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Clicker Question Room Frequency BA

The plastic is in equilibrium so FB = mplasticg = ρplastic V g !

A solid piece of plastic of volume V, and density ρplastic is floating partially submerged in a cup of water. (The density of water is ρwater.) What is the buoyant force on the plastic?

A) ZeroB) ρplastic VC) ρwater VD) ρwater V gE) ρplastic V g

FB

mplasticg

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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

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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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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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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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Continuity Equation

Consider the flow of a fluid through a pipe in which the cross sectional area changes from A1 to A2

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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Continuity EquationTo analyze this mass conservation, we calculate the mass flow rate: Δm

Δt

=mass flowing through a volume

elapsed time

Flow rate in = Δm1

ΔtFlow rate out =

Δm2

Δtmust equal

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Mass Flow Rates

Flow rate in = Δm1

Δt= ρ1ΔV1

Δt

Flow rate out = Δm2

Δt= ρ 2A2v2

=ρ1A1Δl1

Δt=ρ1A1v1

ρ1A1v1 = ρ 2A2v2Continuity Equation:

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Continuity for Incompressible Fluids

If the fluid is incompressible: ρ1 = ρ2 so

Does this make sense?

A1v1 =A2v2

v2 =A1

A2

v1

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“Incompressible” blood flows out of the heart via the aorta at a speed vaorta. The radius of the aorta raorta = 1.2 cm. What is the speed of the blood in a connecting artery whose radius is 0.6 cm?

A) vaorta

B) 2 vaorta

C) (2)1/2 vaorta

D) 4 vaorta

E) 8 vaorta

vartery =Aaoρta

Aaρteρy

⎛⎝⎜

⎞⎠⎟vaoρta =

πρaoρta2

πρaρteρy2

⎛⎝⎜

⎞⎠⎟vaoρta

=ρaoρtaρaρteρy

⎛⎝⎜

⎞⎠⎟2

vaoρta =4vaoρta

Clicker Question Room Frequency BA

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Bernoulli’s Equation: Conservation of Energy

Earlier in the course we learned: PEi +KEi +Wi =PEf +KEf +W f

Applied to fluid flow, we consider energy of pieces of fluid of mass Δm

Δm1gy1 + 12 Δm1v1

2 + (P1A1Δl1) = Δm2gy2 + 12 Δm2v2

2 + (P2A2Δl2 )

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Bernoulli’s Equation: Incompressible Fluids

ρgy1 + 12 ρv1

2 + P1 = ρgy2 + 12 ρv2

2 + P2

Now using Δm1 = ρΔV1 = ρA1Δl1 and Δm2 = ρΔV1 = ρA2Δl2

and the continuity equation you get A1v1 =A2v2

Bernoulli’s Equation:

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Applications of Bernoulli’s Equation

Bernoulli’s Equation is behind many common phenomena!1) Curve balls2) Aerodynamic Lift3) Sailing into the wind4) Transient Ischemic Attacks (“mini-strokes”)5) Light objects getting sucked out your car window6) Shower curtains bowing in7) Flat roofs flying off houses in Boulder!8) Ping pong ball demo

y1 ≈y2 case: 12 ρv12 +P1 =1

2 ρv22 +P2

Static Fluid (v=0): ρgy1 +P1 =ρgy2 +P2

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ρgy1 +12ρv1

2 + P1 = ρgy2 +12ρv2

2 + P2

Wind flows over a flat roof with area A = 240 m2 at a speed of voutside = 35 m/s (125 km/h = 80 mi/h). What net force does the wind apply to the roof?

hh

v = 35 m/s

ρ gh +12ρ vinside

2 + Pinside = ρ gh +12ρvoutside

2 + Poutside

Pinside − Poutside =12ρ voutside

2( ) =FA

F =12ρAvoutside

2 = 0.5(1.29kg / m3)(240m2 )(35m / s)2

= 1.89x105N1lb

4.45N⎛⎝⎜

⎞⎠⎟ ≈ 42,600lbs

h

Inside Outside

Flat Roof Example

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ρgytop + 12ρvtop

2 + Ptop = ρgybot + 12ρvbot

2 + Pbot

12ρvtop

2 + Ptop = 12ρvbot

2 + Pbot

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) FasterB) SlowerC) SameD) Unknown

faster

slower

F = lift=(Pbot-Ptop)(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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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

equilibrium2) The range of the motion (amplitude) is independent of the

frequency3) The position, velocity, and acceleration are all sinusoidal

(harmonic) in time

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Mass and Spring

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A Simple Harmonic Oscillator: Spring and Mass!

F =ma=−kx (Hooke's Law)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 xmax, occurs with v = 0• mass oscillates between x = A & x = -A• maximum speed vmax 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)