velocity, viz. v-i.e. the change in the velocity of the body between a and the horizontal line BB' is irrespective of the path it describes. In explanation of this remarkable fact it may be observed, that at every instant the reaction is perpendicular to the direction in which the body is moving, and therefore cannot accelerate the velocity. The same formula is true of a body suspended by thread, and oscillating; for the tension of the thread will act at each point perpendicularly to the direction of the body's motion, and will neither accelerate nor retard the velocity. Ex. 630. A stone is tied to the end of a string 10 ft. long and describes a vertical circle of which the string is the radius; if at the highest point it is moving at the rate of 25 ft. per second, find its velocity after describing angles of 90°, 180°, and 270° respectively from the highest point. Ans. (1) 35.6. (2) 43.6. (3) 35.6. Ex. 631. Show that if a body oscillate in any are of a circle, the arc of ascent would always equal the arc of descent if there were no passive resistances. Ex. 632. A body is tied to the end of a string 12 ft. long, the other end of which is fastened to a point a; at a distance of 4 ft. vertically below a is a peg B; the body descends through an angle of 30° when the string comes to the peg B; find the angle through which the body will rise. Ans. 36° 58′. Ex. 633. Suppose a body to move in a circle whose radius is rand lowest point ; let v be the velocity it has at a point p and v that which it has at Q; let the chords AP and AQ be denoted by c and c respectively; show that Ex. 634. If a body slide down an arc of a vertical circle to the lowest point of the circle, show that the velocity at that point is proportional to the chord of the arc described. 121. Centrifugal Force. - If a stone is tied to the end of a string and whirled round, there arises a very peculiar case of the action of forces, and one which requires careful consideration. Suppose the string to be r feet long, the stone to weigh w lbs., and to move with a velocity of v feet per second; now, the tendency of the stone at each instant is to move off in the direction of a tangent to the circle it describes, therefore there must be exerted on it at each instant a certain force P (acting along the radius and towards the centre) sufficient to deflect it from the tangent and to keep it in the circle; this force is given in lbs. by the formula where g denotes 32-1912 or approximately 32. In the case supposed this force is supplied by the hand, and gives rise to the same sensation as would be produced if the stone were at rest and pulled outward with a force of P lbs. It must be added, that when any heavy body moves in a circle under the action of any forces whatever, the sum of the resolved parts of the forces along the radius must at each instant equal W. v tinue to move in the circle. g r or the body will not con We have already seen (Art. 118) that if a body whose weight is w lbs. is acted on by a force of P lbs., it would acquire at the end of every second an additional velocity f equal to g; in the present case therefore P The acceleration f is frequently spoken of as the 'centrifugal force.' Ex. 635. A weight of 1 lb. is fastened to the end of a string 3 ft. long and made to perform 50 revolutions in 1 min. with a uniform velocity; the revolutions take place in a horizontal plane: determine the tension of the string. Ans. 2.57 lbs. Ex. 636.-In Ex. 630 determine the tension of the string at the highest and at the other points, supposing the body to weigh 10 lbs. Ans. (1) 9.53 lbs. (2) 39.53 lbs. (3) 69.53 lbs. (4) 39.53 lbs. Ex. 637.- If a body moves in a vertical circle the radius of which is 5 ft. determine the velocity at the highest point that the body may just keep in the circle. Ans. 12.65. [Let T be the tension of the string, then T+w= W will just keep in the circle if T-0. If were less than w the body g would fall within the circle; if it were greater than w there would be a certain tension on the string.] Ex. 638. In the last Example show that the tension of the string at the lowest point will equal 6 times the weight of the body; and that when the body has described a quadrant from the highest point the tension is 3 times the weight of the body. Ex. 639. Show that the centrifugal force at the equator equals 0-11129 or the part of what the accelerating force of gravity would be if the earth were at rest. [See Ex. 569 and Table XV.] Ex. 640. How many revolutions would the earth have to make in 24 hours, if bodies would just stay on her surface at the equator? Ans. 17. Ex. 641. Given that the moon makes one revolution round the earth in about 2,360,000 seconds, and nearly in a circle whose radius is 59-964 times the earth's equatorial radius, show that the accelerative effect of gravity 1 on the moon must equal 112.48 reckoning in feet and seconds: what inference can be deduced from this as to the law of the decrease of the earth's attraction? Ex. 642. A body moves in a circle whose radius is r, the force tending towards the centre necessary to keep it in the circle is p; if u is the work accumulated in the body, show that 2 u=pr. 122. Time of Oscillation of a simple Pendulum. If a small bullet is suspended by a very fine thread, and caused to oscillate in any small arc (e. g. not exceeding 2° or 3° on each side of the lowest point), then the time of each oscillation * is given approximately by the formula where t is the required time in seconds, and I the distance. in feet from the point of suspension to the centre of the bullet. It may be remarked that the above formula would be rigorously true if the bullet were reduced to a point, * i. e. the time of moving from the highest point on one side to the highest on the other. the thread perfectly flexible and without weight, and the arc of vibration indefinitely small: a pendulum possessing these properties (which is of course an abstraction) is called a simple pendulum, and the above formula is said to give the time of a small oscillation of a simple pendulum. Ex. 643-If g=32-2 determine the number of oscillations made in one hour by a pendulum 3 ft. long. Ans. 3754-2. Ex. 644 -It is found that at a certain place a pendulum 39.138 inches long oscillates in one second; determine the accelerative effect of gravity at that place. Ans. 32-1897. Ex. 645. Find the time of 100 oscillations of a pendulum 11 ft. long at a place where the length of the seconds pendulum is 39.047 in. Ans. 183.9 sec. Ex. 646. If L is the length of a seconds pendulum show that g="L Ex. 647. If L is the length of a seconds pendulum at any place, and I the length of a pendulum that oscillates in n seconds at the same place, show that l=n2L. Ex. 648. A pendulum at the average temperature oscillates in one second; it is found that its length is L; after a certain time it is found to lose 50 seconds a day; determine the increase of its length. Ans. 0.00116 L. FIG. 154. A G 123. Centre of Oscillation and Percussion. Let AB represent any body capable of oscillating about an axis passing through s perpendicularly to the plane of the paper, which plane contains the centre of gravity G: let the body be made to oscillate round the axis, and let the time of its small oscillations be noted; determine the length l of the simple pendulum which would make a small oscillation in the same time; insa produced, take o, such that so equals l; then the point o is called the centre of oscillation of the body corresponding to the centre of suspension s. If A B has a definite geometrical form so can be determined by calculation, as will be shown hereafter; but in any case it can be determined by observation as explained. R 0 B * In the plane of the paper draw or at right angles to so; it admits of proof that if A В were struck a blow of any magnitude along the line or, there would be no impulse communicated to the axis; the point o is therefore also called the centre of percussion. Ex. 649. A mass of oak is suspended freely by a horizontal axis; it is observed to make 43 oscillations in one minute; at what distance below the point of support must a shot be fired into it so that there may be no impulsive strain on the point of support ? Ans. 6-313 ft. Ex. 650.-A tilt hammer when allowed to oscillate about its axis is observed to make 35 small oscillations per minute; at what distance from its axis must the point be at which it strikes the object on the anvil in order that no impulse may be communicated to the axis? TABLE XV. Ans. 9.528 ft. THE VALUE OF THE ACCELERATING FORCE OF GRAVITY AT DIFFERENT PLACES. * The body AB is supposed to be symmetrical with reference to the plane of the paper. |