If the web of the beam is divided into areas in. in height (instead of į in.), the value of I obtained will be 23.46 in. If the section is considered to be of the form indi. cated by the dotted lines in Fig. 1, and to have the same area as the original section, then, by the formula for the moment of inertia of an I-beam given in Table V, page 153, the value of 3.50 X 63 — 3.27 X 5.253 I= = 23.57. 12 The true value is almost exactly 23.48 in. Any one of these values would be sufficiently correct for most practical purposes. If it is desired to find the moment of inertia of a body about a given axis with reference to the weight of the body, the process is substantially the same as in the example given for the plane section, except that the weight of each small part of the FIG. 2. body is taken instead of the area of each small part of the section. 0.125 CENTER OF OSCILLATION. The center of oscillation of a pendulum or other body vibrating or rotating about a fixed axis or center is that point at which, if the entire weight of the body were concentrated, the body would continue to vibrate in the same intervals of time. When a pendulum, or other suspended body, is oscillating backward and forward, it is plain that those particles that are farther from the point of suspension travel through greater distances, and therefore move with greater velocities than those particles that are nearer the point of suspension. But there is evidently some point on the pendulum that travels through the same distance and has the same velocity as the average distance and average velocity of all the particles. This point is called the center of oscillation; it is not situated at the center of gravity. It always exists in the ball of a revolving governor or other rotating body. The axis or center around which the body rotates (corresponding to the point of suspension in pendulum) is the axis of rotation. The distance from the axis, or center of rotation, to the center of oscillation is sometimes called the true length of the pendulum; it is also called the radius of oscillation; the latter name is preferable. To find the radius of oscillation: Divide the moment of inertia of the body about the given axis of rotation by the product of the total weight of the body, multiplied by the distance from the given axis to the center of gravity of the body. The centers of oscillation and of rotation (point of suspension) are interchangeable. If the position of a pendulum is reversed, and suspended from its center of oscillation, the pendulum will vibrate in the same intervals of time. EXAMPLE.-It is desired to find the position of the center of oscillation of a wrought-iron bar 1 in. square and 12 in. long, axis of rotation perpendicular to the bar at one end: Weight of Each Sq. of Dist. .281 x 0.52 0.070 161.572 I SOLUTION.- For the purposes of the example it will be sufficiently accurate to find the moment of inertia by considering the bar to be divided into 12 equal cubes, each containing 1 cu. in. of metal, as indicated in the figure, and the weight of each cube to be concentrated at its center of gravity. The weight of 1 cu. in. of wrought iron is .281 lb., and of a bar 1 in. square and 1 ft. long it is .281 x 12 = 3.372 lb. Hence, I = .281 X .52 + .281 X 1.52 + etc. = 161.572. (See page 128.) The exact value of I is 161.856; this shows that the approximate method is very close. According to the rule previously given, if the moment of inertia is divided by the product of the weight of the body, by the distance from the axis of rotation to the center of gravity, the quotient will be the radius of oscillation. Therefore, the distance from the exact center of oscillation of a wrought-iron bar, 1 in. square and 12 in. long, to an axis of rotation perpendicular to the end of the bar, is 161.856 8 in., 3.372 X 6 or two-thirds of the length of the bar. The value of I for a bar of any cross-section, provided it is uniform throughout its length, revolving about an axis perpendicular to it and passing through its end, is W 12 3 in which W is the weight of the bar, and I is its length. W 12 3.372 X 122 Hence, I = 161.856. 3 3 If the axis passes through the center of gravity of the bar, W 12 12 CENTER OF PERCUSSION. The center of percussion with respect to a given axis of rotation may be defined as the point of application of the resultant of the forces that cause the body to rotate. It is that point at which if a force is applied, the force will have no effect at the axis of rotation. Strike anything solid, as an anvil, with a stick. If the end of the stick hits the anvil, the opposite end will sting your hand and will jerk in the direction in which the blow is struck; if the center of the stick hits the anvil it will again sting your hand, but you will jerk it in a direction opposite to the movement of the blow. But somewhere between the end and the center of the stick will be a point where it may hit the anvil and not sting your hand at all. This point is the center of percussion. Level off the surface of some wet sand and lay a strip of board upon it (say 18 in. long and 3 in. wide). Strike or press the board near the center and the entire length of the board will be imprinted in the sand; but press it near one end and the opposite end will be raised up from the sand and will make no imprint. Between the center and the end of the board is a point that if pressed upon will cause no movement in the opposite end, i. e., the end of the board will neither press into the sand nor be lifted from it, but the imprint in the sand will diminish to zero at the end of the board. The point pressed or struck will be the center of percussion. If the board is of uniform width, the center of percussion will be at one-third of the distance from one end of the board. Similarly in the preceding illustration, if the stick is of uni. form size and weight, and your hand grasps it at one end, the point at which it can strike the anvil without affecting your hand will be at one-third the distance from the opposite end. In all cases the center of percussion is identical with the center of oscillation, and its position is found in the same manner. EXAMPLE.- It is desired to find the position of the center of oscillation or percussion of two balls fastened upon a rod. The first, weighing 2 lb., is at a distance of 18 in. from the axis of rotation, and the second, weighing 1 lb., is at a distance of 16 36 in. from the axis. (See figure.) SOLUTION.-For simplicity, the rod will be assumed to have no weight. Consider the weight of each ball to be concentrated at its center of gravity. The moment of inertia is found as follows. Sq. of 1,944 I. The center of gravity of the two balls is found to be at a distance of 6 in. from the larger, or 24 in. from the axis of rotation (see page 124), and the combined weight of the two balls is 2 +1= 3 lb. Therefore, the center of percussion is found to be at a distance of 1,944 = 27 in. from the axis of rotation. 3 X 24 But, in an actual case, the rod would have weight, and its moment of inertia must be considered as well as the moment of inertia of the balls. If we assume that the rod is of steel, f in. in diameter and 32 X.7854 X 36 X .283 = 1.125 lb. .283 lb. is the weight of 1 cu. in. of steel. Using the formula given on page 129, W 12 1.125 X 362 486. 3 3 Adding this result to the former, 1,944 + 486 = 2,430 = moment of inertia of rods and balls. The center of gravity of the combination is found by the formula (see page 124) Pa 1.125 X 6 P + W Substituting, 1.125 + 3 111. 24 – 1/1 distance from end of rod to center of gravity. Applying the rule given for finding the center of oscillation, the distance of the center of percussion from the end of the bar is 2,430 26.34 in., very nearly. (1 + 2 + 1.125) X 2211 36 in. long, it will weigh ( I= 221 in. RADIUS OF GYRATION. The center of gyration is that point in a revolving body at which, if the entire mass of the body were concentrated, the moment of inertia with respect to a given axis would be the same as in the body. An ounce of cork occupies about 94 times as much space as |