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1 Physics 101 Section 3 April 8 th : REVIEW Announcements: n nt Exam #4, today at the same place (Ch ) 18.8) Final Exam: May 11 th (Tuesday), 7:30 AM at Howe- Russell 130 Make up Final: May 15 th (Saturday) 7:30 AM Class Website: 3/

2 m 1 Chapter 13 m Gravitation Newton s law of gravitation, which describes the attractive force between two pointmasses and itsapplicationto to extended objects F G mm = r 1 The acceleration of gravity on the surface of the Earth, above it as well as below it. G = N m / kg 11

3 m Gravitational Potential Energy r The gravitational potential energy is U GmM = r. U = GmM r The negative sign of U expresses the fact that the force is attractive. Assum e: U( r= ) = 0 Note: The gravitational potential energy is not only associated with the mass m but with M as well, i.e., with both objects. If we have three masses m1, m, and m3 positioned as shown in the figure, the potential energy U due to the gravitational forces among the objects is U Gm1 m Gm m Gm m = + + r1 r13 r Wetakeintoaccount each pair once.

4 Gravitational potential energy ΔU = W done by force Conservative force path independent x f ΔU W d g = W done = F g dx x i At Earth s surface, F g ~const. r W = r f r i G mm r dr = GmM r f r i 1 r dr ΔU g = W done = m g y f ( ) dy = mgδy y i If we define U = 0 at infinity, then the work done by taking mass m from R to infinity Note: 1 U U ( r) = W = GmM 0 r GmM U ( r) = r du (r) F (r) = dr d GmM = GmM dr r r 1) As before, U decreases asseparationseparation decreases (more negative) ) Path independent 3) MUST HAVE AT LEAST TWO PARTICLES TO POTENTIAL ENERGY (& force) 4) Knowing potential, you can get force.

5 Escape Velocity v = GM R v = 0 B m The minimum initial speed for which the projectile object m will escape from the gravitational pull of M and will stop at infinity y(point B in the figure). mv GMm EA = K + U = EB = K + U = 0 R E A mv GMm GM = EB = 0 v = R R v A r = m Note: The escape speed does not depend on m.

6 Kepler's First Law. All planets move on elliptical orbits with the Sun at one focus. da Kepler's Second Law. The line that connects a planet to the Sun sweeps out equal areas ΔA in the plane of the orbit in equal da time intervals Δ t : = constant. dt Kepler's second law is equivalent to the law of conservation of angular momentum: da L L= rp = rmv = rmωr = mr ω = = constant dt m

7 Kepler s 3 rd Law The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit. Gm satm E = m sat r sat Circular Orbit πr sat 3 T = GM r sat r sat 4π T = 4π GM T 3 r sat NOTE: r 3 proportional T Elliptical Orbit When used in the equations, r (radius) and a (semimajor axis) are synonymous y

8 SHW#9 problem 3: Assume the orbital of a lunched satellite around the earth is circular. Show that there is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius and it does not depend on the mass of the satellite. What is the speed and height eg H above the earth s surface at which all synchronous ous satellite (regardless (egadessof mass) must be placed in orbit, assuming the earth is spherical? If a synchronous satellite has a mass m. How much additional kinetic energy is needed for it to escape from the earth? M Em v GM E (a) Since G = m is the centripetal force for a satellite with mass m, v =, r r r which is dependent only on radius r but independent on mass m. (b) For a synchronous satellite, π r r 4h = T = = π r = v GM H GM T E /3 = ( ) π R E E π r GM 3/ E GM ET ; r = RE + H = ( ) π /3 1 E (c) Based upon the condition that mvescape = 0 v v escape escape GM E = r = r π GM E 1/3 = ( ) T GM T GmM r E /3 with ( ) for asynchronous satellite, so that π

9 Chapter 14: Fluids In this chapter we will explore the behavior of fluids. In particular we will study the following: Static fluids: Pressure exerted by a static fluid Methods of measuring pressure Pascal s principle Archimedes principle, buoyancy

10 Fluid: Δm ΔV Δm m ρ = Density of mass Δ V Pressure: p = F A Fluid at rest: p = p +ρgh bottom top Can you calculate the pressure at any depth?

11 Pascal's Principle and the Hydraulic Lever: A change in the pressure applied to an enclosed incompressible liquid is transmitted undiminished toevery portion of the fluid and to the walls of the container. F o = A F o i A i Buoyant Force and Archimedes' Principle When a body is fully or partially submerged in a fluid a buoyant force F b is exerted on the body by the surrounding fluid. This force is directed upward and its magnitude is equal to the weight m f g of the fluid that has been displaced by the body. F b F b = ρ gv f

12 Helium Blimps Length 19 feet Width 50 feet Height 59.5 feet Volume 0,700 cubic feet (5740 m 3 ) Maximum Speed 50 mph Cruise Speed 30 mph Powerplant: Two 10 hp fuel injected, air cooled piston engines What is the maximum load weight of blimp (W L ) in order to fly? ρ He = kg/m 3 & ρ air = 1.1 kg/m 3 W At static equilibrium F = 0 : W He + W L = F B W L F B W He The fluid blimp is in is: air W L = m air g m He g W L = F B W He = ρ air V ship g ρ He V ship g = V ship g( ρ air ρ He ) ( ) = (5740 m 3 )(9.8 m /s ) kg/m/ 3 = 58 kn = 13,000 lbs Maximum Gross Weight 1,840 pounds

13 Chapter 15: Oscillations Displacement, velocity, and acceleration of a simple harmonic oscillator Energy of a simple harmonic oscillator Examples of simple harmonic oscillators: spring mass system, simple pendulum, physical pendulum, torsion pendulum Forced oscillations/resonance

14 Simple harmonic Motion: Kinematics Displacement: ( ω φ ) xt ( ) = x cos t+ m Velocity of SHM dx() t d vt () = = xmcos t+ = xmsin t+ dt dt Acceleration of SHM: ( ω φ) ω ( ω φ) dv() t d at () = = ωxmsin( ωt+ φ) = ω xmcosωt dt dt at () = ω xt (). period (symbol T, units: s) vs. frequency (symbol f, unit: Hz): f = 1. T ππ ω = π f = T

15 SHW #10 The figure gives the position function of a 50g block in simple harmonic oscillation on the end of a spring. The period of the oscillation is s. 1) Find the velocity and acceleration function? ) Find the maximum kinetic energy, potential energy, and total energy. 3) Find the maximum force. 4) If you want to stop the oscillation in a quarter of period, what is minimum kinetic friction coefficient required? (Answer: μ k = 0.0) 0) Solutions: π a) x( t) = xmcos( ωt) with ω = = 3.14(1/ s) and xm = 0.07m T vt ( ) = ωx sin( ωt) m at x t ( ) = ω m cos( ω ) ( ωx ) kx mv m kx ( b) Fmax = kxm; Umax = ; K = = = π m 4π m ( c ) T = = π so that t k = ω k T 1 Since W = F xm = μkmgxm = kxm 1 1 4π m π μ k = kxm = ( ) x 0.0 m = x m = mg mg T T g m m m m m

16 Simple Harmonic Motion : Dynamics ( ω ) Newton's second law we get: F = kx = ma = mω x = m x. Simple Pendulum ω = k and T = π m = π m m C k Gravity plays as the restoring force τ = Iα = Lmgsinθ ( ) SHM : θ(t) = θ max cos( ωt + ϕ) mgl I α θ α ω θ ω = mgl I = mgl ml ( ) = g L Independent of amplitude and mass (in small angle approximation)! Dependent only on L and g

17 Physical Pendulums 1) Torsion pendulum τ = Iα = κθ α = κ I θ SHM : θ(t) = θ max cos( ωt + ϕ) ω = κ I τ = Iα = L( mgsinθ) α mgl θ I mgh ω = What is I? I I = I + mh com. T = π I mgh

18 Problem: The pendulum consists of a uniform disk with radius R and mass M attached to a uniform rod of length L and mass m. (a) what is the rotational inertia of the pendulum? (b) What is the distance between the pivot and the COM? (c) What is the period? (a) The disc has an I=MR / about the COM so. MR MR I(Disc) = + Mh = + M r + L ( ) = ml = nl L I(Rod) 1 + mh 1 + m = ml 3 (b) COM d COM = Ml d + ml r M + m With l d and l r being the the COM of the disc and rod. (c) The period T = π I (M + m)gd

19 wave phenomena. Chapter 16: Waves Types of waves Amplitude, phase, frequency, period, propagation speed of a wave Mechanical waves propagating along a stretched t string ti Principle of superposition of waves Wave interference Standing waves, resonance

20 Description of traveling wave: k = π λ Angular wave number ω = π T Angular frequency Phase and Direction i Wave Velocity phase : kx ± ωt kx ωt Wave traveling in + x direction kx + ωtt Wave traveling in x direction Velocity at which the wave crests move : v wave = λ / T = λf Wave on a τ v wave = = λf μ stretched string

21 Standing Wave: The displacement of a standing wave is given by the equation ( ) = [ ] y x, t = y sin kx cos ωtω t. m The position-dependent amplitude is equal to y sin kx. m n a a n n n a Nodes. These are defined as positions where the standing wave amplitude vanishes. They occur when kx = nπ for n = 0,1,, π x= nπ xn = n λ for n= 0,1,,... λ Antinodes. These are defined d as positions i where the standing wave amplitude is maximum. 1 They occur when kx = n+ π for n= 0,1, 01,... π 1 1 λ x= n+ π x n = n+ for n= 0,1,,... λ Note 1 : The distance between adjacent nodes and antinodes is λ/. Note : The distance between a node and an adjacent antinode is λ/4.

22 Standing Waves Summary Notations: Length L Nodes at both ends Tension in String: τ Given mass density: μ Driven at a frequency f This assures nodes at 0 and L λ = L n n th harmonic resonance

23 Problem 16 58: A sting, tied to a sinusodial oscillator at P and running over a support Q, is stretched by a block of mass m. The length is L and the linear density is μ, and the frequency is f. What mass allows the system to set up the fourth harmonic? n = 4 f = n τ = 4 τ = mg L μ L μ L μ

24 Chapter 18 Temperature, Heat, and Thermodynamics Temperature and the zeroth law of thermodynamics Thermometers and temperature scales F vs C vs K Thermal expansion Temperature and heat Specific heat Heat of transformation Heat, work, and the first law of thermodynamics Heat transfer mechanisms First Law of Thermodynamics

25 Temperature and Heat If two objects are NOT in thermal equilibrium, i their temperatures t must be different. To make their temperatures equal (i.e. thermal equilib), HEAT MUST FLOW. HEAT has to do with the transfer of thermal energy. Symbol for HEAT: Q BE VERY CAREFUL ABOUT THE SIGN System particular objecrtor set of objects Environment everythingelse else inthe universe Heat is negative when energy is transferred from the system s thermal energy to its environment (heat is released or lost) Heat is zero when no energy is transferred between the system s thermal energy and its environment (AT THERMAL EQUILIB) Heat is positive when energy is transferred to the system s thermal energy from its environment (heat is absorbed)

26 Thermal expansion Linear Volume ΔL = αl 0 ΔT L = L 0 1+ αδt ( ) ΔV = βv 0 ΔT V = V 0 1+ βδt ( ) Absorption of Heat by solids/liquids (in the same phase) Q = CΔT = C(T f T i ) Change of system energy with change of temperature Heat Transfer Heat Capacity [ J/ºC or cal/ºc ] depends on the material Specific Heat (heat capacity/mass) Q = cmδt = cm(t f T i ) f Water : c water = 1 cal/g. C= 1 Btu /lb. F= 4190 J/kg. C ~ independent of temperature c ~ c(t)

27 Heats of Transformation due to phase transition Heat is transferred in or out of system, but temperature does NOT change: Change of phase Heat of Heat of Fusion Vaporization Q = ±L m ΔT = 0 Amount of heat transferred during phase change depends on L and mass (M) L F F Heat of fusion Solid to liquid (heat is adsorbed : atomic bonds are broken L V Heat of Vaporization Liquid to gas (heat is adsorbed) L S Heat of Sublimation Solid to gas (heat is adsorbed) L F L V H O = 79.5 cal/g = kJ/mol = 333 kj/kg = 539 cal/g = 40.7 kj/mol = 56 kj/kg

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