1 / 23

Lecture 30

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

alima
Download Presentation

Lecture 30

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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

  2. 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)

  3. b/m small steady state amplitude b/m middling b large w  w  w0 Driven SHM with Resistance • Apply a sinusoidal force, F0 cos (wt), and now consider what A andb do, Not Zero!!!

  4. Dramatic example of resonance • In 1940, a steady wind set up a torsional vibration in the Tacoma Narrows Bridge 

  5. Dramatic example of resonance • Eventually it collapsed 

  6. Mechanical Energy of the Spring-Mass System x(t) = A cos( t +  ) v(t) = -A sin( t +  ) a(t) = -2A cos( t +  ) & F(t)=ma(t) Kinetic energy: K = ½ mv2 = ½ m(A)2 sin2(t+f) Potential energy: U = ½ k x2 = ½ k A2 cos2(t + ) And w2 = k / m or k = m w2 K + U = constant

  7. SHM x(t) = A cos( t +  ) v(t) = -A sin( t +  ) a(t) = -2A cos( t +  ) A : amplitude • : angular frequency • =2pf =2p/T  : phase constant xmax = A vmax = A amax = 2A

  8. 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

  9. Common SHMs

  10. SHM: Friction with velocity dependent Drag force -bv b is the drag coefficient; soln is a damped exponential if

  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. Chapter 14 Fluids • Density ρ = m/V • Pressure P = F/A 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

  13. 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

  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 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

  15. 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=-Tcosq + D = 0 SFy=-Tsinq + Fb - Wair = 0 Wair = rair V g= rair (4/3 pr3) g with r=0.49 m Fb= rwater V g= rwater (4/3 pr3) g Variation: What is the maximum wall thickness of a lead balloon filled with He? Fb D T Wair

  16. Pascal’s Principle • Is PA = PB ? Answer: No! Same level, same pressure, only if same fluid density B A

  17. 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

  18. 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)

  19. Key equations • Newton’s Universal “Law” of Gravity Universal Gravitational Constant G = 6.673 x 10-11 Nm2 / kg2 The force points along the line connecting the two objects. On Earth, near sea level, it can be shown that gsurface ≈ 9.8 m/s2. • Gravitational potential energy “Zero” of potential energy defined to be at r = ∞, force  0 • At apogee and perigee:

  20. 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

  21. 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

  22. 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) implies infinite height

  23. We hope everyone does well on Sunday Have a great summer!

More Related