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Lecture 30 • Review (the final is, to a large degree, cumulative) ~50% refers to material in Ch. 1-12 ~50% refers to material in Ch. 13,14,15 -- Chapter 13: Gravitation -- Chapter 14: Newtonian Fluids -- Chapter 15: Oscillatory Motion • Today we will review chapters 13-15 • 1st is short mention of resonance • Order Ch. 15 to 13 Physics 201: Lecture 30, Pg 1 Final Exam Details Sunday, May 13th 10:05am-12:05pm in 125 Ag Hall & quiet room Format: Closed book Up to 4 8½x1 sheets, hand written only Approximately 50% from Chapters 13-15 and 50% 1-12 Bring a calculator Special needs/ conflicts: All requests for alternative test arrangements should be made by today (except for medical emergency) Physics 201: Lecture 30, Pg 2 Driven SHM with Resistance Apply a sinusoidal force, F0 cos (wt), and now consider what A and b do, steady state amplitude d 2 x b dx k F x cos wt 2 dt m dt m m A b/m small Not Zero!!! F0 / m bw 2 (w w ) ( ) m 2 2 2 0 b/m middling b large w w0 w Physics 201: Lecture 30, Pg 3 Dramatic example of resonance In 1940, a steady wind set up a torsional vibration in the Tacoma Narrows Bridge Physics 201: Lecture 30, Pg 4 Dramatic example of resonance Eventually it collapsed Physics 201: Lecture 30, Pg 5 Mechanical Energy of the Spring-Mass System x(t) = A cos( wt + f ) v(t) = -wA sin( wt + f ) a(t) = -w2A cos( wt + f ) & F(t)=ma(t) Kinetic energy: K = ½ mv2 = ½ m(wA)2 sin2(wt+f) Potential energy: U = ½ k x2 = ½ k A2 cos2(wt + f) And w2 = k / m or k = m w2 K + U = constant Physics 201: Lecture 30, Pg 6 SHM x(t) = A cos( wt + f ) v(t) = -wA sin( wt + f ) a(t) = -w2A cos( wt + f ) A : amplitude w : angular frequency w =2p f =2p/T f : phase constant xmax = A vmax = wA amax = w2A Physics 201: Lecture 30, Pg 7 Recognizing the phase constant An oscillation is described by x(t) =A cos(ωt+φ). Find φ for each of the following figures: Answers φ=0 φ = π/2 x(t)= A cos (π) φ=π Physics 201: Lecture 30, Pg 8 Common SHMs w k m w g L w I Physics 201: Lecture 30, Pg 9 SHM: Friction with velocity dependent Drag force -bv b is the drag coefficient; soln is a damped exponential x(t ) A exp ( b t ) cos(wt f ) 2m if wo b / 2m 1.2 1 k b w m 2m 0.8 0.6 0.4 2 A 0.2 0 -0.2 -0.4 b w w 2m 2 o -0.6 -0.8 -1 wt Physics 201: Lecture 30, Pg 10 2 SHM: Friction with velocity dependent Drag force -bv If the maximum amplitude drop 50% in 10 seconds, what will the relative drop be in 30 more seconds? x(t 10 s) 12 x(t ) x(t 30) A exp( 2bm (t 30)) x(t ) x(t 30 s) cx(t ) A exp( b t ) 2m 1.2 b 20) x ( t 10 ) exp ( x(t 30) 2m x(t ) x(t ) b 10)]2 [exp ( x(t 30) 2m x(t ) 2 1 0.8 0.6 0.4 A 0.2 0 -0.2 -0.4 x(t 30) 12 2 1 x(t ) 2 8 -0.6 -0.8 -1 wt Physics 201: Lecture 30, Pg 11 Chapter 14 Fluids Density ρ = m/V P1 atm = 1x105 N/m2 Force is normal to container surface Pressure with Depth/Height P = P0 + ρgh Gauge vs. Absolute pressure Pascal’s Principle: Same depth Same pressure Buoyancy, force, B, is always upwards B = ρfluid Vfluid displaced g (Archimedes’s Principle) Flow Continuity: Q = v2A2 = v1A1 (volume / time or m3/s) Bernoulli’s eqn: P1+ ½ ρv12 + ρgh1 = P2+ ½ ρv22 + ρgh2 Pressure P = F/A Physics 201: Lecture 30, Pg 12 Example problem A piece of iron (ρ=7.9x103 kg/m3) block weighs 1.0 N in air. How much does the scale read in water? Solution: In air T1 = mg = ρiromV g In water: B+T2-mg = 0 T2 = mg-B = mg – ρwaterVg = mg – ( ρwater /ρiron ) ρiron Vg = mg (1-ρwater /ρiron ) = 0.87mg = 0.87 N Physics 201: Lecture 30, Pg 13 Another buoyancy problem A spherical balloon is filled with air (rair 1.2 kg/m3). The radius of the balloon is 0.50 m and the wall thickness of the latex wall is 0.01 m (rlatex 103 kg/m3). The balloon is anchored to the bottom of stream which is flowing from left to right at 2.0 m/s. The massless string makes an angle of 30° from the stream bed. What is the magnitude of the drag force on the balloon? Key physics: Equilbrium and buoyancy. SFx=0 & SFy=0 Physics 201: Lecture 30, Pg 14 Another buoyancy problem A spherical balloon is filled with air (rair 1.2 kg/m3). The radius of the balloon is 0.50 m and the wall thickness of the latex wall is 1.0 cm (rlatex 103 kg/m3). The balloon is anchored to the bottom of stream which is flowing from left to right at 2.0 m/s. The massless string makes an angle of 45° from the stream bed. What is the magnitude of the drag force on the balloon? Key physics: Equilibrium and buoyancy. SFx=0 & SFy=0 SFx=-T cos q + D = 0 SFy=-T sin q + Fb - Wair = 0 Fb Wair = rair V g= rair (4/3 pr3) g with r=0.49 m D Fb= rwater V g= rwater (4/3 pr3) g T Variation: What is the maximum wall Wair thickness of a lead balloon filled with He? Physics 201: Lecture 30, Pg 15 Pascal’s Principle Is PA = PB ? Answer: No! Same level, same pressure, only if same fluid density B A Physics 201: Lecture 30, Pg 16 Power from a river Water in a river has a rectangular cross section which is 50 m wide and 5 m deep. The water is flowing at 1.5 m/s horizontally. A little bit downstream the water goes over a water fall 50 m high. How much power is potentially being generated in the fall? W = Fd = mgh Pavg= W / t = (m/t) gh Q = Av and m/t = rwater Q (kg/m3 m3/s) Pavg= rwater Av gh = 103 kg/m3 x 250 m2 x 1.5 m/s x 10 m/s2 x 50 m = 1.8x108 kg m2/s2/s = 180 MW Physics 201: Lecture 30, Pg 17 Chapter 13 Gravitation Universal gravitation force Always attractive Proportional to the mass ( m1m2 ) Inversely proportional to the square of the distance (1/r2) Central force: orbits conserve angular momentum Gravitational potential energy Always negative Proportional to the mass ( m1m2 ) Inversely proportional to the distance (1/r) Circular orbits: Dynamical quantities (v,E,K,U,F) involve radius K(r) = - ½ U(r) Employ conservation of angular momentum in elliptical orbits No need to derive Kepler’s Laws (know the reasons for them) Energy transfer when orbit radius changes(e.g. escape velocity) Physics 201: Lecture 30, Pg 18 Key equations Newton’s Universal “Law” of Gravity m1m2 Fon 2 by 1 G 2 r̂12 Fon1by 2 r Universal Gravitational Constant G = 6.673 x 10-11 Nm2 / kg2 The force points along the line connecting the two objects. g surface GM 2 RE On Earth, near sea level, it can be shown that gsurface ≈ 9.8 m/s2. Gravitational potential energy U (r ) Gm1m2 / r “Zero” of potential energy defined to be at r = ∞, force 0 At apogee and perigee: L mvr Physics 201: Lecture 30, Pg 19 Dynamics of Circular Orbits For a circular orbit: Force on m: FG= GMm/r2 Orbiting speed: v2 = GM/r (independent of m) Kinetic energy: K = ½ mv2 = ½ GMm/r Potential energy UG= - GMm/r Notice UG = -2 K Total Mech. Energy: E = KE + UG = - ½ GMm/r Physics 201: Lecture 30, Pg 20 Changing orbit A 200 kg satellite is launched into a circular orbit at height h= 200 km above the Earth’s surface. What is the minimum energy required to put it into the orbit ? (ignore Earth’s spin) (ME = 5.97x1024 kg, RE = 6.37x106 m, G = 6.67x10-11 Nm2/kg2) Solution: Initial: h=0, ri = RE Ei = Ki +Ui = 0 + (-GMEm/RE ) = -1.25x1010J In orbit: h = 200 km, rf = RE + 200 km Ef = Kf + Uf = - ½ Uf+ Uf = ½ Uf = - ½ GMEm/(RE+200 km) = -6.06x109J DE = Ef – Ei = 6.4x109 J Physics 201: Lecture 30, Pg 21 Escaping Earth orbit Exercise: suppose an object of mass m is projected vertically upwards from the Earth’s surface at an initial speed v, how high can it go ? (Ignore Earth’s rotation) Ei U G ( RE ) mv 1 2 2 i E f U G ( RE h) 0 2 2 E i Rv h 2 (2GM RE vi ) 2GM RE vi2 0 implies infinite height vEscape 2GM 11.2 km/s RE Physics 201: Lecture 30, Pg 22 We hope everyone does well on Sunday Have a great summer! Physics 201: Lecture 30, Pg 23