mass m. Then the length of each part will be and if the distance of n any part from the axis is taken as the distance of its farther end, the Now, it may be shown that the larger n is taken the more closely does n3 the sum in the parenthesis approach the value and accordingly if the rod is supposed to be divided into an infinite number of parts, The moment of inertia of the bar is therefore the same as though its mass were concentrated at a distance k from the axis, where k2 12 3. The distance k is known as the radius of gyration of the rod about the given axis. 139. Formulas for Moment of Inertia.-In case of bodies of simple figure and having the mass uniformly distributed throughout the volume the moments of inertia may be calculated by the methods of calculus. But in more complicated cases they must be determined by experiment. The following formulas are given for reference: Thin rod, of mass M and length l, having a transverse axis at one end M 12 I = 12 Thin rod, of length 1, having a transverse axis through the center, Rectangular block, of width a and length b and of any thickness whatever, about an axis through the center perpendicular to a and b, Circular disc or cylinder, of any length and of radius r, about an axis through the center and perpendicular to the circular section of the Circular cylinder, of length and radius r, about a transverse axis through its center perpendicular to its length, Sphere, of radius r about an axis through its center, 140. Moment of Inertia about a Parallel Axis.-If the moment of inertia of a body is known about an axis through its center of mass, it may readily be calculated about any parallel axis. For if I is the moment of inertia about the axis through its center of mass and if M is the mass of the body, then the moment of inertia about a parallel axis at a distance h from the center of mass of the body is I = 1, + Mh2; that is, the moment of inertia I about any axis is equal to the moment of inertia which the whole mass would have about that axis if it were concentrated at the center of mass of the body, added to the moment of inertia of the body about the parallel axis through its center of mass. 141. The Compound Pendulum.-In discussing the simple pendulum it was assumed that the oscillating mass was so small that it might be considered as concentrated at a point, and the mass of the suspending system was entirely neglected. A pendulum which has distributed mass and so does not satisfy either of the above simple conditions is said to be a compound or physical pendulum. All actual pendulums belong to this class. Let it be required to find the length of a simple pendulum having the same period of oscillation as a given physical pendulum. Suppose the pendulum to have mass M and let its axis of suspension O be a distance h above its center of gravity C (Fig. 70). Then, when a line joining O and C Mg FIG. 70. m mg makes an angle a with the vertical, the pendulum may be considered as acted upon by a force Mg acting downward through its center of gravity and producing a moment of force about the axis O equal to Mgd or Mgh sin a. If I is the moment of inertia of the pendulum about 0, we have by equation (1) $135, A Mgh sin a But in case of a simple pendulum of length 7 the moment of the force mg about the axis O' is mgl sin a about O' is m l2; therefore mgl sin a tion is, A'= = and the moment of inertia of m =3 ml'A' and the angular accelera g sin a If the two pendulums are to have the same period of vibration their angular accelerations A and A' must be equal when both pendulums make equal angles with the vertical; that is, The length of the equivalent simple pendulum calculated from the above formula will always be greater than h, since the moment of inertia I of the pendulum is always greater than if the whole mass were concentrated at its center of gravity (see §140); that is, I is greater than Mh2 and, consequently, I is greater than h. The point P in line with O and C and at a distance l from O is called the center of oscillation. Each portion of the mass of the pendulum between 2 FIG. 71. P and O is constrained to swing slower than it would if it were free to oscillate by itself about O as a center, while all portions of the pendulum below P have to swing more quickly than if they were free. The mass between P and O therefore tends to quicken the motion of the pendulum while the mass below P tends to retard it, while the mass situated at P is neither hastened nor retarded, but swings exactly as it would if freely suspended from 0. 142. Center of Percussion. If a rod or pendulum is suspended from an axis A (Fig. 71) and if that axis is given a sudden sidewise impulse or if it is moved rapidly back and forth from side to side, the inertia of the rod will cause it to move as though a certain point B was fixed and the rod turned about that point as axis. This instantaneous center of the motion is not the center of gravity C, but is the center of oscillation corresponding to the axis of suspension at A. A marble placed on a little shelf at B is scarcely disturbed by the sudden to-and-fro movements of the axis A, while at any other point it would be instantly thrown off. On the other hand, when the pendulum suspended from the axis A is hanging at rest, if a sudden sidewise impulse is given to the bar at B. as when it is struck a blow at that point, no sidewise impulse is communicated to A in consequence, but the bar simply tends to turn about A as an axis. For this reason the point B is also called the center of percussion corresponding to the axis A. In case of a baseball bat the blow is given to the ball with the least jar to the hands when the ball is struck at the center of percussion of the bat corresponding to an axis at the point where it is grasped. Problems. 1. A cylinder weighing 30 kilos and having a diameter of 1 meter is mounted on an axis and set rotating by a pull of 2 kilograms on a cord wound on an axle 10 cm. in radius. Find the acceleration produced and the speed of rotation 3 sec. from the time of starting. The moment of inertia of a cylinder about its axis is M (§139) M” ($139) 2 or I 30 X 1000 × 502 2 =37,500,000 grm. cm. The force acting is 2 kilos or 2000 grm. or 2000×980 dynes and the moment of the force is 2000 ×980 × 10 dyne-cm. Substitute in the formula L = I A 2000×980 × 10 = 37,500,000 A .. A = 0.523. Hence the system will gain in 1 sec. an angular velocity of a little more than half a radian per sec. = In 3 sec. it will acquire an angular velocity @= ЗА 1.569; where P is the period of revolution. 2. What is the kinetic energy of a wheel which has a moment of inertia 20 ft. lbs. and is rotating at the rate of two turns per sec.? 3. If a 5-lb. weight is raised by means of a rope wound on the axle of the wheel in problem 2, how high will it be raised before the wheel comes to rest? 4. A uniform rod 40 cms. long and weighing 200 grm. can rotate about a transverse axis through its middle point. How many ergs of work will be required to make it revolve at the rate of three turns per sec.? 5. Suppose the rod in question 4 is set in rotation by means of a 200grm. weight attached to a cord wrapped around a cylindrical axle 4 cms. in diameter. How far will the weight have descended in giving a speed of rotation of 3 revolutions per sec. (See Note, over.) Note. First solve neglecting the kinetic energy acquired by the 200-grm. weight as it sinks. Then obtain the more exact solution, taking account of this energy. 6. The fly wheel of an engine weighs 1200 lbs., the bulk of the weight being in the rim of the wheel at a distance of about 3 ft. from the axis. What is approximately its moment of inertia and how many ft. lbs. of work must be done by the engine to set it rotating 3 times per sec.? 7. How much energy will be given out by the fly wheel in problem 6 in slowing down from 3 to 2.5 revolutions per sec. 8. A uniform bar 3 ft. long swings as a pendulum about an axis at one end. Show that the equivalent simple pendulum is 2 ft. long. 9. A uniform spherical steel ball 6 cm. in diameter is hung as a pendulum by a steel wire so that the center of the ball is just 100 cm. below the axis of suspension. Find how far the center of oscillation is below the center of the ball and what is the length of the equivalent simple pendulum, neglecting the mass of the suspending wire. 10. A rectangular bar of steel 1 x 1 x 12 cm. and weighing 90 grm., when suspended in a horizontal position by a wire attached to its middle point, is set oscillating about a vertical axis through its center and makes 4 complete vibrations in 10 sec. Find the moment of force or torque due to the twist in the wire when the bar is at right angles to its equilibrium position. 11. Find the period of oscillation of a solid metal sphere 6 cm. in diameter and weighing 800 grm. when hung by the same wire as the bar in problem 10 and set oscillating about a vertical axis through its center. SOME CASES OF MOTION WITH PARTLY FREE AXIS. 143. Foucault's Pendulum Experiment.-It occurred to the French physicist Foucault that since a pendulum undisturbed by external forces must persist in its original direction of vibration, if one were swung at the north pole by some suspension which could not transmit torsion, its direction of vibration would remain constant while the earth turned around under it, so that to an observer moving with the earth the pendulum would seem to change its direction of vibration at the rate of 15° per hour. At the equator the direction of the meridian remains parallel to itself as the earth rotates, and consequently the plane of vibration of the pendulum would remain unchanged. |