1. Physics 201: Lecture 30, Pg 1 Lecture 30 Review (the final is, to a large degree, cumulative) 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 Today we will review chapters 13-15 1 st is short mention of resonance 1 st is short mention of resonance Order Ch. 15 to 13 Order Ch. 15 to 13

2. Physics 201: Lecture 30, Pg 2 Final Exam Details l Sunday, May 13th 10:05am-12:05pm in 125 Ag Hall & quiet room l Format:  Closed book  Up to 4 8½x1 sheets, hand written only  Approximately 50% from Chapters 13-15 and 50% 1-12  Bring a calculator l Special needs/ conflicts: All requests for alternative test arrangements should be made by today (except for medical emergency)

3. Physics 201: Lecture 30, Pg 3 Driven SHM with Resistance Apply a sinusoidal force, F 0 cos (  t), and now consider what A and b do, Not Zero!!!   b/m small b/m middling b large   steady state amplitude

4. Physics 201: Lecture 30, Pg 4 Dramatic example of resonance l In 1940, a steady wind set up a torsional vibration in the Tacoma Narrows Bridge 

5. Physics 201: Lecture 30, Pg 5 Dramatic example of resonance  l Eventually it collapsed

6. Physics 201: Lecture 30, Pg 6 Mechanical Energy of the Spring-Mass System Kinetic energy: K = ½ mv 2 = ½ m(  A) 2 sin 2 (  t+  ) Potential energy: U = ½ k x 2 = ½ k A 2 cos 2 (  t +  ) And  2 = k / m or k = m  2 K + U = constant x(t) = A cos(  t +  ) v(t) = -  A sin(  t +  ) a(t) = -  2 A cos(  t +  ) & F(t)=ma(t)

7. Physics 201: Lecture 30, Pg 7 SHM x(t) = A cos(  t +  ) v(t) = -  A sin(  t +  ) a(t) = -  2 A cos(  t +  ) A : amplitude  : angular frequency  =2  f =2  /T  : phase constant x max = A v max =  A a max =  2 A

8. Physics 201: Lecture 30, Pg 8 Recognizing the phase constant l 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

9. Physics 201: Lecture 30, Pg 9 Common SHMs

10. Physics 201: Lecture 30, Pg 10 SHM: Friction with velocity dependent Drag force -bv b is the drag coefficient; soln is a damped exponential if

11. Physics 201: Lecture 30, Pg 11 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?

12. Physics 201: Lecture 30, Pg 12 Chapter 14 Fluids l Density ρ = m/V l Pressure P = F/A P 1 atm = 1x10 5 N/m 2 Force is normal to container surface l Pressure with Depth/Height P = P 0 + ρgh l Gauge vs. Absolute pressure l Pascal’s Principle: Same depth  Same pressure l Buoyancy, force, B, is always upwards l B = ρ fluid V fluid displaced g (Archimedes’s Principle) l Flow Continuity: Q = v 2 A 2 = v 1 A 1 (volume / time or m 3 /s) Bernoulli’s eqn: P 1 + ½ ρv 1 2 + ρgh 1 = P 2 + ½ ρv 2 2 + ρgh 2

13. Physics 201: Lecture 30, Pg 13 Example problem l A piece of iron (ρ=7.9x10 3 kg/m 3 ) block weighs 1.0 N in air. How much does the scale read in water? l Solution: l In air T 1 = mg = ρ irom V g l In water: B+T 2 -mg = 0 T 2 = mg-B = mg – ρ water Vg = mg – ( ρ water /ρ iron ) ρ iron Vg = mg (1-ρ water /ρ iron ) = 0.87mg = 0.87 N

14. Physics 201: Lecture 30, Pg 14 Another buoyancy problem A spherical balloon is filled with air (  air 1.2 kg/m 3 ). The radius of the balloon is 0.50 m and the wall thickness of the latex wall is 0.01 m (  latex 10 3 kg/m 3 ). 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. l What is the magnitude of the drag force on the balloon? Key physics: Equilbrium and buoyancy.  F x =0 &  F y =0

15. Physics 201: Lecture 30, Pg 15 Another buoyancy problem A spherical balloon is filled with air (  air 1.2 kg/m 3 ). The radius of the balloon is 0.50 m and the wall thickness of the latex wall is 1.0 cm (  latex 10 3 kg/m 3 ). 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. l What is the magnitude of the drag force on the balloon? Key physics: Equilibrium and buoyancy.  F x =0 &  F y =0  F x =-T cos  + D = 0  F y =-T sin  + F b - W air = 0 W air =  air V g=  air (4/3  r 3 ) g with r=0.49 m F b =  water V g=  water (4/3  r 3 ) g Variation: What is the maximum wall thickness of a lead balloon filled with He? W air FbFb T D

16. Physics 201: Lecture 30, Pg 16 Pascal’s Principle l Is P A = P B ? Answer: No! Same level, same pressure, only if same fluid density A B

17. Physics 201: Lecture 30, Pg 17 Power from a river l 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? l W = Fd = mgh l P avg = W / t = (m/t) gh Q = Av and m/t =  water Q (kg/m 3 m 3 /s) P avg =  water Av gh = 10 3 kg/m 3 x 250 m 2 x 1.5 m/s x 10 m/s 2 x 50 m = 1.8x10 8 kg m 2 /s 2 /s = 180 MW

18. Physics 201: Lecture 30, Pg 18 Chapter 13 Gravitation l Universal gravitation force  Always attractive  Proportional to the mass ( m 1 m 2 )  Inversely proportional to the square of the distance (1/r 2 )  Central force: orbits conserve angular momentum l Gravitational potential energy  Always negative  Proportional to the mass ( m 1 m 2 )  Inversely proportional to the distance (1/r) l Circular orbits: Dynamical quantities (v,E,K,U,F) involve radius  K(r) = - ½ U(r) l Employ conservation of angular momentum in elliptical orbits l No need to derive Kepler’s Laws (know the reasons for them) l Energy transfer when orbit radius changes(e.g. escape velocity)

19. Physics 201: Lecture 30, Pg 19 Key equations l Newton’s Universal “Law” of Gravity Universal Gravitational Constant G = 6.673 x 10 -11 Nm 2 / kg 2 The force points along the line connecting the two objects. On Earth, near sea level, it can be shown that g surface ≈ 9.8 m/s 2. l Gravitational potential energy “Zero” of potential energy defined to be at r = ∞, force  0 l At apogee and perigee:

20. Physics 201: Lecture 30, Pg 20 Dynamics of Circular Orbits For a circular orbit: l Force on m: F G = GMm/r 2 l Orbiting speed: v 2 = GM/r (independent of m)  Kinetic energy: K = ½ mv 2 = ½ GMm/r  Potential energy U G = - GMm/r l Notice U G = -2 K l Total Mech. Energy:  E = KE + U G = - ½ GMm/r

21. Physics 201: Lecture 30, Pg 21 Changing orbit l 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) (M E = 5.97x10 24 kg, R E = 6.37x10 6 m, G = 6.67x10 -11 Nm 2 /kg 2 ) l Solution: Initial: h=0, r i = R E l E i = K i +U i = 0 + (-GM E m/R E ) = -1.25x10 10 J l In orbit: h = 200 km, r f = R E + 200 km E f = K f + U f = - ½ U f + U f = ½ U f = - ½ GM E m/(R E +200 km) = -6.06x10 9 J  E = E f – E i = 6.4x10 9 J

22. Physics 201: Lecture 30, Pg 22 Escaping Earth orbit l 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) implies infinite height

23. Physics 201: Lecture 30, Pg 23 We hope everyone does well on Sunday Have a great summer!