Now let n be increased and a diminished without limit so is 1.t that A is kept unchanged; then the limit of (cos) is Hence, when the polygon becomes a circle, v (cOS a)” = v. So that the speed of the particle is unchanged, when its direction has turned through a finite angle A. Q. E. D. *181. Since the above proposition is true of a circle of any radius, and of an arc of any magnitude, it must be true for any continuous curve. For such a curve may be considered to be made up of successive small arcs of its successive circles of curvature. 182. When a particle slides on a smooth continuously curved surface, under the action of gravity only, its total energy is constant. For, the only forces acting are the pressure of the surface and the weight of the particle; and the pressure of the surface does not alter the kinetic energy of the particle, while the weight of the particle produces its own effect, no matter what other forces may be acting. The effect of the weight is expressed by saying that the total energy of the particle is constant. Example 1. A smooth fine wire, in the form of a circle of radius r ft. is placed in a vertical position, and a particle in the form of a small ring slides on it under the action of gravity. The ring has u velos when at the highest point of the circle; find its velocity at the lowest point. Let the ring contain m lbs. In going from the highest point to the lowest it loses 2 rmg foot-poundals of potential energy. Let its velocity at the lowest point be v velos; it will therefore (§ mv2 — § mu2) foot-poundals have gained of kinetic energy. But on the whole, its energy is unchanged. 2 rmg= (§ mv2 — §mu2), or, § mv2= § mu2 + 2 rmg. + Lock's Higher Trigonometry, Art. 9. Example ii. A stone is fastened by a light inextensible string 1 foot long, to a fixed point, and is describing a vertical circle about the fixed point as centre, under the action of gravity; find the least possible velocity at the lowest point of the circle. The constraint of the string is similar to that of a smooth circular wire; for the tension of the string, being always perpendicular to the direction of motion, does no work; hence, by Art. 179, the total energy of the stone is constant. Therefore, if v velos be the velocity of the stone at the lowest point of the circle, u velos its velocity at a point on the circle whose vertical distance above the lowest point is h feet, then or, mgh=(§ mv2 – 1⁄2mu2) { v2 — § u2=gh...... .(i). Now, in order that the stone may completely describe the circle, the tension of the string (which is least at the highest point) must just vanish at that highest point. Let u velos be the velocity of the particle at the highest point, then its acceleration is The only force acting when the stone is at the highest point of the circle, is the weight, mg poundals vertically downwards; for the only other force available is the tension of the string, which at this moment is zero. Therefore, since is the velocity at the lowest point of the circle, putting h=2 ft. in (i) and combining with (ii), we have v2=u2+48=8+48=58• Hence, the least velocity at the highest point is about 4/2 velos, and the corresponding velocity at the lowest point about 4/10 velos. NOTE. The unit of work in the C. G. S. system of units [see page 65] is the work done by a Dyne working through a Centimetre. This unit work is called an Erg. 1. A particle is describing a parabola under the action of gravity, prove that its total energy is constant. 2. A heavy particle slides down a smooth straight tube from rest and then describes a parabola; find its velocity when it is 100 feet below the point from which it started. 3. A perfectly elastic particle is let fall on a horizontal pavement, and rebounds; prove that the total energy of the particle is constant. 4. Two perfectly elastic spheres impinge directly, prove that their total energy before and after impact is the same. 5. Two perfectly elastic spheres impinge obliquely, prove that their total energy is unaltered by the impact. 6. An imperfectly elastic particle falls on a horizontal plane and rebounds; shew that it loses energy by the impact. 7. Two imperfectly elastic balls m and m', of elasticity e and velocities u velos, u' velos impinge directly; find the energy lost by impact. 8. A multitude of small perfectly elastic spheres are in motion in a closed space surrounded by fixed walls, under the action of no force; prove that the total kinetic energy of the spheres is constant. 9. If the spheres in Question 8 are under the action of gravity, shew that their total energy is constant. NOTE. A simple pendulum consists of a heavy particle fastened to a fixed point by a light rod; the particle moves in the lowest arc of a vertical circle under the action of gravity. 10. A simple pendulum of length I feet swings through an arc 4a, prove that its velocity at the lowest point of its path is 2 sin a √ (gl) velos. 11. A circular wheel, whose rim consists of a heavy uniform wire of m lbs. and whose spokes, (of length feet), and axle are such that their mass may be neglected, revolves about its axis, which is fixed, so as to make n complete revolutions per second; shew that the kinetic energy of the wheel is 2mn222 foot poundals. 12. A fine light string is coiled round the wheel of Question 11, one end being fastened to the rim and the other end supporting a heavy mass m' lbs. This mass descends from rest under the action of gravity, causing the wheel to turn as it descends; prove that when m' has descended 1⁄2 feet from rest its velocity is 13. Two masses m lbs., m' lbs. are fastened to the ends of a light string, one of them m is placed on the slant side of a smooth inclined plane of inclination a, and the other m' hangs freely suspended under the action of gravity by the string which passes over the topmost point of the plane; the masses are moving with uniform velocity; prove that m' = m sin a. 14. A rifle ball moving with 2000 velos has its velocity reduced by 200 velos in passing through a plank; how many such planks would it penetrate, assuming the same amount of work to be performed in overcoming the resistance of each plank. 15. Prove that a foot poundal is 421,394 ergs. 16. A particle, in the form of a ring, slides on a smooth fixed curved wire under the action of gravity; shew that the velocity of the ring at any instant is equal to that due to a fall from a certain horizontal plane to its position at that instant. 17. Assuming that the energy of a projectile is constant, prove that its velocity at any instant is equal to that due to a fall from a certain horizontal line to its position at that instant. 18. A simple pendulum of length / feet just describes a complete revolution; prove that the tensions of the rod in its two vertical positions are in the ratio of 1 to 5. 19. Shew that when a simple pendulum just describes a complete revolution the tension of the rod vanishes at a certain instant; determine the position of the rod at that instant. 20. The velocity which a simple pendulum of length / feet must have at its lowest point that, when it reaches its highest point the rod may have no tension, is (5/g) velos. 21. A regular polygon is made of some rigid material; A, B, C, D... are the middle points of the sides; a smooth perfectly elastic particle under the action of no forces moves along AB with velocity v velos, and impinges at B on that side of the polygon whose middle point is B. Prove that the particle will proceed to describe the polygon A, B, C, D... with uniform speed v velos. Now suppose the polygon to have an infinite number of sides, the length of each side being infinitely small; then it follows that the polygon A, B, C, D... coincides with the material polygon; and we have an independent proof of Art. 180. CHAPTER XVI. POWER. 183. We use the word 'agent' to denote a machine which transforms some stored up chemical or other energy into work. Examples. A navvy transforms the chemical energy of his food into work when he raises an embankment. A steam-engine transforms the energy stored in coal into work when it does work on a train. A windmill transforms the kinetic energy of the air into work when it pumps up water, etc. 184. DEF. The power of an agent varies as the amount of work which the agent can do in a unit of time. The unit power is the power of an agent which can do I foot-poundal per second. Hence the measure of the power of an agent is the number of foot-poundals it can do per second. 185. A Horse-power is the power of an agent which can do 550 foot-pounds per second; that is, 550 × g foot-poundals per second, Example i. At what uniform speed can a horse of 1 horse-power draw a tram-car of 1 ton, supposing the friction etc. to cause a uniform resistance of 50 lbs. weight? Suppose the required speed to be v velos. Then the tram passes over v feet per second. Therefore the horse does v × 50 foot pounds per second. He can do 550 foot pounds per second. Hence VII. That is, he can draw the car at the rate of II velos; or, at the rate of 7 miles per hour. L. D. IO |