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1 Clicker Question Room Frequency BA The plastic is in equilibrium so F B = m plastic g = ρ plastic V g ! A solid piece of plastic of volume V, and density ρ plastic is floating partially submerged in a cup of water. (The density of water is ρ water.) What is the buoyant force on the plastic? A) Zero B) ρ plastic V C) ρ water V D) ρ water V g E) ρ plastic V g FBFB m plastic g

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CAPA assignment #13 is due on Friday at 10 pm. This week in Section: Assignment #6 Start reading Chapter 11 on Vibrations and Waves I will have regular office hours 1:45 – 3:45 in the Physics Helproom today Announcements 2

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3 Fluids in Motion: Fluid Dynamics Many, many different types of motion depending on particular properties of fluid: waves, rivers, geysers, tornados, hurricanes, ocean currents, trade winds, whirlpools, eddies, tsunamis, earthquakes, and on and on! We’ll focus on the simplest motion: flow

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4 Fluids in Motion: Flow Two main types of flow: Laminar and Turbulent Laminar Turbulent We’ll focus on the simplest flow: laminar flow

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5 Analysis of Flow Analyzing flow at the force level is mathematically complex Use conservation laws! 1) Conservation of Mass: the Continuity Equation 2) Conservation of Energy: Bernoulli’s Equation

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6 Continuity Equation Consider the flow of a fluid through a pipe in which the cross sectional area changes from A 1 to A 2 The mass of fluid going in has to equal the mass of fluid coming out: conservation of mass! The speed of the fluids must be different!

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7 Continuity Equation To analyze this mass conservation, we calculate the mass flow rate: Flow rate in = Flow rate out = must equal

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8 Mass Flow Rates Flow rate in = Flow rate out = Continuity Equation:

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9 Continuity for Incompressible Fluids If the fluid is incompressible: ρ 1 = ρ 2 so Does this make sense?

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10 “Incompressible” blood flows out of the heart via the aorta at a speed v aorta. The radius of the aorta r aorta = 1.2 cm. What is the speed of the blood in a connecting artery whose radius is 0.6 cm? A)v aorta B)2 v aorta C)(2) 1/2 v aorta D)4 v aorta E) 8 v aorta Clicker Question Room Frequency BA

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11 Bernoulli’s Equation: Conservation of Energy Earlier in the course we learned: Applied to fluid flow, we consider energy of pieces of fluid of mass Δm

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12 Bernoulli’s Equation: Incompressible Fluids Now using and the continuity equation you get Bernoulli’s Equation:

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13 Applications of Bernoulli’s Equation Bernoulli’s Equation is behind many common phenomena! 1)Curve balls 2)Aerodynamic Lift 3)Sailing into the wind 4)Transient Ischemic Attacks (“mini-strokes”) 5)Light objects getting sucked out your car window 6)Shower curtains bowing in 7)Flat roofs flying off houses in Boulder! 8)Ping pong ball demo

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14 Wind flows over a flat roof with area A = 240 m 2 at a speed of v outside = 35 m/s (125 km/h = 80 mi/h). What net force does the wind apply to the roof? hh v = 35 m/s h Inside Outside Flat Roof Example

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15 For an airplane wing (an air-foil) the upward lift force is derivable from Bernoulli’s equation. How does the air speed over the wing compare to the air speed under the wing? It is…… A) Faster B) Slower C) Same D) Unknown faster slower F = lift=(P bot -P top )(Wing Area) Clicker Question Room Frequency BA On the top side, the air has to travel farther to meet at the back edge of the wing!

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16 Oscillations! Throughout nature things are bound together by forces which allow things to oscillate back and forth. It is important to get a deeper understanding of these phenomena! We’ll focus on the most common and the most simple oscillation: Simple Harmonic Motion (SHM) Requirements for SHM: 1)There is a restoring force proportional to the displacement from equilibrium 2)The range of the motion (amplitude) is independent of the frequency 3)The position, velocity, and acceleration are all sinusoidal (harmonic) in time

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17 Mass and Spring

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18 A Simple Harmonic Oscillator: Spring and Mass! Note: restoring force is proportional to displacement force is not constant, so acceleration isn’t either: a = -(k/m)x “amplitude” A is the maximum displacement x max, occurs with v = 0 mass oscillates between x = A & x = -A maximum speed v max occurs when displacement x = 0 a “cycle” is the full extent of motion as shown the time to complete one cycle is the “period” T frequency is the number of cycles per second: f = 1/T (units Hz)

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