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Physics 1710 Section 004 Mechanics and Thermodynamics Final Review Physics 1710 MWF Session 1 Introduction The “Structure” of this course: Dynamics Fluid Mechanics Gravitation Elasticity Applications Statics Kinematics Oscillations Waves Thermodynamics Physics 1710—Chapter 1 Measurement Summary •Fundamental Dimensions and Units Time, measured in seconds; Length, measured in meters; Mass, measured in kilograms. • Prefixes scale units to convenient size. k =1000, M = 1 000 000 c = 1/100, m = 1/1000, μ =1/1 000 000 • Density is mass per unit volume. ρ = m/V [kg/m3 ] • Avogadro’s number is the number of atoms in a mole of an element. 6.022 x1023 atom/mole Physics 1710 Chapter 2 Motion in One Dimension—II Summary: • The change in the instantaneous velocity is equal to the (constant) acceleration multiplied by its duration. ∆v = at •The displacement is equal to the displacement at constant velocity plus one half of the product of the acceleration and the square of its duration. ∆x = vinitial t + ½ at 2 •The change in the square of the velocity is equal to two times the acceleration multiplied by the distance traveled during acceleration. ∆v 2 = 2a ∆x •The acceleration of falling bodies is 9.8 m/s/s downward. a = - g = - 9.8 m/s/s Physics 1710 Chapter 3 Vectors Summary: •To add vectors, simply add the components separately. •Use the Pythagorean theorem for the magnitude. •Use trigonometry to get the angle. •The vector sum will always be equal or less than the arithmetic sum of the magnitudes of the vectors. Physics 1710 Chapter 4: 2-D Motion—II Summary: • Kinematics in two (or more) dimensions obeys the same 1- D equations in each component independently. – rfinal = rinitial + vinitial t + ½ a t 2 – vfinal = vinitial + a t – vx,final2 = vx,initial 2+ 2 ax /∆x – vy,final2 = vy,initial 2+ 2 ay /∆y • Projectiles follow a parabola [y(x) = A + Bx +Cx2] Physics 1710 Chapter 4: 2-D Motion—II Summary: • In a moving or accelerating Frame of Reference • v ′ = v – vframe of reference • a ′ = a – aframe of reference • The Centripetal acceleration is • a = - ω2 r or |a| = v 2/ |r|, toward the center. Physics 1710 Chapter 5: Laws of Motion—II Summary: • Newton’s Laws of Motion are: (1) Acceleration (or deceleration) occurs if and only if there is a net external force. (2) a = F/m [Note this is a vector eqn.] (3) The force exerted by a first object on a second is always equal and opposite the the force exerted by the second on the first. F12 = - F21 Physics 1710 Chapter 5: Laws of Motion—II Summary (cont’d.) : • Weight is the force of gravity equal to g times the mass of the object. • g =9.80 N/kg • The force of friction is opposed to the motion of a body and proportional to the normal force. • Free body diagrams are sketches of all the forces acting on a body. Physics 1710 Chapter 6—Circular Motion Summary • The net force on a body executing circular motion is equal to the mass times the centripetal acceleration of the body. •acentripedal = v 2/ R [toward the center] • The “centrifugal” force is a fictitious force due to a noninertial frame of reference. Physics 1710 Chapter 7—Work Summary • Work is defined to be the distance traveled multiplied by the distance over which the force acts. • W = ∫ F•d r •[Joules] = [N ][m] Physics 1710 Chapter 7&8—Power & Energy Summary: •The Potential Energy is equal to the negative of the work done on the system to put it in its present state. U = -∫ F•d r • The sum of all energy, potential and kinetic, of a system is conserved, in the absence of dissipation. E=U+K–W • F = - ∇U •P = dE/dt Physics 1710—Chapter 1 Measurement Summary • F = - ∇U = negative gradient of U. • The Potential Energy graph is a complete description of the dynamics of a system. Physics 1710—Chapter 10 Rotating Bodies Summary: •Angular displacement is the angle through which a body has rotated. •Instantaneous angular speed is the time rate of angular displacement. •Instantaneous angular acceleration is the time rate of change in angular speed. Physics 1710—Chapter 10 Rotating Bodies Summary (cont’d): •The moment of inertia is the measure of the (inertial) resistance to angular acceleration and equal to the second moment of the mass distribution. •Torque (“twist”) is the vector product of a force and the “moment” arm. Physics 1710—Chapter 10 Rotating Bodies Summary: •The moment of inertia I is the measure of the (inertial) resistance to angular acceleration and equal to the second moment of the mass distribution about an axis. Physics 1710—Chapter 11 Rotating Bodies Summary: •The total Kinetic energy of a rotating system is the sum of the rotational energy about the Center of Mass and the translational KE of the CM. K = ½ ICM ⍵ 2 + ½ MR 2 ⍵ 2 τ=rxF Physics 1710—Chapter 11 Rotating Bodies Summary: •Angular momentum L is the vector product of the moment arm and the linear momentum. L=rxp • The net externally applied torque is equal to the time rate of change in the angular momentum. ∑ τz = d Lz /dt = Iz ⍺ Physics 1710—Chapters 6-10 Summary • Rotary (circular) motion obeys laws that are analogous to those of translational motion. • Linear Momentum is conserved in absence of external forces. • F = d p/dt • Energy is related to the work done or stored • Work is the cumulative force times distance moved. • Power is the rate of expenditure of work or energy. • Force is the negative of the gradient of the potential. Physics 1710—Chapter 11 Rotating Bodies Summary: • Angular momentum about an axis z is equal to the product of the moment of inertia of the body about that axis and the angular velocity about z. L=I⍵ Lz = Iz ⍵ • In the absence of torques, the angular momentum is conserved. • In the presence of torques the angular moment will change with time. Physics 1710—Chapter 11 App: E & E Summary • Static equilibrium implies that all forces and torques balance. • The center of mass is often the center of gravity. • The moduli of elasticity characterizes the stress-strain relation: • stress= modulus x strain Stress = modulus x strain σ = F/A = Y ε Physics 1710—Chapter 13 Apps: Gravity Summary: • The force of attraction between two bodies with mass M and m respectively is proportional to the product of their masses and inversely proportional to the distance between their centers squared. F = - G M m/ r 2 • The proportionality constant in the Universal Law of Gravitation G is equal to 6.673 x 10 –11 N m2 /kg2 . Physics 1710—Chapter 13 Apps: Gravity Summary: •The gravitational force constant g is equal to G M/(R+h) 2, R is the radius of the planet. • Kepler’s Laws –The orbits of the planets are ellipses. –The areal velocity of a planet is constant. –The cube of the radius of a planet’s orbit is proportional to the square of the period. • The gravitation field is the force divided by the mass. g = Fg / m Physics 1710—Chapter 13 Apps: Gravity Summary: • The force of attraction between two bodies with mass M and m respectively is proportional to the product of their masses and inversely proportional to the distance between their centers squared. F = - G M m/ r 2 • The proportionality constant in the Universal Law of Gravitation G is equal to 6.673 x 10 –11 N m2 /kg2 . Physics 1710—Chapter 13 Apps: Gravity Summary: • The gravitation potential energy for a point mass is proportional to the product of the masses and inversely proportional to the distance between their centers: U = GMm / r • The escape velocity is the minimum speed a projectile must have at the surface of a planet to escape the gravitational field. vescape = √[ 2GM/R] • Total Energy E is conserved for two body geavitational problem; bodies are bound for E ≤ 0 E = L2/2mr 2 – GMm/r Physics 1710—Chapter 14 Fluid Dynamics Summary: •Pressure is the force per unit area. P =F/A • Unit of pressure [Pacal] = [N]/[m2] •The hydrostatic pressure is P = Po + ρgh • Archimedes’ Principle: Fbouyant = ρfluid g V • Equation of Continuity: A1v1 = A2v2 • Bernoulli’s Equation: P + ½ ρv2 + ρgy = constant. Physics 1710—Chapter 15 SHO Summary : • Simple Harmonic Motion is sinusoidal. x = Xo cos(ωt +φ) • The period is the reciprocal of the frequency. • For a mass m on a spring of spring constant k, the period T = 1/ f T = 2π√(m/k) • For Damped SHO, the frequency is decreased and the amplitude decays exponentially. x = Xo e – ½ (b/m)t cos(ωt +φ) with ω = √[k/m – ½ b/m] Physics 1710—Chapter 15 SHO Summary : • For a driven SHO the amplitude is a maximum when the drive frequency is equal to the natural frequency; a condition known as “resonance.” • A simple pendulum oscillates at a frequency of f = (1/2π) √(g/L) •A physical pendulum oscillates at a frequency of f = (1/2π) √(mgL/I) Physics 1710—Chapter 16 Waves Summary : • A traveling wave has the form y(x,t) = Y sin(kx – ωt), with k = 2π/λ, k : wave number, λ : wavelength & ω= 2π f = 2π/ T as previously defined • d 2y/dx 2 = (1/v 2) d 2y/dt 2 is the linear wave equation. • λ f = v, the phase velocity. • For a longitudinal wave on a string v = √(T/μ). T = tension, μ = dm/dx = linear mass density • The time averaged power transmitted on a string is ₧ = ½ μ ω2A2v Physics 1710—Chapter 17 Sound Summary : • Sound is a longitudinal pressure/displacement •V = √B/ρ, the phase velocity is equal to the square root of the ratio of the bulk modulus to the density. • The Doppler effect is a shift in frequency due to the relative motion of the source and observer of a sound. Physics 1710—Chapter 16 Waves Summary : • A traveling wave has the form y(x,t) = Y sin(kx – ωt), with k = 2π/λ, k : wave number, λ : wavelength & ω= 2π f = 2π/ T as previously defined • d 2y/dx 2 = (1/v 2) d 2y/dt 2 is the linear wave equation. • λ f = v, the phase velocity. • For a longitudinal wave on a string v = √(T/μ). T = tension, μ = dm/dx = linear mass density • The time averaged power transmitted on a string is ₧ = ½ μ ω2A2v Physics 1710—Chapter 18 Chapter 18 Superposition and Standing Waves Summary: The propagation of waves is characterized by Reflection — the rebound of the wave. Refraction — the bending of a wave’s direction due to a velocity gradient Diffraction — the bending of a wave around obstacles. Interference — the combination of two or more waves in space. Beats — the combination of two waves in time. Physics 1710—Chapter 18 Chapter 18 Superposition and Standing Waves Summary: • • • • • Angle of incidence = angle of reflection; θi=θr sin θ 1 /v1 = sin θ 2 / v2 fave = ( f1 + f2 )/2; fbeat = ( f1 - f2 ) fn = n /(2L) √(T/μ) A = (Fext /m)/ [ω0 2 - ω2] Physics 1710 Chapter 19 Temperature Summary: •Temperature is a measure of the average kinetic energy of a system of particles. • Thermal Equilibrium means that two bodies are at the same temperature. • The “Zeroth Law of Thermodynamics” states that if system A and B are n thermal equilibrium with system C, then A and B are in thermal Equilibrium with each other. Physics 1710 Chapter 19 Temperature •Kelvin is a unit of temperature where one degree K is 1/279.16 of the temperature of the triple point of water (near freezing). TC = (100/180) (TF – 32 ⁰F) TF = (180/100) TC + 32 ⁰F • ∆L/L = α∆T • PV = n R T = N kT Physics 1710 Chapter 20 Heat & 1st Law of Thermo Summary •The internal energy is the total average energy of the atoms of an object. • Heat is the change in internal energy. • The change in temperature is proportional to the change in internal energy (heat flow) when there is no change of phase and the system does no work. • The first law of thermodynamics states ∆E = ∆Q - W Physics 1710 Chapter 21 Kinetic theory of Gases Summary: • The Ideal Gas Law results from the cumulative action of atoms or molecules. • The average kinetic energy of the atoms or molecules of an ideal gas is equal to 3/2 kT. ½ m<v2> = 3/2 kT • Energy average distributes equally (is equipartitioned) into all available states. •Each degree of freedom contributes 1/2 kT to the energy of a system. Physics 1710 Chapter 21 Kinetic theory of Gases Summary (cont’d.) γ = CP / CV PV γ = constant B=γP • The distribution of particles among available energy states obeys the Boltzmann distribution law. nV = no e –E/kT Physics 1710 Chapter 22 Heat Engines etc Summary: •The work done by a heat engine is equal to the difference in the heat absorbed at the high temperature and expelled at the low. ∆W = ∆Qh – ∆Qc • The thermal efficiency is the work done divided by the heat absorbed. e = 1 - ∆Qc / ∆Qh Physics 1710 Chapter 22 Heat Engines etc Summary: • Kelvin-Planck form of 2 nd Law of Thermo: It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of energy from a reservoir and the performance of an equal amount of work. •Clausius Form of 2 nd Law of Thermo: It is impossible to construct a cyclical machine whose sole effect is the continuous transfer of energy from one object to another at a higher temperature without the input of work. Physics 1710 Chapter 22 Heat Engines etc Summary: •The maximum efficiency is obtained via a Carnot cycle and is equal to the temperature difference divided by the high temperature. eCarnot = 1 - Tc / Th • Entropy S is a measure of the disorder of a system. • ∆S = ∫dQ/T • S≡ k ln N