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j.ir.-.-s. : al an. i ar<>un<i.1 circh- ami ><• oetixed \|]..•:i rt - Dtafmn of winding In alUrnriaf mnwi in th.-n \ ^ S iture UK! iii opposite 7.~>l> ELECTRODYNAMICS directions; so, as the armature is revolved, the induced electro-motive forces are in the same direction in them all, but they are reversed in direction as the coils pass from one magnetic field into the next. If there are n pole pieces, and if the armature makes ra revolutions per minute, the E. M. F. will be reversed nm times each minute. On the armature shaft there are two conducting rings, which are insulated from each other, and to which are joined the terminals of the wire wound on the armature. On them rest the two brushes joined to the external circuit ; so an alternating E. M. F. is applied to it. The pole pieces of the dynamo are magnetized by a separate direct-current dynamo. Oscillatory Discharge of a Condenser. — It has been stated in the description of the discharge of a condenser that under certain conditions it is oscillatory. The reasons for this may now be discussed more fully. When the condenser is charged, the surrounding medium has a certain amount of electrostatic energy. If its two plates are joined by a conductor, a cur- rent will flow in it, and thus, if the resistance is small, so that the energy is not in the main spent in heating the con- ductor, a considerable amount of it will be consumed in pro- ducing a magnetic field. When all the electrostatic energy is thus exhausted, the electro-magnetic energy will flow back into the conductor, continuing the current in the same direc- tion, and thus charging the condenser plates again, but in the opposite manner to their original charges. Finally, all the energy which has not gone into heating the conductor or to producing waves in the ether is again in the form of electrostatic energy ; and the process is repeated in the oppo- site direction ; etc. After a certain number of oscillations the energy is all exhausted, and everything comes to rest. It is evident that the greater the self-induction of the con- ductor joining the condenser plates, so much the more energy goes into
showing that the velocity of short waves along wires is the same as the velocity of light in the free ether, viz., 3 x 1010 cm. per second. These waves may be compared with ordinary mechanical waves along a stretched string or rope. The self-induction of a unit length of the conductor corresponds to the mass per unit length of the string ; the reciprocal of the capacity per unit length of the conductor, to the stiffness of the string ; the resistance of the conductor, to the internal friction of the string. Waves of all lengths decrease in amplitude as they advance along the conductor ; but the long waves decrease least ; so as a complex train of waves advances, it becomes more and more simple, because its shorter components vanish. If the self-induction of the conductor is increased sufficiently, not alone is the attenuation of all waves decreased, but it is the same for waves of all wave lengths ; thus there is no distor- tion of the waves. In Pupin's system of constructing tele- phone lines, this condition is secured by inserting in the line, at regular intervals of every few miles, a helix whose self- induction is large. By this means it is possible to telephone with distinctness over intervals of a thousand miles or more. Electrical Waves in the Ether. — Electrical oscillations pro- duce waves in the surrounding ether, as has been already i.\in < i:i> < n;i;i-:\ i s stated. These waves are identical in their properties with those which produce the sensation of light; and there are many ways by which they may be detected. They will pro- duce oscillations in other conductors : and these may be made evident by sparks or by the heating effects. Again, if a n umber of small metallic particles are put loosely together, hey offer a great resistance to a current : l»nt, when these long ether waves fall upon them, they cohere in such a manner as to have comparatively Mnall resistance. It the particles are jarred slightly, after the v as, their resistance again increases. Such an apparatus is.ailed a "coherer." It is evident that by introducing wires int- two ends of a coherer, and putting it in circuit with a battery and a Lralvanoscope, or a telegraph sounder, one can observe the passage of these " electric waves." There are many m« >i ••• methods, descriptions of which may be found in advanced text-books.
The papers of Henry and Faraday, on the subject of induced currents, have been reprinted in the Seiei. liemoin - Vols. Xi and Xll, New York, CHAPTER XLIX OTHER ELECTRICAL PHENOMENA THERE are several phenomena, showing the connection between electricity and light, which should be mentioned. Faraday Effect. — The fact discovered by Faraday that, if a beam of plane polarized light is passed through a strong magnetic field, parallel to the lines of force, the plane of po- larization is rotated has already been discussed. (See page 564.) The direction of rotation follows the right-handed- screw law ; so that, if the field of force is inside a solenoid, the direction of rotation is that of the current flowing in the coils. This rotation leads one to believe that associated with a line of magnetic force there is a rotational motion in the ether ; so that any vibration in the ether at right angles to the line of force will have its direction changed. The energy of the magnetic field of force should then be considered as due to this motion ; and it is thus seen why it is kinetic. Hall Effect. — This rotational action of a magnetic 'field is shown also in what is called the " Hall effect," a phenome- non discovered in 1879 by E. H. Hall, now of Harvard Uni- versity. If an electric current flows through a thin metallic film, it will so distribute itself that corresponding to any point on one edge of the film there is another on the op- posite edge which has the same potential. Let A and B, in the cut, be two corresponding points ; if they are joined by a wire which includes a galvanometer, no current will flow in it. If now this film is placed between the poles of a mag- net, so that a magnetic field is produced perpendicular to the current sheet, a current will flow from A to B through the 766 KLECTRICAL 757 galvanometer, showing that they are no longer at the same potential. This proves that the lines of flow of the m. Fio. 488. — Lines of flow In a thin metal Htrip. Dotted lines are lines of constant potential have been rotated by the magnetic field. \\\ moving one terminal of the wire slightly along the edge of the film, another point may be found fur which there is again no current. Before the magnetic field is produced, the lines of tlo
was discovered by Du Fay in 1733 ; and the idea that in every process of electrification equal quantities of opposite kinds are produced is due to Symnier, 1759. The phenomena of electrostatic induction and of charging by induction were first investigated by Canton, 1753, and.<Epinus, 1759. We owe many important ideas to Benjamin Franklin. Among other things, he established the identity of atmospheric electricity as manifested in lightning, etc., with electricity as obtained by ordinary means. Many im- portant phenomena were discovered by Cavendish in the last few years of the eighteenth century, but unfortunately his researches were not published for nearly one hundred years. Cavendish was the first to prove the law of the inverse square for electrical forces ; to study the capacity of condensers ; to discover the effect of introducing dielectrics other than air; to propose a law for an electric current, which is practically equivalent to Ohm's law ; and to meas- ure roughly the electrical resistances of many conductors., The most important work done within recent years has been that of Michael Faraday, who was the first to recognize the importance of the surrounding medium in all questions dealing with electrical forces. The varied phenomena dealing with the properties of electric currents and the names of those scientists associated with their discovery are given in the preceding pages. BOOKS OF REFERENCE PERKINS. Outlines of Electricity and Magnetism. New York. 1896. An elementary treatise, in which the phenomena are all explained in terms of tubes of induction, either electrostatic or magnetic. WATSON. A Text-book of Physics. London. 1899. The sections on Electricity are particularly good. J.J.THOMSON. Elements of Electricity and Magnetism. London 1895. This is a text-book which treats the subject from a more mathemati- cal standpoint than most other elementary books; but it may be consulted with advantage by any student. INDEX Aberration, chromatic, 4S9 ; of light, 568 ; spher Attenuation of vibrations and waves, 828, 846, leal, 451. 465,48* Abeotata tamperacwe. M8, M7. Abv.lut.- zero. -.'Jo. :;.•;. A».>..rpti.,n. t,.-ly. MS, DBT; of.-ru-ivy. -.1.7 ; Mir
. Color blindness, 592. Color sensation, 592. Colors, absorption, 587 ; complementary, 524, 587 ; connection between wave number and, 424 ; mixtures of, 587 ; of thin plates, 528 ; polarization, 557 ; surface, 588. Combination, chemical, heat of, 285 ; of lenses, 487 ; of notes, 412. Combustion, 285. Commutator of dynamo and motor, 748. Compensated pendulum, 232. Complex pendulum, 322. Complex vibrations, 320. Complex waves, 347, 754. Composition, of a uniform acceleration and a uniform velocity, 45 ; of displacements, 36 ; of forces, 60, 75 ; of harmonic vibrations, 318, 822; of moments, 91 ; of velocities, 87, 53. Compound microscope, 504. Compound pendulum, 185. Compressed glass, double refraction in, 542. Compressibility, of a gas, 192 ; of a liquid, 171 ; of a solid, 150. Concave grating, 537. Concave mirror, 447. Condensation of vapors, 279. Condensers, electric, 650 ; capacity of, 652, 660 ; discharge of, 654, 752 ; energy of, 656 ; sec- ondary discharge of, 662. Conduction, electric, 668 ; of heat, 227, 287. Conservation, of electricity, 638; of energy. 110, 115, 218, 810; of linear momentum, 78; of mass, 65. Conservative forces, 107. Consonance, 414. Contact, difference of potential, 681 ; electrifi- cation, 625. Continuity of matter, 282. Convection, electric currents, 706 ; of heat, 227, 286. Converging lenses, 470, 475. Convex mirror, 452. Cooling, Newton's law of, 297. Coplanar forces, 96. Cords, vibrations of stretched, 851, 400 ; vocal, 407. Coulomb (unit of electricity), 729. Coulomb's law of electrostatic force, 640. Coulomb's law of magnetic force, 608. Couple, 100 ; theruio-, 294, G7'J. Critical angle, l.">r,. Critical temperature, state, etc., 278. Crookcs, fourth state of matter, 208. Cro
okes, radiometer, 204. Cross hairs of telescope, 506. Cryohydrates, '^t>2. Cryopnorus, 270. Crystalloids, 141. Crystals, biaxal and uniaxal, 542; expansion of, 281. Current, electric, 663. d'Alembert, laws of mechanics, 188. Dalton's law of mixtures of gases, 190, 195. Damping of vibrations, 828. Daniell's cell, CMN d'Arsonval galvanometer, 718. da Vinci, lever, 123, 187. Davy's experiment on the nature of heat, 309. Davy's safety lamp, 289. Declination, magnetic, 618, 622. Density, 146 ; of a gas, 192 ; of a liquid, 147, 176; of a solid, 147, 165; of water, 14G, 14s. Depolarization of light, 560. Depression of the freezing point, 261. Descartes, laws of mechanics, 138 ; theory of rainbow, 517. Deviation, angle of, 459 ; minimum, 460. Dew, 267. Dew point, 268. Dewar flask, 228 ; liquefaction of gases, 280. Diamagnetic bodies, 595. Diaphragms, in optical instruments, 507. Diatonic scale, 416. 615. Dielectric constant, 6*1, 661. Dielectrics, 640. Differential notes, 412. Diffraction, around an edge, 887; through a small opening, 890. Diffraction, grating, 530. Diffuse reflection. 429. Diffusion, 140, 202. Diopter, 488. Dip, magnetic, 618, 620, 622, 739. Direct vision spectroscope, 514. Discord, 414. Dispersion, 423, 491, 508 ; anomalous, 514. Dispersive power, 512. Displacement, angular, 52 ; linear, 86. Dissociation, 201, 284 ; electrolytic, 692 ; heat of, 285. Distortion of waves, 848. Diverging lens, 480. Divided circuits, laws of, 722, 724. Dollond, achromatic lens, 493. Doppler's principle, 845, 898, 584. Double refraction, 541. Ductility, 17.
due to earth, 101, 130; universal, power of, 585. 129. Gravitational waves on liquids, 172, 839. Gravity, centre of, 101, 134; value at different latitudes, 181. Gregory's telescope, 498. Griffith's, mechanical equivalent of heat, 304. Ground ice, 259. Guard ring electrometer, 659. Guericke, von, 167, 759. Hall effect, the, 756. Harmonic motion, rotation, 55 ; translation, 48. Harmonic vibrations, composition of, 318, 322.' Harmonics, 321, 397. Harmony and discord, 414. Heat, flow of, 305; mechanical equivalent of. 226, 804 ; of evaporation, 269 ; of fusion, 260 ; of solution, 283 ; sources of, 215 ; specific, 250 ; transfer of, 227, 286. Heat effects, 216, 303. Heat energy, 219, 226. Heating effect of electric currents, 665, 730. Helmholtz, analysis of sounds, 396 ; explanation Henry, electrical waves, 656 ; induced electric of harmony, 414. currents, 738. Henry, the, unit of inductance, 745. Herschel, fluorescence, 578 ; infra-red radiation, 310. measurement of, 613. HerschePs telescope, 497. Homocentric pencils, 435. Homogeneous waves, 428. Hooke's law, 145. Horizontal intensity of earth's magnetic field, Horn blower, steam engine, 275. Horns, 858, 404. Horse power, 115. Humidity of air, 268. Huygens, clock, 186, 138 ; experiment in double refraction, 547 ; eyepiece, 494 ; fixed points of temperature, 226; impact experiments, 156; principle, 866; reversible pendulum, 136; theory of reflection and refraction, 307, 871 ; variation in "g," 181 ; wave surface for Iceland spar, 548. Hydraulic ram, 206. Hydrostatic press, 163. Ice, lowering of melting point of, by pressure, 258. Ice calorimeter, 250, 252. 789. Iceland spar, 541. linage.-, real and virtual, 435. Impact, 7*. l.V>. Impulse, C.'.t, 106. Incandescent electric lamp, 666. Incidence, angle and plane of, 869, 481
. Inclination, earth's magnetism, 618, 620, 622, Inclined plane, 42, 119, 126. Independence o!' forces, principle of, 60. Index of refraction, 482, 455, 457 ; measure- ment of. 4;.C., 4»SO, 465, 506. Indicator diagram, nil, 274. Induced electric currents, 738. Induction, electro-magnetic, 707 ; electrostatic, Induction coil. 741. Inertia, 18, 14, 15; moment of, 88; principle 631, 645 ; magnetic, 597, 614. of, 59. Infra-red radiation, 293, 424. Insulators, 626. Intensity, of electric field, 641 ; of light, 487 ; of magnetic field, 609 ; of magnetization, 610 ; of sound, 397 ; of waves, 814, 828, 330. Interference of light waves, 374, 424, 519 ; of sound waves, 374 ; of waves on surface of liquid, 374 ; over long paths, 527. Internal energy, 218, •-'»:;. Internal work done when a gas expands, 248, 248, 253. Interval, musical, 416. Inverse square, law of, 180, 608, 640. Inversion, thermoelectric, 680. lonization of gases, 699. Ions, 689, 692. Isoclinal lines, 621, 628. Isogonal lines, 620, 622. Isothermals, 195, 267, 276. Jar, Leyden, 653. Jets of liquid, 189. Joly, steam calorimeter, 252. Joule, determination of mechanical equivalent, 303, 804, 810 ; internal work in a gas, '243. Joule, the unit of energy, 112. Jupiter, occultation of satellites, 421, 5G6, 574. Kaleidoscope, 446. Kathode (see Cathode). Kelvin, absolute scale of temperature, 307; electrometers, 658; expansion of gases through porous plugs, 243 ; thermoelectric- ity, 682. Kepler's laws, 132. Kerr effect, 565. Kinetic theory, of evaporation, 143, 264; of gases, 197, 241, 254; of matter, 142; of vis- Kir
63. Ordinary and extraordinary rays, 542. Orir:in pipes, 355, 401, 402. Oscillations, electrical, 816, 654, 752. Osmosis, 141, 179. Osmotic pressure, 179. Overtones, 897. Parabolic mirror, 408, 454. Parallelogram of forces, 75. Partial vibrations, 322. Particles, reflection by fine, 480, 555, 589. Pascal's law, 1U-J. Path, of molecules, 202. Peltier K.M.F., 681. Pencils, astigmatic and homocentric, 485. Pendulum, compensated, 282; complex, 322; compound, 185 ; magnetic, 610 ; reversible, 136; simple, 74, 91. Penumbra, 426. Period of vibration, 48, 817 ; measurement of, 322. (jnograph, 396, 407. Permeability, magnetic, 609. Phase of simple harmonic motion, 49. — logiston, 309. Phosphorescence, 299, 578, 588. Photo-electric action, 699. Photographic lens, 500. Photometer, 442. Photometry, 437. Physical quantities, 10 et seq. Physics, 7, 9 ; divisions of, 10. Piezometer, 150. Pigment colors, 587. Pile of plates, 554. Pin-hole images, 426, 495. Pitch, of a screw, 127 ; of sounds, 397 ; stand- ard, 418. Plane of incidence, 431 ; of polarization, 554. Plasticity, 17. Plate, refraction through, 458, 466. Plates, colors of thin, 523 ; vibrations of metal, 360. Platinum thermometer, 223. Points, effect of, on electric charge, 633. Polarization, angle of, 552 ; by double refrac- tion, 547; by reflection, 551, 554; colors due to, 557 ; plane of, 554 ; rotation of, 563, 750. Polarized waves, circularly and elliptically, 818, 560, 562; interference of, 556; plane, 313, 555. Polarizer, 560. Porosity, 18. Potential, electric, 642. Potential energy, 109, 117; and force, 114. Pound, «'•('•. Power, 115. Practical
system of electrical units, 728. Pressure, 158; atmospheric, 166, 176; centre of, 174 ; unit of, 165 ; in a bubble or drop, 185 ; in a gas, 158, 198. Pressure in a gas, measurement of, 178, 196. Pressure in liquids, 158; due to cohesion, 162; due to gravity, 163; due to surface-tension, 185. Prevost's theory of exchanges, 297. Primary colors," fi'.i''. Primary electric cell, 676; energy of, 697. Principal axes of rotation, 94. Principal section of a crystal, 548. Prism, angle, edge, face, 458, 466; resolving power of, 510. Projectile, 45. Projection lantern, 501. Projection of a line or area, 28. Pulley, 128. Pulses, 342, 348, 394, 409, 425, 704. Pump, air, -joy ; water, 208. Quadrant electrometer, 658. Quality of sound, 895. Quantity of heat, 226, 809. Quarter- wave-plate, 561. Radian, 24. Radiant energy, 290. Radiation, 290 ; and absorption, 298, 576. Radioactive bodies, 7'H. Radiometer, Crookes', 204, 294. Rainbow, 516. Ram, hydraulic, 206. Ramsden eyepiece, 494. Ratio of specific heatsof a gas, 253, 254, 337. Ray of light, 881, 425. Reaumur temperature scale, 223. Rectilinear motion, 43, 70. Rectilinear propagation of light, 381, 425. Reed pipe, 403. Reflection, 383, 367 ; angle of, 369, 431 ; of,'8b <. ether waves, 867, 429, 430. Reflection of light, concave surface, 447 ; con-. vex surfacju.452 ; diffuse, 429; fine particles, regular, 430 ; total, 428, 456. Reflection of sound, 367, 408; diffuse, •Ir- regular. 430 ; total, 428, 456. Refraction, angle of, 373, 432; double, 4f><> ; in- dex of, 432, 455, 457, 460, 506. Refraction of light, concave surface, 467 ; con- vex surface, It'i!) ; plane surface,
, 154. Young's principle, 874. Zeeinan effect, 758. Zero, absolute, 240, 807.emember that you can only use this equation when acceleration is constant, it is not true otherwise. We also have an equation that deﬁnes average velocity and is true in all cases, v = v0 + v 2. (2.9) Let’s solve this equation for ∆x, and use the new equation for v, v = ∆x t. ∆x = vt, ∆x = ∆x = v0 + v 2 t 1 2 (v0 + v)t, (2.10) (2.11) (2.12) to get an expression for displacement in terms of initial and ﬁnal velocities. We can get an even more useful relationship by eliminating the ﬁnal velocity. If we use Eq. (2.8) to substitute for the ﬁnal velocity, ∆x = 1 2 (v0 + (v0 + at))t ∆x = v0t + 1 2 at2. (2.13) This expression is often handy because it does not contain the ﬁnal velocity. In many cases, we have information about the start of motion, but we rarely have information about the end of it (that’s usually what we are trying to predict). We can get another useful relationship by solving Eq. (2.8) for time, t = v − v0 a, (2.14) and substituting into Eq. (2.12), v − v0 a ∆x = ∆x = (v0 + v) 1 2 v2 − v2 0 2a v2 = v2 0 + 2a∆x. This equation is very handy if information about time is not given in the problem; it still allows us to calculate the ﬁnal velocity without knowing how long the object was accelerating. 19 Example: Car Chase A car traveling at a constant speed of 24.0 m/s passes a trooped hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets oﬀ in chase with a constant acceleration of 3.00 m/s2. (a) How long does it take the trooper to overtake the speeding car? (b) How fast is the trooper
going at that time? Solution: We ﬁrst need to be clear about how we are measuring time and position in this problem. Let’s measure them both relative to the trooper’s motion. That is, x = 0 at the billboard where the trooper starts moving and t = 0 when the trooper starts moving. This means that the speeding car passed the trooper at t = −1 s. (a) Now we need to think about what the problem is asking us to ﬁnd. It asks for a time (“how long”) at which the trooper overtakes the speeding car. At the exact moment that the trooper overtakes the car, their positions on our axis will be equal, i.e. xt = xc. So, we want to ﬁnd the time at which the positions are equal. We have just derived an equation which gives position as a function of time for constant acceleration, ∆x = v0t + 1 2 at2. First check that we can use this equation; are both objects undergoing constant acceleration? Yes, the problem says the trooper has a constant acceleration and the car has constant velocity (a constant acceleration of 0). So let’s ﬁgure out the position of the trooper. The trooper’s initial position is x0t = 0, his initial velocity is v0t = 0, and his acceleration is a = 3.00 m/s2, so we have xt = 1 2 at2. Now we do the same for the car. It’s initial position is x0c = 24.0 m (it had a 1 s head start), it’s initial velocity is v0c = 24.0 m/s, and it’s acceleration is ac = 0, so we have xc − x0c = v0ct xc = x0c + v0ct. To ﬁnd the time at which positions are the same, we set the two expressions equal and solve for time, xt = xc at2 = x0c + v0ct 1 2 at2 − v0ct − x0c = 0. 1 2 You get a quadratic expression for t, so you will need to use the quadratic formula (it’s in your textbook), t = t = v0c ± 24.0 m/s ± 2 a (−x0c) 0
c − 4 1 v2 2 a 2 1 (24.0 m/s)2 − 4 1 2 1 2 · 3.00 m/s2 2 · 3.00 m/s2 (−24.0 m) t = 16.9 s. There is also a negative root, but since we know that the trooper could not overtake the car before it passed him, this root is not physically meaningful. 20 (b) The trooper’s speed at that time is a fairly straightforward problem. We have an equation that gives us ﬁnal velocity if we know the acceleration, time and initial velocity (and we know all those now). v = v0 + at v = (0) + 3.00 m/s2(16.9 s) v = 50.7 m/s. As mentioned earlier, constant acceleration is particularly useful because objects moving due to the gravitational pull of an object will move with a constant acceleration if air resistance is neglected. Objects moving under the inﬂuence of gravity and without air resistance are said to be in free fall. Note that this does not mean that the object necessarily started from rest (v0 = 0). Objects can be moving upward, like when you throw a ball, and still be considered free-falling. The magnitude of the free-fall acceleration is denoted by g and has a value of 9.80 m/s2 on Earth (although it varies slightly depending on latitude). The direction of g is always towards the large object creating the gravitational pull. Because g is a constant, we can use all of the equations we just derived for constant acceleration. Example: Rookie Throw A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof on its way down. Determine (a) the time needed for the ball to reach its maximum height, (b) the maximum height, (c) the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant, (d) the time needed for the ball to reach the ground. Neglect air drag. Solution: You can refer to the ﬁgure in your textbook (Fig. 2.20) to get a visual idea of what’s going on in the problem. We are expressly given an initial velocity and
an initial height in the problem. There are a few other pieces of information that are not expressly stated in the problem. The ball is in free fall, so we know the acceleration, g = 9.80 m/s2, and we know that the velocity of the ball at it’s maximum height will be zero. Let’s set up our coordinate system so that y = 0 corresponds to the top of the building; this means that the bottom of the building is at -50.0 m. (a) The ﬁrst part of the problem is concerned with the upward motion of the ball. We are given an initial velocity, we know acceleration and a ﬁnal velocity, and we want time. We have an equation that contains all these quantities, v = v0 + at v − v0 a t = t = 0 − 20.0 m/s −9.80 m/s2 t = 2.04 s. 21 (b) Now we want to know the maximum height reached by the ball. We have all the information from before and we now have the time at which we reach this maximum height, so we can use y = y0 + v0t + 1 2 at2 ymax = (0) + (20.0 m/s)(2.04 s) + ymax = 20.4 m. 1 2 (−9.80 m/s2)(2.04 s)2 (c) Now the ball begins to move downward and we want to know how long it takes to get back to it’s initial height. We can use the same distance equation, but now use the fact that yf = 0 to solve for time. y = y0 + v0t + 1 2 at2 v0t + t(v0 + 1 2 1 2 at2 = 0 at) = 0. We will have two roots, t = 0 because the ball starts at that height, and the one we really want, v0 + 1 2 at = 0 t = − t = − 2v0 a 2(20.0 m/s) −9.80 m/s2 t = 4.08 s. Note that this is twice the time it takes to get to the maximum height. The velocity of the ball at that time is v = v0 + at v = 20.0 m/s + (−9.80 m/s2)(4.08 s) v =
−20.0 m/s. (d) Now our ﬁnal position is yf = −50.0 m, but this is essentially the same problem we just solved. y = y0 + v0t + 1 2 at2 1 2 at2 + v0t − yf = 0. This leads to a quadratic equation, so it’s a little more eﬀort to solve, t = t = v0 ± 20.0 m/s ± 2 a (−yf ) 0 − 4 1 v2 2 a 2 1 (20.0 m/s).80 m/s2 2 · 9.80 m/s2 (−(−50.0 m)) t = 5.83 s. 22 Chapter 3 Vectors and Two-Dimensional Motion We live in a three-dimensional world and the objects around us move in that three-dimensional space. While some simple examples of motion can be described in one dimension, extend what we’ve learned about motion to more than one dimension will allow us to study many more systems. 3.1 Vector properties Displacement, velocity and acceleration are all vector quantities. That is, they have a magnitude and a direction. In one dimension, there were two options for the direction, left or right, and we could represent the direction of displacement, velocity, or acceleration simply by using a negative or positive sign. When we move to two dimensions, there are many more options for the direction of the vector and we will need to be more formal with the mathematics of vectors. Vectors are represented graphically as arrows with the length of the arrow representing the magnitude of the vector and the direction of the arrow giving the direction of the vector. This graphical representation might help you understand the basic arithmetic of vectors. Mathematically, we denote a vector, A, with an arrow over the variable name. If we use the variable name without the arrow, this means we are referring to the magnitude of the vector. Let’s set up a coordinate system at the starting end of the vector (see Fig. 3.1). Then the vector points to a speciﬁc location in that space. We know that we can give the coordinates of that location using either Cartesian, (x, y), or polar, (r, θ) coordinates. The polar coordinates of this point are quite straightforward; r is the length (magnitude) of the vector and θ
is the angle between the vector and the x-axis. We can also determine the x and y coordinates by ﬁnding the projections of the vector on the x- and y-axes. The projection of a vector A along the x-axis is called the x-component and is represented by Ax. The projection of A along the y-axis is called the y-component and is represented by Ay. We can ﬁnd the x and y components by converting the polar coordinates to rectangular coordinates, Ax = A cos θ Ay = A sin θ. (3.1) Note that we can go from the vector’s x- and y-components back to the polar representation using the Pythagorean theorem and the deﬁnition of the tangent: x + A2 A2 = A2 y Ay Ax. tan θ = 23 (3.2) Figure 3.1: The projections of a vector on the x- and y-axes. Projections give the x- and y components of a vector. Equality of vectors Two vectors are equal only if they have the same magnitude and direction (two arrows are the same only if they are the same length and point in the same direction). This means that you can move a vector around in space as long as you don’t change the length or direction of the vector. Adding vectors Just as when you are adding scalar quantities, you must ensure that the vectors you are trying to add have the same units. Vectors can be added either geometrically (graphically) or algebraically. To graphically add two vectors, A and B, draw the vectors (using the same scale for both) head to tail on a piece of paper. The resultant vector R = A + B is the vector drawn from the unmatched tail to the unmatched head (see Fig. 3.2). You can lay out many vectors head to tail to ﬁnd the result of them all. While graphical addition of vectors is useful for visualizing the addition process. You will most often be adding the vectors algebraically. To add vectors algebraically, resolve the vectors into their x- and y- components. All the x-components are added to get the x-component of the resultant vector. All the y-components are added to get the y-component of the resultant vector. Never add the the x-components to y-components. You can get
the magnitude and direction of the resultant by converting from the Cartesian representation (x and y components) using the equations presented earlier. Negative of a vector The negative of a vector, A, is deﬁned as the vector that gives zero when added to A. If you think about the graphical addition of vectors, this means that the negative of A must have the same magnitude as A, but opposite direction (180◦ diﬀerence). Subtracting vectors Subtraction is simply the addition of a negative quantity. Since we have now deﬁned the negative of a vector, we can ﬁgure out how to subtract. Remember that the negative of a vector points in the opposite direction, so for subtraction we ﬂip the direction of the vector to be subtracted, but otherwise use the same graphical method as for addition. Graphically ﬂipping a vector corresponds to algebraically changing the sign of both the x- and y- components of the vector, so we can use the same algebraic method for subtracting vectors as we do for adding vectors. 24 AAxAyθ Figure 3.2: Graphical addition of vectors. To graphically add vectors, the vectors are drawn head to tail. The resultant vector is the vector that connects the two “loose ends”, drawn from tail to head. Example: Take a Hike A hiker begins a trip by ﬁrst walking 25.0 km 45.0◦ south of east from her base camp. On the second day, she walks 40.0 km in a direction 60.0◦ north of east, at which point she discovers a forest ranger’s tower. (a) Determine the components of the hiker’s displacements in the ﬁrst and second days. (b) Determine the components of the hiker’s total displacement for the trip. (c) Find the magnitude and direction of the displacement from base camp. Solution: (a) Let’s set the origin of our coordinate system at the camp and have the x-axis pointing east and the y-axis pointing north. On the ﬁrst day, the hiker’s displacement, let’s call it A has a magnitude of 25.0 km with a direction θ = −45◦ — it’s negative because she is moving south of the x-axis.
We can use Eqs. 3.1 to ﬁnd the x- and y-components, Ax = A cos θ = (25 km) cos(−45◦) = 17.7 km Ay = A sin θ = (25 km) sin(−45◦) = −17.7 km. On the second day, her displacement, let’s call it B, has a magnitude of 40.0 km with a direction θ = 60.0◦ — positive this time because it is north of east. The x- and y-components of this displacement are Bx = B cos θ = (40 km) cos(60◦) = 20.0 km By = B sin θ = (40 km) sin(60◦) = −34.6 km. (b) The total displacement is the vector sum of A and B. We’ve just found the components of A and B, so we can ﬁnd the components of the total displacement, Rx = Ax + Bx = 17.7 km + 20.0 km = 37.7 km Ry = Ay + By = −17.7 km + 34.6 km = 16.9 km. 25 (c) We just found the components of the total displacement, all we need to do is convert them to a magnitude and direction. The magnitude is found using the Pythagorean theorem, x + R2 y R2 = R2 R = (37.7 km)2 + (16.9 km)2 R = 41.3 km. The direction is found using the tangent function, tan θ = tan θ = Ry Rx 16.9 km 37.7 km 16.9 km 37.7 km θ = tan−1 θ = 24.1◦. Don’t forget to check that the angle is in the correct quadrant. components are both positive, so we are in the ﬁrst quadrant and the angle is correct. In this case, the x- and y- 3.1.1 Displacement, velocity and acceleration in two dimensions In one dimension, we deﬁned the displacement as the diﬀerence between the initial and ﬁnal positions of an object. The position in that case was determined by a single coordinate. We now want to deﬁne displacement in two dimensions where the position
of the object is given by two coordinates. Let’s call the initial position of the object ri and the ﬁnal position of the object rf, where r is a position vector that goes from the origin to the position of the object. The displacement is deﬁned as the vector diﬀerence between the initial and ﬁnal position vectors, ∆r = rf − ri. (3.3) With this generalized deﬁnition of displacement, we can also generalize the average velocity and average acceleration of an object, vav = aav = ∆r ∆t ∆v ∆t. Finally, we can also generalize the instantaneous velocity and instantaneous acceleration v = lim ∆t→0 a = lim ∆t→0 ∆r ∆t ∆v ∆t. (3.4) (3.5) (3.6) (3.7) If you look carefully at these deﬁnitions, you will see that x-component of acceleration is determined by the x-component of velocity which is determined by the x-component of the displacement. The same is true for the y-components of these quantities. This means that the horizontal and vertical components can be treated independently of each other. 26 Example: The swimmer The current in a river is 1.0 m/s. A woman swims across the river to the opposite bank and back. She can swim 2.0 m/s in still water and the river is 300 m wide. She swims perpendicular to the current so she ends up downstream from where she started. Find the time for the round trip Solution: Since the woman swims perpendicular to the current let’s deﬁne the y-axis as parallel to the river. We can treat the x and y motion independently. We are only interested in the motion in the x-direction (across the river) since this will determine how long the trip takes. She is swimming at a constant velocity of 2.0 m/s, so the time to travel a distance of 300 m is vx = t = t = ∆x t ∆x vx 300 m 2.0 m/s t = 150 s. It will take the same amount of time for her to travel back, so the round trip takes 300 s. Note that the current pushing the woman down the river is
completely irrelevant here because it aﬀects her motion in the y-direction (along the river) and this is independent of her motion in the x-direction. 3.2 Motion in two dimensions We have previously studied motion of objects moving in a straight line (one dimension). We will now extend our study to two dimensions. We know that if we break motion up into x and y components, that the motion in the two directions is independent, so that motion in the horizontal direction does not aﬀect motion in the vertical direction and vice versa. This is particularly important when studying something called projectile motion which is the motion of any object thrown in some way. If we throw a ball with some horizontal initial velocity, its motion can be studied by breaking it up into the horizontal and vertical motions. In the vertical direction, the object undergoes acceleration due to gravity just as in free fall. In the horizontal direction, there is no acceleration and the velocity remains constant. The resulting two-dimensional motion is the combination of the two components. 27 Chapter 4 Laws of motion So far, we’ve studied motion by describing what happens without being concerned about what causes the motion. Now, we will start to examine the causes of motion and we will learn the rules that govern changes in motion. 4.1 Newton’s ﬁrst law Isaac Newton developed the laws of motion in the 1600s when he started thinking about why objects close to the Earth tended to fall to Earth unless something was holding them up in some way. His ideas on motion are summed up in three laws that are based on the idea of forces. You probably have an intuitive sense of a force from everyday life. When you push or pull on an object, you are applying an external force on the object. These are examples of contact forces, forces which are caused by one object being in contact with another object. There are also forces that arise without contact of two objects. While this may seem strange (Newton was also uncomfortable with the idea of action-at-adistance), you are very familiar with one such force. Gravity causes all objects near Earth to fall towards the Earth even though the Earth is not touching the object. This is an example of a ﬁeld force, so called because scientists use the idea of a force ﬁeld emanating from an object to explain how it might aﬀect the motion of objects that it hasn’t touched. Essentially a force is something that can
change the state of motion of an object. Note that force (contact or ﬁeld) is a vector — it has both a magnitude and direction. If a force is something that can change the state of motion of an object, will objects move without a force? Obviously, if an object is at rest (not moving), it will just sit there forever unless something pushes or pulls it. Suppose now that the object is given a quick push. It will start moving because of the force that has been applied, but what happens after the initial push? In most cases, the object will start to slow down because there is friction between it and the object on which it moves. But suppose we could eliminate the friction, which is a force and changes the motion of the object? If we completely eliminate friction, then the object would continue moving without speeding up or slowing down. So yes, objects will move without the presence of a force, but with a very speciﬁc type of motion. This is the essence of Newton’s ﬁrst law, “An object moves with a velocity that is constant in magnitude and direction unless a non-zero net force acts on it.” The net force is the vector sum of all the forces acting on the object. So this law can be used in two ways. If we know that there is no net force on the object, then we know that it will continue moving with a constant (possibly zero) velocity. Alternatively, if an object is moving with a constant (possibly zero) velocity, then the net force acting on it must be zero. This law is based on the notion of inertia which is the tendency of an object to continue its state of motion in the absence of a force. The law is sometimes stated as “a body in motion will stay in motion and a body at rest will stay at rest unless acted upon by an outside force.” This is closely related to the idea of 28 mass, which measures an object’s resistance to changes in its velocity due to a force. If the same force acts on two objects with diﬀerent masses, the object with the smaller mass will experience a bigger change in its velocity than the more massive object. 4.2 Newton’s second law In the absence of a force, an object will keep doing whatever it was doing. What happens if a force acts on the object? Clearly it’s velocity will change in some way. If you
push on an object, it will accelerate (change its velocity). If you push harder, it will accelerate faster. So the magnitude of the force is related to the acceleration. In fact, the force is proportional to the acceleration; so if you push twice as hard the acceleration will be twice as large. What other things might aﬀect the acceleration? One quantity that we’ve already discussed is the mass. The same force applied to objects with diﬀerent masses will result in diﬀerent accelerations. In this case the relationship is inversely proportional — if the mass is twice as large, the acceleration is halved. Newton put both of these observations together into his second law “The acceleration a of an object is directly proportional to the net force acting on it and inversely proportional to its mass.” We can write this statement more compactly using mathematics a = F m, (4.1) where a is the acceleration (vector) of the object, F is the vector sum of of all the forces acting on the object and m is the mass of the object. To actually use this equation, we break it up into its x and y (and maybe z) components Fx = max Fy = may. (4.2) Note that the unit of force in the SI system is the newton where 1 N = 1 kg · m/s2. 29 Example: Horses pulling a barge Two horses are pulling a barge with mass 2.0 × 103 kg along a canal. The cable connected to the ﬁrst horse makes an angle of θ1 = 30.0◦ with respect to the direction of the canal, while the cable connected to the second horse makes an angle of θ2 = −45.0◦. Find the initial acceleration of the barge, starting at rest, if each horse exerts a force of magnitude 6.00 × 102 N on the barge. Ignore forces of resistance on the barge. Solution: We’ve been given the mass of an object and the forces acting on it and we’re asked to ﬁnd accelerations. So we want to use Newton’s second law to try to ﬁnd the acceleration. Let’s deﬁne our coordinate system with the x-axis lying along the canal (so the angles are measured relative to the x-axis). Now we can break down the forces into
x and y components, F1x = F cos θ1 = (6.00 × 102 N) cos(30.0◦) = 5.2 × 102 N F1y = F sin θ1 = (6.00 × 102 N) sin(30.0◦) = 3.00 × 102 N F2x = F cos θ2 = (6.00 × 102 N) cos(−45.0◦) = 4.24 × 102 N F2y = F sin θ2 = (6.00 × 102 N) sin(−45.0◦) = −4.24 × 102 N. Newton’s second law tells us to ﬁnd the net force in both the x and y directions, Fx = F1x + F2x = 5.2 × 102 N + 4.24 × 102 N = 9.44 × 102 N Fy = F1y + F2y = 3.00 × 102 N − 4.24 × 102 N = −1.24 × 102 N. Now Newton’s second law says that the net force is related to the acceleration, ax = ay = Fx m Fy m = = 9.44 × 102 N 2.0 × 103 kg −1.24 × 102 N 2.0 × 103 kg = 0.472 m/s2 = −0.062 m/s2. We have the x and y components of the acceleration, so we can ﬁnd the magnitude and acceleration, a = x + a2 a2 y = 0.476 m/s2 θ = tan−1 ay ax = −7.46◦. 4.2.1 Weight The weight of an object is not the same as its mass. The weight of an object is the magnitude of the gravitational force acting on an object, so on Earth the weight is Fw = mg. Because g is the same everywhere on Earth, we can use an object’s weight to determine its mass and so the two terms are used interchangeably in everyday language. While mass is a fundamental property of an object and will not change if you move the object to another location, its weight can change depending on the object’s location. 4.3 Newton’s third law Newton’s third law is perhaps the least intuitive of the three laws of motion. According to Newton, all forces come in
pairs, 30 “If object 1and object 2 interact, the force F12 exerted by object 1 on object 2 is equal in magnitude but opposite in direction to the force F21 exerted by object 2 on object 1” The force exerted by object 1 on object 2 is sometimes called the action force and the force exerted on object 2 by object 1 is called the reaction force. The law is sometimes stated as “every action has an equal and opposite reaction”. Basically anytime two objects interact in some way, there will be two forces, one acting on each object. When you walk, your foot pushes on the ﬂoor and the ﬂoor pushes back on you. When you lean against a wall, the wall pushes back on you. Every time an object falls towards Earth because Earth’s gravity is pulling it, the object also pulls Earth towards it. This may seem strange, but the Earth is so much more massive than other objects that its acceleration due to this force is negligible. One consequence of this law is a force called the normal force. Every object on Earth is being pulled towards the center of the Earth by gravity. Most objects are not moving downward because they are sitting on some surface and this surface is pushing up on the object. The upwards force is the normal force, so named because it is always perpendicular to the surface; this mean it does not always point straight up even though it is a reaction force to the pull of gravity. Example: Standing on a crate (a) A 38 kg crate rests on a horizontal ﬂoor, and a 63 kg person is standing on the crate. Determine the magnitude of the normal force that the crate exerts on the person. (b) Determine the magnitude of the normal force that the ﬂoor exerts on the crate. Solution: (a) Let’s consider the forces acting on the person only. There is, of course a downward force due to gravity Fg = mpg. There is also a normal force from the crate pushing up on the person. The man is not moving so these two forces must be in equilibrium (the net force must be zero), −Fg + Fcp = 0 Fcp = mpg. (b) Now let’s look at the forces acting on the crate. Gravity acts on the crate, Fg = mcg, and the ﬂoor pushes up on the crate through the normal force, Ff c. The person standing on the
crate also pushes down on the crate with a “normal” force that is equal in magnitude and opposite to the force of the crate pushing on the person, Fpc = −Fcp. The crate is not moving,so these forces must be in equilibrium, Ff c − Fg − Fpc = 0 Ff c = mcg + mpg. When we are using Newton’s laws to solve problems, we use several assumptions. We assume that each object is a point mass, or that they are particles without any spatial extent (0-dimensional objects). This means we don’t have to worry about rotation of the objects. If strings or ropes are part of the problem, we assume that their mass is negligible and that any tension in the rope is the same at all points on the rope. When solving force problems, it is useful to draw a free-body diagram. The free body diagram is a drawing of all the forces acting on a particular object. It is very important to only draw the forces acting on the object, any force that the object exerts on its surroundings is not included in the free-body diagram. This diagram helps to isolate the forces of interest for our object and can then be used to apply Newton’s laws. 31 Example: Traﬃc light A traﬃc light weighing 1.00 × 102 N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper cables make angles of 37◦ and 53◦ with the horizontal (see Fig. 4.14 in your textbook). Find the tension in each of the three cables. Solution: We start by drawing free-body diagrams. First for the traﬃc light which has gravity acting downward (the weight, W and tension, T3 from the rope pulling it upward. Because both forces are in the y direction only, this leads to a single equation, T3 − W = 0. Although we now know the value of T3, this does not tell us anything about the tension in the other two ropes. We can also consider the forces acting on the knot. The knot has three ropes pulling on it: tension downwards from T3, tension up to the left from T1 and tension up to the right from T2. Because forces are vectors, we need to break everything into x and y components and apply Newton’s second law along each axis. Note that the knot is in equilibrium, so there is
no acceleration in either direction, x − direction : y − direction : T2 cos(53◦) − T1 cos(37◦) = 0 T1 sin(37◦) + T2 sin(53◦) − T3 = 0. We know T3 from the ﬁrst equation, so we are left T1 and T2 as unknowns. Luckily, we have two equations for our two unknowns. so we can solve one equation, T2 = T1 cos(37◦) cos(53◦) and substitute into the other equation, T1 sin(37◦) + T1 cos(37◦) cos(53◦) sin(53◦) − W = 0 T1(sin(37◦) + cos(37◦) cos(53◦) sin(53◦)) = W T1 = sin(37◦) + cos(37◦) W cos(53◦) sin(53◦) T1 = 60.1 N. We can now ﬁnd T2 as well, T2 = (60.1 N) T2 = 79.9 N. cos(37◦) cos(53◦) 4.4 Friction An object moving on a surface or through some medium encounters resistance as it moves. This resistance is called friction. Friction is an essential force as it allows us to hold objects, drive a car and walk. The strength of the frictional force depends on whether an object is stationary or moving. You’ve undoubtedly had the experience of trying to push a large heavy object; it takes more eﬀort to get the object moving than to keep it moving once it’s going. When an object is stationary, the frictional force is called the force 32 of static friction and when the object is moving, the frictional force is called the force of kinetic friction. Friction points opposite to the direction of motion in the kinetic case or the direction of impending motion in the static case. The force of static friction is larger than the force of kinetic friction. It has been shown experimentally that both kinetic and static friction are proportional to the normal force. The only way this can be true is if the two forces have diﬀerent constants of proportionality. For static friction, we have fs ≤ µsN (4.3) where µs is the
coeﬃcient of static friction. This is an inequality because the force of static friction can take on smaller values if less force is needed to hold an object in place. You will almost always use the ‘=’ sign in problems. For the force of kinetic friction, we have where µk is the coeﬃcient of kinetic friction. µk will almost always be less than µs. Note that the friction coeﬃcients do not have any units. fk = µkN (4.4) 33 Example: Block on a ramp Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coeﬃcient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down? Solution: This is a problem involving forces, so we will need a free body diagram. There are three forces acting on the block: gravity pulls straight down, the normal force acts perpendicular to the surface, and the force of friction acts upwards along the ramp because the block would like to slide down the ramp. We will choose a tilted coordinate system such that the x-axis runs along the ramp. If we do this, the only force that needs to be broken into components is the force of gravity (friction pulls entirely along the x axis and the normal force pulls entirely along the y axis). We can now use Newton’s second law for both axes to get two equations, Fx = 0 mg sin θ − f = 0 Fy = 0 N − mg cos θ = 0. Remember that the force of friction is related to the normal force, f = µsN. I have used the equal sign here because the static force of friction will be largest (and therefore equal) just before the object starts to slip. Let’s use one equation to solve for N, and substitute into the other equation, N = mg cos θ mg sin θ − µsmg cos θ = 0 tan θ = µs θ = 19.3◦. The angle at which the block will slide depends only on the coeﬃcient of static friction between the ramp and block. 34 NWxyf Example: Two blocks A block of mass m = 5.00 kg rides on top of a second block of mass M = 10.0 kg. A person attaches a string to the bottom block
and pulls the system horizontally across a frictionless surface. Friction between the two blocks keeps the 5.0 kg block from slipping oﬀ. If the coeﬃcient of static friction is 0.305, (a) what maximum force can be exerted by the string on the 10.0 kg block without causing the 5.0 kg block to slip? (b) What is the acceleration? Solution: This is a problem involving forces, so we will need a free body diagram. There are ﬁve forces acting on block M : gravity pulls straight down, the normal force from the ﬂoor pushes up, the normal force from block m pushes down, the tension from the rope, and friction from block m opposes the motion (so f points to the left). There are three forces acting on block m: the normal force from block M pushing up, gravity pulling down and the force of friction that opposes the potential motion of block m (so f points to the right). (a) Now let’s use Newton’s second law on both objects. First block M, and for the second block, Fx = M a T − f = M a Fy = 0 N2 − N1 − M g = 0, Fx = ma f = ma Fy = 0 N1 − mg = 0, Remember that the force of friction is related to the normal force, f = µsN1. I have used the equal sign here because the static force of friction will be largest (and therefore equal) just before the object starts to slip. Now we want to ﬁnd T, N1 = mg µsN1 = ma a = µsg T − f = M a T − µsmg = M µsg T = (m + M )µsg 35 T = 51.5 N (b) We already have an expression for the acceleration a = µsg a = 3.43 m/s2. NNTffM2m1W1W2N1 Chapter 5 Work and Energy 5.1 Work You probably deﬁne work as something that expends some of your energy. Typing up a paper, or writing out a homework assignment is considered work as are more physically demanding tasks such as building furniture or moving heavy objects. In physics, work has a very speciﬁc deﬁnition that involves motion and forces. Work is done by a force only if
that force causes a net displacement of the object. There are two key points here: a force is only responsible for motion if it is in the same direction as that motion, so forces that are perpendicular to motion do not result in work, and there must be a net displacement or no work has been done. The mathematical deﬁnition of work done on an object is W = F ∆x cos θ (5.1) where F is the force applied to the object, ∆x is the displacement, θ is the angle between the force and the direction of motion, and W is the work. Work is measured in units of joules (joule ). Note that if the force is doubled, work is doubled or if the object is displaced twice as far, then work is also doubled. Work is a scalar quantity — it has a magnitude, but no direction. Work can, however, be positive or negative; it’s negative when the applied force is opposite to the direction of motion. 36 Example: Work on a block A block of mass m = 2.50 kg is pushed a distance d = 2.20 m along a frictionless horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle θ = 25.0◦ below the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gravity, and (d) the net force on the block. Solution: (a) The applied force has both horizontal and vertical components, but because the motion is entirely horizontal, only the horizontal component contributes to the force. W = F d cos θ W = (16.0 N)(2.20 m) cos(−25.0◦) W = 32 J. (b) The normal force is perpendicular to the motion, so it does not do any work. (c) The force of gravity is also perpendicular to the motion, so it also does no work. (d) We could calculate the net force by vectorially adding the normal force, the applied force and the force of gravity and then ﬁnd the work done by that force. Or we can save ourselves some eﬀort by remembering that only forces along the direction of motion contribute to the work. The only force that has a horizontal component is the applied force, and we found the work due to that force in
part (a). 5.2 Kinetic energy Energy is an indirectly observed quantity that measures an object’s capacity to do work. Energy comes in many diﬀerent forms and can easily change from one form to another, but the total amount of energy in the universe (or in an isolated system) stays the same. That means that energy is a conserved quantity. The concept of energy provides an alternative formulation for Newton’s laws. An object’s energy determines it’s potential to do work and the work it can do is related to the net force exerted by the object. If the force is constant, the acceleration is also constant and we can use kinematics equations, namely, Wnet = Fnet∆x = ma∆x. v2 − V 2 0 = 2a∆x a∆x = v2 − v2 0 2. We can substitute this into the work equation, Wnet = m v2 − v2 0 2 Wnet = 1 2 mv2 − 1 2 mv2 0. (5.2) This equation tells us that the net work done on an object leads to a change in a quantity of the form 1 2 mv2. This term is called the kinetic energy of the object and it is the energy of the motion of the object. The equation tells us that any net work done on an object leads to a change in its kinetic energy and for this reason, the equation is known as the work-energy theorem. It’s important to realize that this is just an alternative formulation of Newton’s second law — two diﬀerent ways of looking at the same process. Both Newton’s second law and the work-energy theorem tell us that interacting with an object in a certain way (by applying a force in Newton’s view, or doing work in the 37 energy view) will lead to changes in the object’s velocity. Why do we need two ways to describe the same process? It is sometimes more convenient to use one formulation over the other when solving problems. The energy formulation uses scalars rather than vectors, which can be easier for calculations, but it is sometimes hard to determine the net work done on an object without considering forces. Example: Stopping a ship A large cruise ship of mass 6.50 × 107 kg has a speed of 12.0 m/s at some instant. (a) What is the ship’s
kinetic energy at this time? (b) How much work is required to stop it? (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of 2.5 km? Solution: (a) We know the ship’s initial speed and we know that the ship’s kinetic energy is determined by its speed, mv2 K0 = 1 2 1 2 K0 = 4.7 × 109 J. K0 = (6.50 × 107 kg)(12.0 m/s)2 (b) We want the ship’s ﬁnal velocity (and also its kinetic energy) to be 0. The work-energy theorem tells us how much work is required to change an object’s kinetic energy, W = Kf − K0 W = −4.7 × 109 J. (c) Remember that work is done by a force acting on an object that travels some distance. The force slowing down the ship in this case is the drag or friction force of the water. Remember that friction is always opposite to the direction of motion so θ = 180◦. W = F ∆x cos θ F = F = W ∆x cos θ −4.7 × 109 J 2500 m cos(180◦) F = 1.9 × 106 N. 5.2.1 Conservative and nonconservative forces Forces can be broken up into two types: conservative and non-conservative. Conservative forces are forces where you can easily get back the energy you put into system. Gravity is one example of a conservative force. If you lift a book, you will be doing work against gravity to raise that book. As you lower the book, the book is now doing work on you (the normal force still points the same way, but the motion is in the opposite direction, so work is negative) and you will recover all the energy you put into the book to lift it. A nonconservative force converts energy of objects into heat or sound — forms of energy that are hard to convert back to motion. Friction is one example of a nonconservative force — you can’t recapture the energy lost to friction simply by moving the object back to where it started (like we did when we lowered the textbook). The proper physics deﬁnition of a conservative force is “A force is conservative if the work it does moving an object between two points is the same no matter
what path is taken.” 38 This is based on the idea that, for conservative forces, we can get back the energy we put in simply by moving the object back to its starting point. We can re-write the work energy theorem to speciﬁcally separate these two types of forces, Wnc + Wc = ∆KE, (5.3) where we’ve separated the work done on the object into two parts: the work done by conservative forces and the work done by nonconservative forces. 5.3 Gravitational potential energy Conservative forces have the nice property that they essentially “store” energy based on their position. When you lift a book, you’ve done some work on that book and put energy into the book. You can get that energy back by lowering the book, or you can convert that energy to something else (like kinetic energy) by letting go of the book. The book is said to have potential energy because it now has the potential to do work on another object. Let’s ﬁgure out how much work is done by gravity as a book of mass m falls from yi to yf. Remember that the formula for work is W = F ∆x cos θ. The force of gravity is Fg = −mg, the displacement is ∆x = yf − yi, and the angle between them is θ = 0. So we have, Wg = −mg(yf − yi). (5.4) Assuming we have no other conservative forces, we can explicitly put the eﬀect of gravity into the work-energy theorem, Wnc + Wg = ∆KE Wnc = ∆KE + mg(yf − yi) Wnc = ∆KE + ∆P E. (5.5) Note that when you are using the work-energy theorem, it does not matter what you choose as your reference point for measuring the height of an object. It is only changes in height (and therefore changes in gravitational potential energy) that matter and the the actual value of the potential energy at an one point. In the absence of any nonconservative forces (which will be the case in most of your homework problems), we have 0 = ∆KE + ∆P E KEi + P Ei = KEf + P Ef. (5.6) This is a conservation law — it tells us that the total
amount of kinetic energy and potential energy (sometimes called mechanical energy) for a particular system stays the same all the time. The amount of kinetic energy and potential energy might change as the object moves, but if you add the energies together, you will always get the same number. A ball sitting on the top of a hill has lots of gravitational potential energy and no kinetic energy. As it starts to roll down, it’s potential energy decreases, but it’s kinetic energy increases. When it gets to the bottom of the hill, it has no more potential energy, but lots of kinetic energy. So the amount of each type of energy changes, but the total will always be the same. 39 Example: Platform diver A diver of mass m drops from a board 10.0 m above the water’s surface. Neglect air resistance. (a) Find his speed 5.0 m above the water’s surface. (b) Find his speed as he hits the water. Solution: (a) When solving problems dealing with gravitational potential energy, we need to set a reference point (it doesn’t matter where the reference point is, we just need to be consistent for all measurements). Let’s choose the bottom of the diving board as y = 0. We are told that the diver “drops” from the board, so v0 = 0 which means that his kinetic energy is KEi = 0. The initial position of the diver is at the top of the board where his gravitational potential energy is P Ei = mgyi. The conservation of mechanical energy tells us, KEi + P Ei = KEf + P Ef 0 + mgyi = mv2 f + mgyf 1 2 2g(yi − yf ) vf = vf = 2(9.8 m/s2)(10.0 m − 5.0 m) vf = 9.90 m/s. (b) We use the same procedure as for part (a), but with a diﬀerent end point where yf = 0. KEi + P Ei = KEf + P Ef 0 + mgyi = 1 2 f + 0 mv2 vf = 2gyi vf = 2(9.8 m/s2)(10.0 m) vf = 14.0 m/s. 40 Example: Waterslides Der Stuka is a waterslide at Six Flags in
Dallas named for the German dive bombers of World War II. It is 21.9 m high. (a) Determine the speed of a 60.0 kg woman at the bottom of such a slide, assuming no friction is present. (b) If the woman is clocked at 18.0 m/s at the bottom of the slide, ﬁnd the work done on the woman by friction. Solution: (a) Let’s choose the bottom of the slide as y = 0. We can assume that the woman starts from rest, so v0 = 0 which means that her kinetic energy is KEi = 0 at the top of the slide. The initial position of the woman is at the top of the board where her gravitational potential energy is P Ei = mgyi. We are interested in the ﬁnal position where her height is yf = 0 and so her gravitational potential energy is P Ef = 0. The conservation of mechanical energy tells us, KEi + P Ei = KEf + P Ef 0 + mgyi = 1 2 f + 0 mv2 vf = 2gyi vf = 2(9.8 m/s2)(21.9 m) vf = 20.7 m/s. (b) In this case we have to use the full work-energy theorem, Wnc = KEf − KEi + P Ef − P Ei f − 0 + 0 − mgyi mv2 Wnc = 1 2 1 2 Wnc = −3.16 × 103 J. Wnc = (60.0 kg)(18.0 m/s)2 − (60.0 kg)(9.8 m/s2)(21.9 m) Note that the work done by friction is negative because the woman is losing her energy to friction. 5.4 Spring potential energy When you compress or stretch a string, you have to apply a force and therefore do some work on the spring. When you move the spring back to its original position, that energy is given back to you. Like gravity, the spring force is a conservative force — any energy you put into the spring when it is stretched or compressed is returned when the spring moves back to its original position. Springs exert a force on an object when they are stretched or compressed and the more you stretch or compress the spring, the larger the force trying to return the spring to its original position. So the force exerted by a
spring is proportional to the displacement, Fs = −k∆x, (5.7) where k is a proportionality constant called the spring constant (units of newtons per meter). This constant is diﬀerent for each spring. This equation is often called Hooke’s law after Robert Hooke who discovered the relationship. In the case of a spring, we measure the displacement from the equilibrium position of the spring. That is x = 0 is the point at which the spring is neither compressed nor stretched. The spring force is sometimes called a restoring force because it tries to return the spring to equilibrium. Calculating the work done by a spring is not as straightforward as calculating the work done by gravity because the size of the spring force changes as the displacement changes (remember the force of gravity is the same no matter the height of the object). The work done by the spring force when an object moves from 41 xi to xf is Ws = −( 1 2 kx2 f − 1 2 kx2 i ). (5.8) We can include this in the work-energy theorem (if there is a spring involved in our system) as part of the work done by conservative forces, Wnc = ∆KE + ∆P Eg + ∆P Es, (5.9) where the potential energy of a spring is 1 2 kx2. Example: Block on a spring A block with mass of 5.00 kg is attached to a horizontal spring with spring constant k = 4.00 × 102 N/m. The surface the block rests upon is frictionless. If the block is pulled out to xi = 0.05 m and released, (a) ﬁnd the speed of the block when it ﬁrst reaches the equilibrium point. (b) Find the speed when x = 0.025 m, and (c) repeat part (a) if friction acts on the block with coeﬃcient µk = 0.150. Solution: (a) In the ﬁrst part of the problem there is no friction, so there aren’t any nonconservative forces and the work-energy theorem can be written as KEi + P Egi + P Esi = KEf + P Egf + P Esf. We can simplify this a little more by realizing that all the action takes place at the same height, so there are no changes in gravitational potential energy and we can remove
that from the equation, KEi + P Esi = KEf + P Esf. The problem states that the block is “released” at a certain point — this means that the initial velocity is vi = 0. So the initial kinetic energy is also 0. The ﬁnal position of the object is at the equilibrium point (xf = 0), so the spring potential energy at this point is also 0. 0 + 1 2 kx2 i = mv2 f + 0 1 2 vf = vf = kx2 i m (4.00 × 102 N/m)(0.05 m)2 5.00 kg vf = 0.45 m/s. (b) This time we’re asked for the velocity at a non-equilibrium point. The method we use is still the same, we just have a non-zero ﬁnal potential energy, 0 + 1 2 kx2 i = 1 2 mv2 f + k(x2 kx2 f 1 2 i − x2 f ) m vf = vf = (4.00 × 102 N/m)[(0.05 m)2 − (0.025 m)2] 5.00 kg vf = 0.39 m/s. 42 (c) Now we add friction, so we will need to consider the energy lost to this nonconservative force. First we need to ﬁnd the magnitude of the frictional force, so we need a free-body diagram of the block. The block has four forces acting on it: gravity pulling down, the normal force pushing up, the spring force pulling towards equilibrium, and the force of friction pulling away from equilibrium. There is no acceleration in the y direction, so we must have Fy = 0 N − mg = 0 N = mg. We know that the frictional force is related to the normal force, The work done by friction then is fk = µkN fk = µkmg. Wf = fk∆x cos θ Wf = −µkmgxi. The work-energy theorem now has to include the nonconservative work, Wnc = ∆KE + ∆P Es −µkmgxi = mv2 f − 1 2 kx2 i 1 2 vf = vf = kx2 i m − 2µkgxi (4.00 × 102 N/m)(
0.05 m)2 5.00 kg − 2(0.150)(9.8 m/s2)(0.05 m) vf = 0.230 m/s. 43 Example: Circus acrobat A 50.0 kg circus acrobat drops from a height of 2.0 m straight down onto a springboard with a force constant of 8.00 × 103N/m By what maximum distance does she compress the spring? Solution: There aren’t any nonconservative forces in this problem, but spring potential, gravitational potential and kinetic energy all play a role. In this problem, we need to be very clear about how we are measuring distances because there are two displacements that are relevant: her change in height and the compression of the spring. Let’s set y = 0 to be the point of maximum spring compression and let’s call the distance that the spring compresses d. We will call the acrobat’s height above the uncompressed springboard h. Now let’s use the work-energy theorem, KEi + P Egi + P Esi = KEf + P Egf + P Esf 0 + mg(h + d kd2 − mgd − mgh = 0. 1 2 kd2 We have a quadratic equation and will need to use the quadratic formula, d = d = mg ± m2g2 + 2kmgh k mg k 1 ± 1 + 2kh mg d = 0.56 m (−0.44). 5.4.1 Power Power is the rate at which energy is transferred from one object to another. Remember that work is the amount of energy transferred from one object to another, so the average power will be the amount of work done over some period of time, P =. (5.10) W ∆t The unit of power is the Watt (W) or joule/second (J/s). We can write the power in another form by using the deﬁnition of work W = F ∆x cos θ, P = F ∆x cos θ ∆t P = F v cos θ. We can generalize this equation (if we use calculus) to get an equation for the instantaneous power, P = F v cos θ, where P and v are the instantaneous power and velocity rather than the average power and velocity. (5.11) (5.12
) 44 Example: Shamu Killer whales are able to accelerate up to 30 mph in a matter of seconds. Neglecting the drag force of water, calculate the average power a killer whale with mass 8.00 × 103 kg would need to generate to reach a speed of 12.0 m/s in 6.00 s. Solution: To ﬁnd the power needed by the whale, we need to ﬁgure out how much work the whale has to do to reach a speed of 12.0 m/s. We do not know the magnitude of the force needed to generate this acceleration, so we can’t use the deﬁnition of work. (We also can’t ﬁnd the acceleration using kinematics because we can’t assume constant acceleration). The other option is to use the work-energy theorem, Wnet = ∆KE mv2 f − 0 Wnet = 1 2 1 2 Wnet = 5.76 × 105 J. Wnet = (8.00 × 103 kg)(12.0 m/s)2 We know the elapsed time, so we can use the equation deﬁning average power, P = P = W ∆t 5.76 × 105 J 6.00 s P = 9.6 × 104 W. You might be wondering why we did not use the equation P = F v cos θ. That equation uses the average velocity and we are given the ﬁnal velocity. We can’t ﬁnd the average velocity without assuming constant acceleration, which we can’t do in this case. 45 Chapter 6 Momentum and Collisions 6.1 Momentum and impulse You probably have an intuitive deﬁnition of momentum. Objects with a large amount of momentum are hard to stop, i.e. a larger force is required to stop an object with lots of momentum. In physics, of course, we like to have precise deﬁnitions for these concepts, so we deﬁne the linear momentum as p = mv. (6.1) The linear momentum is proportional to both mass and velocity. That is the more massive an object, the more momentum it has and the faster an object moves the more momentum it has. The unit of momentum is kilogram meter per second (kg · m/s). Notice that momentum is a vector that points in the same direction as an object�
�s velocity. As usual, when dealing with vectors, you will break up momentum into its x and y components, px = mvx py = mvy. The momentum is related to the kinetic energy of an object, KE = 1 2 mv2 = (mv)2 2m = p2 2m. (6.2) The concept of momentum is closely tied to the idea of inertia and force. A force is required to change the momentum of an object. We can actually restate Newton’s second law in terms of momentum, Fnet = ma = m ∆v ∆t = ∆p ∆t, (6.3) which tells us that the change in momentum over some time is equal to the net force. Two implications arise from this equation. First, if there is no net force, the momentum does not change. Second, to change an object’s momentum you need to continuously apply a force over some time period (however small). We call the change in momentum of an object the impulse, I = ∆p = Fnet∆t. (6.4) This is called the impulse-momentum theorem. Most forces vary over time, making it diﬃcult to use the idea of impulse without calculus. We can, however, replace a time-varying force with an average force which is a constant force that delivers the same impulse in the time ∆t as the real time-varying force. In this case, ∆p = Fav∆t. 46 (6.5) Example: Car crash In a crash test, a car of mass 1.50 × 103 kg collides with a wall and rebounds. The initial and ﬁnal velocities are vi = −15.0 m/s and vf = 2.60 m/s. If the collision lasts for 0.150 s, ﬁnd (a) the impulse delivered to the car due to the collision and (b) the size and direction of the average force exerted on the car. Solution: The impulse is the change in momentum, I = pf − pi I = m(vf − vi) I = (1.50 × 103 kg)(2.60 m/s − (−15.0 m/s)) I = 2.64 × 104 kg · m/s. We know that the impulse is related to the average force, Fav =
I ∆t 2.64 × 104 kg · m/s 0.150 s Fav = 1.76 × 105 N. Fav = 6.2 Conservation of momentum If the two objects are isolated, then we can Let’s think about what happens when two objects collide. consider them as a single system. While the two objects will exert forces on each other during the collision, there are no net external forces acting on the system as a whole. If there are no net forces, then the total momentum of the system stays the same throughout the collision process. This is known as the conservation of momentum. Suppose a particle with mass m1 is travelling with velocity v1i towards a particle with mass m2 which is travelling towards the ﬁrst particle with velocity v2i. These two particles will eventually collide. After they collide, the ﬁrst particle moves away from the second with a velocity v1f and the second particle moves away with velocity v2f. While they are colliding, there will be a contact force between them. The force from particle 2 on particle 1 will change the momentum of particle 1, F21∆t = m1v1f − m1v1i, and similarly, the force from particle 1 on particle 2 will change the momentum of particle 2, F12∆t = m2v2f − m2v2i. (6.6) (6.7) By Newton’s third law, we know that the contact forces are an action/reaction pair and so they must be equal in magnitude and opposite in direction, F21∆t = − F12∆t m1v1f − m1v1i = −(m2v2f − m2v2i) m1v1i + m2v2i = m2v2f + m1v1f. (6.8) This equation tells us that if we add the momenta of the particles before the collision, that will be the same as the sum of all momenta after the collision. 47 Example: Littering ﬁsherman A 75 kg ﬁsherman in a 125 kg boat throws a package of mass 15 kg horizontally with a speed of 4.5 m/s. Neglecting water resistance, and assuming the boat is at rest before the package is thrown, ﬁnd the velocity of the boat after the package is
thrown. Solution: We will treat the ﬁsherman/boat as a single object for the purpose of this problem; the package will be treated as a separate object. Everything is at rest before the package is thrown, so there is no initial momentum. After the package is thrown, the boat/ﬁsherman will have some recoil velocity, 0 = mbVb + mpvp vb = − vb = − mpvp mb (15 kg)(4.5 m/s) 200 kg vb = −0.38 m/s. 6.2.1 Collisions While momentum is conserved during a collision, kinetic energy is not necessarily conserved. It is important to stress that total energy is always conserved, but certain types of energy, like kinetic energy are not always conserved. During a collision, energy is often lost to friction, sound, heat or deformation of the objects. We can classify collisions into several diﬀerent types: Elastic collision: An elastic collision is a collision in which both momentum and kinetic energy are con- served. Inelastic collision: An inelastic collision is a collision in which momentum is conserved, but kinetic energy is not. Perfectly inelastic collision: A perfectly inelastic collision is a collision in which momentum is conserved, kinetic energy is not conserved and the two objects stick together and move with the same velocity after the collision. 48 Example: Truck versus compact A pickup truck with mass 1.80 × 103 kg is travelling eastbound at 15.0 m/s, while a compact car with mass 9.00 × 102 kg is travelling westbound at -15.0 m/s. The vehicles collide head-on, becoming entangled. (a) Find the speed of the entangled vehicles after the collision. (b) Find the change in the velocity of each vehicle. (c) Find the change in the kinetic energy of the system consisting of both vehicles. Solution: (a) This problem involves a collision, so we should use conservation of momentum. Before the collision, we know the velocities of both the truck and compact. After the collision, they have the same velocity. mtvt + mcvc = (mt + mc)v mtvt + mcvc mt + mc v = v = (1.80 × 103 kg)(15.0 m/s) − (9.00 × 102 kg)(15.0 m/s) 1
.80 × 103 kg + 9.00 × 102 kg v = 5.0 m/s. (b) The change in velocity of the truck is ∆v = v − vt = 5.0 m/s − 15.0 m/s = −10 m/s. The change in velocity of the car is ∆v = v − vc = 5.0 m/s + 15.0 m/s = 20 m/s. (c) The change in kinetic energy is ∆KE = KEf − KEi 1 2 ∆KE = −2.7 × 105 J. ∆KE = (mt + mc)v2 − 1 2 mtv2 t − 1 2 mcv2 c 49 Example: Billiard balls Two billiard balls of identical mass move toward each other. Assume that the collision between them is perfectly elastic. If the initial velocities of the balls are 30.0 cm/s and -20.0 cm/s, what are the velocities of the balls after the collision? Assume friction and rotation are unimportant. Solution: This problem involves a collision, so we should use conservation of momentum. Before the collision, we know the velocities of both balls. After the collision, both velocities are unknown. mv1i + mv2i = mv1f + mv2f v1i + v2i = v1f + v2f v1i − v1f = v2f − v2i. We only have one equation with two unknowns, so we will need to ﬁnd another equation. We are told that the collision is perfectly elastic, so we know that kinetic energy is conserved, 1 2 mv2 mv2 mv2 1 1f + 2i = 2 1f + v2 2i = v2 2f 2f − v2 1f = v2 2i (v1i − v1f )(v1i + v1f ) = (v2f − v2i)(v2f + v2i). 1 1i + 2 v2 1i + v2 1i − v2 v2 mv2 2f 1 2 Let’s divide the two equations, (v1i − v1f )(v1i + v1f ) v1i − v1f = (v2f − v2
i)(v2f + v2i) v2f − v2i v1i + v1f = v2f + v2i. This equation is easier to deal with than the one with squared velocities, so let’s solve it for v1f and substitute into the conservation of momentum equation v1f = v2f + v2i − v1i v1i − v2f − v2i + v1i = v2f − v2i v2f = v1i v2f = 30.0 cm/s v1f = v1i + v2i − v1i v1f = v2i v1f = −20.0 cm/s. The balls have swapped their velocities as if they had passed through each other. The linear equation derived in the above example v1i − v2i = −(v1f − v2f ) (6.9) is actually true for any elastic collision, even if the masses are not equal. Instead of using the conservation of kinetic energy, which has quadratic terms and is hard to use, you can use the above linear equation for an elastic collision. 50 6.2.2 Collisions in two dimensions Most collisions are not head-on collisions with both objects travelling along a single line before and after the collision. Although most collisions occur in three dimensions, we will limit ourselves to two-dimensional collisions. When doing two-dimensional momentum problems, we break up momentum into x and y components, just as we did for Newton’s second law. The conservation of momentum tells us that momentum will be conserved in the x direction and in the y direction. That is,we now have two equations m1v1ix + m2v2ix = m1v1f x + m2v2f x m1v1iy + m2v2iy = m1v1f y + m2v2f y. (6.10) There are now three subscripts on the velocity: one telling us which object, one telling us which direction, and one telling us whether it was before or after the collision. Example 6.8: Collision at an intersection A car with mass 1.50 × 103 kg travelling east at a speed of 25.0 m/s collides at an intersection with a 2.50 × 103 kg van travelling north at a speed of 20.0
m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming friction between the vehicles and the road can be neglected. Solution: Before the collision, the car has a velocity only in the x direction and the van has a velocity only in the y direction. After the collision, the velocity of the combined object will have components in both directions, mcvc + 0 = (mc + mv)vf x 0 + mvvv = (mc + mv)vf y. We can use these equations to ﬁnd the x and y components of the ﬁnal velocity vf x = vf y = mc mc + mv mv mc + mv vc = vv = 1.50 × 103 kg 1.50 × 103 kg + 2.50 × 103 kg 2.50 × 103 kg 1.50 × 103 kg + 2.50 × 103 kg (25.0 m/s) = 9.38 m/s (20.0 m/s) = 12.5 m/s, from which we can ﬁnd the magnitude and direction f x + v2 v2 f y v = v = (9.38 m/s)2 + (12.5 m/s)2 v = 15.6 m/s. vf y vf x 12.5 m/s 9.38 m/s tan θ = tan θ = θ = 53◦. 51 Example P54: Colliding blocks Consider a frictionless track as shown in Figure P6.54 of your textbook. A block of mass m1 = 5.00 kg is released from height h = 5.00 m. It makes a head-on elastic collision with a block of mass m2 = 10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision. Solution: Let’s think about what happens in this problem. The block m1 starts from some height with no initial velocity and will travel to the bottom of the ramp, gaining speed as it falls. At the bottom of the ramp, it will hit block 2 and transfer some of its speed to block 2. We want to know if block 1 will travel backward after the collision and, if so, how far back up the ramp it travels. Let’s consider each step separately. First, the
block falls from a height h. How fast is it moving at the bottom of the ramp? This is an energy conservation problem since all of the block’s initial potential energy is converted to kinetic energy (no friction), KEi + P Ei = KEf + P Ef 0 + mgh = 1 2 f + 0 m1v2 vf = 2gh vf = 2(9.8 m/s2)(5.00 m) vf = 9.9 m/s. Now we can look at the collision. Block 1 has initial velocity v1i = 9.9 m/s and an unknown ﬁnal velocity v1f. The second block is initially at rest v2i = 0 and has some unknown ﬁnal velocity v2f. Use conservation of momentum because this is a collision m1v1i + 0 = m1v1f + m2v2f. We have one equation with two unknowns, but we also know that the collision is elastic will give us a second equation. We’ll use the linear equation derived earlier instead of the quadratic equation to make the calculation easier, v1i − 0 = −(v1f − v2f ). We’re actually not too interested in what block 2 does, so let’s solve for its ﬁnal velocity ﬁrst and substitute into the other equation to ﬁnd v1f v2f = v1i + v1f m1v1i = m1v1f + m2(v1i + v1f ) v1f (m1 + m2) = v1i(m1 − m2) v1f = v1f = v1i m1 − m2 m1 + m2 5.00 kg − 10.0 kg 5.00 kg + 10.0 kg v1f = −3.3 m/s. (9.9 m/s) Finally, we do the reverse of what we did in the ﬁrst part to see how high the block will go KEi + P Ei = KEf + P Ef 1 2 1f + 0 = 0 + mgh m1v2 h = 52 h = v2 1f 2g (3.3 m/s)2 2(9.8 m/s2) h = 0.56 m
. Chapter 7 Rotational Motion All the motion that we have studied to this point is linear motion. All the objects travelled in a straight line (or a series of straight lines). Objects do not always move in a straight line, they often rotate or move in circles. Luckily, many of the concepts you have learned for linear motion have analogues in rotational motion. 7.1 Angular displacement, speed and acceleration When describing linear motion, the important quantities are displacement ∆x, velocity v and acceleration, a. For rotational motion we use the angular displacement ∆θ, angular velocity ω, and angular acceleration α. For linear motion, the displacement measured the change in linear position of the object. For rotational motion, we want to measure the net change in angle as the object moves around the circle. You are used to measuring angles in degrees, but a more natural unit for measuring angles is the radian. Remember that the circumference of a circle of radius r is s = 2πr. Rearranging this equation a little gives s/r = 2π. This quantity is dimensionless, but it tells us that the displacement around any circle is 2π. A displacement around half the circle is π; a quarter circle is π/2. This forms the basis of the unit of radians. Note that we can convert from degrees to radians using the relation 180◦ = π. Angular quantities in physics must be expressed in radians, so be sure to set your calculators to radian mode when doing rotation problems. We can ﬁnd the angle travelled by an object rotating at a distance r through an arc length s through θ = s r You might wonder why we bother deﬁning a new type of displacement if we can just measure the arc length and use our linear displacement equations. Consider a solid object, called a rigid body, like a rotating compact disc where the entire object rotates as a unit. As the object rotates, all points on the CD will have the same angular displacement, but the arc length travelled by points on the disc will vary depending on the radius. Using the notion of angular displacement, we can easily describe the motion of the entire CD, but using linear displacement makes it diﬃcult to treat the CD as a single object. (7.1). Let’s properly deﬁne the angular properties. Suppose an object starts at an angle θi and ends at an angle θ
f after some time ∆t. The angular displacement is determined by the initial and ﬁnal angles, Note that for a rigid body the angular displacement is the same for all points on the object. The unit for angular displacement is the radian (rad). The average angular velocity of an object is the angular displacement divided by the time, ∆θ = θf − θi. (7.2) ω = ∆θ ∆t. 53 (7.3) For a rigid body, again, all points will have the same angular velocity. The units of angular velocity are radian per second (rad/s). We will use the term angular speed when we are not concerned with the direction, but just using the magnitude of the velocity. A positive angular speed denotes counterclockwise rotation and a negative angular speed denotes clockwise rotation. Angular velocity is a vector and the direction is speciﬁed by the right-hand rule. Take your right hand, curl your ﬁngers in the direction of the motion and your thumb will give the direction of the vector. This rule speciﬁes the rotation axis of the spinning object. Example: Spinning wheel A wheel has a radius of 2.0 m. (a) How far does a point on the circumference travel if the wheel is rotated through an angle of 30 rad. (b) If this occurs in 2 s, what is the average angular speed of the wheel? Solution: (a) We are ﬁrst asked to ﬁnd the arc length travelled by a point on the edge of the wheel when the wheel rotates through 30 rad. We use the equation relating the arc length to the angle, s = rθ s = (2.0 m)(30 rad) s = 60 m. (b) Angular speed is the angular displacement divided by time, ω = ω = ∆θ ∆t 30 rad 2 s ω = 15 rad/s. We deﬁne the instantaneous angular speed by taking the limit (as we did for velocity), We use this instantaneous angular speed to deﬁne the average angular acceleration, ω = lim ∆t→0 ∆θ ∆t. α = ∆ω ∆t. (7.4) (7.5) The units of angular acceleration are radian per second squared (rad/s2). A
rigid body will have the same angular acceleration at all points on the body. The direction of angular acceleration is in the same direction as angular velocity if the object is accelerating, otherwise it is in the opposite direction. 7.1.1 Constant angular acceleration For linear motion, we developed a number of useful equations when we could assume constant linear acceleration. We can derive similar equations using the angular quantities under the assumption of constant angular acceleration. Linear motion Rotational motion 2 v = vf +vi vf = vi + at ∆x = vit + 1 f = v2 v2 2 at2 i + 2a∆x ω = ωf +ωi ωf = ωi + αt 2 ∆θ = ωit + 1 f = ω2 ω2 2 αt2 i + 2α∆θ 54 Example 7.2: Spinning wheel II A wheel rotates with a constant angular acceleration of 3.5 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions. (b) What is the angular speed of the wheel at 2.00 s? (c) What angular displacement (in revolutions) results while the angular speed of part (b) doubles? Solution: (a) We are told that the angular acceleration is constant, so we can use the equations above. We are given angular acceleration, initial angular speed and a time and we want to ﬁnd angular displacement, ∆θ = ωit + 1 2 αt2 ∆θ = (2.00 rad/s)(2.00 s) + ∆θ = 11.0 rad. 1 2 (3.5 rad/s2)(2.00 s)2 To convert this to revolutions, we remember that one revolution is 2π rad, 11.0 rad/2π = 1.75. (b) We know want the ﬁnal angular speed, ωf = ωi + αt ωf = (2.00 rad/s) + (3.5 rad/s2)(2.00 s) ωf = 9.00 rad/s. (c) In this case, we have an initial and ﬁnal angular
speed and we want displacement, f = ω2 ω2 i + 2α∆θ f − ω2 ω2 i 2α ∆θ = ∆θ = (18.00 rad/s)2 − (9.00 rad/s)2 2(3.5 rad/s2) ∆θ = 34.7 rad = 5.52. 7.1.2 Relations between angular and linear quantities Remember that we could relate the distance travelled along an arced path to the angular displacement, ∆s = r∆θ. We can use this relationship to ﬁnd relationships between other angular and linear quantities. For example, we can divide both sides of this equation by time to get the relationship between angular and linear velocity, = r ∆θ ∆s ∆t ∆t vt = rω. (7.6) I’ve used the subscript t here to denote the tangential velocity. The instantaneous velocity of the rotating object is always tangent to the circle. That is, if the object were no longer forced to rotate, it would continue in a straight line tangent to the circle, as dictated by Newton’s ﬁrst law. So at every point on the path of a rotating object, the velocity is tangent to the path. 55 We can similarly derive a relationship between the angular and linear accelerations, where again, we use the subscript t to denote the tangential acceleration of the object. In this case, the distinction is very important because, as we shall see shortly, there is a second type of acceleration that is important during rotational motion. at = rα, (7.7) Example 7.3: Germy compact disc A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. (a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform? (b) Through what angle does the disc turn while coming up to speed? (c) If the radius of the disc is 4.45 cm, ﬁnd the tangential speed of a microbe riding on the rim of the disc when t = 0.892 s. (d) What is the magnitude of the tangential acceleration of the microbe at the given time? Solution: (a) We are told to assume uniform (constant
) angular acceleration, so we can use the equations from the previous section. ωf = ωi + αt ωf − ωi t α = α = 31.4 rad/s 0.892 s α = 35.2 rad/s2. (b) Now we want to ﬁnd the angular displacement, f = ω2 ω2 i + 2α∆θ f − ω2 ω2 i 2α ∆θ = ∆θ = (31.4 rad/s)2 2(35.2 rad/s2) ∆θ = 14.0 rad. (c) Now we use the relationship relating angular velocity and tangential velocity, vt = rω vt = (0.0445 m)(31.4 rad/s) vt = 1.4 m/s. (d) This time we need to relate the angular acceleration and the tangential acceleration, at = rα at = (0.0445 m)(35.2 rad/s2) at = 1.57 m/s2. 7.2 Centripetal acceleration We just discussed the relationship between the tangential acceleration and the angular acceleration. The tangential acceleration of an object is determined by changes in how fast an object is spinning. There is 56 another type of acceleration that is present in all rotational motion, even when the rate of rotation is not changing. Recall that Newton’s ﬁrst law tells us that objects will continue to move in a straight line unless acted upon by an outside force. In order for an object to rotate, it must be continuously pulled from the straight line path that it wants to take. The direction of motion of the object is constantly changing, which means that there is some kind of acceleration. This acceleration is called the centripetal acceleration. This acceleration is towards the center of rotation and is responsible for changing the direction of motion. The tangential acceleration is responsible for changing the speed of rotation. The centripetal acceleration is given by ac = v2 r, where v is the tangential velocity. We can re-write this formula in terms of the angular speed, ac = (rω)2 r = rω2. (7.8) (7.9) The total acceleration consists of both the tangential and the centripetal acceleration. Since the two are always perpendicular, the magnitude of the
total acceleration is given by a = c + a2 a2 t. (7.10) 57 Example 7.5: At the racetrack A race car accelerates uniformly from a speed of 40.0 m/s to a speed of 60.0 m/s in 5.00 s while travelling counterclockwise around a circular track of radius 4.00 × 102 m. When the car reaches a speed of 50.0 m/s, ﬁnd (a) the magnitude of the car’s centripetal acceleration, (b) the angular speed, (c) the magnitude of the tangential acceleration, and (d) the magnitude of the total acceleration. Solution: (a) The centripetal acceleration can be determined from the tangential speed, ac = ac = v2 r (50.0 m/s)2 4.00 × 102 m ac = 6.25 m/s2. (b) The angular speed is related to the tangential speed, ω = v r ω = 50.0 m/s 4.00 × 102 m ω = 0.125 rad/s. (c) The acceleration is uniform, so we can use the constant acceleration equations, at = at = vf − vi ∆t 60.0 m/s − 40.0 m/s 5.00 s at = 4.00 m/s2. (d) The total acceleration is found from the centripetal and tangential accelerations a2 t + a2 c a = a = (4.00 m/s2)2 + (6.25 m/s2)2 a = 7.42 m/s2. Centripetal force Since there is an acceleration that is directed towards the center, there must be some force pulling objects towards the center of rotation. This is often called the centripetal force, but this is not actually a new force. The centripetal force is one of the forces you are already familiar with (friction, normal force, tension, gravity) that happens to be pulling an object towards the center of rotation. In the case of planets orbiting the sun, the centripetal force is gravity. In the case of a yo-yo being swung in a circle, the centripetal force is tension. When solving rotational motion problems, it is often useful to set up a coordinate system based on the radial and tangential
directions of motion. In this case, the net force in the radial direction is the centripetal 58 force and gives rise to the centripetal acceleration, Fc = mac = m v2 r, (7.11) where we have used the formula for centripetal acceleration. If the centripetal force were to disappear, the spinning object would continue travelling in a straight line tangent to the circle. Centrifugal force Many people refer to the centrifugal force when discussing rotational motion. The centrifugal force is a ﬁctitious force; it does not exist. When we experience rotational motion, we often ﬂee like we are being pushed out from the center of rotation, but this feeling comes about because your body would like to continue in a straight line, tangent to the circle, and you have to exert a centripetal force to continue the rotation. Example 7.6: Car in a turn A car travels at a constant speed of 13.4 m/s on a level circular turn of radius 50.0 m. What minimum coeﬃcient of static friction between the tires and the roadway will allow the car to make the circular turn without sliding? Solution: There are three forces acting on the car: gravity, the normal force and friction. There is a frictional force pointing opposite to the direction of motion of the car, but we are not concerned with that kinetic frictional force. There is another frictional force, static friction in this case, due to the circular path of the car. Due to inertia, the car would like to move tangent to the circle, but friction prevents it from doing so. In this example, friction is the centripetal force. fs = m µsN = m µsmg = m v2 r v2 r v2 r µs = µs = v2 gr (13.4 m/s)2 (50.0 m)(9.8 m/s2) µs = 0.366. 59 Example 7.8: Riding the tracks A roller coaster car moves around a frictionless circular loop of radius R. (a) What speed must the car have so that it will just make it over the top without any assistance from the track? (b) What speed will the car subsequently have at the bottom of the loop? (c) What will be the normal force on a passenger at the bottom of the loop if the loop has
a radius of 10.0 m? Solution: (a) There are two forces acting on the car: gravity and the normal force, both acting downwards when the car is at the top of the loop. Since we want the car to make it through the loop without the assistance of the track, we set the normal force to 0. mg + N = m Fy = mac v2 t R v2 t mg = m R vt = gR. (b) We can use conservation of energy to ﬁnd the car’s speed at the bottom. At the top, the car has both potential and kinetic energy; at the bottom it only has kinetic energy, KEi + P Ei = KEf + P Ef 1 2 mv2 b + 0 mv2 t + mg(2R) = 1 2 1 2 5 v2 b = 2 vb = 5gR. gR (c) At the bottom of the loop the normal force and the force of gravity are in opposite directions, N − mg = m Fy = mac v2 b R 5gR R N = 6mg. N = m + mg 7.3 Gravitation One of the reasons we are interested in circular motion is because we know that astronomical bodies move in (roughly) circular orbits. The centripetal force that causes this rotation is the gravitational force ﬁrst written by Isaac Newton. His law of universal gravitation states that any two objects will exert an attractive force because of their mass. If two particles with masses m1 and m2 are separated by a distance r, a gravitational force F acts along a line joining them, with magnitude given by where G = 6.673 × 10−11 kg−1 · m3 · s−2 is a constant of proportionality called the constant of universal gravitation. F = G m1m2 r2 (7.12) 60 It is important to notice that both objects feel an attractive force; gravitational forces form an action-reaction pair. Example 7.10: Ceres An astronaut standing on the surface of Ceres, the largest asteroid, drops a rock from a height of 10.0 m. It takes 8.06 s to hit the ground. (a) Calculate the acceleration of gravity on Ceres. (b) Find the mass of Ceres, given that the radius of Ceres is RC = 5.1 × 102 km. (c) Calculate the gravitational acceleration 50.0
km from the surface of Ceres. Solution: (a) We can use kinematics to ﬁnd the acceleration, ∆y = v0t + 1 2 at2 a = a = 2∆y t2 2(10.0 m) (8.06 s)2 a = 0.308 m/s2. (b) This acceleration is caused by the force of gravity pulling on the rock, mM R2 C aR2 C G ma = G M = M = (0.308 m/s2)(5.1 × 102 km)2 6.673 × 10−11 kg−1 · m3 · s−2 M = 1.2 × 1021 kg. (c) From the previous part, we have ma = G a = G mM R2 M R2 a = (6.673 × 10−11 kg−1 · m3 · s−2) 1.2 × 1021 kg (5.1 × 102 km + 50 km)2 a = 0.255 m/s2. 7.3.1 Kepler’s Laws Before Isaac Newton discovered the law of gravity, another astronomer, Johannes Kepler discovered laws that helped him describe the motion of the planets in our solar system. Kepler’s three laws state 1. All planets move in elliptical orbits with the Sun at one of the focal points. 2. A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals, 3. The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun. It turns out that these laws are a consequence of Newton’s gravitational force. 61 First Law It turns out that any object moving under the inﬂuence of an inverse-square force (force varies as 1/r2) will move in an elliptical orbit. Ellipses are curves drawn such that the sum of the distances from any point on the curve to the two foci is constant. For planets in our solar system, the Sun is always at one focus, so a planets distance from the Sun will vary as the planet orbits. Second Law Kepler’s second law has an interesting consequence. Since the planet is closer to the Sun at some points of its orbit than at other points, in order to sweep out equal area in equal time, it must change its speed as it moves around the Sun. The planet will move more slowly when it
is far from the Sun and more quickly when it is close to the Sun. Third Law We can derive Kepler’s third law from the law of gravity. Suppose that an object (Mp) is moving in a circular orbit around an object (Ms) with a constant velocity (Is this possible given the second law?). We know that the object undergoes centripetal acceleration and that the force causing this acceleration is the gravitational force, Mpac = Mpv2 r = GMpMs r2. The speed of the object is simply the circumference divided by the time required for one revolution (period) Substituting, we ﬁnd v = 2πr T. Mp(4π2r2 rT 2 = GMpMs r2 T 2 = 4π2 GMs r3. (7.13) (7.14) 62 Example 7.13: Geosynchronous orbit From a telecommunications point of view, it’s advantageous for satellites to remain at the same location relative to a location on Earth. This can occur only if the satellite’s orbital period is the same as the Earth’s period of rotation, 24.0 h. (a) At what distance from the center of the Earth can this geosynchronous orbit be found? (b) What’s the orbital speed of the satellite? Solution: (a) We can use Kepler’s third law to ﬁnd the radius of the orbit, T 2 = 4π2 r3 GMs GME 4π2 r = 3 T 2 (6.673 × 10−11 kg−1 · m3 · s−2)(5.98 × 1024 kg (86400 s)2 r = 3 4(3.14)2 r = 4.23 × 107 m. (b) The speed is simply the circumference divided by the period, v = 2πr T 2(3.14)(4.23 × 107 m) 86400 s v = 3.08 × 103 m/s. v = 7.4 Torque When we studied forces, we treated all objects as point masses and assumed that it didn’t matter where on the object a force was applied. The force would simply move linearly in response to the applied force. This is an extreme simpliﬁcation of reality, it actually does matter at which point on the object the force is applied. Forces applied near the edges of an extended object
will tend to rotate the object rather than move it forward. If you lay your textbook on the table and push it with a force applied near the center of the book, it will slide forward. If you push the book with a force applied near the edge, it will rotate rather than move forward. Remember that forces cause an acceleration. Similarly, we deﬁne something called a torque which causes an angular acceleration. Forces and torques are not completely independent — forces cause torques, but torques also depend on where the force is applied and on the angle at which the force is applied. Let F be a force acting on an object, and let r be a position vector from the point of rotation to the point of application of the force. The magnitude of the torque τ exerted by the force F is τ = rF sin θ (7.15) where r is the length of the position vector, F is the magnitude of the force and θ is the angle between r and F. The unit of torque is newton-meter. Torque is a vector with the direction given by the right hand rule. Point your ﬁngers in the direction of r and curl them toward the direction of F. Your thumb will point in the direction of the torque. This will be perpendicular to the plane that contains both r and F. When your ﬁngers point in the direction of the torque, your ﬁngers will curl in the direction of rotation that the torque will cause. 63 Example 8.2: Swinging door (a) A man applies a force of F = 3.00 × 102 N at an angle of 60.0◦ to the door 2.00 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) Suppose a wedge is placed at 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door? Solution: (a) We use the above equation to calculate the torque on the door τ = F r sin θ τ = (F = 3.00 × 102 N)(2.00 m) sin(60.0◦) τ = 520 N · m. The direction of torque is out of the board/page (towards top of door). (b) We don’t want the door to rotate
, so there must be no net torque on the door. We need to identify all the torques/forces acting on the door. There are three forces acting on the door: the applied force, the force of the wedge and the force of the hinges. Although the hinges apply a force to the door, they do not exert a torque because they act at the point of rotation. τhinges + τwedge + τ = 0 0 + Fwedgerwedge sin(−90◦) + τ = 0 Fwedge = Fwedge = τ rwedge sin(−90◦) 520 N · m (1.50 m) sin(−90◦) Fwedge = 347 N. 7.4.1 Equilibrium An object that is in equilibrium must satisfy two conditions, 1. The net force must be zero, i.e. no linear acceleration ( F = 0). 2. The net torque must be zero, i.e. no angular acceleration ( τ = 0). This does not mean that the object is not moving or rotating, it can be moving at a constant velocity or rotating at a constant angular speed. If an object is in equilibrium, we can choose the axis of rotation so we should choose one which makes the calculation convenient. An axis where at least one torque is zero makes calculations easier. 64 Example 8.3: Balancing act A woman of mass m = 55.0 kg sits of the left side of a seesaw — a plank of length L = 4.00 m, pivoted in the middle. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 75.0 kg sit if the system is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of mp = 12.0 kg. (c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank. Solution: (a) We need to identify all the forces acting on the plank and their point of application. The woman and the man will push down on the plank with a force equal to their respective weights. We know the woman’s distance from the center of the plank, but the man’s distance is unknown. Gravity also acts on the plank itself, and will pull down on the center of the plank as if all the plank’s mass was
concentrated at that one point. Finally, there is a normal force that pushes upwards from the pivot point. We want the seesaw to be in equilibrium, so the sum of all torques about some axis of rotation (the pivot point in this case) must be zero, τ = 0 τN + τg + τw + τm = 0 0 + 0 + mg L 2 − M gx = 0 x = x = mL 2M (55.0 kg)(4.00 m) 2(75.0 kg) x = 1.47 m. (b) We now want to ﬁnd the normal force. We can use the ﬁrst condition of equilibrium to do this, F = 0 −M g − mg − mpg + N = 0 N = (M + m + mp)g N = (75.0 kg + 55.0 kg + 12.0 kg)(9.8 m/s2) N = 1.39 × 103 N. (c) We will now use the left end of the plank (where the woman sits) as the axis of rotation, τ = 0 τN + τg + τw + τm = 0 −N L 2 + mpg L 2 + 0 + M g(N − mpg − M g) L 2 M g (M g + mg + mpg − mpg − M g) L 2 M g mL 2M x = 1.47 m. 65 7.4.2 Center of gravity As we saw in the previous problem, we treat gravity as if it acts on a single point of an extended body. This point is called the center of gravity. Suppose we have an object with some arbitrary shape. We can treat this object as if it is divided into very small pieces of weights m1g, m2g... at locations (x1,y1), (x2,y2).... Each piece contributes some torque about the axis of rotation due to its weight. For example, τ1 = m1gx1 and so forth. The center of gravity is the point where we apply a single force of magnitude F = i mig which has the same eﬀect on the rotation of the object as all the individual little pieces. F xcg = ( i mig)xcg = xcg = i τi migxi i i mixi i mi. (7.16) This gives us the x coordinate
of the center of gravity. We can ﬁnd the y and z coordinates in a similar fashion, ycg = zcg = i miyi i mi i mizi i mi. (7.17) If an object is symmetric, the center of gravity will lie on the axis of symmetry, so it is sometimes possible to guess where the center of gravity is for such objects (like we did for the plank in the example problem). Example 8.4: Center of gravity Three objects are located on the x-axis as follows: a 5.00 kg mass sits at x = −0.500 m, a 2.00 kg mass sits at the origin, and a 4.00 kg mass sits at x = 1.00 m. Find the center of gravity. (b) How does the answer change if the object on the left is displaced upward by 1.00 m and the object on the right is displaced downward by 0.50 m? Solution: (a) We simply apply the formula for center of gravity that we just derived, xcg = xcg = i mixi i mi (5.00 kg)(−0.500 m) + (2.00 kg)(0) + (4.00 kg)(1.00 m) 5.00 kg + 2.00 kg + 4.00 kg xcg = 0.136 m. (b) The x coordinate of the center of gravity will not change since the masses have not been moved along the x-axis. We will, however, have to consider the y axis now, ycg = ycg = i miyi i mi (5.00 kg)(1.00 m) + (2.00 kg)(0) + (4.00 kg)(−0.500 m) 5.00 kg + 2.00 kg + 4.00 kg ycg = 0.273 m. 66 7.5 Torque and angular acceleration When an object is subjected to a torque, it undergoes an angular acceleration. We can derive a law similar to Newton’s second law for the eﬀect of a torque. Suppose we have an object of mass m connected to a very light rod of length r. The rod is pivoted about the end opposite the mass and its movement is conﬁned to a horizontal frictionless table. Suppose a tangential force Ft acts on the mass. This will cause a tangential acceleration
, Multiplying both sides of the equation by r, Ft = mat. Ftr = mrat, and substituting for the at = rα for the tangential acceleration gives The left side is simply the torque, Ftr = mr2α. τ = (mr2)α. (7.18) (7.19) This tells us that the torque is proportional to the angular acceleration. The constant of proportionality is mr2 and is called the moment of inertia of the object. Moment of inertia has units of kg · m2 and is denoted by I. So we can write, τ = Iα. (7.20) This is the rotational analog of Newton’s second law. 7.5.1 Moment of inertia The formula for moment of inertia that we just derived is true for a single point mass only. For extended objects, the moment of inertia will be diﬀerent. Consider a solid object rotating about its axis. We can break this object up into many little pieces like we did to ﬁnd the center of gravity. The net torque on the object will be the sum of the torques caused by all the small pieces, i τi = ( mir2 i )α. i The moment of inertia of the whole object then is I = mir2 i. i (7.21) (7.22) We can ﬁnd the moment of inertia of any object or any collection of objects by adding the moments of inertia of its constituents. Notice that the moment of inertia depends not just on the mass of an object, but on how the mass is distributed within the object. Importantly, it matters how the mass is distributed relative to the axis of rotation. 67 Table 7.1: Moments of inertia for various rigid objects of uniform composition Object Hoop or cylindrical shell Solid sphere Solid cylinder or disk Thin spherical shell Long thin rod Long thin rod Axis of Rotation Moment of inertia center center center center center end I = M R2 I = 2 5 M R2 I = 1 2 M R2 I = 2 3 M R2 I = 1 12 M L2 I = 1 3 M L2 Example 8.9: Baton twirler In an eﬀort to be the star of the halftime show, a majorette twirls an unusual baton made up of four spheres fastened to the ends of very light rods. Each rod is 1.0 m long. Two of
the spheres have a mass of 0.20 kg and the other two spheres have a mass of 0.30 kg. Spheres of equal masses are placed across from each other. (a) Find the moment of inertia of the baton through the point where the rods cross. (b) The majorette tries spinning her strange baton about the rod holding the 0.2 kg spheres. Calculate the moment of inertia of the baton about this axis. Solution: (a) When the baton is spinning around the point where the rods cross, all four spheres contribute to the moment of inertia. We can treat the spheres as point masses since their radius is small compared to the length of the rods. I = mir2 i i I = 2m1r2 + 2m2r2 I = 2(0.20 kg)(0.5 m)2 + 2(0.30 kg)(0.5 m)2 I = 0.25 kg · m2. (b) In this case only the 0.3 kg spheres contribute to the moment of inertia because the 0.2 kg spheres lie along the axis of rotation (so r = 0). I = mir2 i i I = 2m2r2 I = (0.30 kg)(0.5 m)2 I = 0.15 kg · m2. The moment of inertia for solid extended objects can be calculated using calculus. The moment of inertia for some common objects is given in Table 7.1. Note that the assumption is that the mass is distributed uniformly throughout these objects. Parallel axis theorem A useful property of the moment of inertia is that it is fairly easy to calculate the moment of inertia about an axis parallel to the axis through the center of gravity of the object. This result is called the parallel axis theorem and is as follows, Iz = Icm + M d2, (7.23) 68 where Icm is the moment of inertia of the object rotating about the center of mass, M is the mass of the object and d is the distance between the two parallel axes. 69 Example 8.11: Falling bucket A solid uniform frictionless cylindrical reel of mass M = 3.00 kg and radius R = 0.400 m is used to draw water from a well. A bucket of mass m = 2.00 kg is attached to a cord that is wrapped around the cylinder. (a) Find the tension T in the cord and acceleration a of the bucket. (b) If the
bucket starts from rest at the top of the well and falls for 3.00 s before hitting the water, how far does it fall? Solution: (a) We will need free body diagrams for both the wheel and the bucket. The bucket has two forces acting on it: tension pulling up and gravity pulling down. Note that we don’t care where these forces act on the bucket because this object is not rotating. The cylinder has three forces acting on it: gravity acting at the center and pulling down, a normal force (from the bar holding the cylinder) also acting at the center and pushing up, and tension acting at a distance R and pulling down. We know that the bucket is accelerating and the cylinder has an angular acceleration. We can use Newton’s second law on the bucket, F = ma mg − T = ma. We can use the rotational analog of Newton’s second law on the cylinder. In this case, we don’t get to choose the point of rotation because the object is rotating about a speciﬁc axis, τ = Iα R2α M Rα. Gravity and the normal force don’t contribute to the torque because they act at the axis of rotation. We now have two equations and three unknowns (a, α and T ). We will need one more equation to solve this problem. Remember that the tangential acceleration is related to the angular acceleration of a rotating object. In this case, the rope is causing the tangential acceleration of the cylinder and we know that the acceleration of the rope is the same as that of the bucket, Using this relationship, we ﬁnd a = Rα. T = 1 2 M a, which we can use to substitute into the ﬁrst equation, M a = ma mg − a(m + 1 2 1 2 M ) = mg a = 2 M ) mg (m + 1 (2.00 kg)(9.8 m/s2) 2.00 kg + 1 2 (3.00 kg) a = 5.60 m/s2. a = Now we can also ﬁnd the tension 70 (3.00 kg)(5.60 m/s2) T = 8.4 N. (b) This is a kinematics problem, ∆y = v0t + 1 2 at2 ∆y = 1 2 (5.60 m/s2)(3.0 s)2 �
�y = 25.2 m. 7.6 Rotational kinetic energy Recall that an object moving through space has kinetic energy. Similarly, a rotating object will have rotational kinetic energy. Consider the mass connected to a light rod rotating on a horizontal frictionless table. The kinetic energy of the mass is KE = mv2. 1 2 We know that the velocity is related to the angular speed, KE = 1 2 m(rω)2 = 1 2 (mr2)ω2 = 1 2 Iω2. (7.24) Notice that again the equations for translational (linear) kinetic energy and rotational kinetic energy are quite similar with moment of inertia replacing mass and angular speed replacing linear velocity. In the case of the rotating mass on a rod, either expression can be used to describe it’s energy because it only undergoes rotational motion. There are cases, however when both expressions are used such as when balls or wheels are rolling. In this case, there is rotation about the center of mass while the center of mass itself is moving through space. The translational kinetic energy refers to the energy of the center of mass’ motion while the rotational kinetic energy refers to the energy of the rotation. This new type of energy needs to be included in the work-energy theorem, Wnc = ∆KEt + ∆KEr + ∆P E. (7.25) 71 Example 8.12: Ball on an incline A ball of mass M and radius R starts from rest at a height of 2.00 m and rolls down a 30◦ slope. What is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping. Solution: We can use energy to solve this problem. Let’s consider the energy at the top and bottom of the ramp, remembering that the ball rolls (rotates) down the ramp, KEti + KEri + P Eg = KEtf + KErf + P E 0 + 0 + M gh = 1 2 M v2 + 1 2 Iω2 + 0. Remember that there is a relationship between the translational velocity and the angular speed. Note that this relationship will only hold if the object “rolls without slipping.” If the ball slips then the center of mass moves while the object is not rotating and the relationship does not hold. 2 5 1 2 M R2 2 v R M gh = gh = v = v = 1 5
v2 M v2 + 1 2 1 v2 + 2 10 7 10 7 gh (9.8 m/s2)(2.00 m) v = 5.29 m/s. The velocity is smaller than the velocity of a block sliding down the incline because some of the gravitational potential energy goes into rotational kinetic energy. 7.7 Angular momentum When we apply a torque to an object, we change its angular acceleration — and we have an equation relating the two. Just as we re-wrote Newton’s second law in terms of momentum, we can re-write the rotational equivalent in terms of angular momentum. τ = Iα = I ∆ω ∆t = I∆ω ∆t = ∆L ∆t, (7.26) where we have deﬁned L = Iω as the angular momentum of an object. If there is no net torque, then the total angular momentum of a system does not change, Li = Lf. The law of conservation of angular momentum explains why ﬁgure skaters spin faster when they bring their arms closer to their bodies. As their arms move in, their moment of inertia decreases, so their angular speed must increase to compensate. Note that angular momentum is a vector with the direction determined by the direction of angular velocity. Changes in direction of angular momentum (changes in the direction of the axis of rotation) are also subject to conservation of momentum. If the axis of rotation changes, there must be a change in the angular momentum of some other part of the system to compensate. 72 Example 8.14: Merry-go-round A merry-go-round modeled as a disk of mass M = 1.00 × 102 kg and radius R = 2.00 m is rotating in a horizontal plane about a frictionless vertical axle. (a) After a student with mass m = 60.0 kg jumps on the rim of the merry-go-round, the system’s angular speed decreases to 2.00 rad/s. If the student walks slowly from the edge toward the center, ﬁnd the angular speed of the system when she reaches a point 5.00 m from the center. (b) Find the change in the system’s rotational kinetic energy caused by her movement to r = 0.500 m. (c) Find the work done on the student as she walks to r = 0.500 m. Solution: (a)
There are two parts to the moment of inertia of the system, the moment of inertia of the disk and the moment of inertia of the person. It is the moment of inertia of the student that changes as she walks towards the center — the moment of inertia of the disk remains the same, ωf = Li = Lf (Id + Ipi)ωi = (Id + Ipf )ωf (Id + Ipi)ωi Id + Ipf ( 1 2 M R2 + mR2)ωi 1 2 M R2 + mr2 ( 1 2 (1.00 × 102 kg)(2.00 m)2 + (60.0 kg)(2.00 m)2)(2.00 rad/s) 1 2 (1.00 × 102 kg)(0.500 m)2 ωf = ωf = ωf = 4.09 rad/s. (b) The change in rotational kinetic energy is ∆KEr = ( ∆KEr = 1 2 1 2 ∆KEr = 920 J. (( − 1 2 1 2 1 2 (Id + Ipf )ω2 f − 1 2 (Id + Ipi)ω2 i (1.00 × 102 kg)(0.500 m)2 + (60.0 kg)(2.00 m)2)(4.09 rad/s)2 (1.00 × 102 kg)(2.00 m)2 + (60.0 kg)(2.00 m)2)(2.00 rad/s))2 (c) When calculating the work done by the student we need to use the change in kinetic energy of the student only, ∆KEr = ∆KEr = 1 2 1 2 Ipf ω2 f − 1 2 Ipiω2 i (60.0 kg)(2.00 m)2(4.09 rad/s)2 − 1 2 (60.0 kg)(2.00 m)2)(2.00 rad/s))2 ∆KEr = −355 J. 73 Chapter 8 Vibrations and Waves We have now studied linear motion and circular motion. There is one more very important type of motion that arises in many aspects of physics. Periodic motion, such as waves or vibrations underlies sound and light and many other physical phenomena. 8.1 Return of springs One simple type of periodic motion is an object attached to a spring.
Remember that the force of a spring is given by Hooke’s law, F = −kx. (8.1) This force is sometimes called a restoring force because it likes to pull the object back to the equilibrium position. The negative sign ensures that the force is pulling opposite to the direction of displacement. Suppose we pull the object so that the spring is stretched and let go. The spring force will cause an acceleration back towards the equilibrium position. The object will pick up speed as it moves back towards equilibrium and will overshoot the equilibrium position. Once it passes the equilibrium position the object starts to compress the spring and the force changes direction. The force now decelerates the object, eventually causing the object to stop. When the object stops, the spring is compressed and the force still points towards the equilibrium. So the object will accelerate towards the center again. In this way and object will move back and forth endlessly. This is an example of simple harmonic motion. Simple harmonic motion occurs when the net force along the direction of motion obeys Hooke’s Law. Not all periodic motion is simple harmonic motion. Two people tossing a ball back and forth is not simple harmonic motion even though it is periodic. The force causing the motion of the ball is not of the form of Hooke’s Law, so it cannot be simple harmonic motion. The acceleration of an object undergoing simple harmonic motion can be found using Newton’s second law, F = ma −kx = ma k m a = − x. (8.2) 8.1.1 Energy of simple harmonic motion Let’s consider the energy of an object attached to a spring. Suppose that we pull the object and stretch the spring then release it. Just before the object is released, the spring is at it’s maximum stretch. This is called the amplitude. The energy at this point is E = kA2, (8.3) 1 2 74 where A is the amplitude. Now we release the spring and as it moves it picks up speed. The object now has both potential energy and kinetic energy, so E = 1 2 kx2 + 1 2 mv2. We can use this to ﬁnd the velocity at any position, 1 2 kA2 = 1 2 1 2 mv2 kx2 + k m v = ± (A2 − x2). (8.4) (8.5) The ± appears because of the square root. The usual convention
is that if the object moves to the right, the velocity is positive; if it moves to the left, it is negative. 8.1.2 Connecting simple harmonic motion and circular motion When the object on the spring moves back and forth, it’s similar to an object moving with constant angular velocity around a circle. The object moving around the circle will come back to its original position at regular time intervals, just like the mass on a spring. Remember that the period of a rotating object is T = 2πr v. (8.6) For the rotating object, r is the size of the spatial displacement and corresponds to the amplitude of the mass on a spring, T = 2πA v. The velocity of the mass after it has travelled a distance A can be found from Eq. (8.5) by setting x = 0, Now we can put this into the equation for the period, v = A k m. T = 2π m k. (8.7) This represents the time it takes for a mass on a spring to return to its starting position. A larger mass gives a longer period, while a larger spring constant (stiﬀer spring) gives a shorter period. The frequency is the inverse of the period f = 1 T = 1 2π k m. (8.8) The units of frequency are cycles per second or Hz. This is related the angular frequency (which is in radians per second), k m. (8.9) ω = 2πf = 75 Example 13.5: Shock absorbers A 1.3 × 103 kg car is constructed on a frame supported by four springs. Each spring has a spring constant of 2.00 × 104 N/m. If two people riding in the car have a combined mass of 1.6 × 102 kg, ﬁnd the frequency of vibration of the car when it is driven over a pothole. Find also the period and the angular frequency. Assume the weight is evenly distributed. Solution: First we need to ﬁnd the total mass, mt = mc + mp = 1.3 × 103 kg + 1.6 × 102 kg = 1.46 × 103 kg Each spring will hold up one quarter of the total mass. The frequency is f = f = 1 2π 1 2π k m 2.00 × 104 N/m 365 kg f = 1.18 Hz. The period is the inverse of frequency and the angular
frequency is T = 1 f = 1 1.18 Hz = 0.847 s, ω = 2πf = 2π(1.18 Hz) = 7.41 rad/s. 8.2 Position, velocity and acceleration Suppose a mass is moving on a circle with constant angular velocity. If we look at it’s x position as it moves around the circle, we see that it oscillates somewhat like a mass on a spring. The x position of the mass is given by We know that the mass is moving with constant angular speed so x = A cos θ. x = A cos(ωt) = A cos(2πf t). (8.10) (8.11) This equation describes the position of an object undergoing simple harmonic motion as a function of time. We can substitute this into Eq. (8.5) v = ± (A2 − x2) k m k v = ± m v = Aω1 − cos2(2πf t) v = Aω sin(2πf t). (A2 − (A cos(2πf t))2 (8.12) The velocity also oscillates, but it is 90◦ out of phase with the displacement. When the displacement is a maximum or minimum, velocity is zero and vice versa. The maximum value (amplitude) of velocity is Aω 76 (when sin(πf t) = 1). We can also derive an expression for the acceleration a = − k m x a = −Aω2 cos ω(2πf t). (8.13) The acceleration is also sinusoidal and 180◦ out of phase with the displacement. When the displacement is a maximum, acceleration is a minimum and vice versa. The maximum acceleration (amplitude) is Aω2. 77 Example 13.6: Vibrating system (a) Find the amplitude, frequency, and period of motion for an object vibrating at the end of a horizontal spring if the equation for its position as a function of time is x = (0.250 m) cos π 8.00 t. (b) Find the maximum magnitude of the velocity and acceleration. (c) What are the position, velocity, and acceleration of the object after 1.00 s has elapsed? Solution: (a) Compare the given function to the standard function for simple harmonic motion and we can just read oﬀ the amplitude, and
the frequency x = A cos(2πf t) A = 0.250 m, 2πf = f = π 8.00 1 16 = 0.0625 Hz. The period is the inverse of frequency (b) The maximum velocity is T = 1 f = 1 0.0625 Hz = 16 s. vmax = Aω vmax = (0.250 m)(2π)(0.0625 Hz) vmax = 0.098 m/s, and the maximum acceleration is amax = Aω2 amax = (0.250 m)(2π)2(0.0625 Hz)2 amax = 0.039 m/s2. (c) We simply substitute into the given equation and for the velocity and for the acceleration x = (0.250 m) cos π 8.00 = 0.231 m, v = −(0.098 m) cos π 8.00 = −0.038 m/s, a = −(0.039 m) cos π 8.00 = −0.036 m/s2. 78 8.3 Motion of a pendulum Another type of periodic motion that you may have observed is that of a pendulum swinging back and forth. To determine if it is simple harmonic motion, we need to ﬁgure out whether there is a Hooke’s law type force causing the pendulum to move. There are two forces acting on the pendulum: the force of gravity pulls down and tension pulls towards the center of rotation. If the mass is pulled away from the equilibrium position, then the force trying to pull it back towards equilibrium is F = −mg sin θ, (8.14) where θ gives the angular displacement of the pendulum. We know that the linear displacement is s = Lθ where L is the length of the pendulum (radius of the circle on which the pendulum moves). So the force pulling along the path towards equilibrium is F = −mg sin. s L (8.15) This does not look like Hooke’s law, so in general the motion of a pendulum is not simple harmonic. At small angles, however, the sine of an angle is approximately the same as the angle itself (as long as it’s measured in radians). So for small angles, we can write F = −mg s L = − mg L s. (
8.16) Now the equation looks like Hooke’s law. The force is propotional to the linear displacement with the “spring constant” given by k = mg/L. Remember, however, that this is only valid for small angles. Recall that the angular frequency for an object undergoing simple harmonic motion is ω = k m. We have an expression for k for a pendulum, so we can substitute, ω = mg/L m = g L. From that we can ﬁnd the frequency and the period g L f = T = 1 2π ω = 1 f = 2π 1 2π L g. (8.17) (8.18) (8.19) Note that the period depends only on the length of the pendulum and not on its mass or the amplitude of the motion. 79 Example 13.7: Measuring g Using a small pendulum of length 0.171 m, a geophysicist counts 72.0 complete swings in a time of 6.00 s. What is the value of g in this location? Solution: We ﬁrst need to determine the period of oscillation, T = T = time # of oscillations 6.00 s 72.0 T = 0.833 s. We can use this to ﬁnd g, T = 2π L g g = 4π2 L T 2 g = 4π2 0.171 m (0.833 s)2 g = 9.73 m/s2. 8.4 Damped oscillations So far we have assumed that the objects will continue oscillating forever. In the real world energy losses due to friction will cause the oscillating object to slow down. In this case the motion is said to be damped. If we consider the mass on the spring, we know that the mass will oscillate for some time but that the amplitude will decrease over time. This scenario is an underdamped oscillation. Suppose now we put the mass on a spring into a liquid. If the liquid is thick enough it will prevent the oscillations and simply allow the spring to come back to its equilibrium position. If the object returns to equilibrium rapidly without oscillating, then the motion is critically damped. If the object returns to equilibrium slowly, the motion is overdamped. 8.5 Waves A wave is typically thought of as a disturbance moving through a medium. When a wave passes through a medium, the individual components of that
medium oscillate about some equilibrium point, but they do not move with the wave. Imagine a leaf ﬂoating in a pond. You throw a pebble into the pond near the leaf. This creates a wave in the water. When the wave reaches the leaf, it causes the leaf to bo up and down, but it does not carry the leaf with it. The leaf was temporarily disturbed, but once the wave passes it goes back to its original state. 8.5.1 Types of waves Suppose you ﬁx one end of a string to a wall and you hold the other end. If you quickly move your hand up and down, you will create a wave (in this case a pulse) that travels down the string. This is a traveling wave. In the case of the pulse on the string, the individual bits of string move up and down as the pulse goes through. They do not move in the direction of pulse. When the disturbed medium moves perpendicular to the direction of the wave, the wave is called a transverse wave. A longitudinal wave occurs when the disturbed medium oscillates along the direction of travel of the wave. A good example of this is to alternately stretch 80 and compress a spring. The stretched and compressed regions will move down the spring with each coil oscillating along the direction of the wave. While the coils oscillate along this direction they still do not actually move with the wave. It turns out that each point in the medium undergoes simple harmonic motion as the wave passes through. 8.5.2 Velocity of a wave The frequency of a wave is determined by the frequency of the individual oscillating points. The amplitude of the wave is the amplitude of the oscillations. The wavelength of a wave is the distance between two successive points that behave identically (peak to peak, for example). The wavelength is denoted by λ. From these quantities we can determine the speed of the wave. The wave speed is the speed at which a particular part of the wave (like the peak) travels through the medium. Remember that speed is the displacement over time, v = ∆x ∆t. We know that the wave moves a distance of one wavelength in the time it takes for one point on the wave to move through a single cycle (it’s period), v = λ T = f λ. (8.20) 8.5.3 Interference of waves One interesting aspect of waves is how they interact with other waves. Two travelling
waves will pass right through each other when they meet. When you throw two pebbles into the water near each other they will each create waves rippling from the point of entry. When those two waves meet they don’t destroy each other. Each wave comes out of the interaction undisturbed. At the point(s) where the two waves meet, they interact with each other in a process called interference. At these points, the motion of the points in space is determined by the principle of superposition: When two or more travelling waves encounter each other while moving through a medium, the resultant wave is found by adding together the displacements of the individual waves point by point. If the peaks and troughs of two waves occur at the same place at the same time, the waves are in phase and the resulting interference is constructive interference. If the peak of one wave occurs at the same time and place as the trough of another wave then the waves are inverted and the resulting interference is destructive interference. In this case the waves completely cancel each other in the region where they interact (they will re-appear once they pass through each other). 8.5.4 Reﬂection of waves Waves cannot travel in a particular medium indeﬁnitely. Eventually the waves will reach a boundary. When the waves reach the boundary, some of the wave will be reﬂected and some of the wave will be transmitted. The reﬂected wave can sometimes be inverted with respect to the incoming wave. Consider a wave on a string that approaches a wall where the string is ﬁxed. The string will pull on the wall as the wave hits. The wall will exert an equal and opposite force on the string (Newton’s third law), pulling the string in the opposite direction. This causes the reﬂected wave to be inverted. If the end of the string is free to move, the reﬂected wave will have the same orientation as the original wave. 81 8.6 Sound waves Sound waves are longitudinal waves that are caused by vibrating objects. When an object vibrates, it pushes the air near it causing alternating compression and stretching of the spacing between molecules (density) in the air. This vibration is picked up by our ears and is interpreted by our brains as sound. In a sound wave, the air molecules oscillate along the direction of travel of the wave (think of the longitudinal wave on a spring). Sound waves can
have a range of frequencies. The audible waves have frequencies between 20 and 20000 Hz. Infrasonic waves have frequencies below the audible range while ultrasonic waves have frequencies above the audible range. 8.6.1 Energy and intensity of sound waves Sound waves are created because a vibrating object pushes air molecules. The vibrating object exerts a force on the air and so is doing work on the air. The sound wave carries that energy away from the vibrating object. For waves, we don’t typically measure the total energy in the wave, but instead measure the ﬂow of energy. The average intensity I of a wave on a given surface is deﬁned as the rate at which energy ﬂows through the surface, divided by the surface area, I = 1 A ∆E∆t, (8.21) where the direction of energy ﬂow is perpendicular to the surface. The rate of energy transfer is power, so we can also write this as. (8.22) The units of intensity are W/m2. I = P A The faintest sounds a human ear can hear have an intensity of about 1 × 10−12 W/m2. This is the threshold of hearing. The loudest sounds the ear can tolerate have an intensity of about 1 W/m2. This is the threshold of pain. You’ll notice that the intensities that a human ear can detect vary over a very wide range. The quietest sounds don’t seem to us to be 1×1012 times quieter than the loudest sounds because our brains us an approximately logarithmic scale to determine loudness. This is measured by the intensity level deﬁned by β = 10 log, (8.23) I I0 where the constant I0 = 1 × 10−12 W/m2 is the reference intensity. β is measured in decibels (dB). 82 Example 14.2: Noisy grinding machine A noisy grinding machine in a factory produces a sound intensity of 1.0 × 10−5 W/m2. Calculate (a) the decibel level of this machine and (b) the new intensity level when a second, identical machine is added to the factory. (c) A certain number of additional machines are put into operation alongside these two. The resulting decibel level is 77.0 dB. Find the sound intensity. Solution: (a) For a single grinder
β = 10 log I I0 β = 10 log 1.0 × 10−5 W/m2 1.0 × 10−12 W/m2 β = 70.0 dB. (b) With a second grinder the total intensity is 2.0 × 10−5 W/m2. The decibel level is β = 10 log I I0 β = 10 log 2.0 × 10−5 W/m2 1.0 × 10−12 W/m2 β = 73.0 dB. (c) In this case we are given the decibel level and want the intensity β = 10 log I I0 77 dB = 10 log I 1.0 × 10−12 W/m2 I 1.0 × 10−12 W/m2 I = 5.01 × 10−5 W/m2. 107.7 = There are ﬁve machines in all. Many sound waves can be thought of as coming from a point source. A point source is small compared to the waves and emits waves symmetrically. The waves emitted by a point source are spherical waves; they spread in a uniform sphere. Suppose that the average power emitted by the source is Pav. Then the intensity at a distance r is I = Pav A = Pav 4πr2. The average power always remains the same, no matter the distance so we can write Since the average power is the same, we get I1 = I2 = Pav 4πr2 1 Pav 4πr2 2. I1 I2 = r2 2 r2 1. 83 (8.24) (8.25) 8.6.2 The doppler eﬀect When a moving object is making a sound, the frequency of the sound changes as as the object moves towards or away from the observer. Think of a train blowing its whistle — the whistle changes in pitch as the train approaches and as it moves away. This is known as the Doppler eﬀect. Suppose that a source is moving through the air with velocity vs towards a stationary observer. Since the source is moving towards the observer, in the same direction as the sound wave, the waves emitted by the source get “squished”. The wavelength measured by the observer is shorter than the actual wavelength emitted by the source. During a single vibration, which lasts a time T (the period), the source moves a distance vsT = vs/fs. The
wavelength detected by the observer is shortened by this amount, The frequency heard by the observer is λo = λs − vs fs. Rearranging, we get fo = v λo = v λs − vs fs = v − vs fs. v fs fo = fs v v − vs. (8.26) (8.27) The observed frequency increases when the source move towards the observer and decreases when it moves away (vs becomes negative). We can do a similar analysis for the case when the source is stationary and the observer is moving. In fact, both source and observer can be moving and this is covered by the general equation fo = fs v + vo v − vs. (8.28) The sign convention is that velocities are positive when source and observer move towards each other and negative when source and observer move away from each other. Example 14.4: Train whistle A train moving at a speed of 40.0 m/s sounds its whistle, which has a frequency of 5.00 × 102 Hz. Determine the frequency heard by a stationary observer as the train approaches the observer. Solution: The velocity of sound is 345 m/s. We use the equation for the doppler eﬀect, keeping in mind that the train is approaching the observer, fo = fs v + vo v − vs fo = (500 Hz) fo = 566 Hz. 331 m/s 331 m/s − 40 m/s 8.7 Standing waves Suppose we connect one end of a string to a stationary clamp and the other end to a vibrating object. The vibrating object will move down to the end of the string and will be reﬂected. The reﬂected wave will interact with the wave originating from the object and the two waves will combine according to the principle If the string vibrates at exactly the right frequency the wave appears to stand still so of superposition. 84 it is called a standing wave. A node occurs when the two travelling waves have the same amplitude but opposite displacement, so the net displacement is zero. Halfway between two nodes there will be an antinode where the string vibrates with the largest amplitude. Note that the distance between two nodes is half the wavelength dN N = λ/2. Suppose we ﬁx both ends of the string, then both ends must be nodes. We can then pluck the string so that we get a single antinode over the length of
the string. This is the fundamental or ﬁrst harmonic and we have half a wavelength on the string. Alternatively, we could set up our wave so that there is another node in the middle of the string. This is the second harmonic and we now have a full wavelength on the string. In fact there are many node/antinode patterns we can set up on the string always keeping in mind that the ends must be nodes. In general, we can set up waves whose wavelengths satisfy the condition λn = 2L n, (8.29) where n is the harmonic of the wave and can be any positive integer. The frequency of the harmonic is fn = v λn = vn 2L. (8.30) The velocity of a wave on a string depends on the tension in the string F and on the mass density of the string µ, F µ. v = This allows us to write fn = F µ. n 2L This series of frequencies forms a harmonic series. Example 14.8: Harmonics of a stretched wire (8.31) (8.32) (a) Find the frequencies of the fundamental and second harmonics of a steel wire 1.00 m long with a mass per unit length of 2.00 × 10−3 kg/m and under a tension of 80.0 N. (b) Find the wavelengths of the sound waves created by the vibrating wire. Assume the speed of sound is 345 m/s. Solution: (a) We use the formula for frequency with n = 1 and n = 2 for the fundamental and second harmonics, f1 = f1 = F µ F µ 1 2L 2 2L = = 1 2(1.00 m) 80.0 N 2.00 × 10−3 kg/m = 100 Hz 1 (1.00 m) 80.0 N 2.00 × 10−3 kg/m = 200 Hz. (b) The frequency of the vibrating string will be transferred to the air. The wave will then move at the speed of sound, λ1 = λ2 = v f1 v f2 = = 345 m/s 100 Hz 345 m/s 200 Hz = 3.45 m = 1.73 m 85 Standing waves can also be set up with sound waves in a pipe. Even if the end of the pipe is open, some of the sound wave will be reﬂected back into the pipe by the edges
of the pipe. The reﬂected wave will interfere with the original wave and, if the frequency is right, a standing wave can be established. For pipes, the possible standing wave frequencies will depend on whether one end of the pipe is closed or if both ends are open. If both ends are open, then there must be antinodes at either end of the pipe. The ﬁrst harmonic will have a single node in middle. If the length of the pipe is L, then the wavelength is λ1 = 2L. The second harmonic will have two nodes and a wavelength of λ2 = L. In general, the wavelength will be λn = 2L/n. The frequency will be fn = v λn nv 2L =, (8.33) where v is the speed of sound. For a pipe that is closed at one end, there must be a node at the closed end and an antinode at the open end. In this case, the ﬁrst harmonic has only a quarter wavelength inside the pipe, λ1 = 4L. The next possible harmonic will have one node inside the pipe, but not at the center. In this case, the wavelength is λ3 = 4L/3. There are no even harmonics in a pipe with one end closed; only the odd harmonics are possible. In general, the wavelength is λn = 4L/n where n is odd integers only. The frequency is fn = nv 4L. (8.34) 8.8 Beats Standing waves are created by interference between two waves of the same frequency, More often, waves of diﬀerent frequencies will interfere with each other. When waves of diﬀerent frequencies interfere, then they will alternately go in and out of phase causing periods of constructive interference followed by periods of destructive interference. If you are listening to two sound waves of diﬀerent frequencies, you will hear a sound that alternates between loud and soft. This loud/soft pattern is known as beats and is also wave. The frequency of the beats is determined by the frquency diﬀerence of the two waves fb = \|f2 − f1\|. (8.35) 86 Example: Out of tune pipes Two pipes of equal length are each open at one end. Each has a fundamental frequency of 480 Hz when the speed of sound is 347 m/s. In
one pipe the air temperature is increased so that the speed of sound is now 350 m/s. If the two pipes are sounded together, what beat frequency results? Solution: We will need to ﬁnd the new fundamental frequency of the second pipe. In order to do this, we need to know the length of the pipe. We know the frequency of the unheated pipe, so we can ﬁnd the length, f1 = L = L = v 2L v 2f 347 m/s 2(480 Hz L = 0.36 m. Now we can ﬁnd the new fundamental frequency of the second pipe, f1 = f1 = v 2L 350 m/s 2(0.36 m f1 = 486 Hz. The frequency of beats is the diﬀerence between those two frequencies, fb = \|f1 − f2\| fb = \|486 Hz − 480 Hz\| fb = 6 Hz. 87 Chapter 9 Solids and Fluids 9.1 States of matter The matter you interact with every day is typically classiﬁed as being in one of three states: solid, liquid or gas. There is also a fourth state of matter that you will not ordinarily encounter, plasma. Matter consists of molecules, which are groups of atoms. The properties of particular states of matter are determined by how molecules of a substance interact. At low temperatures, most substances are solid. Macroscopically, solids have a deﬁnite volume and shape. In a solid, the molecules are held in (relatively) ﬁxed positions relative to each other. There are usually electrical bonds between molecules that make it diﬃcult for molecules to move away from each other. As temperature increases, substances change from solid to liquid. A liquid has a deﬁnite volume, but no ﬁxed shape. In a liquid, the molecules are weakly bound to each other and so can move within the substance with some freedom. Further increases in temperature completely break the bonds between molecules, changing the liquid into a gas. A gas has no deﬁnite shape and no deﬁnite volume. The molecules of a gas are far from each other (relative to the size of the molecules) and very rarely interact. This means that the gas can expand to ﬁll an volume. At extremely high temperatures (such as those encountered inside stars)
the molecules and atoms of the substance are torn apart. Positive and negatively charged particles are free to move around within the substance creating long-range electrical and magnetic forces. This is a plasma. 9.1.1 Characterizing matter Even though substances may be in the same state and will have some broad general characteristics in common, they are by no means identical. For example, two equal masses of diﬀerent substances may not take up the same volume. This property is the density of a substance and is deﬁned as the mass divided by its volume, ρ =. (9.1) M V The SI unit for density is kg/m3. The density of a liquid or solid varies slightly with temperature and pressure. The density of a gas is very sensitive to temperature and pressure. Liquids are generally, but not always, less dense than solids. Gasses are about 1000 times less dense than liquids or solids. It is sometimes convenient to standardize density by comparing it to some standard. The speciﬁc gravity of a substance is the ratio of its density to the density of water at 4 ◦C, which is 1.0 × 103 kg/m3. 88 9.2 Deformation of solids While a solid tends to have a deﬁnite shape, the shape can often be altered with the application of a force. While a strong enough force will permanently alter the shape, often when the force is removed, the substance will return to its original shape. This is called elastic behaviour. Diﬀerent substances have diﬀerent elastic properties, so we will need some way to quantify or characterize this. The stress on a material is the force per unit area that is causing some deformation. The strain is a measure of the amount of deformation in the material. For small stresses, stress is proportional to the strain with the constant of proportionality depending on the properties of the material. The proportionality constant is called the elastic modulus. The equation for the elasticity of a substance is similar to Hooke’s law, F = −k∆x and the elastic modulus can be thought of as a spring constant. It is a measure of the stiﬀness of a material. A substance with a large elastic modulus is hard to deform. 9.2.1 Young’s modulus Suppose we have a long bar of cross-sectional A and length L0 that
is clamped at one end. When we apply an external force F along the bar, we can change the length of the bar. At this new length, the external force is balanced by internal forces that resist the stretch. The bar is said to be stressed. The tensile stress is the magnitude of the external force divided by the cross-sectional area. The tensile strain is the ratio of the change in length to the original length. Since we know that stress and strain are proportional, we have In this equation the proportionality constant is called Young’s modulus. We can re-write this equation as F A = Y ∆L L0. (9.2) F = Y A L0 ∆L, so that it looks like Hooke’s law with a spring constant of k = Y A/L0. The Young’s modulus depends on whether the material is being stretched or compressed. Many materials are easier to stretch than compress. The elastic response is also not quite linear and substances have an elastic limit. The elastic limit is the point at which the stress is no longer proportional to the strain. If stretched (compressed) beyond this limit, substances will not return to their original shape once the force is released. The ultimate strength is the largest stress that the substance can endure and any force beyond that reaches the substance’s breaking point. 9.2.2 Shear modulus Suppose we have a rectangular block of some substance. One side of the rectangle is held in a ﬁxed position. A force is applied to the other side, parallel to the side (think of sliding the cover of a book that is sitting on a table). This is called shear stress. The shear stress is the ratio of the magnitude of the applied force to the area of the face being sheared. The shear strain is the ratio of the horizontal distance moved to the height of the object. F A ∆x h = S, (9.3) where S is the shear modulus of the substance. In this case, the “spring constant” is k = SA/h. A substance with a large shear modulus is diﬃcult to bend. 89 9.2.3 Bulk modulus Suppose we have a block of some substance and we squeeze it uniformaly with a perpendicular force from all sides. This type of squeezing is common when a substance is immersed in a ﬂuid. The volume
stress is the ratio of the change in the magnitude of the applied force to the surface area. The volume strain is the ratio of the change in volume to the original volume. ∆F A = −B ∆V V, (9.4) where B is the bulk modulus. The negative sign appears so that B is positive. An increase in the external force (more squeezing) results in a decrease of the volume. Materials with a large bulk modulus are diﬃcult to compress. Example: Shear stress on the spine Between each pair of vertebrae of the spine is a disc of cartilage of thickness 0.5 cm. Assume the disc has a radius of 0.04 m. The shear modulus of cartilage is 1 × 107 N/m2. A shear force of 10 N is applied to one end of the disc while the other end is held ﬁxed. (a) What is the resulting shear strain? (b) How far has one end of the disc moved with respect to the other end? Solution: (a) The shear strain is caused by the shear force, strain = strain = F AS 10 N π(0.04 m)2(1 × 107 N/m2) strain = 1.99 × 10−4. (b) A shear strain is deﬁned as the displacement over the height, strain = ∆x h ∆x = h × strain ∆x = (0.5 cm)(1.99 × 10−4) ∆x = 0.99 µm. 9.3 Pressure and ﬂuids While a force can deform or break a solid, forces applied to a ﬂuid have a diﬀerent result. When a ﬂuid is at rest, all parts of the ﬂuid are in static equilibrium. This means that the forces are balanced for every point in the ﬂuid. If there was some kind of a force imbalance at one point, then that part of the ﬂuid would move. When discussing ﬂuids, we often don’t consider a force directly, but rather use the pressure which is the force per unit area, since ﬂuids (and the forces that act on them) tend to be extended over some region of space. Mathematically, the pressure is given by the formula, P = F A.
(9.5) The units of pressure are newton per meter2 or pascal (Pa). Suppose we have some ﬂuid sitting in equilibrium in a large container. Consider the forces acting on a piece of the ﬂuid extending from y1 to y2 (y = 0 is the top of the ﬂuid) and having a cross-sectional area A. There are three forces acting on this piece of ﬂuid: the force of gravity (M g), the force caused by the 90 pressure of the ﬂuid above this piece pushing down (P1A), and the force caused by the pressure from the ﬂuid below pushing up (P2A). This piece of ﬂuid is not moving, so the forces must balance, P2A − P1A − M g = 0. (9.6) We can ﬁnd the mass of the water from the density M = ρV = ρA(y1 − y2). Substituting into our equation, we get P2 = P1 + ρg(y1 − y2). (9.7) You’ll notice that the force of the ﬂuid pushing upward is larger than the force of the ﬂuid pushing down (the diﬀerence being the weight of the ﬂuid we’re considering). For a liquid near the surface of the earth exposed to the earth’s atmosphere, this equation can give us the pressure at any depth h, P = P0 + ρgh, (9.8) where P0 = 1.013 × 105 Pa is the atmospheric pressure at sea level. The atmospheric pressure arises because air is also a ﬂuid and the large column of air over the surface of any point on earth will exert a downward pressure on the earth. This equation also suggests that if you change the pressure at the surface of a ﬂuid, then the change is transmitted to every point in the ﬂuid. This is known as Pascal’s principle. The change in pressure will also be transmitted to the containers enclosing the ﬂuid. Example: Oil and Water In a huge oil tanker, salt water has ﬂooded an oil tank to a depth h2 = 5.00 m. On top of the water is a layer of oil h1 = 8.00 m deep. The oil has a density
of 700 kg/m3 and salt water has a density of 1025 kg/m3. Find the pressure at the bottom of the tank. Solution: The surface of the oil is exposed to air, so the pressure at that point will be atmospheric pressure. We can ﬁnd the pressure at the bottom of the oil layer using our equation, P1 = P0 + ρoilgh1 P1 = 1.01 × 105 Pa + (700 kg/m3)(9.8 m/s2)(8.00 m) P1 = 1.56 × 105 Pa. This is the pressure at the surface of the water layer. At the bottom of that layer, the pressure is, P2 = P1 + ρwatergh2 P2 = 1.56 × 105 Pa + (1025 kg/m3)(9.8 m/s2)(5.00 m) P2 = 2.06 × 105 Pa. 9.4 Buoyant forces The idea of buoyancy was discovered by the Greek mathematician Archimedes and is known as Archimedes’ principle: Any object completely or partially submerged in a ﬂuid is buoyed up by a force with magnitude equal to the weight of the ﬂuid displaced by the object. Basically, buoyancy is the pressure diﬀerence between the ﬂuid below and above an object. We know that the pressure from ﬂuid below is larger than pressure from the ﬂuid above, so the object will feel lighter if we try to lift it (since the ﬂuid is helping us to move the object upward). 91 Suppose we replace the little piece of ﬂuid in our container with a piece of lead of the same volume. The pressure above and below the lead will not change — their diﬀerence is still equal to the mass of the ﬂuid. The lead is denser than water, so the piece of lead is heavier and the downward force of gravity is now larger. Since the forces are no longer in equilibrium, the lead will sink. The buoyant force is due to pressure diﬀerences in the surrounding ﬂuid and will not change if a new substance is introduced. The buoyant force is given by, B = ρﬂuidVﬂuidg, (9.9) where Vﬂuid is the volume of ﬂuid displaced by the
object. 9.4.1 Fully submerged object For a fully submerged object, the buoyant force pushes upwards while the force of gravity pulls the object downwards. M g − B = M a ρﬂuidVﬂuidg − ρobjectVobjectg = ρobjectVobjecta a = (ρﬂuid − ρobject) g ρobject. (9.10) The acceleration will be positive (upwards) if the density of the ﬂuid is larger than the density of the object. It will be negative (downwards) if the density of the object is larger than the density of the ﬂuid. 9.4.2 Partially submerged object In this case, the object is in equilibrium since it is ﬂoating in the ﬂuid and not moving either up or down. This means that the forces must be in equilibrium. B = M g ρﬂuidVﬂuidg = ρobjectVobjectg ρobject ρﬂuid = Vﬂuid Vobject. (9.11). 92 Example 9.8: Weighing a crown A bargain hunter purchases a “gold” crown at a ﬂea market. After she gets home, she hangs it from a scale and ﬁnds its weight to be 7.48 N. She then weighs the crown while it is immersed in water and now the scale reads 6.86 N. Is the crown made of pure gold? Solution: To determine whether the crown is actually made of gold, we need to ﬁnd the density of the crown. For this, we will need both the volume and mass of the crown. When the crown is weighed in the air, we have, When the crown is in water, the buoyant force needs to be included, Tair − mg = 0. Given these two equations, then, we must have that Twater + B − mg = 0. B = Tair − Twater B = 7.48 N − 6.86 N B = 0.980 N. The buoyant force is equal to the weight of the water displaced, B = ρwaterVobjg Vobj = Vobj = B ρwaterg 0.980 N (1.0 × 103 kg/m3)(9.8 m/s2) Vobj = 1.0 × 10−4 m
3. We can easily get the mass from the ﬁrst equation, m = Tair g 7.48 N 9.8 m/s2 m = 0.800 kg. m = The density of the crown is ρcrown = ρcrown = m Vobj 0.800 kg 1.0 × 10−4 m3 ρcrown = 8.0 × 103 kg/m3. The density of gold is 19.3 × 103 kg/m3, so this crown is deﬁnitely not solid gold. 93 9.5 Fluids in motion When ﬂuids move, there are two broad categories for the type of motion. Laminar or streamline motion occurs when every particle that passes a particular point moves along the same smooth path followed by previous particles passing that point. The path itself is called a streamline. During laminar motion, diﬀerent streamlines will not cross. When the motion of the ﬂuid becomes irregular, or turbulent, the streamlines disappear and neighbouring particles can end up moving in very diﬀerent directions. In turbulent ﬂow, you tend to see eddy currents (little whirlpools) and other non-linear patterns. We will only study laminar motion in this course — turbulent motion is very complicated — and we will only consider the motion of an ideal ﬂuid. The ideal ﬂuid has the following properties: • The ﬂuid is non-viscous, which means there is no internal friction between adjacent particles. (The viscosity of a ﬂuid is a measure of the amount of internal friction.) • The ﬂuid is incompressible, which means the density is constant. • The ﬂuid motion is steady, meaning that the velocity and pressure at each point does not change in time. • The ﬂuid moves without turbulence. This condition means that the particles have no rotational motion and no angular velocity — they only move in straight lines. 9.5.1 Equation of continuity Suppose a ﬂuid ﬂows in a pipe whose cross-sectional area increases from left to right, going from A1 at one end to A2 at the other end. Suppose the ﬂuid enters the pipe with a velocity v1. The ﬂuid entering the pipe moves a distance ∆x1 = v1
∆t in a time ∆t. The mass of water contained in this region is ∆M1 = ρwaterA1∆x1 = ρwaterA1v1∆t. We can write a similar equation for the mass ﬂowing out of the other end of the pipe, ∆M2 = ρwaterA2v2∆t. Since mass is conserved (the ﬂuid is incompressible), we must have the same amount of mass going in as is coming out, ∆M1 = ∆M2, or A1v1 = A2v2. (9.12) This equation is known as the equation of continuity. It tells us that ﬂuid will speed up or slow down as the area through which they ﬂow changes. Fluid ﬂows faster through a pipe of small area than through a pipe of large area. The product Av is also known as the ﬂow rate. 94 Example 9.12: Garden hose A water hose 2.5 cm in diameter is used by a gardener to ﬁll a 30.0 L bucket. The gardener notices that it takes 1.0 min to ﬁll the bucket. A nozzle with an opening of cross-sectional are 0.500 cm2 is then attached to the hose. The nozzle is held so the water is projected horizontally from a point 1.0 m above the ground. Over what horizontal distance will the water be projected? Solution: We ﬁrst need to determine the ﬂow rate for the water in the absence of the nozzle. We can ﬁgure this out from how long it takes to ﬁll the bucket, A1v1 = 1000 cm3 1 L A1v1 = 5.0 × 10−4 m3/s. 30 L 1.00 min 1 m 100 cm 3 1 min 60 s The ﬂow rate remains constant when the new nozzle is attached, v2 = A1v1 = A2v2 A1v1 A2 5.0 × 10−4 m3/s 0.5 × 10−4 m2 v2 = v2 = 10.0 m/s. This is the initial horizontal velocity of the water. Once the water leaves the hose, it is a projectile undergoing acceleration in the vertical direction (but not horizontally). We can �
�nd how long it takes for the water to hit the ground, 1 2 gt2 y = v0yt − 2y g t = t = 2(1.0 m) 9.8 m/s2 t = 0.452 s. Now we can ﬁnd the horizontal distance travelled, x = v0xt x = (10 m/s)(0.452 s) x = 4.52 m. 9.6 Bernoulli’s equation Suppose that the pipe with varying diameter is now angled upwards. Let’s consider the work done on the ﬂuid in the pipe. The ﬂuid at the lower end is pushed by the ﬂuid behind it. The ﬂuid at the upper end is pushed by the ﬂuid in front of it. The net work done on the ﬂuid in the pipe then is W = F1∆x1 − F2∆x2 W = P1A1∆x1 − P2A2∆x2 W = P1V − P2V. 95 (9.13) The work done on the ﬂuid can do one of two things: it can change the kinetic energy of the ﬂuid, and it can change the gravitational potential energy, W = ∆KE + ∆P E. Combining these two equations we have, Now we can put in expressions for the kinetic and potential energies, P1V − P2V = ∆KE + ∆P E. We can re-write this as P1 − P2 = 1 2 ρv2 2 − 1 2 ρv2 1 + ρgy2 − ρgy1. P1 + 1 2 ρv2 1 + ρgy1 = P2 + 1 2 ρv2 2 + ρgy2. (9.14) (9.15) (9.16) (9.17) This is known as Bernoulli’s equation. Note that this is not really a new concept, it is just the conservation of energy applied to a ﬂuid. This equation is only true, however, for laminar ﬂow. One consequence of this equation is that faster moving ﬂuids exert less pressure than slowly moving ﬂuids. We can see this by considering the pipe with a varying diameter when it is horizontally
level. In this case, Bernoulli’s equation simpliﬁes to P1 + 1 2 ρv2 1 = P2 + 1 2 ρv2 2. We know that the ﬂuid moves faster in the narrow region, so the pressure in that region must be lower than in the wide region in order for the two sides of the equation to balance. Example 9.13: Shootout A nearsighted sheriﬀ ﬁres at a cattle rustler with his trust six-shooter. Fortunately for the rustler, the bullet misses him and penetrates the town water tank, causing a leak. If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the hole when the water level is 0.500 m above the hole. Solution: We can use Bernoulli’s equation to ﬁnd the velocity. Let’s choose the ﬁrst point to be the top of the tank and the second point will be the hole. The pressure at both of these points is just P0, the standard atmospheric pressure. We assume that the water level drops very slowly, so the velocity of the ﬂuid at the top of the tank is zero. Putting these into Bernoulli’s equation, P1 + 1 2 ρv2 1 + ρgy1 = P2 + 1 2 ρv2 2 + ρgy2 1 2 ρgy1 = ρv2 2 + ρgy2 v = 2g(y1 − y2) v = 2(9.8 m/s2)(0.500 m) v = 3.13 m/s. 96 Chapter 10 Thermal physics There is a form of energy that we have to date neglected to consider in our description of objects. This is primarily because this form of energy is not typically involved in macroscopic motion. All objects have thermal energy, a type of energy which we intuitively detect as the object being hot or cold. 10.1 Temperature Determining whether an object is hot or cold is rather inexact and we would like to ﬁnd a more quantitative way of measuring thermal energy. We say that two objects are in thermal contact if energy (particularly thermal energy) can be exchanged between them. Two objects are in thermal equilibrium if they are in thermal contact but there is no net exchange of energy between them. The zer
oth law of thermodynamics (law of equilibrium) states: If objects A and B are separately in thermal equilibrium with a third object C, then A and B are in thermal equilibrium with each other. This law allows us to use a thermometer to compare the thermal energies of two objects. Suppose we want to compare the thermal energies of two objects A and B, we could put them into thermal contact and try to detect the direction of energy ﬂow. Suppose, however, that we cannot put the two objects into thermal contact directly. We can then use a third object, C, to compare the thermal energies of A and B. We ﬁrst put the thermometer into thermal contact with A until it reaches thermal equilibrium at which point we read the thermometer. We then put the thermometer into thermal contact with B until it reaches thermal equilibrium and read the temperature again. If the two readings are the same then A and B are also in thermal equilibrium. This property allows us to deﬁne temperature — if two objects are in thermal equilibrium, then they have the same temperature. The thermometer used to measure thermal energy must have some physical property that changes with temperature. Most thermometers use the fact that substances (solids, gasses, liquids) expand as temperature increases. This physical change can usually be measured visually allowing us to put a number on the temperature. We must ﬁrst, however, calibrate the thermometer. That is, we must agree on a measurement scale for temperature. The Celsius temperature scale is deﬁned by measuring the freezing point of water which is set to be 0 ◦C. and the boiling point of water which is set to be 100 ◦C. The scale most commonly used in the US is the Fahrenheit scale. On this scale, the freezing point is at 32◦F and the boiling point is at 212◦F. We can convert between the two using the formula The temperature scale most often used by scientists is the Kelvin scale. One of the problems with both the Celsius and Fahrenheit scales is that the freezing point and boiling point of water depend not only on the TF = 9 5 TC + 32. (10.1) 97 temperature but on the pressure. Scientists removed the pressure dependence by observing that the pressure of all substances goes to zero at a temperature of −273.15 ◦C. This temperature is known as absolute zero and is deﬁned to be 0 K. The second point used to
deﬁne the Kelvin scale is the triple point of water. This is the temperature and pressure at which water, water vapour, and ice exist in equilibrium. This point occurs at 0.01 ◦C and 4.58 mm of mercury. This temperature is deﬁned to be 273.16 K. This means that the unit size of both the celsius and kelvin scales are the same. We convert between the two using, TC = TK − 273.15. (10.2) 10.2 Thermal expansion Most substances increase in volume as their temperature (thermal energy) increases. The thermal energy of an object is actually a measure of the average velocity of the constituent atoms. As temperature increases, atoms move faster. In solids and liquids, these atoms cannot actually leave the substance, so their vibrational motion increases leading to an increased separation between atoms. Macroscopically, we see this as an increase in volume. If the expansion is small compared to the object’s original size, the expansion in one dimension is approximately linear with temperature, where ∆L is the change in length (not volume), L is the original length of the object, and α is the coeﬃcient of linear expansion for a particular substance. ∆L = αL0∆T, (10.3) Example 10.3: Expansion of a railroad track (a) A steel railroad track has a length of 30.0 m when the temperature is 0 ◦C. What is the length on a hot day when the temperature is 40.0 ◦C? (b) What is the stress caused by this expansion? Solution: (a) The change in length due to the temperature change, ∆L = αL0∆T ∆L = (11 × 10−6 /◦C)(30.0 m)(40.0 ◦C) ∆L = 0.013 m. So the new length is 30.013 m. (b) The railroad undergoes a linear expansion, so this is a tensile strain, = Y ∆L L = (2.0 × 1011 Pa = 8.7 × 107 Pa. F A F A F A 0.013 m 30.0 m Since their is a linear expansion of objects with temperature, there must also be a change in their area and volume. Suppose we have a square of material with a length of L0. Each dimension
of the square will undergo linear expansion and the new area is A = L2 A = (L0 + αL0∆T )(L0 + αL0∆T ) A = L2 0α∆T + (αL0∆T )2. 0 + 2L2 98 The last term in that equation will be very small, so we will ignore it, We can re-write this so that it looks like the linear expansion equation, A = A0 + 2αA0∆T. ∆A = 2αA0∆T. (10.4) (10.5) We deﬁne a new coeﬃcient γ = 2α as the coeﬃcient of area expansion. We can perform the same type of derivation and show that the increase in volume of a substance is given by ∆V = βV0∆T, (10.6) where β is the coeﬃcient of volume expansion and is given by β = 3α. 10.3 Ideal gas law The eﬀect of temperature change on a gas is somewhat more complex than in solids and liquids. A gas will expand to ﬁll a particular container no matter what the temperature. What will change instead as the temperature increases is the pressure. There is usually a fairly complex relationship between the pressure, volume and temperature of gasses, but for an ideal gas, we can derive a simple relationship. An ideal gas is a gas that is maintained at low density or pressure. In an ideal gas, particles of the gas are so far apart that they rarely interact and so we can assume there are no forces acting on any of the particles and no collisions take place. Each particle of the gas moves randomly. Since gases contain large numbers of particles, we usually count the number of particles in moles where one mole is 6.02×1023 gas particles. The number 6.02×1023 is known as Avogadro’s number and is denoted by NA. Avogadro’s number was chosen so that the mass in grams of one mole of an element is numerically the same as the atomic mass units of the element. Carbon 12 has an atomic mass of 12 amu, so one mole of carbon 12 weighs 12 g. For an ideal gas, the relationship between pressure, volume, and temperature is P V = nRT,
(10.7) where n is the number of moles of the substance, and R is the universal gas constant with a value of R = 8.31 J/mol · K. The ideal gas law tells us that the pressure is linearly proportional to temperature and inversely proportional to the volume. As temperature increases, pressure increases. As volume increases, pressure decreases. 99 Example 10.6: Expanding gas An ideal gas at 20.0 ◦C and a pressure of 1.50 × 105 Pa is in a container having a volume of 1.0 L. (a) Determine the number of moles of gas in the container. (b) The gas pushes against a piston, expanding to twice its original volume, while the pressure falls to atmospheric pressure. Find the ﬁnal temperature. Solution: (a) We need to be careful that all quantities are in SI units. We will need to convert the temperature to kelvins: T = 20 + 273 = 293 K. And we need to convert the volume to m3: V = 1.0 L = 1.0 × 10−3 m3. Now we can go ahead and plug the values into the gas law, n = P V = nRT P V RT (1.50 × 105 Pa)(1.0 × 10−3 m3) (8.31 J/mol · K)(293 K) n = n = 6.16 × 10−2 mol. (b) We can ﬁnd the new temperature from the gas law, T = T = P V nR (1.01 × 105 Pa)(2.0 × 10−3 m3) (6.16 × 10−2 mol)(8.31 J/mol · K) T = 395 K. 100ed by classical physics. One reason for this is that they are small enough to travel at great speeds, near the speed of light. Figure 1.5 Using a scanning tunneling microscope (STM), scientists can see the individual atoms that compose this sheet of gold. (Erwinrossen) At particle colliders (Figure 1.6), such as the Large Hadron Collider on the France-Swiss border, particle physicists can make subatomic particles travel at very high speeds within a 27 kilometers (17 miles) long superconducting tunnel. They can then study the properties of the particles at high speeds, as well as collide them with each other to see how they exchange energy. This
has led to many intriguing discoveries such as the Higgs-Boson particle, which gives matter the property of mass, and antimatter, which causes a huge energy release when it comes in contact with matter. Figure 1.6 Particle colliders such as the Large Hadron Collider in Switzerland or Fermilab in the United States (pictured here), have long tunnels that allows subatomic particles to be accelerated to near light speed. (Andrius.v ) Physicists are currently trying to unify the two theories of modern physics, relativity and quantum mechanics, into a single, comprehensive theory called relativistic quantum mechanics. Relating the behavior of subatomic particles to gravity, time, and space will allow us to explain how the universe works in a much more comprehensive way. Application of Physics You need not be a scientist to use physics. On the contrary, knowledge of physics is useful in everyday situations as well as in nonscientific professions. For example, physics can help you understand why you shouldn’t put metal in the microwave (Figure 1.7), why a black car radiator helps remove heat in a car engine, and why a white roof helps keep the inside of a house cool. The operation of a car’s ignition system, as well as the transmission of electrical signals through our nervous system, are much easier to understand when you think about them in terms of the basic physics of electricity. Figure 1.7 Why can't you put metal in the microwave? Microwaves are high-energy radiation that increases the movement of electrons in Access for free at openstax.org. 1.1 • Physics: Definitions and Applications 11 metal. These moving electrons can create an electrical current, causing sparking that can lead to a fire. (= MoneyBlogNewz) Physics is the foundation of many important scientific disciplines. For example, chemistry deals with the interactions of atoms and molecules. Not surprisingly, chemistry is rooted in atomic and molecular physics. Most branches of engineering are also applied physics. In architecture, physics is at the heart of determining structural stability, acoustics, heating, lighting, and cooling for buildings. Parts of geology, the study of nonliving parts of Earth, rely heavily on physics; including radioactive dating, earthquake analysis, and heat transfer across Earth’s surface. Indeed, some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines. Physics also describes the chemical processes that power the human body. Physics is involved in medical diagn
ostics, such as xrays, magnetic resonance imaging (MRI), and ultrasonic blood flow measurements (Figure 1.8). Medical therapy Physics also has many applications in biology, the study of life. For example, physics describes how cells can protect themselves using their cell walls and cell membranes (Figure 1.9). Medical therapy sometimes directly involves physics, such as in using X-rays to diagnose health conditions. Physics can also explain what we perceive with our senses, such as how the ears detect sound or the eye detects color. Figure 1.8 Magnetic resonance imaging (MRI) uses electromagnetic waves to yield an image of the brain, which doctors can use to find diseased regions. (Rashmi Chawla, Daniel Smith, and Paul E. Marik) Figure 1.9 Physics, chemistry, and biology help describe the properties of cell walls in plant cells, such as the onion cells seen here. (Umberto Salvagnin) BOUNDLESS PHYSICS The Physics of Landing on a Comet On November 12, 2014, the European Space Agency’s Rosetta spacecraft (shown in Figure 1.10) became the first ever to reach and orbit a comet. Shortly after, Rosetta’s rover, Philae, landed on the comet, representing the first time humans have ever landed a space probe on a comet. 12 Chapter 1 • What is Physics? Figure 1.10 The Rosetta spacecraft, with its large and revolutionary solar panels, carried the Philae lander to a comet. The lander then detached and landed on the comet’s surface. (European Space Agency) After traveling 6.4 billion kilometers starting from its launch on Earth, Rosetta landed on the comet 67P/ChuryumovGerasimenko, which is only 4 kilometers wide. Physics was needed to successfully plot the course to reach such a small, distant, and rapidly moving target. Rosetta’s path to the comet was not straight forward. The probe first had to travel to Mars so that Mars’s gravity could accelerate it and divert it in the exact direction of the comet. This was not the first time humans used gravity to power our spaceships. Voyager 2, a space probe launched in 1977, used the gravity of Saturn to slingshotover to Uranus and Neptune (illustrated in Figure 1.11), providing the first pictures ever taken of these planets. Now, almost 40 years after its launch, Voyager 2 is at the very edge of our solar
system and is about to enter interstellar space. Its sister ship, Voyager 1 (illustrated in Figure 1.11), which was also launched in 1977, is already there. To listen to the sounds of interstellar space or see images that have been transmitted back from the Voyager I or to learn more about the Voyager mission, visit the Voyager’s Mission website (https://openstax.org/l/28voyager). Figure 1.11 a) Voyager 2, launched in 1977, used the gravity of Saturn to slingshot over to Uranus and Neptune. NASA b) A rendering of Voyager 1, the first space probe to ever leave our solar system and enter interstellar space. NASA Both Voyagers have electrical power generators based on the decay of radioisotopes. These generators have served them for almost 40 years. Rosetta, on the other hand, is solar-powered. In fact, Rosetta became the first space probe to travel beyond the asteroid belt by relying only on solar cells for power generation. At 800 million kilometers from the sun, Rosetta receives sunlight that is only 4 percent as strong as on Earth. In addition, it is very cold in space. Therefore, a lot of physics went into developing Rosetta’s low-intensity low-temperature solar cells. In this sense, the Rosetta project nicely shows the huge range of topics encompassed by physics: from modeling the movement of gigantic planets over huge distances within our solar systems, to learning how to generate electric power from low-intensity light. Physics is, by far, the broadest field of science. GRASP CHECK What characteristics of the solar system would have to be known or calculated in order to send a probe to a distant planet, such as Jupiter? Access for free at openstax.org. 1.1 • Physics: Definitions and Applications 13 a. b. c. d. the effects due to the light from the distant stars the effects due to the air in the solar system the effects due to the gravity from the other planets the effects due to the cosmic microwave background radiation In summary, physics studies many of the most basic aspects of science. A knowledge of physics is, therefore, necessary to understand all other sciences. This is because physics explains the most basic ways in which our universe works. However, it is not necessary to formally study all applications of physics. A knowledge of the basic laws of physics will be most useful to you, so that you can use them to solve some everyday problems
. In this way, the study of physics can improve your problem-solving skills. Check Your Understanding 1. Which of the following is notan essential feature of scientific explanations? a. They must be subject to testing. b. They strictly pertain to the physical world. c. Their validity is judged based on objective observations. d. Once supported by observation, they can be viewed as a fact. 2. Which of the following does notrepresent a question that can be answered by science? a. How much energy is released in a given nuclear chain reaction? b. Can a nuclear chain reaction be controlled? c. Should uncontrolled nuclear reactions be used for military applications? d. What is the half-life of a waste product of a nuclear reaction? 3. What are the three conditions under which classical physics provides an excellent description of our universe? a. b. c. d. 1. Matter is moving at speeds less than about 1 percent of the speed of light 2. Objects dealt with must be large enough to be seen with the naked eye. 3. Strong electromagnetic fields are involved. 1. Matter is moving at speeds less than about 1 percent of the speed of light. 2. Objects dealt with must be large enough to be seen with the naked eye. 3. Only weak gravitational fields are involved. 1. Matter is moving at great speeds, comparable to the speed of light. 2. Objects dealt with are large enough to be seen with the naked eye. 3. Strong gravitational fields are involved. 1. Matter is moving at great speeds, comparable to the speed of light. 2. Objects are just large enough to be visible through the most powerful telescope. 3. Only weak gravitational fields are involved. 4. Why is the Greek word for nature appropriate in describing the field of physics? a. Physics is a natural science that studies life and living organism on habitable planets like Earth. b. Physics is a natural science that studies the laws and principles of our universe. c. Physics is a physical science that studies the composition, structure, and changes of matter in our universe. d. Physics is a social science that studies the social behavior of living beings on habitable planets like Earth. 5. Which aspect of the universe is studied by quantum mechanics? a. objects at the galactic level b. objects at the classical level c. objects at the subatomic level d. objects at all levels, from subatomic to galactic 14 Chapter 1 • What is Physics? 1.2 The Scientific Methods Section Learning Objectives By the end
of this section, you will be able to do the following: • Explain how the methods of science are used to make scientific discoveries • Define a scientific model and describe examples of physical and mathematical models used in physics • Compare and contrast hypothesis, theory, and law Section Key Terms experiment hypothesis model observation principle scientific law scientific methods theory universal Scientific Methods Scientists often plan and carry out investigations to answer questions about the universe around us. Such laws are intrinsic to the universe, meaning that humans did not create them and cannot change them. We can only discover and understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle, triumph, and disappointment inherent in any creative effort. The cornerstone of discovering natural laws is observation. Science must describe the universe as it is, not as we imagine or wish it to be. We all are curious to some extent. We look around, make generalizations, and try to understand what we see. For example, we look up and wonder whether one type of cloud signals an oncoming storm. As we become serious about exploring nature, we become more organized and formal in collecting and analyzing data. We attempt greater precision, perform controlled experiments (if we can), and write down ideas about how data may be organized. We then formulate models, theories, and laws based on the data we have collected, and communicate those results with others. This, in a nutshell, describes the scientific method that scientists employ to decide scientific issues on the basis of evidence from observation and experiment. An investigation often begins with a scientist making an observation. The scientist observes a pattern or trend within the natural world. Observation may generate questions that the scientist wishes to answer. Next, the scientist may perform some research about the topic and devise a hypothesis. A hypothesis is a testable statement that describes how something in the natural world works. In essence, a hypothesis is an educated guess that explains something about an observation. Scientists may test the hypothesis by performing an experiment. During an experiment, the scientist collects data that will help them learn about the phenomenon they are studying. Then the scientists analyze the results of the experiment (that is, the data), often using statistical, mathematical, and/or graphical methods. From the data analysis, they draw conclusions. They may conclude that their experiment either supports or rejects their hypothesis. If the hypothesis is supported, the scientist usually goes on to test another hypothesis related to the first. If their hypothesis is rejected, they will often then test a new and different hypothesis in their effort
to learn more about whatever they are studying. Scientific processes can be applied to many situations. Let’s say that you try to turn on your car, but it will not start. You have just made an observation! You ask yourself, "Why won’t my car start?" You can now use scientific processes to answer this question. First, you generate a hypothesis such as, "The car won’t start because it has no gasoline in the gas tank." To test this hypothesis, you put gasoline in the car and try to start it again. If the car starts, then your hypothesis is supported by the experiment. If the car does not start, then your hypothesis is rejected. You will then need to think up a new hypothesis to test such as, "My car won’t start because the fuel pump is broken." Hopefully, your investigations lead you to discover why the car won’t start and enable you to fix it. Modeling A model is a representation of something that is often too difficult (or impossible) to study directly. Models can take the form of physical models, equations, computer programs, or simulations—computer graphics/animations. Models are tools that are especially useful in modern physics because they let us visualize phenomena that we normally cannot observe with our senses, such as very small objects or objects that move at high speeds. For example, we can understand the structure of an atom using models, despite the fact that no one has ever seen an atom with their own eyes. Models are always approximate, so they are simpler to consider than the real situation; the more complete a model is, the more complicated it must be. Models put the Access for free at openstax.org. 1.2 • The Scientific Methods 15 intangible or the extremely complex into human terms that we can visualize, discuss, and hypothesize about. Scientific models are constructed based on the results of previous experiments. Even still, models often only describe a phenomenon partially or in a few limited situations. Some phenomena are so complex that they may be impossible to model them in their entirety, even using computers. An example is the electron cloud model of the atom in which electrons are moving around the atom’s center in distinct clouds (Figure 1.12), that represent the likelihood of finding an electron in different places. This model helps us to visualize the structure of an atom. However, it does not show us exactly where an electron will be within its cloud at any one particular time. Figure 1.12 The
electron cloud model of the atom predicts the geometry and shape of areas where different electrons may be found in an atom. However, it cannot indicate exactly where an electron will be at any one time. As mentioned previously, physicists use a variety of models including equations, physical models, computer simulations, etc. For example, three-dimensional models are often commonly used in chemistry and physics to model molecules. Properties other than appearance or location are usually modelled using mathematics, where functions are used to show how these properties relate to one another. Processes such as the formation of a star or the planets, can also be modelled using computer simulations. Once a simulation is correctly programmed based on actual experimental data, the simulation can allow us to view processes that happened in the past or happen too quickly or slowly for us to observe directly. In addition, scientists can also run virtual experiments using computer-based models. In a model of planet formation, for example, the scientist could alter the amount or type of rocks present in space and see how it affects planet formation. Scientists use models and experimental results to construct explanations of observations or design solutions to problems. For example, one way to make a car more fuel efficient is to reduce the friction or drag caused by air flowing around the moving car. This can be done by designing the body shape of the car to be more aerodynamic, such as by using rounded corners instead of sharp ones. Engineers can then construct physical models of the car body, place them in a wind tunnel, and examine the flow of air around the model. This can also be done mathematically in a computer simulation. The air flow pattern can be analyzed for regions smooth air flow and for eddies that indicate drag. The model of the car body may have to be altered slightly to produce the smoothest pattern of air flow (i.e., the least drag). The pattern with the least drag may be the solution to increasing fuel efficiency of the car. This solution might then be incorporated into the car design. Snap Lab Using Models and the Scientific Processes Be sure to secure loose items before opening the window or door. In this activity, you will learn about scientific models by making a model of how air flows through your classroom or a room in your house. • One room with at least one window or door that can be opened • Piece of single-ply tissue paper 1. Work with a group of four, as directed by your teacher. Close all of the windows and doors in the room you are working in. Your teacher may assign you
a specific window or door to study. 16 Chapter 1 • What is Physics? 2. Before opening any windows or doors, draw a to-scale diagram of your room. First, measure the length and width of your room using the tape measure. Then, transform the measurement using a scale that could fit on your paper, such as 5 centimeters = 1 meter. 3. Your teacher will assign you a specific window or door to study air flow. On your diagram, add arrows showing your hypothesis (before opening any windows or doors) of how air will flow through the room when your assigned window or door is opened. Use pencil so that you can easily make changes to your diagram. 4. On your diagram, mark four locations where you would like to test air flow in your room. To test for airflow, hold a strip of single ply tissue paper between the thumb and index finger. Note the direction that the paper moves when exposed to the airflow. Then, for each location, predict which way the paper will move if your air flow diagram is correct. 5. Now, each member of your group will stand in one of the four selected areas. Each member will test the airflow Agree upon an approximate height at which everyone will hold their papers. 6. When you teacher tells you to, open your assigned window and/or door. Each person should note the direction that their paper points immediately after the window or door was opened. Record your results on your diagram. 7. Did the airflow test data support or refute the hypothetical model of air flow shown in your diagram? Why or why not? Correct your model based on your experimental evidence. 8. With your group, discuss how accurate your model is. What limitations did it have? Write down the limitations that your group agreed upon. GRASP CHECK Your diagram is a model, based on experimental evidence, of how air flows through the room. Could you use your model to predict how air would flow through a new window or door placed in a different location in the classroom? Make a new diagram that predicts the room’s airflow with the addition of a new window or door. Add a short explanation that describes how. a. Yes, you could use your model to predict air flow through a new window. The earlier experiment of air flow would help you model the system more accurately. b. Yes, you could use your model to predict air flow through a new window. The earlier experiment of air flow is not useful for modeling the new system. c. No, you

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