Conceptual Physics Matter Liquids...

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Conceptual Physics Matter Liquids Gases Lana Sheridan De Anza College July 19, 2016

Transcript of Conceptual Physics Matter Liquids...

Conceptual PhysicsMatterLiquidsGases

Lana Sheridan

De Anza College

July 19, 2016

Last time

• the atom

• history of our understanding of the atom

• solids

• density

Overview

• elasticity

• liquids

• pressure

• buoyancy and Archimedes’ Principle

• Pascal’s principle

• gases

• the atmosphere

• Bernoulli’s principle

Elasticity

Elasticity is a property of some solids: they can be deformed andreturn on their own to their original shape.

A spring is a device that makes use of this effect: it is a coil ofmetal.

It can be stretched to a longer length or compressed to a shorterone, but when released will return to its original length.

Elasticity

The force that the spring exerts to restore itself to its originallength is proportional to how much it is compressed or stretched.

This is called Hooke’s Law:

F = −kx

where x is the distance that the spring is stretched or compressedby and k is a constant that depends on the spring itself. (The“spring constant”).

If a very large force is put on the spring eventually it will break: itwill not return to its original shape. The elastic limit is themaximum distance the spring can be stretched so that it stillreturns to its original shape.

Spring example

If a 2 kg painting is hung from a spring, the spring stretches 10cm. What if instead a 4 kg painting is hung from the spring? Howfar will it stretch?

We don’t know the spring constant, but we can work it out fromthe information about the first 2 kg painting. The force on thespring is just the weight of the painting.

k =Fgx

=(2 kg)g

0.1 m≈ 200 N/m

x =F

k=

40

200= 0.2 m

If you put on twice the force, you stretch the spring twice as far!

Spring example

If a 2 kg painting is hung from a spring, the spring stretches 10cm. What if instead a 4 kg painting is hung from the spring? Howfar will it stretch?

We don’t know the spring constant, but we can work it out fromthe information about the first 2 kg painting. The force on thespring is just the weight of the painting.

k =Fgx

=(2 kg)g

0.1 m≈ 200 N/m

x =F

k=

40

200= 0.2 m

If you put on twice the force, you stretch the spring twice as far!

Spring example

If a 2 kg painting is hung from a spring, the spring stretches 10cm. What if instead a 4 kg painting is hung from the spring? Howfar will it stretch?

We don’t know the spring constant, but we can work it out fromthe information about the first 2 kg painting. The force on thespring is just the weight of the painting.

k =Fgx

=(2 kg)g

0.1 m≈ 200 N/m

x =F

k=

40

200= 0.2 m

If you put on twice the force, you stretch the spring twice as far!

Spring example

If a 2 kg painting is hung from a spring, the spring stretches 10 cm.

Now suppose a 6 kg painting is hung from the same spring. Howfar does it stretch?

x = 0.3 m.

[Also see prob. 4 from the textbook.]

Spring example

If a 2 kg painting is hung from a spring, the spring stretches 10 cm.

Now suppose a 6 kg painting is hung from the same spring. Howfar does it stretch?x = 0.3 m.

[Also see prob. 4 from the textbook.]

Liquids

liquid

a state of matter where the constituent particles are free to movearound relative to one another, but the substance is nearlyincompressible. Volume is fixed, but shape is not.

A solid that is heated will melt to form a liquid.

If that liquid is further heated it will boil to form a gas.

Both liquids and gases can flow, but the density of a substance inits liquid state is close to that in its solid state. For gases thedensity can vary.

Liquids

Liquids require a fairly narrow range of temperatures and pressuresto exist.

Earth has the right temperature and pressure on its surface forwater to be liquid.

Examples of liquids at room temperature:

• water

• oil

• alcohol

• mercury

Pressure

Most liquids maintain nearly constant volume whatever thepressure.

Pressure = ForceArea

P =F

A

An illustration: a hammer and a nail pounded by the samehammer exert the same force on a board. The pressure at the tipof the nail is much higher.

Pressure

Pressure is measured in Pascals, Pa.

1 Pa = 1 N/m2.

Atmospheric pressure is 101.3 kPa, or 1.013× 105 Pa.

Since the Pascal is a small unit (even 1 atmosphere is ∼ 105)sometimes atmospheres are also used as a unit of pressure.

1 atmosphere = 1.013× 105 Pa

Fluid Pressure

Pressure is a scalar quantity.

We understand that its underlying cause is molecular collisions:

1Diagram by Brant Carlson, on Wikipedia.

Liquid Pressure

Liquids also apply pressure onto submerged objects.

Liquid pressure depends on depth.

liquid pressure =weight density

of fluid× depth

Pliq = ρgh

where ρ = m/V is the mass density of the fluid and h is the depth.

It does not depend on the total amount of water involved, just thedepth of water.

Liquid Pressure

The force exerted downward by the column of water isF = mg = ρVg .

Pressure, Pliq = ρVg/A = ρgh.

Liquid Pressure

The force exerted downward by the column of water isF = mg = ρVg .

Pressure, Pliq = ρVg/A = ρgh.

Total Pressure

The liquid pressure only expresses the pressure due to the weight ofthe fluid above.

However, this is not the total pressure in most circumstances,eg. diving on earth.

The total pressure is the sum of the pressure due to the liquid andthe pressure due to the atmosphere.

Ptotal = ρgh + Patm

where Patm = 1.013× 105 Pa.

Total Pressure

The liquid pressure only expresses the pressure due to the weight ofthe fluid above.

However, this is not the total pressure in most circumstances,eg. diving on earth.

The total pressure is the sum of the pressure due to the liquid andthe pressure due to the atmosphere.

Ptotal = ρgh + Patm

where Patm = 1.013× 105 Pa.

Pressure varies with Depth

Pascal’s Barrel

0Illustration from ‘The forces of nature’ by Amedee Guillemin, 1872.

Question

Plug and Chug # 2, page 243.

Calculate the water pressure at the base of the Hoover Dam. Thedepth of water behind the dam is 220 m.

Surface Tension

The H2O molecules that make up water are slightly polar andattract one another.

This means that water droplets tend to arrange themselves tominimize their surface area.

cohesion

The attraction between like molecules

adhesion

The attraction between molecules of different substances

Water in air behaves as if it is covered with an elastic membranebecause the cohesive forces between the water molecules arestronger than the adhesive forces between water and air.

Surface Tension

Mensicus

A meniscus, or curvature of a liquid surface happens when a waterdroplet is in air, or when water is in a glass beaker.

Adhesive forces between water and glass are stronger than cohesiveforces within the water, so the water tends to stick to the glass.

This effect can also be used to draw water up in a glass tube.

Capillary Action

Siphoning

1Figure from Wikipedia.

Buoyancy

Astronauts training in their spacesuits:

The total mass of NASA’s EMU (extravehicular mobility unit) is178 kg. Why does training underwater make maneuvering in thesuits easier?

1Picture from Hubblesite.org.

Buoyancy

The apparent weight of submerged objects is less than its fullweight.

For an object that would float, but is held underwater, its apparentweight is negative!

There is an upward force on an object in a fluid called the buoyantforce.

Buoyancy and Archimedes’ Principle

Archimedes’ Principle

The buoyant force on an object is equal to the weight of the fluidthat the object displaces.

Logically, if a brick falls to the bottom of a pool it must push anamount water equal to its volume up and out of the way.

Buoyancy and Archimedes’ Principle

For a fully submerged object the buoyant force is:

Fbuoy = ρfVobj g

where ρf is the mass density of the fluid and Vobj is the volume ofthe object.

ρf Vobj is the mass of the water moved aside by the object.

Buoyancy and Archimedes’ Principle

An object that floats will displace less fluid than its entire volume.

For a floating object:

Fbuoy = ρfVsub g

where Vsub is the volume of the part of the object underneath thefluid level only.

Sinking and Floating

Will a particular object sink or float in a particular fluid?

• If the object is less dense than the fluid it will float.

• If the object is more dense than the fluid it will sink.

• If the object and the fluid have the same density if will neitherfloat or sink, but drift at equilibrium.

Sinking and Floating

A floating object displaces a mass of water equal to its own mass!

(Equivalently, a weight of water equal to its own weight.)

This also means that ρfVsub = mobj.

Questions

Military ships are often compared by their displacements, theweight (or mass, depending on context) of water they displace.

The USS Enterprise was an aircraft carrier (now decommissioned).

Displacement: 94,781 tonnes (metric tons), fully loaded.

1 tonne = 1000 kg

What is the mass of the fully loaded USS Enterprise in kgs?

Problem # 8, page246.

Your friend of mass 100 kg can just barely float in fresh water.Calculate her approximate volume.

Questions

Military ships are often compared by their displacements, theweight (or mass, depending on context) of water they displace.

The USS Enterprise was an aircraft carrier (now decommissioned).

Displacement: 94,781 tonnes (metric tons), fully loaded.

1 tonne = 1000 kg

What is the mass of the fully loaded USS Enterprise in kgs?

Problem # 8, page246.

Your friend of mass 100 kg can just barely float in fresh water.Calculate her approximate volume.

Pascal’s Principle

(Or, how hydraulic systems work.)

Liquids are (mostly) incompressible.

If you push on a piston at one end of a pipe you will move.

Pascal’s Principle

A change in pressure at any point in an enclosed fluid at rest istransmitted undiminished to all points in the fluid.

Pascal’s Principle

Since the pressures at the left end and the right end are the same:

P1 = P2

F1A1

=F2A2

Since A2 > A1, F2 > F1.

Pascal’s Principle

This has applications:

1Figure from hyperphysics.phy-astr.gsu.edu.

Question

If a pair of pistons are connected on either end of a hydraulic tube.The first has area 0.2 m2 and the second has an area of 4 m2.

A force of 30 N is applied the first piston. What is the forceexerted by the second piston on a mass that rests on it?

If the first piston is depressed a distance of 1 m by the 30 N force,how far does the second piston rise?

Question

If a pair of pistons are connected on either end of a hydraulic tube.The first has area 0.2 m2 and the second has an area of 4 m2.

A force of 30 N is applied the first piston. What is the forceexerted by the second piston on a mass that rests on it?

If the first piston is depressed a distance of 1 m by the 30 N force,how far does the second piston rise?

Gases

gas

a state of matter where the constituent particles are vastlyseparated from one another and free to move around.

A liquid that is heated will boil to form a gas.

A gas heated much further will form a plasma.

Both liquids and gases can flow, but the density of gases can varyand they are readily compressible.

The large distances between individual molecules or atoms is whymost gases are transparent.

Elemental Gases

• hydrogen (H2),

• nitrogen (N2)

• oxygen (O2)

• fluorine (F2)

• chlorine (Cl2)

• helium (He)

• neon (Ne)

• argon (Ar)

• krypton (Kr)

• xenon (Xe)

• radon (Rn)

The Atmosphere

We are living in a sea of air.

It is held in place by Earth’s gravity and made up of:

• nitrogen, N2, ∼ 78%

• oxygen, O2, ∼ 21%

• argon, Ar, ∼ 1%

• carbon dioxide, CO2, ∼ 0.4%

It also contains less that 1% water vapor.

Layers of the Atmosphere

1Picture shot from the ISS, courtesy of NASA Earth Observatory.

Air Pressure

Just like liquids, gases also exert pressure on their surroundings.

Atmospheric pressure at sea level is 101.3 kPa.

Atmospheric pressure falls with increasing altitude.

Atmospheric pressure fluctuates locally from place to place, day today.

BarometersBarometers are devices for measuring local atmospheric pressure.

Typically, simple barometers are filled with mercury, which is verydense.

The weight of the mercury in the tube exerts the same pressure asthe surrounding atmosphere. On low pressure days, the level of themercury drops. On high pressure days it rises.

Boyle’s Law

Suppose you seal a fixed mass of gas in a container.

If you compress it, you will decrease the volume, but increase thepressure. If you let it expand, you increase the volume but decreasethe pressure.

For a fixed mass of gas at a constant temperature:

P1V1 = P2V2

This is Boyle’s Law.

Buoyancy in Air

Buoyancy in air works the same way as in liquids:

Fbuoy = ρfVobj g

If an object is less dense than air, it will float upwards.

However, in the atmosphere, the density of air varies with height.

Bernoulli’s Principle

A law discovered by the 18th-century Swiss scientist, DanielBernoulli.

Bernoulli’s Principle

As the speed of a fluid’s flow increases, the pressure in the fluiddecreases.

This leads to a surprising effect: for liquids flowing in pipes, thepressure drops as the pipes get narrower.

Bernoulli’s Principle

Why should this principle hold? Where does it come from?

Actually, it just comes from the conservation of energy, and anassumption that the fluid is incompressible.1

Consider a fixed volume of fluid, V .

In a narrower pipe, this volume flows by a particular point 1 intime ∆t.

However, it must push the same volume of fluid past a point 2 inthe same time. If the pipe is wider at point 2, it flows more slowly.

1Something similar can be argued for compressible fluids also.

Bernoulli’s Principle

Why should this principle hold? Where does it come from?

Actually, it just comes from the conservation of energy, and anassumption that the fluid is incompressible.1

Consider a fixed volume of fluid, V .

In a narrower pipe, this volume flows by a particular point 1 intime ∆t.

However, it must push the same volume of fluid past a point 2 inthe same time. If the pipe is wider at point 2, it flows more slowly.

1Something similar can be argued for compressible fluids also.

Bernoulli’s Principle

V = A1v1∆t

also, V = A2v2∆t

This meansA1v1 = A2v2

The “Continuity equation”.

Bernoulli’s Principle

V = A1v1∆t

also, V = A2v2∆t

This meansA1v1 = A2v2

The “Continuity equation”.

Bernoulli’s Equation

Bernoulli’s equation is just the conservation of energy for this fluid.Let the volume of fluid, V , have mass m.

Remember:

• kinetic energy:

• gravitational potential energy:

• work:

Bernoulli’s Equation

Bernoulli’s equation is just the conservation of energy for this fluid.Let the volume of fluid, V , have mass m.

Remember:

• kinetic energy: 12mv2

• gravitational potential energy: mgh

• work: W = Fd

Bernoulli’s Equation

Bernoulli’s equation is just the conservation of energy for thisfluid.The system here is all of the fluid in the pipe shown.

Both light blue cylinders of fluid have the same volume, V , andsame mass m.

We imagine that in a time ∆t, volume V of fluid enters the leftend of the pipe, and another V exits the right.

Bernoulli’s EquationIt makes sense that the energy of the fluid might change: the fluidis moved along, and some is lifted up.

428 Chapter 14 Fluid Mechanics

The path taken by a fluid particle under steady flow is called a streamline. The velocity of the particle is always tangent to the streamline as shown in Figure 14.15. A set of streamlines like the ones shown in Figure 14.15 form a tube of flow. Fluid particles cannot flow into or out of the sides of this tube; if they could, the stream-lines would cross one another. Consider ideal fluid flow through a pipe of nonuniform size as illustrated in Fig-ure 14.16. Let’s focus our attention on a segment of fluid in the pipe. Figure 14.16a shows the segment at time t 5 0 consisting of the gray portion between point 1 and point 2 and the short blue portion to the left of point 1. At this time, the fluid in the short blue portion is flowing through a cross section of area A1 at speed v1. During the time interval Dt, the small length Dx1 of fluid in the blue portion moves past point 1. During the same time interval, fluid at the right end of the segment moves past point 2 in the pipe. Figure 14.16b shows the situation at the end of the time interval Dt. The blue portion at the right end represents the fluid that has moved past point 2 through an area A2 at a speed v2. The mass of fluid contained in the blue portion in Figure 14.16a is given by m1 5 rA1 Dx1 5 rA1v1 Dt, where r is the (unchanging) density of the ideal fluid. Similarly, the fluid in the blue portion in Figure 14.16b has a mass m2 5 rA2 Dx2 5 rA2v2 Dt. Because the fluid is incompressible and the flow is steady, however, the mass of fluid that passes point 1 in a time interval Dt must equal the mass that passes point 2 in the same time interval. That is, m1 5 m2 or rA1v1 Dt 5 rA2v2 Dt, which means that

A1v1 5 A2v2 5 constant (14.7)

This expression is called the equation of continuity for fluids. It states that the product of the area and the fluid speed at all points along a pipe is constant for an incompressible fluid. Equation 14.7 shows that the speed is high where the tube is constricted (small A) and low where the tube is wide (large A). The product Av, which has the dimensions of volume per unit time, is called either the volume flux or the flow rate. The condition Av 5 constant is equivalent to the statement that the vol-ume of fluid that enters one end of a tube in a given time interval equals the volume leaving the other end of the tube in the same time interval if no leaks are present. You demonstrate the equation of continuity each time you water your garden with your thumb over the end of a garden hose as in Figure 14.17. By partially block-

Equation of Continuity Xfor Fluids

Figure 14.17 The speed of water spraying from the end of a garden hose increases as the size of the opening is decreased with the thumb.©

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Lea

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Sem

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vS

At each point along its path, the particle’s velocity is tangent to the streamline.

Figure 14.15 A particle in laminar flow follows a streamline.

v2

v1

At t ! 0, fluid in the blueportion is moving pastpoint 1 at velocity v1.

After a time interval "t,the fluid in the blue portion is moving past point 2 at velocity v2.

"x1

"x2

Point 2

Point 1

A1

A2

a

S

S

S

S

b

Figure 14.16 A fluid moving with steady flow through a pipe of varying cross-sectional area. (a) At t 5 0, the small blue-colored portion of the fluid at the left is moving through area A1. (b) After a time interval Dt, the blue-colored portion shown here is that fluid that has moved through area A2.

How does it change? Depends on the work done:

W = ∆K + ∆U

Bernoulli’s Equation

430 Chapter 14 Fluid Mechanics

14.6 Bernoulli’s EquationYou have probably experienced driving on a highway and having a large truck pass you at high speed. In this situation, you may have had the frightening feeling that your car was being pulled in toward the truck as it passed. We will investigate the origin of this effect in this section. As a fluid moves through a region where its speed or elevation above the Earth’s surface changes, the pressure in the fluid varies with these changes. The relationship between fluid speed, pressure, and elevation was first derived in 1738 by Swiss physicist Daniel Bernoulli. Consider the flow of a segment of an ideal fluid through a nonuniform pipe in a time interval Dt as illustrated in Figure 14.18. This figure is very similar to Figure 14.16, which we used to develop the continuity equation. We have added two features: the forces on the outer ends of the blue portions of fluid and the heights of these portions above the reference position y 5 0. The force exerted on the segment by the fluid to the left of the blue portion in Figure 14.18a has a magnitude P1A1. The work done by this force on the segment in a time interval Dt is W1 5 F1 Dx1 5 P1A1 Dx1 5 P1V, where V is the volume of the blue portion of fluid passing point 1 in Figure 14.18a. In a similar manner, the work done on the segment by the fluid to the right of the segment in the same time interval Dt is W2 5 2P2A2 Dx2 5 2P2V, where V is the volume of the blue portion of fluid passing point 2 in Figure 14.18b. (The volumes of the blue portions of fluid in Figures 14.18a and 14.18b are equal because the fluid is incompressible.) This work is negative because the force on the segment of fluid is to the left and the displace-ment of the point of application of the force is to the right. Therefore, the net work done on the segment by these forces in the time interval Dt is

W 5 (P1 2 P2)V

Finalize The time interval for the element of water to fall to the ground is unchanged if the projection speed is changed because the projection is horizontal. Increasing the projection speed results in the water hitting the ground farther from the end of the hose, but requires the same time interval to strike the ground.

y1

y2

The pressure atpoint 1 is P1.

P1A1 i

The pressure atpoint 2 is P2. v2

v1!x1

!x2

Point 2

Point 1a

S

S

"P2A2 i

ˆ

ˆ

b

Figure 14.18 A fluid in laminar flow through a pipe. (a) A segment of the fluid at time t 5 0. A small portion of the blue-colored fluid is at height y1 above a reference position. (b) After a time interval Dt, the entire segment has moved to the right. The blue-colored por-tion of the fluid is that which has passed point 2 and is at height y2.

▸ 14.7 c o n t i n u e d

Daniel BernoulliSwiss physicist (1700–1782)Bernoulli made important discoveries in fluid dynamics. Bernoulli’s most famous work, Hydrodynamica, was published in 1738; it is both a theoreti-cal and a practical study of equilibrium, pressure, and speed in fluids. He showed that as the speed of a fluid increases, its pressure decreases. Referred to as “Bernoulli’s principle,” Bernoulli’s work is used to produce a partial vacuum in chemical laboratories by connecting a vessel to a tube through which water is running rapidly.

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The work done is the sum of thework done on each end of thefluid by more fluid that is oneither side of it:

W = F1∆x1 − F2∆x2

= P1A1∆x1 − P2A2∆x2

(The “environment fluid” just tothe right of the system fluid doesnegative work on the system as itmust be pushed aside by thesystem fluid.)

1Diagram from Serway & Jewett.

Bernoulli’s EquationNotice that V = A1∆x1 = A2∆x2

W = P1A1∆x1 − P2A2∆x2

= (P1 − P2)V

Conservation of energy:

W = ∆K + ∆U

(P1 − P2)V =1

2m(v22 − v21 ) +mg(h2 − h1)

Dividing by V :

P1 − P2 =1

2ρv22 + ρg(h2 − h1)

P1 +1

2ρv21 + ρgh1 = P2 +

1

2ρv22 + ρgh2

Bernoulli’s EquationNotice that V = A1∆x1 = A2∆x2

W = P1A1∆x1 − P2A2∆x2

= (P1 − P2)V

Conservation of energy:

W = ∆K + ∆U

(P1 − P2)V =1

2m(v22 − v21 ) +mg(h2 − h1)

Dividing by V :

P1 − P2 =1

2ρv22 + ρg(h2 − h1)

P1 +1

2ρv21 + ρgh1 = P2 +

1

2ρv22 + ρgh2

Bernoulli’s EquationNotice that V = A1∆x1 = A2∆x2

W = P1A1∆x1 − P2A2∆x2

= (P1 − P2)V

Conservation of energy:

W = ∆K + ∆U

(P1 − P2)V =1

2m(v22 − v21 ) +mg(h2 − h1)

Dividing by V :

P1 − P2 =1

2ρv22 + ρg(h2 − h1)

P1 +1

2ρv21 + ρgh1 = P2 +

1

2ρv22 + ρgh2

Bernoulli’s Equation

P11

2ρv21 + ρgh1 = P2 +

1

2ρv22 + ρgh2

This expression is true for any two points along a streamline.

Therefore,

P +1

2ρv2 + ρgh = const

is constant along a streamline in the fluid.

This is Bernoulli’s equation.

Bernoulli’s Equation

P +1

2ρv2 + ρgh = const

Even though we derived this expression for the case of anincompressible fluid, this is also true (to first order) forcompressible fluids, like air and other gases.

The constraint is that the densities should not vary too much fromthe ambient density ρ.

Bernoulli’s Principle from Bernoulli’s Equation

For two different points in the fluid, we have:

1

2ρv21 + ρgh1 + P1 =

1

2ρv22 + ρgh2 + P2

Suppose the height of the fluid does not change, so h1 = h2:

1

2ρv21 + P1 =

1

2ρv22 + P2

If v2 > v1 then P2 < P1.

Bernoulli’s Principle from Bernoulli’s Equation

For two different points in the fluid, we have:

1

2ρv21 + ρgh1 + P1 =

1

2ρv22 + ρgh2 + P2

Suppose the height of the fluid does not change, so h1 = h2:

1

2ρv21 + P1 =

1

2ρv22 + P2

If v2 > v1 then P2 < P1.

Bernoulli’s Principle from Bernoulli’s Equation

For two different points in the fluid, we have:

1

2ρv21 + ρgh1 + P1 =

1

2ρv22 + ρgh2 + P2

Suppose the height of the fluid does not change, so h1 = h2:

1

2ρv21 + P1 =

1

2ρv22 + P2

If v2 > v1 then P2 < P1.

Bernoulli’s Principle

However, rom the continuity equation A1v1 = A2v2 we can seethat if A2 is smaller than A1, v2 is bigger than v1.

So the pressure really does fall as the pipe contracts!

Implications of Bernoulli’s Principle

Bernoulli’s principle can help explain why airplanes making use ofairfoil-shaped wings can fly.

Air must travel further over the top of the wing, reducing pressurethere.

That means the air beneath the wing pushes upward on the wingmore strongly than the air on the top of the wing pushes down.This is called lift.

Air Flow over a Wing

In fact, the air flows over the wing much faster than under it: notjust because it travels a longer distance than over the top.

This is the result of circulation of air around the wing.

1Diagram by John S. Denker, av8n.com.

Airflow at different Angles of Attack

1Diagram by John S. Denker, av8n.com.

Implications of Bernoulli’s Principle

A stall occurs when turbulence behind the wing leads to a suddenloss of lift.

The streamlines over the wing detach from the wing surface.

This happens when the plane climbs too rapidly and can bedangerous.

1Photo by user Jaganath, Wikipedia.

Implications of Bernoulli’s Principle

Spoilers on cars reduce lift and promote laminar flow.

1Photo from http://oppositelock.kinja.com.

Implications of Bernoulli’s Principle

Wings on racing cars are inverted airfoils that produce downforceat the expense of increased drag.

This downforce increases the maximum possible static frictionforce ⇒ turns can be taken faster.

1Photo from http://oppositelock.kinja.com.

Implications of Bernoulli’s Principle

A curveball pitch in baseballalso makes use of Bernoulli’sprinciple.

The ball rotates as it movesthrough the air.

Its rotation pulls the airaround the ball, so the airmoving over one side of theball moves faster.

This causes the ball to deviatefrom a parabolic trajectory.

1Diagram by user Gang65, Wikipedia.

Implications of Bernoulli’s Principle

Bernoulli’s principle also explains why in a tornado, hurricane, orother extreme weather with high speed winds, windows blowoutward on closed buildings.

The high windspeed outside the building corresponds to lowpressure.

The pressure inside remains higher, and the pressure difference canbreak the windows.

It can also blow off the roof!

It makes sense to allow air a bit of air to flow in or out of abuilding in extreme weather, so that the pressure equalizes.

Implications of Bernoulli’s Principle

Bernoulli’s principle also explains why in a tornado, hurricane, orother extreme weather with high speed winds, windows blowoutward on closed buildings.

The high windspeed outside the building corresponds to lowpressure.

The pressure inside remains higher, and the pressure difference canbreak the windows.

It can also blow off the roof!

It makes sense to allow air a bit of air to flow in or out of abuilding in extreme weather, so that the pressure equalizes.

Summary• elasticity

• liquids

• pressure

• buoyancy

• gases

• the atmosphere

• Bernoulli’s principle

Talks! Aug 2-3.

Homework Hewitt,

• Ch 12, onward from page 225. Problem: 3

• Ch 13, onward from page 243. Plug and Chug: 1, 3;Exercises: 11, 15, 19, 25; Problems: 1, 3, 5, 7

• Ch 14, onward from page 262. Ranking 1; Exercises: 9, 17,19, 40, 53; Problems: 1