Substituting for ƒ its value, taken from (132), we have,

But, from the triangle ABC, we have, lcosph, which

That is, the time of a revolution is equal to the time of a double vibration of a pendulum whose length is h.

The Governor.

142. The principle of the conical pendulum is employed in the governor, a machine attached to engines, to regulate the motive force.

H

K

F

AB is a vertical axis connected with the machine near its working point, and revolving with a velocity proportional to that of the working point; FE and GD are two arms turning freely about AB, and bearing heavy balls D and E, at their extremities; these bars are united by hinge-joints with two other bars at G and F, which latter bars are attached to a ring at I, free to slide up and down the shaft.

(D

Fig. 125.

E

The governor is so constructed, that the figure GCFH is always a parallelogram. The ring at His connected with a lever HK, which may be made to act upon the valve that admits steam to the cylinder.

When the shaft revolves, the centrifugal force developed in the balls, causes them to recede from the axis, and the ring H is depressed; and when the velocity has become sufficiently great, the lever begins to act in closing the valve. If the velocity slackens, the balls approach the axis, and the ring H ascends, opening the valve again. In any given case, if we know the velocity required at the working point, we can from it compute the required angular velocity of the shaft, and, consequently, the value of t. This value of t being substituted in Equation (135), makes known the value of h. We may, therefore, make the proper adaptation of the ring, and of the lever HK.

EXAMPLES.

1. A ball weighing 10 lbs. is whirled around in a circle whose radius is 10 feet, with a velocity of 30 feet per second. What is the acceleration of the centrifugal force?

Ans. 90 ft.

2. In the preceding example, what is the tension upon the cord which restrains the ball?

SOLUTION.

Denote the tension in pounds, by t; then, since the pressures produced by two forces are proportional to their accelerations, we shall have,

10 t 9: 90,

.*. t = 28 lbs., nearly. Ans.

3. A body is whirled around in a circular path whose radius is 5 feet, and it is observed that the pressure due to the centrifugal force is just equal to the weight of the body. What is the velocity of the moving body?

SOLUTION.

Denoting the velocity by v, we have the acceleration

due to the centrifugal force equal to

v2

; but, by the condi

5

tions of the problem, this is equal to the acceleration due to the weight of the body. Hence,

g= 32

.. v = 12.9 ft. Ans.

5

4. In how many seconds must the earth revolve on its axis in order that the centrifugal force at the equator may exactly counterbalance the force of gravity, the radius of the equator being taken equal to 3962.8 miles?

SOLUTION.

Reducing the miles to feet, and denoting the required velocity, by v, we have,

v2 20923584

= 321,

v= √321 × 20923584.

But the time of revolution is equal to the circumference of the equator, divided by the velocity. Denoting the time by t, we have,

and, substituting the value of v, taken from the preceding equation, we have, after reduction,

But the earth actually revolves in 86400 sideral, or in about 86164 mean solar seconds. Hence, the earth would have to revolve 17 times as fast as at present, in order that the centrifugal force at the equator might be equal to the force of gravity.

5. A body is placed on a horizontal plane, which is made to revolve about a vertical axis, with an angular

velocity of 2 feet. How far must the body be situated from the axis that it may be on the point of sliding outwards, the coefficient of friction between the body and plane being equal to .6?

SOLUTION.

Denote the required distance by r; then will the velocity of the body be equal to 2r, and the acceleration due to the centrifugal force will be equal to 4r. But the acceleration due to the force of friction is equal to 0.6 × g = 19.3 ft. From the conditions of the problem, these two are equal, hence,

4r 19.3 ft.,

r = 4.825 ft. Ans.

6. What must be the elevation of the outer rail of a railroad track, the radius of curvature being 3960 ft., the distance between the rails 5 feet, and the velocity of the car 30 miles per hour, in order that the centrifugal force may be exactly counterbalanced by the component of the weight parallel to the line joining the rails?

Ans. 0.076 ft., or 0.9 in., nearly.

7. The distance between the rails is 5 feet, the radius of the curve 600 feet, and the height of the centre of gravity of the car 5 feet. What velocity must be given to the car that it may be on the point of being overturned by the centrifugal force, the rails being on the same level?

143. By the term work, in mechanics, is meant the effect produced by a force in overcoming a resistance, such as weight, inertia, &c. The idea of work implies that a force is continually exerted, and that the point at which it is applied moves through a certain space. Thus, when a weight is raised through a vertical height, the power which

overcomes the resistance offered by the weight is said to work, and the amount of work performed evidently depends, first, upon the weight raised, and, secondly, upon the height through which it is raised. All kinds of work may be assimilated to the raising of a weight. Hence it is, that this kind of work is assumed as a standard to which all other kinds of work are referred.

The unit of work most generally adopted in this country, is the effort required to raise one pound through a height of one foot. The number of units of work required to raise any weight to any height will, therefore, be equal to the product obtained by multiplying the number of pounds in the weight by the number of feet in the height. If we take the weight of the body as it would be at the equator, for the sake of uniformity in notation, we may regard the weight and the mass as identical (Art. 11). If we denote the quantity of work expended in raising a body, by Q, the mass of the body, by m, and the height, by h, we shall have,

Q mh.

When very large quantities of work are to be estimated, as in the case of steam-engines and other powerful machines, a different unit is sometimes employed, called a horse power. When this unit is employed, time enters as an element. A horse power is a power which is capable of raising 33,000 lbs. through a height of one foot in one minute; that is, it is a power capable of performing 33,000 units of work in a minute of time, or 550 units of work in one second. When an engine, then, is spoken of as being of 100 horse power, it is to be understood that it is capable of performing 55,000 units of work in a second.

In general, if a force acts to overcome a resistance of m pounds, through a distance of n feet, whatever may be the cause of the resistance, or whatever may be the direction of the motion, the quantity of work will be measured by a unit of work taken mn times.