Physics I Overview-Review Corrections and updated Fri. May 20 1:00PM
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1 1. Physical Dimensions Physics I Overview-eview (Length, L,m(meter)), (Time,T, s (sec)), (mass,m,kg), (area, L 2, m 2 ), (volume, L 3, m 3 ), (velocity,l/t, m/s), (acceleration, L/T2, m/s 2 ), (orce,ml/t 2, N(Newton)), (Work&Energy, ML 2 /T 2, J(Joule)), Linear Momentum(p), ML/T ), Torque ( O ) (Nm), 2. Scientific Notation Example: x10 12 =3.467 e+12 significant figure 3.467, order of magnitude(power of 10) 10 12,e Measurements Uncertainty, accuracy, precision. 4. Conversion of Units the pseudo program, the program, checking 5. Vectors (The trip as the quintessential model) Magnitude and direction, addition and subtraction, components, Methods: Graphical (ruler and protractor), Analytical (trigonometry) 6. Motion - translational (x, v, a;t) (translational displacement, translational velocity, translational acceleration; time) v= x/ t = slope of x vs t graph; x = v t = area under the v vs t graph x f =x i +v i t+½at 2 ; for constant a. a= v/ t = slope of v vs t graph; v = a t = area under the a vs t graph v f =v i +at; for constant a. Motion rotational (,, :t) (rotational displacement, rotational velocity, rotational acceleration: time) = t = slope of vs t graph; = t = area under the vs t graph f = i + i t+½ t 2 ; for constant. = / t = slope of vs t graph; = t = area under the vs t graph f = i + t; for constant. a. Types of motion ree all: vertical direction, a = a vertical = a v = g down (9.8m/s 2 down) Projectile Motion: 1. In vertical direction - ree all. 2. In horizontal direction - ideally a = a h = 0, so v= v h = constant. otation without slipping a = r, v = r, x = r half turn half a revolution rad x String unwinding on a spool without slipping full turn one revolution rad one revolution rad x = 2 Wheel rolling without slipping x = 2 May 12, 2011 Page 1 of 12
2 7. Newton s Laws of Motion (abbreviated) orces,, - a push or pull, a vector, lie at the base of these laws. Units N (Newtons) Torque is, O, a force acting at a displacement r O from the axis O that is perpendicular plane of the and r, where, O = r O * ^ = r O * sin. = r O^ * = r O sin *. Units mn O =r = r sin O o O ^ O ro Two Points of View of O 1 st Law If net = 0 on a body, then translational motion remains constant. If the torque Onet = 0 on a body, then rotational motion remains constant. r O ^ = sin ro ro O = r = r sin O^ O ro O o r = r sin O O ro ^ r O Line of Action of ro axis of rotation O o r O slide position physical position 2 nd Law If there is a net acting on a body of mass m, that body has a translational acceleration a, where net = ma. If there is a torque Onet acting on a body of moment of inertia about axis I O, that body has a rotational acceleration where Onet = I O. 3 rd Law or every action, there is an opposite and equal reaction. The 3 rd Law permits the replacement of physical connections with action-reaction pairs, then the creation of ree Body Diagrams (BD) so that the first two laws can be used (note they talk about forces and torques, not connections). Inertia the tendency to resist change in motion. mass, m - measure of inertia for translational motion. Units kg, g ((gm) Moment of Inertia, I O, measure of inertia for rotational motion. I O = m i r io 2, Units kg m 2, May 12, 2011 Page 2 of 12
3 a. The Laws of riction 1. = N, where is the friction force, N the normal to the surface, the coefficient of friction. 2. riction always acts to oppose actual motion (velocity) or pending motion. 3. riction never exceeds the active net force, net active (the pushes and pulls) or active net torque on the body. 4. There are two types of friction, (1) static (not moving ) where the maximum static friction smax = smax N. (2) kinetic (moving) k = k N. (3) s > k, once net active exceeds smax, k takes over causing a net = smax, k =ma. Newton s Law of Gravity The absolute gravitational is G = -Gm 1 m 2 /r where m 1 and m 2 are masses with their m centers separated by distance r 12 and 1 G = 6.67x10-11 Nm 2 /kg 2, is the Universal Gravitational Constant. b. Problem Methodology See 1. List given information and results wanted. 2. Write all appropriate definitions 3. Draw ree Body Diagrams of All bodies 4. Set up Newton s 1 st, 2 nd Laws for each body in symbols 5. Get solution in symbols (symbolic equations) See Phy210_Problem_C4_36.pdf 8. Work, Energy and Power a. Mechanical Work units Nm = J Joules Work translational = parallel d = d parallel, where d is the displacement and parallel is the component of the force parallel to the displacement or equivocally, is the force and d parallel is the component of the displacement parallel to the force. Work rotational = torque parallel = Oparallel G G m2 O 1 O 2 r12 b. Mechanical Energy the ablility to do work c. Kinetic Energy KE the ablility to do work because of motion KE translational = ½ mv 2, KE rotational = ½I O 2, Work-Energy Theorem May 12, 2011 Page 3 of 12
4 net x + Onet = = KE final KE initial d. Potential Energy, PE, is the ability to do work because of position. Local Gravitational PE = PE Local gravity = mgh = WH, H=height Absolute Gravitational PE = PE Absolute Gravity = - MmG/r, where m and M are masses with their centers separated by distance r and G = 6.67x10-11 Nm 2 /kg 2, is the Universal Gravitational Constant. Elastic or Spring PE = PE Spring = ½ kx 2. k=stiffness, x = stretch Hooke s Law =kx e. Conservative orces (a) The work they do depends only on the end points and is independent of the path. (b)they produce Potential Energies. (c ) Gravity and Elastic (Spring) forces are examples. f. Non Conservative orces (a) These are all forces that are not Conservative orce. (b) Their work is accounted for with Work other in Conservation of Energy, below. (c) riction is an example of a non conservative force. Its work depends on the path, its work is always negative because it always opposes the motion, hence consumes ( reduces) energy. g. Conservation of Energy (a) Two Stage from Initial to inal KE initial +PE initial = KE final +PE final Work other Initial to inal See f (b) just above. General including otational Motion KE i Trans + KE i ot + PE initial = KE f Trans + KE f ot +PE final Work other Init to inal KE translationa l = ½ mv 2, KE rotational = ½ I O w O 2. (b) Multi Stage from 1 to 2 to 3 KE 1 +PE 1 = KE 2 + PE 2 Work other 1 to 2 = KE 3 +PE 3 Work other 1 to 3 May 12, 2011 Page 4 of 12
5 KE s include both translational and rotational as they occur. h. Power = Work/Time = parallel v v is velocity and parallel is the component of the force parallel to the velocity. (See 8a Work in this Section) Power otational = Work otational /Time = Oparallel. 9. Linear Momentum p a. Definition p = mv. m is the mss of the body, and v its velocity. units kg * m/s b. Impulse I = t. units Ns kg * m/s is the action-reaction force acting between the interacting bodies and t the time (duration) of the interaction. c. Impulse Momentum Theorem p = I = t. d. Conservation of Linear Momentum p final = p initial, p finals = p initials p 1initial p 2initial p 1final p 2final p 1initial p2initial p 3initial p 1final p2final p initial = p final p initial = p final p 3final 10. Angular Momentum L O L O = I O O units kg * m 2 /s Conservation of Angular Momentum If O = 0 then L O = I O O = constant or I O1 O1 = I O2 O2 May 12, 2011 Page 5 of 12
6 Solids and luids Physics I Overview-eview Density = mass/volume = m/vol Units kg/m 2 Stress = orce/area = /A Units 1 N/m 2 = 1 Pa(Pascal)\ Strain is a deformation, a change in size or shape. Deformations Elasticity Stress = Elastic Modulus * strain A axial Young s Modulus - Elasticity, a change in length along central axis Stress Axial = Y L /L o /A = Young s Modulus * L /L o Examples, tension and compression L L axial Shear Modulus Elasticity, a change in Shape Shear = S x /h = S*tan /A = Shear Modulus S x /h = S tan Bulk Modulus Volume Elasticity a change in volume L L P = - B V/V o Change in Pressure = Bulk Modulus*change in volume/original volume Compressibility luid Pressure P P = orce/ Area = /A In a static fluid the pressure is always perpendicular to the surface with which it is in contact and has the same value at the same depth below top surface of the fluid. Pressure Variation with depth H P(H) = fluid g H 1. H is depth of the top below the surface of the fluid 2. fluid is density fluid and must be independent of H in the model (formula) 3. g is the acceleration due to gravity P top = fluid ghsubmerged h H submerged h Buoyant orce why do bodies float and sink in a fluid? May 12, 2011 Page 6 of 12 P Bottom = fluid g(h submerged + h) A submerged body
7 luids In Motion Ideal luids non-viscous, incompressible, steady motion, no turbulence. low rate = Volume/time = A x/ t = Av Q = V/ t = Av = Area*velocity units m 3 /s m/ t = ( V)/ t = V/ t = Av Units kg/s Equation of Continuity Conservation of matter Av = constant, O equivalently, A 1 v 1 = A 2 v 2. Bernoulli s Equation (or Princple) - Conservation of Energy P/ + ½v 2 + gh = constant or equivalently P 1 / + ½v gh 1 = P 2 / + ½v gh 2. Notes on units: Note the term ½v 2. It is ½mv 2, which is KE. So, ½v 2 is energy/mass, hence units of J/kg urther, therefore, P/ = PE/Mass due to pressure, ½v 2 = KE/Mass and, gh is PE/Mass due to gravity Units J/kg Thermal Physics Temperature T Common Arbitrary Scales Celsius Centigrade T o C C T ahrenheit o Boiling water 100 o C 212 o Ice-water 0 o C 32 o BOILING WATE CELSIUS or CENTIGADE 100 o C AHENHEIT 212 o o 212 Absolute Scales 180 o Kelvin K ankine 0K = o C, K = 1 C 0 o C = K 0 = o, = 1 ICE WATE MIXTUE 0C o 32 o o C 100 o C May 12, 2011 Page 7 of 12
8 T = (9/5) T C, T C = (5/9) T Oth (Zeroth) Law of Thermodynamics One Statement -If objects A and B are separately in thermal equilibrium with a third object C, then they are in thermal equilibrium with each other. Another Statement Two bodies in thermal equilibrium with each other are at the same temperature. Another Statement Two bodies in thermal contact with each other will come to a common temperature. Thermal Expansion of Solids and Liquids Linear Thermal Expansion T HIGH > T LOW L = (L f - L o ) = L o T = L o (T f - T o ) Coefficient of Linear Thermal Expansion ( o C -1 ) T LOW T HIGH L 0 Thermal Expansion L Areal Thermal Expansion = (A f - A o ) = A o T = A o (T f - T o ) Coefficient of Areal Thermal Expansion ( o C -1 ) Approximation 2 Volumetric Thermal Expansion V = (V f - V o ) = V o T = V o (T f - T o ) Coefficient of Volumetric Thermal Expansion ( o C -1 ) Approximation 3 May 12, 2011 Page 8 of 12
9 The Ideal Gas Macroscopic Description Physics I Overview-eview PV=nT This is the defining equation for the Ideal Gas. It is experimental behave of gases at high temperature and low density. O 2, N 2 the maim gases of air around room and normal pressure (1 atmosphere). This an Empirical Law. P pressure, V volume, T Absolute Temperature K or. DO NOT USE o C or o!!!! n number of moles Universal Gas Constant = 8.31 J/(mol*K) Other Important Numbers Avogadro s Number N A = 6.02x10 23 particles/mole Boltzmann s Constant k B = /N A = 1.38x10-23 J/K The Kinetic Theory of Gases This is a theoretical model that attempts to explain the behavior of the ideal gas in terms of molecular motion. It starts by examining the elastic collision of a molecule with the walls of the container. With a few assumptions yields 1.Pressure is due to the molecules making elastic collisions with the container walls. 2. Temperature is Energy in Thermal Processes Internal Energy U Energy associated with the kinetic energy of all the atoms and molecules in the system and all of interactions between, i.e. electrical, magnetic, and gravitational potential energies and collisions. Heat Q The transfer of energy between a system and its environment due to a temperature difference between them. Units 1 calorie =1 cal = energy required to raise the temperature of one gram of water from 14.5 to 15.5 o C 1 British Thermal Units = 1Btu = = energy required to raise the temperature of one pound (lb) of water from 63 to 64 o. 1 Joule J, now by definition 1cal = J. Specific Heat C = heat required to raise the temperature of 1 gm of substance by 1 o C cal/(gm o C) = heat required to raise the temperature of 1 kg of substance by 1 o C J/(kg o C) May 12, 2011 Page 9 of 12
10 C = Q/(m T) so Q = m C T Latent Heat of usion L f = Energy required to transition a unit mass of a substance between the solid and liquid phases. SOLID -> LIQUID melting LIQUID -> SOLID fusing or freezing. Units: cal/gm J/kg Latent Heat of Vaporization L v = Energy require to transition a unit mass of a substance between the liquid and gas phases. LIQUID -> GAS vaporizing or boiling GAS->LIQUID condensing Units: cal/gm J/kg Energy Transfer Thermal Conduction Transfer by molecular collisions T h > T l P = ka T/x = ka(t h - T l )/ x units watts, Thermal conductivity k J/(s m o C) = Watt/(m o C) T h A T l P Convection transfer of energy by mass move, literally by the movement of matter in convection currents. x adiation transfer of energy by electromagnetic waves. It does not need a medium to travel through as do conduction and convection. Stefan s Law P = Ae(T 4 T o 4 ) watts A is the area of the radiating surface in m 2, Stefan-Boltzmann constant. = 5.669x10-8 watts/(m 2 K 4 ). e emissivity e=1 for black body, the perfect absober. It absorbs all radiation and re-emits it in all directions e=0 for white body, the perfect reflector. It reflects all radiation back into environment from which it came. The Laws of Thermodynamics Here T is always Absolute Temperature, K or The irst Law of Thermodynamics The Internal Energy,U. (See the beginning of Energy in Thermal Processes) U = Q( energy transferred to the system by heat) + W(work done on the system) May 12, 2011 Page 10 of 12
11 Q( energy transferred to the system by heat) = U + W(work done by the system) U = 0 for a isolated - NO contact with the outside world, Q=W=0. U = 0 for any system that goes through a closed cycle. Thermodynamic or Thermal Processes Adiabatic - Q = 0. So, U = W Isochoric or Isovolumetric - Constant Volume V =0. W =0, U = Q. Isobaric - Constant Pressure, P = 0. W = P V. W = U. Isothermal Constant Temperature, T = 0. U for an ideal gas. So Q = -W Heat Engines and The Second Law of Thermodynamics The figure Engine shows a schematic model of a heat engine. It takes in heat Q H from a high temperature reservoir at temperature T H. It does work on the environment and dumps heat Q C at cool environment at temperature T C. T H Q H Work or a gasoline car engine, the high temperature reservoir is the combustion chamber where the gasoline fuel burns. The cool temperature reservoir is the T C Q C engine outside atmosphere perhaps more accurately, the atmosphere between the engine exhaust and the muffler. A refrigerator is a heat engine run backwards absorbs heat Q C at T C, dumps heat Q H at T H, has work W done on it. (Hence your electric bill $$$) Thermal Efficiency, e is defined as e = Work/ Q H = ( Q H Q C ) / Q H = 1 + ( Q C / Q H ). Note efficiency in general is e = What you want/what it costs. Here you want work, Q H is from what you pay for the fuel. P Adiabatic compression A Isothermalexpansion Isothermal compression D Q H T H Adiabatic expansion B W Q C C TC The Carnot Cycle Carnot Cycle The Carnot Cycle shown in the figure is the cycle of an ideal gas heat engine. 1. It operates at two temperatures, the high temperature reservoir at T H where it absorbs heat Q H,and the low temperature reservoir at T C where it dumps heat Q C. 2. The cycle has four processes. 1. A->B An Isothermal Expansion at temperature T H where the gas absorbs Q H ; 2. B -> C An Adiabatic Expansion (Q=0) from T H to T C ; 3. C -> D An Isothermal Compression at temperature T C where the gas dumps Q C ; 4. D -> A An Adiabatic Compression (Q=0) from T C to T H. V May 12, 2011 Page 11 of 12
12 A result is that Q H / Q C = T H / T C, or, Q H / T H = Q C / T C = Q / T for all Temperatures. This leads to concept of Entropy S. The efficiency becomes e = 1 - T C /T H for the Carnot Cycle The Second Law of Thermodynamics This law has many statements. or example: 1. No heat engine can be more efficient than the Carnot Cycle. 2. Heat cannot spontaneously flow from a low temperature reservoir to a high temperature reservoir. May 12, 2011 Page 12 of 12
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