scribed are principally used for astronomical clocks, where great accuracy and great uniformity in the measure of time is indispensable. Basis of a system of Weights and Measures. 128. The pendulum is of further importance, in a practical point of view, in furnishing the standard of comparison which has been made use of as a basis of the English system of weights and measures. The length of the seconds pendulum at any place, can always be found, and it must always be the same at that place. We have seen that this length was determined, with great accuracy, in the Tower of London, to be 3.2616 ft. It has been decreed by the British Government, that the th part of the length of the simple seconds pendulum, in the Tower of London, shall be regarded as a standard foot. From this, by multiplication and division, every other unit of lineal measure may be derived. By constructing squares and cubes upon, the linear units, we at once arrive at the units of area and of volume. 16 It has further been decreed, that a cubic foot of distilled water, at the temperature of maximum density, shall be regarded as weighing 1000 standard ounces. This fixes the ounce; and by multiplication and division, all other units of weight may be derived. This system enables us to refer to the original standard, when, from any circumstances, doubt may exist as to the accuracy of standard measures. Even should every vestige of a standard be swept from existence, they might be perfectly restored, by the process above indicated. The American system of weights and measures is adopted from that of Great Britain, and is, in all respects, the same as that above described. EXAMPLES. 1. The length of a seconds pendulum is 39.13921 in. If it be shortened 0.130464 in., how many vibrations will be gained in a day of 24 hours? SOLUTION. The times of vibration of two pendulums at the same place, are to each other as the square roots of their lengths (Art. 97). Hence, the number of vibrations made in any given time, are inversely proportional to the square roots of their lengths. If, therefore, we denote the number of vibrations gained in 24 hours, or 86400 seconds, by x, we shall have, 86400 86400 + x:: √39.008747 : or, 86400 86400+x:: 6.2457: 6.2561. Whence, x = 144, nearly. Ans. 39.13921; 2. A seconds pendulum being carried to the top of a mountain, was observed to lose 5 vibrations per day of 86400 seconds. Required the height of the mountain, reckoning the radius of the earth at 4000 miles. SOLUTION. The times of vibration of the same pendulum, at any two points, are inversely proportional to the forces of gravity at those points (Art. 98). But the forces of gravity at the same points are inversely as the squares of their distances from the centre of the earth. Hence, the times of vibration are proportional to the distances of the points from the centre of the earth; and, consequently, the number of vibrations in any given time, as 24 hours, for example, will be inversely as those distances. If, therefore, we denote the height of the mountain in miles by x, we shall have, 86400 86405: 4000 4000 + x. Whence, x= 20000 = 0.2315 miles, or, 1222 feet. Ans. 86400 3. What is the time of vibration of a pendulum whose length is 60 inches, when the force of gravity is reckoned at 32 ft? Ans. 1.2387 sec. 4. How many vibrations will a pendulum 36 inches in length make in one minute, the force of gravity being the same as before? Ans. 62.53. 5. A pendulum is found to make 43170 vibrations in 12 hours. How much must it be shortened that it may beat seconds? SOLUTION. We shall have, as in Example 1st, 43170 43200 :: √39.13921 : /39.13921 + x. Whence, x 0.0548 in. Ans. 6. In a certain latitude, the length of a pendulum vibrating seconds is 39 inches. What is the length of a pendulum vibrating seconds, in the same latitude, at the height of 21000 feet above the first station, the radius of the earth being 3960 miles? Ans. 38.9195 in. 7. If a pendulum make 40000 vibrations in 6 hours, at the level of the sea, how many vibrations will it make in the same time, at an elevation of 10560 feet above the same point, the radius of the earth being 3960 miles? Ans. 39979.8. Centre of Percussion. a 129. The point O, Fig. 108, is a point at which, if the entire mass were concentrated, and the impressed forces applied to it, the effect produced would be in nowise different from what actually obtains. Were an impulse applied at this point, capable of generating a quantity of motion equal and directly opposed to the resultant of all the quantities of motion of the particles of the body, at any instant, the body would evidently be brought to a state of rest without imparting any shock to the axis of suspension. The direction of the impulse remaining the same as before, Fig. 108. no matter what may be its intensity, there will still be no shock on the axis. This point is, therefore, called the centre of percussion. We may then define the centre of percussion to be that point of a body restrained by an axis, at which, if the body be struck in a direction perpendicular to a plane passed through this point and the axis of suspension, no shock will be imparted to the axis. It is a matter of common observation that, if a rod held in the hand be struck at a certain point, the hand will not feel the blow, but if it be struck at any other point. of its length, there will be a shock felt, the intensity of which will depend upon the intensity of the blow, and upon the distance of its point of application from the first point. Moment of Inertia. 130. The MOMENT OF INERTIA of a body with respect to an axis, is the algebraic sum of the products obtained by multiplying the mass of each elementary particle by the square of its distance from the axis. Denoting the moment of inertia with respect to any axis, by K, the mass of any element of the body, by m, and its distance from the axis, by r, we have, from the definition, The moment of inertia evidently varies, in the same body, according to the position of the axis. To investigate the law of variation, let AB represent any section of the body by a plane perpendicular to the axis; C, the point in which this plane cuts the axis; and G, the point in which it cuts a parallel axis through the centre of gravity. Let P be any element of the body, whose mass is m, and denote PC by r, PG by s, and CG by k. CH Fig. 109. From the triangle CPG, according to a principle of Trigonometry, we have, Substituting in (104), and separating the terms, we have, K = 2(ms2) + (mk2) — 22(mskcos CGP). Or, since is constant, and (m) = M, the mass of the K = 2(ms2) + M2 2k (mscos CGP). But scos CGP = GH, the lever arm of the mass m, with respect to the axis through the centre of gravity. Hence, (ms cos CGP), is the algebraic sum of the moments of all the particles of the body with respect to the axis through the centre of gravity; but from the principle of moments, this is equal to 0. Hence, The first term of the second member of (105), is the expression for the moment of inertia, with respect to the axis through the centre of gravity. Hence, the moment of inertia of a body with respect to any axis, is equal to the moment of inertia with respect to a parallel axis through the centre of gravity, plus the mass of the body into the square of the distance between the two axes. The moment of inertia is, therefore, the least possible, when the axis passes through the centre of gravity. If any number of parallel axes be taken at equal distances from the centre of gravity, the moment of inertia with respect to each, will be the same. The moment of inertia of a body with respect to any axis, may be determined experimentally as follows. Make the axis horizontal, and allow the body to vibrate about it, as a compound pendulum. Find the time of a single vibration, and denote it by t. This value of t, in Equation (95), makes known the value of 7. Determine the centre of gravity by some one of the methods given, and denote its |