21.1 Hidrostática: definições 21.2 Pressão hidrostática 21.3 … · 28/Maio/2018 – Aula 21...

28/Maio/2018 – Aula 21

30/Maio/2018 – Aula 22 22 Relatividade (introdução) 22.1 Descrição do movimento 22.2 Relatividade de Galileu 22.3 Velocidade da luz no vácuo 22.4 Relatividade de Einstein 23.4.1 Dilatação do tempo 23.4.2 Contração do espaço 22.5 Transformações de Lorentz 22.6 Momento linear 22.7 Energia

21 Fluidos 21.1 Hidrostática: definições 21.2 Pressão hidrostática 21.3 Princípio de Pascal 21.4 Princípio de Arquimedes 21.5 Hidrodinâmica: equação da continuidade 21.6 Equação de Bernoulli

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12.2 Pressure in a Fluid 375

12.2 Pressure in a FluidWhen a fluid (either liquid or gas) is at rest, it exerts a force perpendicular to anysurface in contact with it, such as a container wall or a body immersed in thefluid. This is the force that you feel pressing on your legs when you dangle themin a swimming pool. While the fluid as a whole is at rest, the molecules that makeup the fluid are in motion; the force exerted by the fluid is due to molecules col-liding with their surroundings.

If we think of an imaginary surface within the fluid, the fluid on the twosides of the surface exerts equal and opposite forces on the surface. (Otherwise,the surface would accelerate and the fluid would not remain at rest.) Considera small surface of area dA centered on a point in the fluid; the normal forceexerted by the fluid on each side is (Fig. 12.2). We define the pressure pat that point as the normal force per unit area—that is, the ratio of to dA(Fig. 12.3):

(definition of pressure) (12.2)

If the pressure is the same at all points of a finite plane surface with area A, then

(12.3)

where is the net normal force on one side of the surface. The SI unit of pressureis the pascal, where

We introduced the pascal in Chapter 11. Two related units, used principally inmeteorology, are the bar, equal to and the millibar, equal to 100 Pa.

Atmospheric pressure is the pressure of the earth’s atmosphere, the pres-sure at the bottom of this sea of air in which we live. This pressure varies withweather changes and with elevation. Normal atmospheric pressure at sea level(an average value) is 1 atmosphere (atm), defined to be exactly 101,325 Pa. Tofour significant figures,

= 1.013 bar = 1013 millibar = 14.70 lb>in.21pa2av = 1 atm = 1.013 * 105 Pa

pa

105 Pa,

1 pascal = 1 Pa = 1 N>m2

F!

p =F!

A

p =dF!

dA

dF!

dF!

sea level, but the density varies negligibly over the room’s 3.0-mheight; see Section 12.2.) We use Eq. (12.1) to relate the mass to the room’s volume V (which we’ll calculate) and the air density

(given in Table 12.1).

EXECUTE: We have sofrom Eq. (12.1),

wair = mairg = 172 kg219.8 m>s22 = 700 N = 160 lb

mair = rairV = 11.20 kg>m32160 m32 = 72 kg

V = 14.0 m215.0 m213.0 m2 = 60 m3,

rair

m air

The mass and weight of an equal volume of water are

EVALUATE: A roomful of air weighs about the same as an averageadult. Water is nearly a thousand times denser than air, so its massand weight are larger by the same factor. The weight of a roomfulof water would collapse the floor of an ordinary house.

= 5.9 * 105 N = 1.3 * 105 lb = 66 tons

wwater = mwaterg = 16.0 * 104 kg219.8 m>s22mwater = rwaterV = 11000 kg>m32160 m32 = 6.0 * 104 kg

Test Your Understanding of Section 12.1 Rank the following objectsin order from highest to lowest average density: (i) mass 4.00 kg, volume

(ii) mass 8.00 kg, volume (iii) mass 8.00 kg,volume (iv) mass 2560 kg, volume (v) mass 2560 kg, volume ❙1.28 m3.

0.640 m3;3.20 * 10-3 m3;1.60 * 10-3 m3;1.60 * 10-3 m3;

The surface does not accelerate, so thesurrounding fluid exerts equal normal forceson both sides of it. (The fluid cannot exert anyforce parallel to the surface, since that wouldcause the surface to accelerate.)

dAdF!

dF!

A small surface of areadA within a fluid at rest

12.2 Forces acting on a small surfacewithin a fluid at rest.

... but the pressure onthem (force magnitudedivided by area) is thesame (and is a scalar).

These surfaces differin area and orientation ...

dA2dAdF!dF!

2dF!

2dF!

12.3 The pressure on either side of asurface is force divided by area. Pressure isa scalar with units of newtons per squaremeter. By contrast, force is a vector withunits of newtons.


12.2 Pressure in a FluidWhen a fluid (either liquid or gas) is at rest, it exerts a force perpendicular to anysurface in contact with it, such as a container wall or a body immersed in thefluid. This is the force that you feel pressing on your legs when you dangle themin a swimming pool. While the fluid as a whole is at rest, the molecules that makeup the fluid are in motion; the force exerted by the fluid is due to molecules col-liding with their surroundings.

If we think of an imaginary surface within the fluid, the fluid on the twosides of the surface exerts equal and opposite forces on the surface. (Otherwise,the surface would accelerate and the fluid would not remain at rest.) Considera small surface of area dA centered on a point in the fluid; the normal forceexerted by the fluid on each side is (Fig. 12.2). We define the pressure pat that point as the normal force per unit area—that is, the ratio of to dA(Fig. 12.3):

(definition of pressure) (12.2)

If the pressure is the same at all points of a finite plane surface with area A, then

(12.3)

where is the net normal force on one side of the surface. The SI unit of pressureis the pascal, where

We introduced the pascal in Chapter 11. Two related units, used principally inmeteorology, are the bar, equal to and the millibar, equal to 100 Pa.

Atmospheric pressure is the pressure of the earth’s atmosphere, the pres-sure at the bottom of this sea of air in which we live. This pressure varies withweather changes and with elevation. Normal atmospheric pressure at sea level(an average value) is 1 atmosphere (atm), defined to be exactly 101,325 Pa. Tofour significant figures,

= 1.013 bar = 1013 millibar = 14.70 lb>in.21pa2av = 1 atm = 1.013 * 105 Pa

pa

105 Pa,

1 pascal = 1 Pa = 1 N>m2

F!

p =F!

A

p =dF!

dA

dF!

dF!

sea level, but the density varies negligibly over the room’s 3.0-mheight; see Section 12.2.) We use Eq. (12.1) to relate the mass to the room’s volume V (which we’ll calculate) and the air density

(given in Table 12.1).

EXECUTE: We have sofrom Eq. (12.1),

wair = mairg = 172 kg219.8 m>s22 = 700 N = 160 lb

mair = rairV = 11.20 kg>m32160 m32 = 72 kg

V = 14.0 m215.0 m213.0 m2 = 60 m3,

rair

m air

The mass and weight of an equal volume of water are

EVALUATE: A roomful of air weighs about the same as an averageadult. Water is nearly a thousand times denser than air, so its massand weight are larger by the same factor. The weight of a roomfulof water would collapse the floor of an ordinary house.

= 5.9 * 105 N = 1.3 * 105 lb = 66 tons

wwater = mwaterg = 16.0 * 104 kg219.8 m>s22mwater = rwaterV = 11000 kg>m32160 m32 = 6.0 * 104 kg

Test Your Understanding of Section 12.1 Rank the following objectsin order from highest to lowest average density: (i) mass 4.00 kg, volume

(ii) mass 8.00 kg, volume (iii) mass 8.00 kg,volume (iv) mass 2560 kg, volume (v) mass 2560 kg, volume ❙1.28 m3.

0.640 m3;3.20 * 10-3 m3;1.60 * 10-3 m3;1.60 * 10-3 m3;

The surface does not accelerate, so thesurrounding fluid exerts equal normal forceson both sides of it. (The fluid cannot exert anyforce parallel to the surface, since that wouldcause the surface to accelerate.)

dAdF!

dF!

A small surface of areadA within a fluid at rest

12.2 Forces acting on a small surfacewithin a fluid at rest.

... but the pressure onthem (force magnitudedivided by area) is thesame (and is a scalar).

These surfaces differin area and orientation ...

dA2dAdF!dF!

2dF!

2dF!

12.3 The pressure on either side of asurface is force divided by area. Pressure isa scalar with units of newtons per squaremeter. By contrast, force is a vector withunits of newtons.

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21.1 Definições

Quando um fluido está em contacto com uma superfície sólida, exerce uma força normal (perpendicular) em cada ponto da superfície. Ao quociente entre essa força (F) e a área de contacto (A), chama-se pressão.

p =dF⊥dA

p⎡⎣ ⎤⎦= N/m2 = Pascal (Pa)

15-2 PressureThe same force applied over a smaller area results in greater pressure – think of poking a balloon with your finger and then with a needle.

By Dr. Weining man 5

421

Matter is normally classified as being in one of three states: solid, liquid, or gas. Fromeveryday experience, we know that a solid has a definite volume and shape. A brickmaintains its familiar shape and size day in and day out. We also know that a liquid hasa definite volume but no definite shape. Finally, we know that an unconfined gas hasneither a definite volume nor a definite shape. These descriptions help us picture thestates of matter, but they are somewhat artificial. For example, asphalt and plastics arenormally considered solids, but over long periods of time they tend to flow like liquids.Likewise, most substances can be a solid, a liquid, or a gas (or a combination of any ofthese), depending on the temperature and pressure. In general, the time it takes a par-ticular substance to change its shape in response to an external force determineswhether we treat the substance as a solid, a liquid, or a gas.

A fluid is a collection of molecules that are randomly arranged and held togetherby weak cohesive forces and by forces exerted by the walls of a container. Both liquidsand gases are fluids.

In our treatment of the mechanics of fluids, we do not need to learn any new physi-cal principles to explain such effects as the buoyant force acting on a submerged ob-ject and the dynamic lift acting on an airplane wing. First, we consider the mechanicsof a fluid at rest—that is, fluid statics. We then treat the mechanics of fluids in motion—that is, fluid dynamics. We can describe a fluid in motion by using a model that is basedupon certain simplifying assumptions.

14.1 Pressure

Fluids do not sustain shearing stresses or tensile stresses; thus, the only stress that canbe exerted on an object submerged in a static fluid is one that tends to compress theobject from all sides. In other words, the force exerted by a static fluid on an object isalways perpendicular to the surfaces of the object, as shown in Figure 14.1.

The pressure in a fluid can be measured with the device pictured in Figure 14.2.The device consists of an evacuated cylinder that encloses a light piston connected to aspring. As the device is submerged in a fluid, the fluid presses on the top of the pistonand compresses the spring until the inward force exerted by the fluid is balanced bythe outward force exerted by the spring. The fluid pressure can be measured directly ifthe spring is calibrated in advance. If F is the magnitude of the force exerted on thepiston and A is the surface area of the piston, then the pressure P of the fluid at thelevel to which the device has been submerged is defined as the ratio F/A:

(14.1)

Note that pressure is a scalar quantity because it is proportional to the magnitude ofthe force on the piston.

If the pressure varies over an area, we can evaluate the infinitesimal force dF on aninfinitesimal surface element of area dA as

P ! FA

Figure 14.1 At any point on thesurface of a submerged object, theforce exerted by the fluid is per-pendicular to the surface of the ob-ject. The force exerted by the fluidon the walls of the container is per-pendicular to the walls at all points.

Definition of pressure

F

Vacuum

A

Figure 14.2 A simple device formeasuring the pressure exerted bya fluid.

Aula anterior


(pressure in a fluid of uniform density) (12.5)

It’s often convenient to express Eq. (12.5) in terms of the depth below the sur-face of a fluid (Fig. 12.5). Take point 1 at any level in the fluid and let p representthe pressure at this point. Take point 2 at the surface of the fluid, where the pres-sure is (subscript zero for zero depth). The depth of point 1 below the surfaceis and Eq. (12.5) becomes


The pressure p at a depth h is greater than the pressure at the surface by anamount Note that the pressure is the same at any two points at the same levelin the fluid. The shape of the container does not matter (Fig. 12.6).

Equation (12.6) shows that if we increase the pressure at the top surface,possibly by using a piston that fits tightly inside the container to push down onthe fluid surface, the pressure p at any depth increases by exactly the sameamount. This fact was recognized in 1653 by the French scientist Blaise Pascal(1623–1662) and is called Pascal’s law.

Pascal’s law: Pressure applied to an enclosed fluid is transmitted undiminishedto every portion of the fluid and the walls of the containing vessel.

The hydraulic lift shown schematically in Fig. 12.7 illustrates Pascal’s law. Apiston with small cross-sectional area exerts a force on the surface of a liq-uid such as oil. The applied pressure is transmitted through the con-necting pipe to a larger piston of area The applied pressure is the same in bothcylinders, so

(12.7)

The hydraulic lift is a force-multiplying device with a multiplication factor equalto the ratio of the areas of the two pistons. Dentist’s chairs, car lifts and jacks,many elevators, and hydraulic brakes all use this principle.

For gases the assumption that the density is uniform is realistic only overshort vertical distances. In a room with a ceiling height of 3.0 m filled with air ofuniform density the difference in pressure between floor and ceiling,given by Eq. (12.6), is

or about 0.00035 atm, a very small difference. But between sea level and thesummit of Mount Everest (8882 m) the density of air changes by nearly a factorof 3, and in this case we cannot use Eq. (12.6). Liquids, by contrast, are nearlyincompressible, and it is usually a very good approximation to regard their den-sity as independent of pressure. A pressure of several hundred atmospheres willcause only a few percent increase in the density of most liquids.

Absolute Pressure and Gauge PressureIf the pressure inside a car tire is equal to atmospheric pressure, the tire is flat.The pressure has to be greater than atmospheric to support the car, so the signif-icant quantity is the difference between the inside and outside pressures. Whenwe say that the pressure in a car tire is “32 pounds” (actually equal to220 kPa or we mean that it is greater than atmospheric pressure2.2 * 105 Pa),

32 lb>in.2,

rgh = 11.2 kg>m3219.8 m>s2213.0 m2 = 35 Pa

1.2 kg>m3,

r

p =F1

A1=

F2

A2 and F2 =

A2

A1F1

A2.p = F1>A1

F1A1

p0

rgh.p0

p = p0 + rgh

p0 - p = -rg1y2 - y12 = -rgh or

h = y2 - y1,p0

p2 - p1 = -rg1y2 - y12

Pressure difference between levels 1 and 2:

The pressure is greater at the lower level.p2 2 p1 5 2rg(y2 2 y1)

At a depth h, thepressure p equalsthe surface pressurep0 plus the pressurergh due to theoverlying fluid:p 5 p0 1 rgh.

p2 5 p0

p1 5 p

y1

y2

y2 2 y1 5 h

2

1

Fluid, density r

12.5 How pressure varies with depth ina fluid with uniform density.

The pressure at the bottom of each liquidcolumn has the same value p.

The pressure at the top of each liquidcolumn is atmospheric pressure, p0.

The difference between p and p0 is rgh, whereh is the distance from the top to the bottom ofthe liquid column. Hence all columns have thesame height.

12.6 Each fluid column has the sameheight, no matter what its shape.

F2

pA2

F1

pA1

A small force is applied to a small piston.

Because the pressure p is thesame at all pointsat a givenheight inthe fluid ...

... a piston of larger area at the sameheight experiences a larger force.

12.7 The hydraulic lift is an applicationof Pascal’s law. The size of the fluid-filledcontainer is exaggerated for clarity.

3

Variação da pressão com a altura:

dpdy

= −ρg p = p0 + ρgh

21.2 Pressão hidrostática

Pressão vs altura

simulação

15-3 Static Equilibrium in Fluids: Pressure and Depth


Aula anterior









(12.7)





32 lb>in.2,

rgh = 11.2 kg>m3219.8 m>s2213.0 m2 = 35 Pa

1.2 kg>m3,

r

p =F1

A1=

F2

A2 and F2 =

A2

A1F1

A2.p = F1>A1

F1A1

p0

rgh.p0

p = p0 + rgh

p0 - p = -rg1y2 - y12 = -rgh or

h = y2 - y1,p0

p2 - p1 = -rg1y2 - y12




p2 5 p0

p1 5 p

y1

y2

y2 2 y1 5 h

2

1

Fluid, density r






F2

pA2

F1

pA1













(12.7)





32 lb>in.2,

rgh = 11.2 kg>m3219.8 m>s2213.0 m2 = 35 Pa

1.2 kg>m3,

r

p =F1

A1=

F2

A2 and F2 =

A2

A1F1

A2.p = F1>A1

F1A1

p0

rgh.p0

p = p0 + rgh

p0 - p = -rg1y2 - y12 = -rgh or

h = y2 - y1,p0

p2 - p1 = -rg1y2 - y12




p2 5 p0

p1 5 p

y1

y2

y2 2 y1 5 h

2

1

Fluid, density r






F2

pA2

F1

pA1





4

A pressão dentro de um fluido é transmitida a todos os pontos desse fluido, de modo uniforme: princípio (ou lei) de Pascal.

21.3 Princípio de Pascal

p =F1A1=F2A2

⇒ F2 =A2A1F1

Aula anterior

5

Um objeto, completa ou parcialmente imerso num fluido, sofre uma força flutuante, que é igual e contrária à força da gravidade, no fluido que foi deslocado (peso do fluido deslocado): princípio de Arquimedes.

21.4 Princípio de Arquimedes

simulação

14.4 Buoyant Forces and Archimedes’s Principle

Have you ever tried to push a beach ball under water (Fig. 14.7a)? This is extremelydifficult to do because of the large upward force exerted by the water on the ball. Theupward force exerted by a fluid on any immersed object is called a buoyant force. Wecan determine the magnitude of a buoyant force by applying some logic. Imagine abeach ball!sized parcel of water beneath the water surface, as in Figure 14.7b. Becausethis parcel is in equilibrium, there must be an upward force that balances the down-ward gravitational force on the parcel. This upward force is the buoyant force, and itsmagnitude is equal to the weight of the water in the parcel. The buoyant force is theresultant force due to all forces applied by the fluid surrounding the parcel.

Now imagine replacing the beach ball!sized parcel of water with a beach ball of thesame size. The resultant force applied by the fluid surrounding the beach ball is the same,regardless of whether it is applied to a beach ball or to a parcel of water. Consequently, wecan claim that the magnitude of the buoyant force always equals the weight of thefluid displaced by the object. This statement is known as Archimedes’s principle.

With the beach ball under water, the buoyant force, equal to the weight of a beachball-sized parcel of water, is much larger than the weight of the beach ball. Thus, thereis a net upward force of large magnitude—this is why it is so hard to hold the beach ballunder the water. Note that Archimedes’s principle does not refer to the makeup of theobject experiencing the buoyant force. The object’s composition is not a factor in thebuoyant force because the buoyant force is exerted by the fluid.

S E C T I O N 1 4 . 4 • Buoyant Forces and Archimedes’s Principle 427

Quick Quiz 14.3 Several common barometers are built, with a variety offluids. For which of the following fluids will the column of fluid in the barometer bethe highest? (a) mercury (b) water (c) ethyl alcohol (d) benzene

Quick Quiz 14.4 You have invented a spacesuit with a straw passing throughthe faceplate so that you can drink from a glass while on the surface of a planet. Outon the surface of the Moon, you attempt to drink through the straw from an open glassof water. The value of g on the Moon is about one sixth of that on Earth. Compared tothe difficulty in drinking through a straw on Earth, you find drinking through a strawon the Moon to be (a) easier (b) equally difficult (c) harder (d) impossible.

(a)

B

Fg

(b)

Figure 14.7. (a) A swimmer attempts to push a beach ball underwater. (b) The forceson a beach ball–sized parcel of water. The buoyant force B on a beach ball that replacesthis parcel is exactly the same as the buoyant force on the parcel.

ArchimedesGreek Mathematician, Physicist, and Engineer(287–212 B.C.)

Archimedes was perhaps thegreatest scientist of antiquity. Hewas the first to computeaccurately the ratio of a circle’scircumference to its diameter,and he also showed how tocalculate the volume and surfacearea of spheres, cylinders, andother geometric shapes. He iswell known for discovering thenature of the buoyant force andwas also a gifted inventor. Oneof his practical inventions, still inuse today, is Archimedes’sscrew, an inclined, rotating,coiled tube used originally to liftwater from the holds of ships. Healso invented the catapult anddevised systems of levers,pulleys, and weights for raisingheavy loads. Such inventionswere successfully used todefend his native city, Syracuse,during a two-year siege byRomans.

Aula anterior

De acordo com o princípio de Arquimedes, um objeto pode flutuar quando a força flutuante for igual ao seu peso, ou seja, quando o volume de fluido deslocado representar um peso igual ao do objeto.

21.4 Princípio de Arquimedes

15-5 Applications of Archimedes’ Principle

An object made of material that is denser than water can float only if it has indentations or pockets of air that make its average density less than that of water.


Wood

Metal

Metal

Condição de equilíbrio:

mg = ρobjetoVobjeto g

mg = Fflutuante

Fflutuante = ρfluidoVfluido g

ρobjetoVobjeto= ρfluidoVfluido

15-5 Floating objectThe fraction of an object that is submerged when it is floating depends on the densities of the object and of the fluid.

By Dr. Weining man

18

1, Analyze forces:2, Fb=mg;

Ice density = 0.9 water densityVsub/V = Uice/Uwater =0.9That’s why “The tip of a iceberg”means a very small portion

15-5 Applications of Archimedes’ Principle

An object floats, when the buoyancy force is equal to its weight.

An object floats, when it displaces an amount of fluid equal to its weight.


Aula anterior

7

O fluxo (do fluido) através de uma superfície A é a taxa a que essa superfície é atravessada por um dado volume de fluido.

(taxa de) Fluxo:

Volume de fluido deslocado num certo intervalo de tempo:

vΔt

Volume = A(vΔt)

A

V = AvΔt

R = VΔt

= Av

Aula anterior 21.5 Hidrodinâmica: equação da continuidade

8

Consideremos um fluido que se move num tubo cilíndrico, de secção variável A; se o movimento for laminar, todos os elementos do fluido têm velocidade v1 quando passam pela secção A1, e velocidade v2 quando passam por A2.

A conservação da massa do fluido requer que o fluxo seja constante (equação da continuidade, para fluidos incompressíveis – que têm densidade constante):

A1v1 = A2v2 = constante

Eq. continuidade simulação

m1 =m2⇒ ρ1V1 = ρ2V2⇒ ρ1A1v1Δt = ρ2 A2 v2Δt

Aula anterior 21.5 Hidrodinâmica: equação da continuidade

12.5 Bernoulli’s Equation 385

12.5 Bernoulli’s EquationAccording to the continuity equation, the speed of fluid flow can vary along thepaths of the fluid. The pressure can also vary; it depends on height as in the staticsituation (see Section 12.2), and it also depends on the speed of flow. We canderive an important relationship called Bernoulli’s equation that relates the pres-sure, flow speed, and height for flow of an ideal, incompressible fluid. Bernoulli’sequation is an essential tool in analyzing plumbing systems, hydroelectric gener-ating stations, and the flight of airplanes.

The dependence of pressure on speed follows from the continuity equation,Eq. (12.10). When an incompressible fluid flows along a flow tube with varyingcross section, its speed must change, and so an element of fluid must have anacceleration. If the tube is horizontal, the force that causes this acceleration has tobe applied by the surrounding fluid. This means that the pressure must be differ-ent in regions of different cross section; if it were the same everywhere, the netforce on every fluid element would be zero. When a horizontal flow tube narrowsand a fluid element speeds up, it must be moving toward a region of lower pres-sure in order to have a net forward force to accelerate it. If the elevation alsochanges, this causes an additional pressure difference.

Deriving Bernoulli’s EquationTo derive Bernoulli’s equation, we apply the work–energy theorem to the fluid ina section of a flow tube. In Fig. 12.22 we consider the element of fluid that atsome initial time lies between the two cross sections a and c. The speeds at thelower and upper ends are and In a small time interval dt, the fluid that isinitially at a moves to b, a distance and the fluid that is initially at cmoves to d, a distance The cross-sectional areas at the two ends are

and as shown. The fluid is incompressible; hence by the continuity equa-tion, Eq. (12.10), the volume of fluid dV passing any cross section during time dtis the same. That is,

Let’s compute the work done on this fluid element during dt. We assumethat there is negligible internal friction in the fluid (i.e., no viscosity), so theonly nongravitational forces that do work on the fluid element are due to thepressure of the surrounding fluid. The pressures at the two ends are and the force on the cross section at a is and the force at c is The network dW done on the element by the surrounding fluid during this displace-ment is therefore

(12.13)

The second term has a negative sign because the force at c opposes the displace-ment of the fluid.

The work dW is due to forces other than the conservative force of gravity, so itequals the change in the total mechanical energy (kinetic energy plus gravita-tional potential energy) associated with the fluid element. The mechanical energyfor the fluid between sections b and c does not change. At the beginning of dtthe fluid between a and b has volume mass and kinetic energy . At the end of dt the fluid between c and d has kinetic energy

. The net change in kinetic energy dK during time dt is

(12.14)

What about the change in gravitational potential energy? At the beginning ofdt, the potential energy for the mass between a and b is Atdm gy1 = r dV gy1.

dK = 12r dV1v2

2 - v122

12r1A2 ds22v2

2

12r1A1 ds12v1

2rA1 ds1,A1 ds1,

dW = p1A1 ds1 - p2A2 ds2 = 1p1 - p22dV

p2A2.p1A1,p2;p1

dV = A1 ds1 = A2 ds2.

A2,A1

ds2 = v2 dt.ds1 = v1 dt,

v2.v1

p2A2

ds2

ds1

ab

cd

dV

dV

v1

v2

Flow

y1

y2

A1

A2

p1A1

12.22 Deriving Bernoulli’s equation.The net work done on a fluid element bythe pressure of the surrounding fluidequals the change in the kinetic energyplus the change in the gravitational potentialenergy.

9

O primeiro termo representa a energia cinética por unidade de volume, o segundo é interpretado como a energia potencial por unidade de volume, associada ao gradiente de pressão, e o terceiro é a energia potencial por unidade de volume, devido à força gravitacional.

A chamada equação de Bernoulli expressa a conservação da energia num fluido e relaciona a pressão, a velocidade e a altura do fluxo de um fluido incompressível:

12ρv2 + p+ ρgz = constante

Aula anterior 21.6 Equação de Bernoulli

10

22.1 Descrição do movimento

A mecânica de Newton

•  descreve bem o movimento dos objetos que se desloquem com velocidades muito menores do que a velocidade da luz

•  não descreve suficientemente bem o movimento de objetos que se desloquem com velocidades próximas da velocidade da luz (demonstrado por experiências recentes)

•  não impõe restrições (limite superior) ao valor das velocidades.

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22.2 Relatividade de Galileu

As leis da mecânica têm que ser iguais em todos os referenciais de inércia (princípio da relatividade de Galileu).

Para se descrever um acontecimento físico, é preciso estabelecer um referencial (sistema de coordenadas).

Não há um referencial de inércia absoluto

•  por exemplo, os resultados de uma experiência realizada num veículo, que se mova com velocidade constante, têm que ser idênticos aos da mesma experiência realizada quando o veículo estiver parado.

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Exemplo

Um automóvel desloca-se com velocidade constante, relativamente ao chão.

Um observador no carro atira uma bola, na vertical.

Para esse observador no carro, o movimento da bola

•  é vertical

•  tem aceleração constante (g).

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Exemplo

Um automóvel desloca-se com velocidade constante, relativamente ao chão.

Um observador no carro atira uma bola, na vertical.

Para um observador no chão, o movimento da bola

•  é parabólico

•  tem velocidade para a direita, igual à velocidade do carro.

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Os dois observadores detetam trajetórias diferentes.

O tempo que a bola demorou no ar é o mesmo.

O movimento da bola é bem descrito pelas equações do movimento, nos dois casos.

Conclusão: o movimento é bem descrito nos dois referenciais (não há um referencial “preferido” ou preferencial).

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A posição de um acontecimento (fenómeno físico) pode ser especificada pelas coorde-denadas (x, y, z, t) num dos referenciais (S).

Se o referencial S’ se mover com velocidade constante v, relativamente ao referencial S, a transformação de coordenadas é

''''

x x vty yz zt t

= −

=

=

=

dx 'dt

=dxdt− v ⇔ u 'x = ux − v

(u e u’ dizem respeito à velocidade do objeto e v é a velocidade relativa entre os dois referenciais)

S E C T I O N 3 9 . 1 • The Principle of Galilean Relativity 1247

Suppose that some physical phenomenon, which we call an event, occurs and isobserved by an observer at rest in an inertial reference frame. The event’s location andtime of occurrence can be specified by the four coordinates (x, y, z, t). We would liketo be able to transform these coordinates from those of an observer in one inertialframe to those of another observer in a frame moving with uniform relative velocitycompared to the first frame. When we say an observer is “in a frame,” we mean that theobserver is at rest with respect to the origin of that frame.

Consider two inertial frames S and S! (Fig. 39.2). The frame S! moves with aconstant velocity v along the common x and x! axes, where v is measured relative to S.We assume that the origins of S and S! coincide at t " 0 and that an event occursat point P in space at some instant of time. An observer in S describes the eventwith space–time coordinates (x, y, z, t), whereas an observer in S! uses the coordinates(x!, y!, z!, t!) to describe the same event. As we see from the geometry in Figure 39.2,the relationships among these various coordinates can be written

x! " x # vt y! " y z! " z t! " t (39.1)

These equations are the Galilean space–time transformation equations. Note thattime is assumed to be the same in both inertial frames. That is, within the frameworkof classical mechanics, all clocks run at the same rate, regardless of their velocity, sothat the time at which an event occurs for an observer in S is the same as the time forthe same event in S!. Consequently, the time interval between two successive eventsshould be the same for both observers. Although this assumption may seem obvious, itturns out to be incorrect in situations where v is comparable to the speed of light.

Now suppose that a particle moves through a displacement of magnitude dx alongthe x axis in a time interval dt as measured by an observer in S. It follows fromEquations 39.1 that the corresponding displacement dx! measured by an observer in S!is dx! " dx # v dt, where frame S! is moving with speed v in the x direction relative toframe S. Because dt " dt!, we find that

or

u!x " ux # v (39.2)

where ux and u!x are the x components of the velocity of the particle measured byobservers in S and S!, respectively. (We use the symbol u for particle velocity ratherthan v, which is used for the relative velocity of two reference frames.) This is theGalilean velocity transformation equation. It is consistent with our intuitive notionof time and space as well as with our discussions in Section 4.6. As we shall soon see,however, it leads to serious contradictions when applied to electromagnetic waves.

dx !

dt !"

dxdt

# v

y

0 x

y′

0′ x ′

xvt x ′

P (event)

v

S ′S

Figure 39.2 An event occurs at apoint P. The event is seen by twoobservers in inertial frames S andS!, where S! moves with a velocity vrelative to S.

Galilean transformationequations

▲ PITFALL PREVENTION 39.1 The Relationship

Between the S andS! Frames

Many of the mathematical repre-sentations in this chapter are trueonly for the specified relationshipbetween the S and S! frames. Thex and x! axes coincide, exceptthat their origins are different.The y and y! axes (and the z andz! axes), are parallel, but do notcoincide due to the displacementof the origin of S! with respect tothat of S. We choose the timet " 0 to be the instant at whichthe origins of the two coordinatesystems coincide. If the S! frameis moving in the positive x direc-tion relative to S, v is positive;otherwise it is negative.

Quick Quiz 39.2 A baseball pitcher with a 90-mi/h fastball throws a ballwhile standing on a railroad flatcar moving at 110 mi/h. The ball is thrown in thesame direction as that of the velocity of the train. Applying the Galilean velocitytransformation equation, the speed of the ball relative to the Earth is (a) 90 mi/h(b) 110 mi/h (c) 20 mi/h (d) 200 mi/h (e) impossible to determine.

The Speed of Light

It is quite natural to ask whether the principle of Galilean relativity also applies toelectricity, magnetism, and optics. Experiments indicate that the answer is no.Recall from Chapter 34 that Maxwell showed that the speed of light in free space isc " 3.00 $ 108 m/s. Physicists of the late 1800s thought that light waves moved through amedium called the ether and that the speed of light was c only in a special, absolute frame

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!vpassageiro, chão =!vpassageiro, comboio +

!vcomboio, chão


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22.3 Velocidade da luz no vácuo

Ole Rømer, 1676: 3220 10 km/sc ≈ ×

3 8300 10 km/s 3 10 m/sc ≈ × = ×

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22.3 Velocidade da luz no vácuo

3 8300 10 km/s 3 10 m/sc ≈ × = ×

Experiência de Michelson-Morley (1881): a velocidade da luz no vácuo é constante.

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22.4 Relatividade de Einstein

Elimina a contradição entre a relatividade de Galileu e o facto de a velocidade da luz ser a mesma para todos os observadores.

Postulados:

•  as leis físicas são as mesmas em todos os referenciais de inércia (Princípio da Relatividade)

•  o valor da velocidade da luz no vácuo é c = 3x108 m/s em todos os referenciais de inércia, independentemente da velocidade do observador ou da velocidade da fonte que emite a luz (a velocidade da luz é constante)

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22.4.1 Dilatação do tempo

Exemplo

Um espelho está colocado no teto de um vagão, que se desloca para a direita, com velocidade v.

Um observador segura uma lanterna, à distância d do espelho.

A lanterna emite um impulso luminoso na direção do espelho (acontecimento 1), esse impulso é refletido no espelho e regressa à lanterna (acontecimento 2).

2p

dtc

Δ =

Para este observador no vagão,

•  os dois acontecimentos ocorreram no mesmo local

•  o tempo entre os dois acontecimentos é

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Exemplo

Para um observador no chão,

•  o espelho e o ponto O’ movem-se, para a direita, com velocidade v

•  quando a luz regressa à lanterna, esta já se deslocou para a direita

•  a trajetória do impulso luminoso é maior, relativamente a O, do que relativamente a O’.

Dilatação do tempo simulação

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Para os dois observadores, a velocidade da luz tem que ser c.

A distância que a luz percorre é maior para o observador em O.

O intervalo Δt, relativamente a O, tem que ser maior do que Δtp, relativamente a O’.

2 22

2 2v t c tdΔ Δ⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2

2 1

1

dtc v

c

Δ =

−

Pt γ= Δ

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γ >1 ⇒ Δt > Δt p

Δtp é o chamado tempo próprio (intervalo de tempo medido por um observador que vê os acontecimentos ocorrerem no mesmo ponto do espaço).

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22.4.1 Dilatação do tempo Os muões são partículas com a mesma carga do eletrão, mas com massas 207 vezes maiores. São criados pela colisão de raios cósmicos com átomos nas camadas superiores da atmosfera. Os muões têm tempos de semi-vida

Δtp = 2,2 µs. Em relação a um observador na Terra, o tempo de semi-vida deve ser γ Δtp , o que tem sido comprovado experimentalmente.

1256 C H A P T E R 3 9 • Relativity

high in the atmosphere where they are produced. However, experiments show that alarge number of muons do reach the surface. The phenomenon of time dilationexplains this effect. As measured by an observer on Earth, the muons have a dilatedlifetime equal to ! "tp. For example, for v # 0.99c, ! ! 7.1 and ! "tp ! 16 $s. Hence,the average distance traveled by the muons in this time as measured by an observeron Earth is approximately (0.99)(3.0 % 108 m/s)(16 % 10&6 s) ! 4.8 % 103 m, as indi-cated in Figure 39.8b.

In 1976, at the laboratory of the European Council for Nuclear Research (CERN)in Geneva, muons injected into a large storage ring reached speeds of approximately0.999 4c. Electrons produced by the decaying muons were detected by counters aroundthe ring, enabling scientists to measure the decay rate and hence the muon lifetime.The lifetime of the moving muons was measured to be approximately 30 times as longas that of the stationary muon (Fig. 39.9), in agreement with the prediction of relativityto within two parts in a thousand.

Muonat rest

Muon movingat 0.999 4c

50 100 150

0.5

1.0

Frac

tion

of m

uons

rem

aini

ng

t( s)µ

Figure 39.9 Decay curves for muonsat rest and for muons traveling at aspeed of 0.999 4c.

Example 39.1 What Is the Period of the Pendulum?

The period of a pendulum is measured to be 3.00 s in thereference frame of the pendulum. What is the period whenmeasured by an observer moving at a speed of 0.950c rela-tive to the pendulum?

Solution To conceptualize this problem, let us changeframes of reference. Instead of the observer moving at 0.950c,we can take the equivalent point of view that the observer is atrest and the pendulum is moving at 0.950c past the stationaryobserver. Hence, the pendulum is an example of a clockmoving at high speed with respect to an observer and we cancategorize this problem as one involving time dilation.

To analyze the problem, note that the proper timeinterval, measured in the rest frame of the pendulum, is"tp # 3.00 s. Because a clock moving with respect to anobserver is measured to run more slowly than a stationaryclock by a factor !, Equation 39.7 gives

To finalize this problem, we see that indeed a movingpendulum is measured to take longer to complete a period

9.60 s# (3.20)(3.00 s) #

"t # ! "tp #1

√1 &(0.950c)2

c2

"tp #1

√1 & 0.902 "tp

than a pendulum at rest does. The period increases by afactor of ! # 3.20. We see that this is consistent with Table39.1, where this value lies between those for ! for v/c # 0.94and v/c # 0.96.

What If? What if we increase the speed of the observer by5.00%? Does the dilated time interval increase by 5.00%?

Answer Based on the highly nonlinear behavior of ! as afunction of v in Figure 39.7, we would guess that theincrease in "t would be different from 5.00%. Increasing vby 5.00% gives us

vnew # (1.050 0)(0.950c) # 0.997 5c

(Because ! varies so rapidly with v when v is this large, we willkeep one additional significant figure until the final answer.)If we perform the time dilation calculation again, we find that

Thus, the 5.00% increase in speed has caused over a 300%increase in the dilated time!

# (14.15)(3.00 s) # 42.5 s

"t # ! "tp #1

√1 &(0.997 5c)2

c 2

"tp #1

√1 & 0.995 0 "tp

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22.4.2 Contração do espaço

A distância medida entre dois pontos depende do referencial do observador.

O comprimento próprio Lp de um objeto é o comprimento desse objeto medido por um observador em repouso, relativamente ao objeto.

O comprimento do objeto, L, medido num referencial que se esteja a mover relativamente ao objeto, é sempre menor do que o comprimento próprio.

A este efeito chama-se contração do espaço.

Nem sempre o tempo próprio e o comprimento próprio são medidos pelo mesmo observador:

•  o comprimento próprio é medido pelo observador para o qual as extremidades do objeto estão fixas no espaço

•  o tempo próprio entre dois acontecimentos é medido pelo observador que vê os dois acontecimentos darem-se na mesma posição do espaço.

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A medida do comprimento de um objeto, que se mova com velocidade v, paralelamente ao seu comprimento, é mais pequena do que a medida feita por um observador em repouso.

Só existe contração do comprimento na direção do movimento.

L =Lpγ= Lp 1− v

2

c2

Contração do espaço simulação

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L =Lpγ= Lp 1− v

2

c2

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22.5 Transformações de Lorentz

Dois acontecimentos, nos pontos P e Q, são detetados por dois observadores.

Um dos observadores está em repouso, no referencial S, e o outro está no referencial S’, que se move para a direita, com velocidade v.

O observador em S descreve os acontecimentos nas coordenadas (x, y, z, t) e o outro descreve-os nas coordenadas (x’, y’, z’, t’ ).

Para transformar as coordenadas de S para S’, usam-se as equações de transformação de Lorentz:

x ' = γ x − v t( ) ; y ' = y ; z ' = z ; t ' = γ t − v

c2x

⎛

⎝⎜

⎞

⎠⎟

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Para transformar as coordenadas de S’ para S:

x = γ x '+ v t( ) ; y = y ' ; z = z '

t = γ t '+ v

c2x '

⎛

⎝⎜

⎞

⎠⎟

Para transformar a velocidade: ux' =

ux − v

1−ux v

c2

; ux =ux' + v

1+ux' v

c2

No limite v << c, as transformações de Lorentz são iguais às de Galileu.

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Invariância da velocidade da luz:

' velocidade da luz no vácuo.'

x xc ct t= ⇒ =

Invariância do intervalo espaço-tempo (intervalo do universo):

( ) ( )22 2

2 2 2 222 2

2

' ' ' '( ) ' ( ')

' ' ( ) ' '

x x Vt x x Vtx ct x ctV Vt t x ct ct xc c

γ γ

γ γ

⎧⎧ = + = +⎪⎪

⇒ ⇒ − = −⎨ ⎨⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟⎪ ⎪ ⎜ ⎟⎝ ⎠⎩ ⎝ ⎠⎩

(x)2 − (ct)2 = (x ')2 − (ct ')2 = ΔI 2 = constante (intervalo do universo)

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Consequências da invariância do intervalo do espaço-tempo:

•  para um acontecimento do tipo espaço,

•  para um raio luminoso, 0IΔ =

2 0IΔ >Para um acontecimento do tipo espaço, a luz não consegue percorrer a distância nesse intervalo de tempo. Não pode haver causalidade.

•  para um acontecimento do tipo espaço,

(x)2 − (ct)2 = (x ')2 − (ct ')2 = ΔI 2 = constante (intervalo do universo)

ΔI 2 < 0Para um acontecimento do tipo tempo, a luz consegue percorrer a distância nesse intervalo de tempo. Pode haver causalidade.

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22.6 Momento linear

O momento linear de um sistema isolado deve conservar-se, em todas as colisões.

O valor relativista do momento p de uma partícula, com massa m e velocidade u, deve aproximar-se do valor clássico mu, quando u se aproxima de zero.

p = mu

1− u2

c2

= γmu

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22.7 Energia

A energia total de uma partícula relativista é definida como a soma da energia cinética (Ecin) com a energia em repouso (mc2), e é igual a γmc2 :

E = γmc2 = Ecin +mc2

Por sua vez, a energia cinética é Ecin = γ −1( )mc2

A energia total pode ser expressa em termos do momento linear, como

E2 = p2 c2 + mc2( )2

Partículas com massa em repouso nula, como os fotões, têm energia

E = pc

21.1 Hidrostática: definições 21.2 Pressão hidrostática 21.3 … · 28/Maio/2018 – Aula 21...

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