SUMMARY Phys 2113 (General Physcs I) Compled by Prof. Erckson Poston Vector (m): r = xˆx + yŷ + zẑ Average Velocty (m/s): v = r Instantaneous Velocty (m/s): v = lm 0 r = ṙ Lnear Momentum (kg m/s): p = mv Average Acceleraton (m/s 2 ): a = v Instantaneous Acceleraton (m/s 2 ): a = lm 0 v = v = r Centrpetal Acceleraton. At any nstant, the moton of an object can be vewed as beng tangent to a crcle. The acceleraton can represent ether a change n speed of the object (tangental acceleraton), a change n drecton of the object (radal acceleraton), or both. The radal acceleraton s the centrpetal acceleraton. The magntude of the centrpetal acceleraton s a c = v2 r where r s the radus of that nstantaneous crcle. Its drecton s toward the center of that nstantaneous crcle, a c = v2 r ˆr = v2 r. (Centrpetal means center-seekng.) Note that for straght-lne r 2 moton, r, and a c 0. For unform crcular moton the tangental speed v s just the crcumference of the crcle dvded by the perod (T ) or tme to complete one revoluton,.e., v = 2π/T = 2πf, where f = 1/T s the frequency. Force (N=kg m/s 2 ): F = ma = p Impulse (N s): J = F = p = p f p
Work (J=N m): W = F s = (F cos θ)s Power (W=J/s): P = W t = F v; generally, P = F v Knetc Energy (J): KE = 1 2 mv2 Potental Energy (J): If the work s ndependent of path, then the force s conservatve and can be derved from a potental energy P E, where F = lm 0 (P E) s Example: Gravtatonal potental energy s P E g = mgh = mg(y y ), where y s the vertcal drecton. (The correspondng force of gravty s F g = mg.) Example: Potental energy of a sprng s P E s = 1 2 kx2, where x s the dsplacement from ts equlbrum poston. (The correspondng force s F s = kx, where k s the sprng constant, and x s the dsplacement from ts equlbrum poston.) Non-conservatve forces (e.g., frcton) are those for whch the work depends on the path. Newton s Frst Law states that n the absence of a net external force, when vewed from an nertal frame, v of an object s constant. An nertal frame of reference s one for whch a = 0. Gallean Transformatons. Poston, velocty and acceleraton measured n frame S are related to those quanttes measured n a frame S, movng wth constant velocty v S relatve to S, by r = r v S t; v = v v S ; a = a Gallean transformaton s only vald for knematc stuatons n whch veloctes are much less than the speed of lght (c = 3 10 8 m/s). Gallean transformatons are naccurate for electromagnetc phenomena, quantum phenomena, and knematc stuatons n whch v/c s a sgnfcant fracton of unty. Fcttous Forces. Fcttous forces appear when observng from a non-nertal reference frame,.e., observng from a reference frame undergong acceleraton. Examples are the centrfugal force (responsble for the sensaton of gravty) and Corols force (whch causes hurrcanes and whrlpools). Newton s Second Law states that F = F net = ma,.e., the acceleraton of an object s proportonal to the net force on the object and nversely proportonal to the (nertal) mass. (Note that f F net = 0, then a = 0, and the object s n equlbrum.) 2
F Equaton of Moton: a = r = m = F net m Example: If a = constant, then the soluton s r = r o + v o t + 1 2 at2 ; v = v o + at Note that tme can be elmnated from these equatons to yeld for each component: v 2 = v 2 + 2a r, where = x, y, z. Example: Sprng, F = ma x = kx or pendulum, F = mgx/l ( ) k x = x max cos m t g ; x = x max cos L t A free-body dagram allow one to vsualze the forces on an object wth the help of Newton s laws. The equatons of moton can then be used to descrbe the motons. Frctonal Forces. Resstance to moton s offered on surfaces or n a vscous medum. On a surface, statc frcton, f s, opposes a force appled parallel to the surface, F, to try to keep the object motonless w.r.t. the surface. Whle motonless, f s = F, and the magntude of the frcton force f s µ s n where µ s s the coeffcent of statc frcton, and n s the magntude of the normal force appled by the surface on the object. If the statc frctonal force s exceeded, the object wll accelerate. If the object slps, the frctonal force s reduced to that of knetc frcton ( ) v f k = µ k nˆv = µ k n v When an object moves through a vscous meda, t colldes wth the ndvdual consttuents of that medum. Lke a fly httng the wndsheld of a car, the object apples an mpulse,.e., a force on the ndvdual consttuents, and there s a reactve force (by Newton s thrd law) on the object from each of the consttuents. We expect a resstve force, R = bv. An object fallng through the medum feels a net force Solvng for ts speed where v T = mg b ma = g bv v = v T ( 1 e t/τ ) s the termnal speed of the object. In the case of an object travelng at hgh speeds n ar, the magntude of the ar drag s R = 1 2 DρAv2, where ρ s the densty of ar, A s the cross-sectonal area of the object, and D s the 3
drag coeffcent. Thje object wll feel an acceleraton a = g velocty v T = ( 2mg DρA) 1/2. ( ) DρA 2m v 2 untl t reaches ts termnal Newton s Thrd Law states that for every acton there s an equal and opposte reacton,.e., F 12 = F 21. An solated force cannot exst n nature. Newton s law of unversal gravtaton states that every partcle n the Unverse attracts every other partcle wth a force that s drectly proportonal to the product of ther masses and nversely proportonal to the square of the dstance between them. Specfcally, the gravtatonal force exerted by partcle 1 on partcle 2 s F 12 (r) = G m 1m 2 r 2 ˆr 12 where r s the dstance between them, and the unversal gravtatonal constant G = 6.673 10 11 N m 2 /kg 2. The gravtatonal potental energy assocated wth a par of partcles separated by a dstance r s P E g (r) = Gm 1m 2 r Note that gravtaton s a conservatve force. The gravtatonal force exerted by a fnte-sze, sphercally symmetrc mass dstrbuton on a partcle outsde the dstrbuton s the same as f the entre mass of the dstrbuton were concentrated at the center. For example, the force exerted by the Earth on a partcle of mass m s F g (r) = G M Em ˆr where M E s the Earth s mass and r s measured from Earth s center. The gravtatonal potental energy s r 2 P E g (r) = G M Em r Kepler s laws: 1. All planets move n ellptcal orbts wth the Sun at one focus. 2. The radus vector drawn from the Sun to a planet sweeps out equal areas n equal tme ntervals. 3. The square of the orbtal perod of any planet s proportonal to the cube of the semmajor axs of the ellptcal orbt. Conservaton of Energy states that energy can nether be created or destroyed energy s always conserved. If the total energy of a system E system changes, t can only be due to the fact that energy has crossed the system boundary by a transfer mechansm T E system = T 4
E system s the total energy of the system, ncludng E mechancal = KE + P E and E nternal, whch ncludes several knetc and potental energes nternal to the system, such as the rest-mass energy, chemcal energy, vbratonal energy, etc., that we have not addressed yet except for frcton whch changes mechancal energy nto nternal energy. T represents mechansms by whch energy s transferred across the system boundary, ncludng heat, work, electromagnetc radaton, etc. Energy can nether be created nor destroyed, so conservaton of energy s a bookkeepng problem. Thnk of energy as currency and follow the money. Work transforms energy from one form to another. At ths pont, we have dscussed only knetc energy, KE, potental energy P E, the work done by frcton, f k s (at ths pont, frcton s the only non-conservatve force dscussed, W nc = f k s), and the work done by other stated forces, W otherforces. These yeld the followng conservaton of energy relaton KE f + P E f = KE o + P E o f k s + W otherforces Conservaton of Momentum. A drect consequence of Newton s thrd law s that the total momentum of an solated system (.e., a system subject to no external forces) at all tmes equals ts ntal momentum m v f = m v An elastc collson s one n whch the knetc energy s conserved 1 2 m vf 2 = 1 2 m v 2 (elastc collsons) 5
Extended Mass Dstrbutons The dscusson so far has dealt wth the moton of a pont partcle or mass. Everythng dscussed to here s vald for an extended mass dstrbuton as well. The moton of an extended object can be descrbed as the moton of the center of mass (CM) plus the moton about the center of mass. Newton s law stll apply to all partcles of the system or extended body. Forces may be external or nternal to the system. However, because of Newton s thrd law all nternal forces occur n equal and opposte pars and hence do not contrbute to the moton of the center of mass. Moton of the Center of Mass If M s the total mass of the extended object, then Poston Vector of CM (m): r cm = 1 M m r Average Velocty of CM (m/s): v cm = r cm Instantaneous Velocty of CM (m/s): v cm = lm 0 r cm = ṙ cm Total Lnear Momentum of Body (kg m/s): p total = Mv cm Average Acceleraton of CM (m/s 2 ): a cm = v cm Instantaneous Acceleraton of CM (m/s 2 ): a cm = lm 0 v cm = v cm = r cm Force on CM (N=kg m/s 2 ): Fext = Ma cm = p cm 6
Impulse on CM (N s): I cm = F = p cm Work on CM (J=N m): W cm = F s cm = (F cos θ)s cm Power on CM (W=J/s): P cm = W cm t = F v cm ; generally, P cm = F v cm Knetc Energy of CM (J): KE cm = 1 2 Mv2 cm All the laws and notons pertanng to the pont partcle descrbed earler apples to the total mass (M) as f appled to the center of mass (CM) of the extended body. Ths ncludes mpulse, work, knetc energy, power, potental energy, etc. of the extended body as t pertans to the translatonal moton of the center of mass. As the noton of potental energy comes from work performed by an external conservatve force, there can be potental energy, as before, assocated wth M at the poston of the center of mass. For example, the gravtatonal potental energy s Mgh cm. Rotatons Newton s laws stll apply to every mass element of the extended body, so there s moton, momentum, work, energy (potental and knetc), etc. assocated to the motons about the center of mass of the body. For a rgd body, that moton takes the form of rotatons. Relaxng the rgdty assumpton, objects may deform, there may be random moton of molecules that comprse the body (heat and temperature), and there may be motons of the ndvdual consttuents of the molecules wth respect to each other (vbraton). There s work and energy assocated wth each of these motons. When dealng wth lquds, gases, or plasmas (a gas comprsed of charges partcles) each element obeys Newton s laws and has an energy and momentum. Such systems wll be descrbed by a dstrbuton functon descrbng the poston and veloctes of the consttuents. By ntegraton, propertes of the dstrbuton, such as pressure (force per unt area), bulk velocty (akn to the velocty of the center of mass), etc. can be computed. At ths pont we wll consder rgd bodes and rotaton. (The notons, f properly appled pertan generally, except for those nvolvng the moment of nerta whch apples to the rgd body.) Angular Poston (radan): θ Angular Dsplacement (radan): θ = θ f θ 7
Average Angular Velocty (rad/s = s 1 ): ω = θ Average Angular Acceleraton (rad/s 2 =s 2 ): α = ω If α = constant, then we have θ = α = constant n analogy to r = a = constant for lnear moton. The soluton s ω = ω 0 + αt; θ = θ 0 + ω 0 t + 1 2 αt2 Note that tme can be elmnated from these equatons to yeld: ω 2 f = ω 2 + 2αθ Arc Length (m): s = rθ Angular dsplacement s measured n radans. Do not use degrees. When s = r, θ = 1 radan. Thus, 2π radans s the same as 360. Note: θ does not behave as a vector. However, the nfntesmal angular dsplacement does behave as a vector, as does the angular velocty, angular acceleraton, as well as other quanttes derved from these quanttes. Its drecton obeys the rght-hand rule, where you curl your fngers n the drecton that the angle changes, and your thump ponts n the drecton of the vector. For counterclockwse rotatons n a horzontal plane, the thumb ponts up; for clockwse rotatons, the thumb ponts down. For rgd body rotaton, θ, ω, and α all pont along the axs of rotaton followng the rght-hand rule. Tangental Velocty (m/s): v t = ω r Tangental Acceleraton (m/s 2 ): a t = α r There may also be a radal velocty and acceleraton. The total lnear velocty (acceleraton) s the vector sum of the tangental velocty (acceleraton) and the radal velocty (acceleraton). If r represents the radus vector of the crcle on whch the partcle (or mass element) s at that nstant movng, then the radal acceleraton s just the centrpetal acceleraton, and a = a 2 t + a2 c = r α 2 + ω 4. 8
Moment of Inerta (kg m 2 ): I = m r 2 Parallel-Axs Theorem: Suppose the moment of nerta about an axs through the center of mass s I CM. The parallel-axs theorem states that the moment of nerta about any axs parallel to and a dstance D away from ths axs s I = I CM + MD 2 Angular Momentum (kg m 2 /s): L = r p = mvr sn φ For a rgd body rotatng about an axs L = I ω If a force s appled to a rgd object pvoted about an axs, then the object tends to rotate about that axs. The tendency of a force to rotate an object s the torque. Torque (N m): L τ = r F = lm 0 = rf sn φ; r sn φ = l = moment or lever arm Torque should not be confused wth force. Forces can cause a change n lnear moton, as descrbed by Newton s second law. Forces can also cause a change n rotatonal moton, but the effectveness of the forces n causng ths change depends on both the forces and the moment arms of the forces, n the combnaton that s called torque. Do not confuse torque wth work, whch have the same unts but are very dfferent concepts. However, t s the torque that changes the angular momentum and results n angular acceleraton of a rgd body τ ext = I α = lm L tot 0 Rotatonal Work (J): W R = τ θ Rotatonal Power (J/s): P R = lm W R 0 = τ ω 9
The total work done by the net external force causng a system to rotate s WR = 1 2 Iω2 f 1 2 Iω2 Knetc Energy of Rotaton (J): KE R = 1 2 Iω2 The total knetc energy of a rgd body s the sum of the knetc energes of all the partcles. Ths s the knetc energy of the translatonal moton of the total mass M located at ts center of mass plus the knetc energy of rotaton about the center of mass KE = 1 2 I cmω 2 + 1 2 Mv2 cm Condton for Rollng Moton: v cm = ωr Conservaton of Angular Momentum states that the total angular momentum of an solated system (.e., a system subject to no external torque) at all tmes equals ts ntal momentum. Thus, so far we have that, for an solated system, the total energy, momentum, and angular momentum each reman constant. The two necessary condtons for an object to be n equlbrum s that (1) the net external force must equal zero F = 0 (translatonal equlbrum) and (2) that the net external torque about any axs be zero τ = 0 (rotatonal equlbrum) Note that f an object s n translatonal equlbrum and the net torque s zero about one axs, then the net torque must be zero about any other axs. Hooke s Law and Smple Harmonc Moton. Any physcal system whch undergoes a small dsplacement from a stable equlbrum poston s subject to a restorng force proportonal to the dsplacement from ts equlbrum poston known as Hooke s law: F = kx In one dmenson, a = ẍ = ω 2 x. The soluton s the condton for smple harmonc moton; the system wll oscllate about ts equlbrum poston accordng to x = A cos(ωt φ) 10
where A s the ampltude of the dsplacement, ω = 2πf = 2π/T s the angular frequency of the oscllatons, f and T = 1/f are the frequency and perod of the oscllaton, and φ s a possble phase shft of the wave. Example systems obeyng Hooke s law nclude the sprng (ω = k/m) and the smple pendulum (ω = g/l). Wave Speed: v = fλ where f s the frequency and λ s the wavelength of the wave. Speed of Sound: v = γp B ρ, or v = ρ where γ s the rato of specfc heats (adabatc ndex), p s the pressure, B s the bulk modulus, and ρ s the mass densty. Waves on a Strng: v = T µ where T s the tenson on the strng, and µ s the mass per unt length of the strng. Standng Waves: f n = nv, where n = 1, 2, 3,... (fxed ends) 2L f n = (2n 1)v, where n = 1, 2, 3,... (one open end) 4L The lowest frequency, correspondng n = 1, s called the fundamental frequency. correspondng to n = 2 s the frst harmonc, n = 3 s the second harmonc, etc... The frequency 11