, the angle at A is equal to 7. Hence,

From (128), we conclude that the tangential component is