Phys. 131 -11/26/2012

DerekStampone

Phys. 131 - 11/26/2012

Derek Stampone

Binghamton University

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Outline

1 ReviewPascal’s and Archimedes’ Principle

2 Fluids - Cont.Ideal Fluids in MotionThe Equation of ContinuityBernoulli’s Equation

3 Test 3 - ReviewChapter 10Chapter 11Chapter 12Chapter 13Chapter 15

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Pascal’s and Archimedes’ Principle

Definition (Pascal’s Principle)

A change in the pressure applied to an enclosed incompressiblefluid is transmitted undiminished to every portion of the fluidand the walls of its container.

Definition (Archimedes’ Principle)

When a body is fully or partially submerged in a fluid, abuoyant force ~Fb from the surrounding fluid acts on the body.The force is directed upward and has a magnitude equal to theweight mfg of the fluid that has been displaced by the body.

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Pascal’s and Archimedes’ Principle

Definition (Pascal’s Principle)

A change in the pressure applied to an enclosed incompressiblefluid is transmitted undiminished to every portion of the fluidand the walls of its container.

Definition (Archimedes’ Principle)

When a body is fully or partially submerged in a fluid, abuoyant force ~Fb from the surrounding fluid acts on the body.The force is directed upward and has a magnitude equal to theweight mfg of the fluid that has been displaced by the body.

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Ideal Fluids in Motion

37114-8 I DEAL FLU I DS I N MOTIONPART 2

14-8 Ideal Fluids in MotionThe motion of real fluids is very complicated and not yet fully understood.Instead, we shall discuss the motion of an ideal fluid, which is simpler to handlemathematically and yet provides useful results. Here are four assumptions thatwe make about our ideal fluid; they all are concerned with flow:

1. Steady flow In steady (or laminar) flow, the velocity of the moving fluid at anyfixed point does not change with time. The gentle flow of water near the center ofa quiet stream is steady; the flow in a chain of rapids is not. Figure 14-12 shows atransition from steady flow to nonsteady (or nonlaminar or turbulent) flow for arising stream of smoke. The speed of the smoke particles increases as they riseand, at a certain critical speed, the flow changes from steady to nonsteady.

2. Incompressible flow We assume, as for fluids at rest, that our ideal fluid is incompressible; that is, its density has a constant, uniform value.

3. Nonviscous flow Roughly speaking, the viscosity of a fluid is a measure of howresistive the fluid is to flow. For example, thick honey is more resistive to flowthan water, and so honey is said to be more viscous than water. Viscosity is thefluid analog of friction between solids; both are mechanisms by which the kineticenergy of moving objects can be transferred to thermal energy. In the absence offriction, a block could glide at constant speed along a horizontal surface. In thesame way, an object moving through a nonviscous fluid would experience no vis-cous drag force—that is, no resistive force due to viscosity; it could move at con-stant speed through the fluid.The British scientist Lord Rayleigh noted that in anideal fluid a ship’s propeller would not work, but, on the other hand, in an idealfluid a ship (once set into motion) would not need a propeller!

4. Irrotational flow Although it need not concern us further, we also assumethat the flow is irrotational. To test for this property, let a tiny grain of dustmove with the fluid.Although this test body may (or may not) move in a circu-lar path, in irrotational flow the test body will not rotate about an axis throughits own center of mass. For a loose analogy, the motion of a Ferris wheel is ro-tational; that of its passengers is irrotational.

We can make the flow of a fluid visible by adding a tracer. This might bea dye injected into many points across a liquid stream (Fig. 14-13) or smoke parti-cles added to a gas flow (Fig. 14-12). Each bit of a tracer follows a streamline,which is the path that a tiny element of the fluid would take as the fluid flows.Recall from Chapter 4 that the velocity of a particle is always tangent to thepath taken by the particle. Here the particle is the fluid element, and its velocity

is always tangent to a streamline (Fig. 14-14). For this reason, two streamlinescan never intersect; if they did, then an element arriving at their intersectionwould have two different velocities simultaneously—an impossibility.

v:

Fig. 14-12 At a certain point, therising flow of smoke and heated gaschanges from steady to turbulent.(Will McIntyre/Photo Researchers)

Fig. 14-13 The steadyflow of a fluid around acylinder, as revealed by adye tracer that was injectedinto the fluid upstream ofthe cylinder. (Courtesy D.H.Peregrine,University ofBristol)

Streamline

Fluidelement

v

Fig. 14-14 A fluid element tracesout a streamline as it moves.The ve-locity vector of the element is tan-gent to the streamline at every point.

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 371

Four Assumptions of Ideal Fluid

• Steady

• Incompressible - ConstantDensity

• Nonviscous - Frictional LossesNegligible

• Irrotational

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Ideal Fluids in Motion

37114-8 I DEAL FLU I DS I N MOTIONPART 2

14-8 Ideal Fluids in MotionThe motion of real fluids is very complicated and not yet fully understood.Instead, we shall discuss the motion of an ideal fluid, which is simpler to handlemathematically and yet provides useful results. Here are four assumptions thatwe make about our ideal fluid; they all are concerned with flow:

1. Steady flow In steady (or laminar) flow, the velocity of the moving fluid at anyfixed point does not change with time. The gentle flow of water near the center ofa quiet stream is steady; the flow in a chain of rapids is not. Figure 14-12 shows atransition from steady flow to nonsteady (or nonlaminar or turbulent) flow for arising stream of smoke. The speed of the smoke particles increases as they riseand, at a certain critical speed, the flow changes from steady to nonsteady.

2. Incompressible flow We assume, as for fluids at rest, that our ideal fluid is incompressible; that is, its density has a constant, uniform value.

3. Nonviscous flow Roughly speaking, the viscosity of a fluid is a measure of howresistive the fluid is to flow. For example, thick honey is more resistive to flowthan water, and so honey is said to be more viscous than water. Viscosity is thefluid analog of friction between solids; both are mechanisms by which the kineticenergy of moving objects can be transferred to thermal energy. In the absence offriction, a block could glide at constant speed along a horizontal surface. In thesame way, an object moving through a nonviscous fluid would experience no vis-cous drag force—that is, no resistive force due to viscosity; it could move at con-stant speed through the fluid.The British scientist Lord Rayleigh noted that in anideal fluid a ship’s propeller would not work, but, on the other hand, in an idealfluid a ship (once set into motion) would not need a propeller!

4. Irrotational flow Although it need not concern us further, we also assumethat the flow is irrotational. To test for this property, let a tiny grain of dustmove with the fluid.Although this test body may (or may not) move in a circu-lar path, in irrotational flow the test body will not rotate about an axis throughits own center of mass. For a loose analogy, the motion of a Ferris wheel is ro-tational; that of its passengers is irrotational.

We can make the flow of a fluid visible by adding a tracer. This might bea dye injected into many points across a liquid stream (Fig. 14-13) or smoke parti-cles added to a gas flow (Fig. 14-12). Each bit of a tracer follows a streamline,which is the path that a tiny element of the fluid would take as the fluid flows.Recall from Chapter 4 that the velocity of a particle is always tangent to thepath taken by the particle. Here the particle is the fluid element, and its velocity

is always tangent to a streamline (Fig. 14-14). For this reason, two streamlinescan never intersect; if they did, then an element arriving at their intersectionwould have two different velocities simultaneously—an impossibility.

v:

Fig. 14-12 At a certain point, therising flow of smoke and heated gaschanges from steady to turbulent.(Will McIntyre/Photo Researchers)

Fig. 14-13 The steadyflow of a fluid around acylinder, as revealed by adye tracer that was injectedinto the fluid upstream ofthe cylinder. (Courtesy D.H.Peregrine,University ofBristol)

Streamline

Fluidelement

v

Fig. 14-14 A fluid element tracesout a streamline as it moves.The ve-locity vector of the element is tan-gent to the streamline at every point.

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 371

Four Assumptions of Ideal Fluid

• Steady

• Incompressible - ConstantDensity

• Nonviscous - Frictional LossesNegligible

• Irrotational

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Ideal Fluids in Motion

37114-8 I DEAL FLU I DS I N MOTIONPART 2

14-8 Ideal Fluids in MotionThe motion of real fluids is very complicated and not yet fully understood.Instead, we shall discuss the motion of an ideal fluid, which is simpler to handlemathematically and yet provides useful results. Here are four assumptions thatwe make about our ideal fluid; they all are concerned with flow:

1. Steady flow In steady (or laminar) flow, the velocity of the moving fluid at anyfixed point does not change with time. The gentle flow of water near the center ofa quiet stream is steady; the flow in a chain of rapids is not. Figure 14-12 shows atransition from steady flow to nonsteady (or nonlaminar or turbulent) flow for arising stream of smoke. The speed of the smoke particles increases as they riseand, at a certain critical speed, the flow changes from steady to nonsteady.

2. Incompressible flow We assume, as for fluids at rest, that our ideal fluid is incompressible; that is, its density has a constant, uniform value.

3. Nonviscous flow Roughly speaking, the viscosity of a fluid is a measure of howresistive the fluid is to flow. For example, thick honey is more resistive to flowthan water, and so honey is said to be more viscous than water. Viscosity is thefluid analog of friction between solids; both are mechanisms by which the kineticenergy of moving objects can be transferred to thermal energy. In the absence offriction, a block could glide at constant speed along a horizontal surface. In thesame way, an object moving through a nonviscous fluid would experience no vis-cous drag force—that is, no resistive force due to viscosity; it could move at con-stant speed through the fluid.The British scientist Lord Rayleigh noted that in anideal fluid a ship’s propeller would not work, but, on the other hand, in an idealfluid a ship (once set into motion) would not need a propeller!

4. Irrotational flow Although it need not concern us further, we also assumethat the flow is irrotational. To test for this property, let a tiny grain of dustmove with the fluid.Although this test body may (or may not) move in a circu-lar path, in irrotational flow the test body will not rotate about an axis throughits own center of mass. For a loose analogy, the motion of a Ferris wheel is ro-tational; that of its passengers is irrotational.

We can make the flow of a fluid visible by adding a tracer. This might bea dye injected into many points across a liquid stream (Fig. 14-13) or smoke parti-cles added to a gas flow (Fig. 14-12). Each bit of a tracer follows a streamline,which is the path that a tiny element of the fluid would take as the fluid flows.Recall from Chapter 4 that the velocity of a particle is always tangent to thepath taken by the particle. Here the particle is the fluid element, and its velocity

is always tangent to a streamline (Fig. 14-14). For this reason, two streamlinescan never intersect; if they did, then an element arriving at their intersectionwould have two different velocities simultaneously—an impossibility.

v:

Fig. 14-12 At a certain point, therising flow of smoke and heated gaschanges from steady to turbulent.(Will McIntyre/Photo Researchers)

Fig. 14-13 The steadyflow of a fluid around acylinder, as revealed by adye tracer that was injectedinto the fluid upstream ofthe cylinder. (Courtesy D.H.Peregrine,University ofBristol)

Streamline

Fluidelement

v

Fig. 14-14 A fluid element tracesout a streamline as it moves.The ve-locity vector of the element is tan-gent to the streamline at every point.

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 371

Four Assumptions of Ideal Fluid

• Steady

• Incompressible - ConstantDensity

• Nonviscous - Frictional LossesNegligible

• Irrotational

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Ideal Fluids in Motion

37114-8 I DEAL FLU I DS I N MOTIONPART 2

14-8 Ideal Fluids in MotionThe motion of real fluids is very complicated and not yet fully understood.Instead, we shall discuss the motion of an ideal fluid, which is simpler to handlemathematically and yet provides useful results. Here are four assumptions thatwe make about our ideal fluid; they all are concerned with flow:

1. Steady flow In steady (or laminar) flow, the velocity of the moving fluid at anyfixed point does not change with time. The gentle flow of water near the center ofa quiet stream is steady; the flow in a chain of rapids is not. Figure 14-12 shows atransition from steady flow to nonsteady (or nonlaminar or turbulent) flow for arising stream of smoke. The speed of the smoke particles increases as they riseand, at a certain critical speed, the flow changes from steady to nonsteady.

2. Incompressible flow We assume, as for fluids at rest, that our ideal fluid is incompressible; that is, its density has a constant, uniform value.

3. Nonviscous flow Roughly speaking, the viscosity of a fluid is a measure of howresistive the fluid is to flow. For example, thick honey is more resistive to flowthan water, and so honey is said to be more viscous than water. Viscosity is thefluid analog of friction between solids; both are mechanisms by which the kineticenergy of moving objects can be transferred to thermal energy. In the absence offriction, a block could glide at constant speed along a horizontal surface. In thesame way, an object moving through a nonviscous fluid would experience no vis-cous drag force—that is, no resistive force due to viscosity; it could move at con-stant speed through the fluid.The British scientist Lord Rayleigh noted that in anideal fluid a ship’s propeller would not work, but, on the other hand, in an idealfluid a ship (once set into motion) would not need a propeller!

4. Irrotational flow Although it need not concern us further, we also assumethat the flow is irrotational. To test for this property, let a tiny grain of dustmove with the fluid.Although this test body may (or may not) move in a circu-lar path, in irrotational flow the test body will not rotate about an axis throughits own center of mass. For a loose analogy, the motion of a Ferris wheel is ro-tational; that of its passengers is irrotational.

We can make the flow of a fluid visible by adding a tracer. This might bea dye injected into many points across a liquid stream (Fig. 14-13) or smoke parti-cles added to a gas flow (Fig. 14-12). Each bit of a tracer follows a streamline,which is the path that a tiny element of the fluid would take as the fluid flows.Recall from Chapter 4 that the velocity of a particle is always tangent to thepath taken by the particle. Here the particle is the fluid element, and its velocity

is always tangent to a streamline (Fig. 14-14). For this reason, two streamlinescan never intersect; if they did, then an element arriving at their intersectionwould have two different velocities simultaneously—an impossibility.

v:

Fig. 14-12 At a certain point, therising flow of smoke and heated gaschanges from steady to turbulent.(Will McIntyre/Photo Researchers)

Fig. 14-13 The steadyflow of a fluid around acylinder, as revealed by adye tracer that was injectedinto the fluid upstream ofthe cylinder. (Courtesy D.H.Peregrine,University ofBristol)

Streamline

Fluidelement

v

Fig. 14-14 A fluid element tracesout a streamline as it moves.The ve-locity vector of the element is tan-gent to the streamline at every point.

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 371

Four Assumptions of Ideal Fluid

• Steady

• Incompressible - ConstantDensity

• Nonviscous - Frictional LossesNegligible

• Irrotational

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Ideal Fluids in Motion

37114-8 I DEAL FLU I DS I N MOTIONPART 2

14-8 Ideal Fluids in MotionThe motion of real fluids is very complicated and not yet fully understood.Instead, we shall discuss the motion of an ideal fluid, which is simpler to handlemathematically and yet provides useful results. Here are four assumptions thatwe make about our ideal fluid; they all are concerned with flow:

1. Steady flow In steady (or laminar) flow, the velocity of the moving fluid at anyfixed point does not change with time. The gentle flow of water near the center ofa quiet stream is steady; the flow in a chain of rapids is not. Figure 14-12 shows atransition from steady flow to nonsteady (or nonlaminar or turbulent) flow for arising stream of smoke. The speed of the smoke particles increases as they riseand, at a certain critical speed, the flow changes from steady to nonsteady.

2. Incompressible flow We assume, as for fluids at rest, that our ideal fluid is incompressible; that is, its density has a constant, uniform value.

3. Nonviscous flow Roughly speaking, the viscosity of a fluid is a measure of howresistive the fluid is to flow. For example, thick honey is more resistive to flowthan water, and so honey is said to be more viscous than water. Viscosity is thefluid analog of friction between solids; both are mechanisms by which the kineticenergy of moving objects can be transferred to thermal energy. In the absence offriction, a block could glide at constant speed along a horizontal surface. In thesame way, an object moving through a nonviscous fluid would experience no vis-cous drag force—that is, no resistive force due to viscosity; it could move at con-stant speed through the fluid.The British scientist Lord Rayleigh noted that in anideal fluid a ship’s propeller would not work, but, on the other hand, in an idealfluid a ship (once set into motion) would not need a propeller!

4. Irrotational flow Although it need not concern us further, we also assumethat the flow is irrotational. To test for this property, let a tiny grain of dustmove with the fluid.Although this test body may (or may not) move in a circu-lar path, in irrotational flow the test body will not rotate about an axis throughits own center of mass. For a loose analogy, the motion of a Ferris wheel is ro-tational; that of its passengers is irrotational.

We can make the flow of a fluid visible by adding a tracer. This might bea dye injected into many points across a liquid stream (Fig. 14-13) or smoke parti-cles added to a gas flow (Fig. 14-12). Each bit of a tracer follows a streamline,which is the path that a tiny element of the fluid would take as the fluid flows.Recall from Chapter 4 that the velocity of a particle is always tangent to thepath taken by the particle. Here the particle is the fluid element, and its velocity

is always tangent to a streamline (Fig. 14-14). For this reason, two streamlinescan never intersect; if they did, then an element arriving at their intersectionwould have two different velocities simultaneously—an impossibility.

v:

Fig. 14-12 At a certain point, therising flow of smoke and heated gaschanges from steady to turbulent.(Will McIntyre/Photo Researchers)

Fig. 14-13 The steadyflow of a fluid around acylinder, as revealed by adye tracer that was injectedinto the fluid upstream ofthe cylinder. (Courtesy D.H.Peregrine,University ofBristol)

Streamline

Fluidelement

v

Fig. 14-14 A fluid element tracesout a streamline as it moves.The ve-locity vector of the element is tan-gent to the streamline at every point.

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 371

Four Assumptions of Ideal Fluid

• Steady

• Incompressible - ConstantDensity

• Nonviscous - Frictional LossesNegligible

• Irrotational

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

The Equation of Continuity

372 CHAPTE R 14 FLU I DS

14-9 The Equation of ContinuityYou may have noticed that you can increase the speed of the water emergingfrom a garden hose by partially closing the hose opening with your thumb.Apparently the speed v of the water depends on the cross-sectional area Athrough which the water flows.

Here we wish to derive an expression that relates v and A for the steady flowof an ideal fluid through a tube with varying cross section, like that in Fig. 14-15.The flow there is toward the right, and the tube segment shown (part of a longertube) has length L.The fluid has speeds v1 at the left end of the segment and v2 atthe right end. The tube has cross-sectional areas A1 at the left end and A2 at theright end. Suppose that in a time interval !t a volume !V of fluid enters the tubesegment at its left end (that volume is colored purple in Fig. 14-15).Then, becausethe fluid is incompressible, an identical volume !V must emerge from the rightend of the segment (it is colored green in Fig. 14-15).

We can use this common volume !V to relate the speeds and areas. To do so,we first consider Fig. 14-16, which shows a side view of a tube of uniform cross-sec-tional area A. In Fig. 14-16a, a fluid element e is about to pass through the dashedline drawn across the tube width.The element’s speed is v, so during a time interval!t, the element moves along the tube a distance !x " v !t. The volume !V of fluidthat has passed through the dashed line in that time interval !t is

!V " A !x " Av !t. (14-22)

Applying Eq. 14-22 to both the left and right ends of the tube segment inFig. 14-15, we have

!V " A1v1 !t " A2v2 !t

or A1v1 " A2v2 (equation of continuity). (14-23)

This relation between speed and cross-sectional area is called the equation ofcontinuity for the flow of an ideal fluid. It tells us that the flow speed increaseswhen we decrease the cross-sectional area through which the fluid flows.

Equation 14-23 applies not only to an actual tube but also to any so-calledtube of flow, or imaginary tube whose boundary consists of streamlines. Such atube acts like a real tube because no fluid element can cross a streamline; thus, allthe fluid within a tube of flow must remain within its boundary. Figure 14-17shows a tube of flow in which the cross-sectional area increases from area A1 toarea A2 along the flow direction. From Eq. 14-23 we know that, with the increasein area, the speed must decrease, as is indicated by the greater spacing betweenstreamlines at the right in Fig. 14-17. Similarly, you can see that in Fig. 14-13 thespeed of the flow is greatest just above and just below the cylinder.

Fig. 14-16 Fluid flows at a constantspeed v through a tube. (a) At time t, fluidelement e is about to pass the dashed line.(b) At time t # !t, element e is a distance!x " v !t from the dashed line.

ve

ve

(a) Time t

(b) Time t + !t

!x

Fig. 14-15 Fluid flows from left toright at a steady rate through a tube seg-ment of length L. The fluid’s speed is v1 atthe left side and v2 at the right side.Thetube’s cross-sectional area is A1 at the leftside and A2 at the right side. From time t in(a) to time t # !t in (b), the amount offluid shown in purple enters at the left sideand the equal amount of fluid shown ingreen emerges at the right side.

L

v1

A1

A2

v2

(a) Time t

L

(b) Time t + !t

The volume flow persecond here mustmatch ...

... the volume flowper second here.

A1

A2

The volume flow persecond here must match ...

... the volume flowper second here.

Fig. 14-17 A tube of flow is defined bythe streamlines that form the boundary ofthe tube.The volume flow rate must be thesame for all cross sections of the tube of flow.

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 372

• Comes from Conservation ofMass

A1v1 = A2v2

• Mass Flow Rate:Rm = ρAv = constant

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

The Equation of Continuity

372 CHAPTE R 14 FLU I DS

14-9 The Equation of ContinuityYou may have noticed that you can increase the speed of the water emergingfrom a garden hose by partially closing the hose opening with your thumb.Apparently the speed v of the water depends on the cross-sectional area Athrough which the water flows.

Here we wish to derive an expression that relates v and A for the steady flowof an ideal fluid through a tube with varying cross section, like that in Fig. 14-15.The flow there is toward the right, and the tube segment shown (part of a longertube) has length L.The fluid has speeds v1 at the left end of the segment and v2 atthe right end. The tube has cross-sectional areas A1 at the left end and A2 at theright end. Suppose that in a time interval !t a volume !V of fluid enters the tubesegment at its left end (that volume is colored purple in Fig. 14-15).Then, becausethe fluid is incompressible, an identical volume !V must emerge from the rightend of the segment (it is colored green in Fig. 14-15).

We can use this common volume !V to relate the speeds and areas. To do so,we first consider Fig. 14-16, which shows a side view of a tube of uniform cross-sec-tional area A. In Fig. 14-16a, a fluid element e is about to pass through the dashedline drawn across the tube width.The element’s speed is v, so during a time interval!t, the element moves along the tube a distance !x " v !t. The volume !V of fluidthat has passed through the dashed line in that time interval !t is

!V " A !x " Av !t. (14-22)

Applying Eq. 14-22 to both the left and right ends of the tube segment inFig. 14-15, we have

!V " A1v1 !t " A2v2 !t

or A1v1 " A2v2 (equation of continuity). (14-23)

This relation between speed and cross-sectional area is called the equation ofcontinuity for the flow of an ideal fluid. It tells us that the flow speed increaseswhen we decrease the cross-sectional area through which the fluid flows.

Equation 14-23 applies not only to an actual tube but also to any so-calledtube of flow, or imaginary tube whose boundary consists of streamlines. Such atube acts like a real tube because no fluid element can cross a streamline; thus, allthe fluid within a tube of flow must remain within its boundary. Figure 14-17shows a tube of flow in which the cross-sectional area increases from area A1 toarea A2 along the flow direction. From Eq. 14-23 we know that, with the increasein area, the speed must decrease, as is indicated by the greater spacing betweenstreamlines at the right in Fig. 14-17. Similarly, you can see that in Fig. 14-13 thespeed of the flow is greatest just above and just below the cylinder.

Fig. 14-16 Fluid flows at a constantspeed v through a tube. (a) At time t, fluidelement e is about to pass the dashed line.(b) At time t # !t, element e is a distance!x " v !t from the dashed line.

ve

ve

(a) Time t

(b) Time t + !t

!x

Fig. 14-15 Fluid flows from left toright at a steady rate through a tube seg-ment of length L. The fluid’s speed is v1 atthe left side and v2 at the right side.Thetube’s cross-sectional area is A1 at the leftside and A2 at the right side. From time t in(a) to time t # !t in (b), the amount offluid shown in purple enters at the left sideand the equal amount of fluid shown ingreen emerges at the right side.

L

v1

A1

A2

v2

(a) Time t

L

(b) Time t + !t

The volume flow persecond here mustmatch ...

... the volume flowper second here.

A1

A2

The volume flow persecond here must match ...

... the volume flowper second here.

Fig. 14-17 A tube of flow is defined bythe streamlines that form the boundary ofthe tube.The volume flow rate must be thesame for all cross sections of the tube of flow.

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 372

• Comes from Conservation ofMass

A1v1 = A2v2

• Mass Flow Rate:Rm = ρAv = constant

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

The Equation of Continuity

372 CHAPTE R 14 FLU I DS

14-9 The Equation of ContinuityYou may have noticed that you can increase the speed of the water emergingfrom a garden hose by partially closing the hose opening with your thumb.Apparently the speed v of the water depends on the cross-sectional area Athrough which the water flows.

Here we wish to derive an expression that relates v and A for the steady flowof an ideal fluid through a tube with varying cross section, like that in Fig. 14-15.The flow there is toward the right, and the tube segment shown (part of a longertube) has length L.The fluid has speeds v1 at the left end of the segment and v2 atthe right end. The tube has cross-sectional areas A1 at the left end and A2 at theright end. Suppose that in a time interval !t a volume !V of fluid enters the tubesegment at its left end (that volume is colored purple in Fig. 14-15).Then, becausethe fluid is incompressible, an identical volume !V must emerge from the rightend of the segment (it is colored green in Fig. 14-15).

We can use this common volume !V to relate the speeds and areas. To do so,we first consider Fig. 14-16, which shows a side view of a tube of uniform cross-sec-tional area A. In Fig. 14-16a, a fluid element e is about to pass through the dashedline drawn across the tube width.The element’s speed is v, so during a time interval!t, the element moves along the tube a distance !x " v !t. The volume !V of fluidthat has passed through the dashed line in that time interval !t is

!V " A !x " Av !t. (14-22)

Applying Eq. 14-22 to both the left and right ends of the tube segment inFig. 14-15, we have

!V " A1v1 !t " A2v2 !t

or A1v1 " A2v2 (equation of continuity). (14-23)

This relation between speed and cross-sectional area is called the equation ofcontinuity for the flow of an ideal fluid. It tells us that the flow speed increaseswhen we decrease the cross-sectional area through which the fluid flows.

Equation 14-23 applies not only to an actual tube but also to any so-calledtube of flow, or imaginary tube whose boundary consists of streamlines. Such atube acts like a real tube because no fluid element can cross a streamline; thus, allthe fluid within a tube of flow must remain within its boundary. Figure 14-17shows a tube of flow in which the cross-sectional area increases from area A1 toarea A2 along the flow direction. From Eq. 14-23 we know that, with the increasein area, the speed must decrease, as is indicated by the greater spacing betweenstreamlines at the right in Fig. 14-17. Similarly, you can see that in Fig. 14-13 thespeed of the flow is greatest just above and just below the cylinder.

Fig. 14-16 Fluid flows at a constantspeed v through a tube. (a) At time t, fluidelement e is about to pass the dashed line.(b) At time t # !t, element e is a distance!x " v !t from the dashed line.

ve

ve

(a) Time t

(b) Time t + !t

!x

Fig. 14-15 Fluid flows from left toright at a steady rate through a tube seg-ment of length L. The fluid’s speed is v1 atthe left side and v2 at the right side.Thetube’s cross-sectional area is A1 at the leftside and A2 at the right side. From time t in(a) to time t # !t in (b), the amount offluid shown in purple enters at the left sideand the equal amount of fluid shown ingreen emerges at the right side.

L

v1

A1

A2

v2

(a) Time t

L

(b) Time t + !t

The volume flow persecond here mustmatch ...

... the volume flowper second here.

A1

A2

The volume flow persecond here must match ...

... the volume flowper second here.

Fig. 14-17 A tube of flow is defined bythe streamlines that form the boundary ofthe tube.The volume flow rate must be thesame for all cross sections of the tube of flow.

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 372

• Comes from Conservation ofMass

A1v1 = A2v2

• Mass Flow Rate:Rm = ρAv = constant

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Continuity Equation: Example Problem

37314-9 TH E EQUATION OF CONTI N U ITYPART 2

We can rewrite Eq. 14-23 as

RV ! Av ! a constant (volume flow rate, equation of continuity), (14-24)

in which RV is the volume flow rate of the fluid (volume past a given point perunit time). Its SI unit is the cubic meter per second (m3/s). If the density r of thefluid is uniform, we can multiply Eq. 14-24 by that density to get the mass flowrate Rm (mass per unit time):

Rm ! rRV ! rAv ! a constant (mass flow rate). (14-25)

The SI unit of mass flow rate is the kilogram per second (kg/s). Equation 14-25says that the mass that flows into the tube segment of Fig. 14-15 each second mustbe equal to the mass that flows out of that segment each second.

CHECKPOINT 3

The figure shows a pipe and gives the volume flow rate (in cm3/s) and the di-rection of flow for all but one section. What are the volume flow rate and thedirection of flow for that section?

4 8

2 56

4

Sample Problem

A water stream narrows as it falls

Figure 14-18 shows how the stream of water emerging froma faucet “necks down” as it falls. This change in the horizontalcross-sectional area is characteristic of any laminar (non-turbulant) falling stream because the gravitational forceincreases the speed of the stream. Here the indicatedcross-sectional areas are A0 ! 1.2 cm2 and A ! 0.35 cm2.The two levels are separated by a vertical distance h ! 45 mm.What is the volume flow rate from the tap?

KEY I DEA

Fig. 14-18 As water falls from a tap, its speedincreases. Because the volume flow rate must bethe same at all horizontal cross sections of thestream, the stream must “neck down” (narrow).

h

A0

A

The volume flow persecond here mustmatch ...

... the volume flowper second here.

The volume flow rate through the higher cross section mustbe the same as that through the lower cross section.

Calculations: From Eq. 14-24, we have

A0v0 ! Av, (14-26)

where v0 and v are the water speeds at the levels correspond-ing to A0 and A. From Eq. 2-16 we can also write, because thewater is falling freely with acceleration g,

v2 ! v20 " 2gh. (14-27)

Eliminating v between Eqs. 14-26 and 14-27 and solving forv0, we obtain

! 0.286 m/s ! 28.6 cm/s.

From Eq. 14-24, the volume flow rate RV is thenRV ! A0v0 ! (1.2 cm2)(28.6 cm/s)

! 34 cm3/s. (Answer)

! A (2)(9.8 m/s2)(0.045 m)(0.35 cm2)2

(1.2 cm2)2 # (0.35 cm2)2

v0 ! A 2ghA2

A20 # A2

Additional examples, video, and practice available at WileyPLUS

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 373

• What is the volume flow rate ifA0 = 1.2 cm2, A = 0.35 cm2 andh = 45 cm

• Continuity equation tells us A0v0 = Av

• We can relate velocity between twoparts of stream: v2 = v20 + 2gh

• Combining the two equations, we get

v0 =

√2ghA2

A20 −A2

= 28.6 cm/s

• We get that the volume flow rate isRV = A0v0 = 34 cm3/s

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Continuity Equation: Example Problem

37314-9 TH E EQUATION OF CONTI N U ITYPART 2

We can rewrite Eq. 14-23 as

RV ! Av ! a constant (volume flow rate, equation of continuity), (14-24)

in which RV is the volume flow rate of the fluid (volume past a given point perunit time). Its SI unit is the cubic meter per second (m3/s). If the density r of thefluid is uniform, we can multiply Eq. 14-24 by that density to get the mass flowrate Rm (mass per unit time):

Rm ! rRV ! rAv ! a constant (mass flow rate). (14-25)

The SI unit of mass flow rate is the kilogram per second (kg/s). Equation 14-25says that the mass that flows into the tube segment of Fig. 14-15 each second mustbe equal to the mass that flows out of that segment each second.

CHECKPOINT 3

The figure shows a pipe and gives the volume flow rate (in cm3/s) and the di-rection of flow for all but one section. What are the volume flow rate and thedirection of flow for that section?

4 8

2 56

4

Sample Problem

A water stream narrows as it falls

Figure 14-18 shows how the stream of water emerging froma faucet “necks down” as it falls. This change in the horizontalcross-sectional area is characteristic of any laminar (non-turbulant) falling stream because the gravitational forceincreases the speed of the stream. Here the indicatedcross-sectional areas are A0 ! 1.2 cm2 and A ! 0.35 cm2.The two levels are separated by a vertical distance h ! 45 mm.What is the volume flow rate from the tap?

KEY I DEA

Fig. 14-18 As water falls from a tap, its speedincreases. Because the volume flow rate must bethe same at all horizontal cross sections of thestream, the stream must “neck down” (narrow).

h

A0

A

The volume flow persecond here mustmatch ...

... the volume flowper second here.

The volume flow rate through the higher cross section mustbe the same as that through the lower cross section.

Calculations: From Eq. 14-24, we have

A0v0 ! Av, (14-26)

where v0 and v are the water speeds at the levels correspond-ing to A0 and A. From Eq. 2-16 we can also write, because thewater is falling freely with acceleration g,

v2 ! v20 " 2gh. (14-27)

Eliminating v between Eqs. 14-26 and 14-27 and solving forv0, we obtain

! 0.286 m/s ! 28.6 cm/s.

From Eq. 14-24, the volume flow rate RV is thenRV ! A0v0 ! (1.2 cm2)(28.6 cm/s)

! 34 cm3/s. (Answer)

! A (2)(9.8 m/s2)(0.045 m)(0.35 cm2)2

(1.2 cm2)2 # (0.35 cm2)2

v0 ! A 2ghA2

A20 # A2

Additional examples, video, and practice available at WileyPLUS

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 373

• What is the volume flow rate ifA0 = 1.2 cm2, A = 0.35 cm2 andh = 45 cm

• Continuity equation tells us A0v0 = Av

• We can relate velocity between twoparts of stream: v2 = v20 + 2gh

• Combining the two equations, we get

v0 =

√2ghA2

A20 −A2

= 28.6 cm/s

• We get that the volume flow rate isRV = A0v0 = 34 cm3/s

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Continuity Equation: Example Problem

37314-9 TH E EQUATION OF CONTI N U ITYPART 2

We can rewrite Eq. 14-23 as

RV ! Av ! a constant (volume flow rate, equation of continuity), (14-24)

in which RV is the volume flow rate of the fluid (volume past a given point perunit time). Its SI unit is the cubic meter per second (m3/s). If the density r of thefluid is uniform, we can multiply Eq. 14-24 by that density to get the mass flowrate Rm (mass per unit time):

Rm ! rRV ! rAv ! a constant (mass flow rate). (14-25)

The SI unit of mass flow rate is the kilogram per second (kg/s). Equation 14-25says that the mass that flows into the tube segment of Fig. 14-15 each second mustbe equal to the mass that flows out of that segment each second.

CHECKPOINT 3

The figure shows a pipe and gives the volume flow rate (in cm3/s) and the di-rection of flow for all but one section. What are the volume flow rate and thedirection of flow for that section?

4 8

2 56

4

Sample Problem

A water stream narrows as it falls

Figure 14-18 shows how the stream of water emerging froma faucet “necks down” as it falls. This change in the horizontalcross-sectional area is characteristic of any laminar (non-turbulant) falling stream because the gravitational forceincreases the speed of the stream. Here the indicatedcross-sectional areas are A0 ! 1.2 cm2 and A ! 0.35 cm2.The two levels are separated by a vertical distance h ! 45 mm.What is the volume flow rate from the tap?

KEY I DEA

Fig. 14-18 As water falls from a tap, its speedincreases. Because the volume flow rate must bethe same at all horizontal cross sections of thestream, the stream must “neck down” (narrow).

h

A0

A

The volume flow persecond here mustmatch ...

... the volume flowper second here.

The volume flow rate through the higher cross section mustbe the same as that through the lower cross section.

Calculations: From Eq. 14-24, we have

A0v0 ! Av, (14-26)

where v0 and v are the water speeds at the levels correspond-ing to A0 and A. From Eq. 2-16 we can also write, because thewater is falling freely with acceleration g,

v2 ! v20 " 2gh. (14-27)

Eliminating v between Eqs. 14-26 and 14-27 and solving forv0, we obtain

! 0.286 m/s ! 28.6 cm/s.

From Eq. 14-24, the volume flow rate RV is thenRV ! A0v0 ! (1.2 cm2)(28.6 cm/s)

! 34 cm3/s. (Answer)

! A (2)(9.8 m/s2)(0.045 m)(0.35 cm2)2

(1.2 cm2)2 # (0.35 cm2)2

v0 ! A 2ghA2

A20 # A2

Additional examples, video, and practice available at WileyPLUS

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 373

• What is the volume flow rate ifA0 = 1.2 cm2, A = 0.35 cm2 andh = 45 cm

• Continuity equation tells us A0v0 = Av

• We can relate velocity between twoparts of stream: v2 = v20 + 2gh

• Combining the two equations, we get

v0 =

√2ghA2

A20 −A2

= 28.6 cm/s

• We get that the volume flow rate isRV = A0v0 = 34 cm3/s

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Continuity Equation: Example Problem

37314-9 TH E EQUATION OF CONTI N U ITYPART 2

We can rewrite Eq. 14-23 as

RV ! Av ! a constant (volume flow rate, equation of continuity), (14-24)

in which RV is the volume flow rate of the fluid (volume past a given point perunit time). Its SI unit is the cubic meter per second (m3/s). If the density r of thefluid is uniform, we can multiply Eq. 14-24 by that density to get the mass flowrate Rm (mass per unit time):

Rm ! rRV ! rAv ! a constant (mass flow rate). (14-25)

The SI unit of mass flow rate is the kilogram per second (kg/s). Equation 14-25says that the mass that flows into the tube segment of Fig. 14-15 each second mustbe equal to the mass that flows out of that segment each second.

CHECKPOINT 3

The figure shows a pipe and gives the volume flow rate (in cm3/s) and the di-rection of flow for all but one section. What are the volume flow rate and thedirection of flow for that section?

4 8

2 56

4

Sample Problem

A water stream narrows as it falls

Figure 14-18 shows how the stream of water emerging froma faucet “necks down” as it falls. This change in the horizontalcross-sectional area is characteristic of any laminar (non-turbulant) falling stream because the gravitational forceincreases the speed of the stream. Here the indicatedcross-sectional areas are A0 ! 1.2 cm2 and A ! 0.35 cm2.The two levels are separated by a vertical distance h ! 45 mm.What is the volume flow rate from the tap?

KEY I DEA

Fig. 14-18 As water falls from a tap, its speedincreases. Because the volume flow rate must bethe same at all horizontal cross sections of thestream, the stream must “neck down” (narrow).

h

A0

A

The volume flow persecond here mustmatch ...

... the volume flowper second here.

The volume flow rate through the higher cross section mustbe the same as that through the lower cross section.

Calculations: From Eq. 14-24, we have

A0v0 ! Av, (14-26)

where v0 and v are the water speeds at the levels correspond-ing to A0 and A. From Eq. 2-16 we can also write, because thewater is falling freely with acceleration g,

v2 ! v20 " 2gh. (14-27)

Eliminating v between Eqs. 14-26 and 14-27 and solving forv0, we obtain

! 0.286 m/s ! 28.6 cm/s.

From Eq. 14-24, the volume flow rate RV is thenRV ! A0v0 ! (1.2 cm2)(28.6 cm/s)

! 34 cm3/s. (Answer)

! A (2)(9.8 m/s2)(0.045 m)(0.35 cm2)2

(1.2 cm2)2 # (0.35 cm2)2

v0 ! A 2ghA2

A20 # A2

Additional examples, video, and practice available at WileyPLUS

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 373

• What is the volume flow rate ifA0 = 1.2 cm2, A = 0.35 cm2 andh = 45 cm

• Continuity equation tells us A0v0 = Av

• We can relate velocity between twoparts of stream: v2 = v20 + 2gh

• Combining the two equations, we get

v0 =

√2ghA2

A20 −A2

= 28.6 cm/s

• We get that the volume flow rate isRV = A0v0 = 34 cm3/s

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Continuity Equation: Example Problem

37314-9 TH E EQUATION OF CONTI N U ITYPART 2

We can rewrite Eq. 14-23 as

RV ! Av ! a constant (volume flow rate, equation of continuity), (14-24)

in which RV is the volume flow rate of the fluid (volume past a given point perunit time). Its SI unit is the cubic meter per second (m3/s). If the density r of thefluid is uniform, we can multiply Eq. 14-24 by that density to get the mass flowrate Rm (mass per unit time):

Rm ! rRV ! rAv ! a constant (mass flow rate). (14-25)

The SI unit of mass flow rate is the kilogram per second (kg/s). Equation 14-25says that the mass that flows into the tube segment of Fig. 14-15 each second mustbe equal to the mass that flows out of that segment each second.

CHECKPOINT 3

The figure shows a pipe and gives the volume flow rate (in cm3/s) and the di-rection of flow for all but one section. What are the volume flow rate and thedirection of flow for that section?

4 8

2 56

4

Sample Problem

A water stream narrows as it falls

Figure 14-18 shows how the stream of water emerging froma faucet “necks down” as it falls. This change in the horizontalcross-sectional area is characteristic of any laminar (non-turbulant) falling stream because the gravitational forceincreases the speed of the stream. Here the indicatedcross-sectional areas are A0 ! 1.2 cm2 and A ! 0.35 cm2.The two levels are separated by a vertical distance h ! 45 mm.What is the volume flow rate from the tap?

KEY I DEA

Fig. 14-18 As water falls from a tap, its speedincreases. Because the volume flow rate must bethe same at all horizontal cross sections of thestream, the stream must “neck down” (narrow).

h

A0

A

The volume flow persecond here mustmatch ...

... the volume flowper second here.

The volume flow rate through the higher cross section mustbe the same as that through the lower cross section.

Calculations: From Eq. 14-24, we have

A0v0 ! Av, (14-26)

where v0 and v are the water speeds at the levels correspond-ing to A0 and A. From Eq. 2-16 we can also write, because thewater is falling freely with acceleration g,

v2 ! v20 " 2gh. (14-27)

Eliminating v between Eqs. 14-26 and 14-27 and solving forv0, we obtain

! 0.286 m/s ! 28.6 cm/s.

From Eq. 14-24, the volume flow rate RV is thenRV ! A0v0 ! (1.2 cm2)(28.6 cm/s)

! 34 cm3/s. (Answer)

! A (2)(9.8 m/s2)(0.045 m)(0.35 cm2)2

(1.2 cm2)2 # (0.35 cm2)2

v0 ! A 2ghA2

A20 # A2

Additional examples, video, and practice available at WileyPLUS

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 373

• What is the volume flow rate ifA0 = 1.2 cm2, A = 0.35 cm2 andh = 45 cm

• Continuity equation tells us A0v0 = Av

• We can relate velocity between twoparts of stream: v2 = v20 + 2gh

• Combining the two equations, we get

v0 =

√2ghA2

A20 −A2

= 28.6 cm/s

• We get that the volume flow rate isRV = A0v0 = 34 cm3/s

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Bernoulli’s Equation

374 CHAPTE R 14 FLU I DS

14-10 Bernoulli’s EquationFigure 14-19 represents a tube through which an ideal fluid is flowing at a steadyrate. In a time interval !t, suppose that a volume of fluid !V, colored purple inFig. 14-19, enters the tube at the left (or input) end and an identical volume,colored green in Fig. 14-19, emerges at the right (or output) end. The emergingvolume must be the same as the entering volume because the fluid is incompress-ible, with an assumed constant density r.

Let y1, v1, and p1 be the elevation, speed, and pressure of the fluid entering atthe left, and y2, v2, and p2 be the corresponding quantities for the fluid emergingat the right. By applying the principle of conservation of energy to the fluid, weshall show that these quantities are related by

(14-28)

In general, the term is called the fluid’s kinetic energy density (kinetic energyper unit volume).We can also write Eq. 14-28 as

(Bernoulli’s equation). (14-29)

Equations 14-28 and 14-29 are equivalent forms of Bernoulli’s equation,after Daniel Bernoulli, who studied fluid flow in the 1700s.* Like the equation ofcontinuity (Eq. 14-24), Bernoulli’s equation is not a new principle but simply the reformulation of a familiar principle in a form more suitable to fluid mechanics. As a check, let us apply Bernoulli’s equation to fluids at rest, byputting v1 " v2 " 0 in Eq. 14-28.The result is

p2 " p1 # rg(y1 $ y2),

which is Eq. 14-7.A major prediction of Bernoulli’s equation emerges if we take y to be a

constant (y " 0, say) so that the fluid does not change elevation as it flows.Equation 14-28 then becomes

(14-30)which tells us that:

p1# 12%v2

1 " p2 # 12%v2

2,

p # 12%v2 # %gy " a constant

12%v2

p1 # 12%v2

1 # %gy1 " p2 # 12%v2

2 # %gy2.

*For irrotational flow (which we assume), the constant in Eq. 14-29 has the same value for allpoints within the tube of flow; the points do not have to lie along the same streamline. Similarly,the points 1 and 2 in Eq. 14-28 can lie anywhere within the tube of flow.

If the speed of a fluid element increases as the element travels along a horizontalstreamline, the pressure of the fluid must decrease, and conversely.

Put another way, where the streamlines are relatively close together (where thevelocity is relatively great), the pressure is relatively low, and conversely.

The link between a change in speed and a change in pressure makes sense ifyou consider a fluid element that travels through a tube of various widths. Recallthat the element’s speed in the narrower regions is fast and its speed in the widerregions is slow. By Newton’s second law, forces (or pressures) must cause thechanges in speed (the accelerations). When the element nears a narrow region,the higher pressure behind it accelerates it so that it then has a greater speed inthe narrow region. When it nears a wide region, the higher pressure ahead of itdecelerates it so that it then has a lesser speed in the wide region.

Bernoulli’s equation is strictly valid only to the extent that the fluid is ideal. Ifviscous forces are present, thermal energy will be involved. We take no accountof this in the derivation that follows.

Fig. 14-19 Fluid flows at a steady ratethrough a length L of a tube, from the inputend at the left to the output end at the right.From time t in (a) to time t # !t in (b), theamount of fluid shown in purple enters theinput end and the equal amount shown ingreen emerges from the output end.

p1

L

Input

v1

y1

(a)

(b)

y

v2

p2

y2

y

x

t

t + !t

x

Output

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 374

• Bernoulli’s Equation tells us how tolook at Conservation of Energy foran ideal fluid.

• Let y1, v1 and p1 be the elevation,speed, and pressure of the fluid atone point and let y2, v2 and p2 bethe elevation, speed, and pressure ofthe fluid at another point along astream line.

• p1+ 12ρv

21 +ρgy1 = p2+ 1

2ρv22 +ρgy2

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Bernoulli’s Equation

374 CHAPTE R 14 FLU I DS

14-10 Bernoulli’s EquationFigure 14-19 represents a tube through which an ideal fluid is flowing at a steadyrate. In a time interval !t, suppose that a volume of fluid !V, colored purple inFig. 14-19, enters the tube at the left (or input) end and an identical volume,colored green in Fig. 14-19, emerges at the right (or output) end. The emergingvolume must be the same as the entering volume because the fluid is incompress-ible, with an assumed constant density r.

Let y1, v1, and p1 be the elevation, speed, and pressure of the fluid entering atthe left, and y2, v2, and p2 be the corresponding quantities for the fluid emergingat the right. By applying the principle of conservation of energy to the fluid, weshall show that these quantities are related by

(14-28)

In general, the term is called the fluid’s kinetic energy density (kinetic energyper unit volume).We can also write Eq. 14-28 as

(Bernoulli’s equation). (14-29)

Equations 14-28 and 14-29 are equivalent forms of Bernoulli’s equation,after Daniel Bernoulli, who studied fluid flow in the 1700s.* Like the equation ofcontinuity (Eq. 14-24), Bernoulli’s equation is not a new principle but simply the reformulation of a familiar principle in a form more suitable to fluid mechanics. As a check, let us apply Bernoulli’s equation to fluids at rest, byputting v1 " v2 " 0 in Eq. 14-28.The result is

p2 " p1 # rg(y1 $ y2),

which is Eq. 14-7.A major prediction of Bernoulli’s equation emerges if we take y to be a

constant (y " 0, say) so that the fluid does not change elevation as it flows.Equation 14-28 then becomes

(14-30)which tells us that:

p1# 12%v2

1 " p2 # 12%v2

2,

p # 12%v2 # %gy " a constant

12%v2

p1 # 12%v2

1 # %gy1 " p2 # 12%v2

2 # %gy2.

*For irrotational flow (which we assume), the constant in Eq. 14-29 has the same value for allpoints within the tube of flow; the points do not have to lie along the same streamline. Similarly,the points 1 and 2 in Eq. 14-28 can lie anywhere within the tube of flow.

If the speed of a fluid element increases as the element travels along a horizontalstreamline, the pressure of the fluid must decrease, and conversely.

Put another way, where the streamlines are relatively close together (where thevelocity is relatively great), the pressure is relatively low, and conversely.

The link between a change in speed and a change in pressure makes sense ifyou consider a fluid element that travels through a tube of various widths. Recallthat the element’s speed in the narrower regions is fast and its speed in the widerregions is slow. By Newton’s second law, forces (or pressures) must cause thechanges in speed (the accelerations). When the element nears a narrow region,the higher pressure behind it accelerates it so that it then has a greater speed inthe narrow region. When it nears a wide region, the higher pressure ahead of itdecelerates it so that it then has a lesser speed in the wide region.

Bernoulli’s equation is strictly valid only to the extent that the fluid is ideal. Ifviscous forces are present, thermal energy will be involved. We take no accountof this in the derivation that follows.

Fig. 14-19 Fluid flows at a steady ratethrough a length L of a tube, from the inputend at the left to the output end at the right.From time t in (a) to time t # !t in (b), theamount of fluid shown in purple enters theinput end and the equal amount shown ingreen emerges from the output end.

p1

L

Input

v1

y1

(a)

(b)

y

v2

p2

y2

y

x

t

t + !t

x

Output

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 374

• Bernoulli’s Equation tells us how tolook at Conservation of Energy foran ideal fluid.

• Let y1, v1 and p1 be the elevation,speed, and pressure of the fluid atone point and let y2, v2 and p2 bethe elevation, speed, and pressure ofthe fluid at another point along astream line.

• p1+ 12ρv

21 +ρgy1 = p2+ 1

2ρv22 +ρgy2

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Bernoulli’s Equation

374 CHAPTE R 14 FLU I DS

14-10 Bernoulli’s EquationFigure 14-19 represents a tube through which an ideal fluid is flowing at a steadyrate. In a time interval !t, suppose that a volume of fluid !V, colored purple inFig. 14-19, enters the tube at the left (or input) end and an identical volume,colored green in Fig. 14-19, emerges at the right (or output) end. The emergingvolume must be the same as the entering volume because the fluid is incompress-ible, with an assumed constant density r.

Let y1, v1, and p1 be the elevation, speed, and pressure of the fluid entering atthe left, and y2, v2, and p2 be the corresponding quantities for the fluid emergingat the right. By applying the principle of conservation of energy to the fluid, weshall show that these quantities are related by

(14-28)

In general, the term is called the fluid’s kinetic energy density (kinetic energyper unit volume).We can also write Eq. 14-28 as

(Bernoulli’s equation). (14-29)

Equations 14-28 and 14-29 are equivalent forms of Bernoulli’s equation,after Daniel Bernoulli, who studied fluid flow in the 1700s.* Like the equation ofcontinuity (Eq. 14-24), Bernoulli’s equation is not a new principle but simply the reformulation of a familiar principle in a form more suitable to fluid mechanics. As a check, let us apply Bernoulli’s equation to fluids at rest, byputting v1 " v2 " 0 in Eq. 14-28.The result is

p2 " p1 # rg(y1 $ y2),

which is Eq. 14-7.A major prediction of Bernoulli’s equation emerges if we take y to be a

constant (y " 0, say) so that the fluid does not change elevation as it flows.Equation 14-28 then becomes

(14-30)which tells us that:

p1# 12%v2

1 " p2 # 12%v2

2,

p # 12%v2 # %gy " a constant

12%v2

p1 # 12%v2

1 # %gy1 " p2 # 12%v2

2 # %gy2.

*For irrotational flow (which we assume), the constant in Eq. 14-29 has the same value for allpoints within the tube of flow; the points do not have to lie along the same streamline. Similarly,the points 1 and 2 in Eq. 14-28 can lie anywhere within the tube of flow.

If the speed of a fluid element increases as the element travels along a horizontalstreamline, the pressure of the fluid must decrease, and conversely.

Put another way, where the streamlines are relatively close together (where thevelocity is relatively great), the pressure is relatively low, and conversely.

The link between a change in speed and a change in pressure makes sense ifyou consider a fluid element that travels through a tube of various widths. Recallthat the element’s speed in the narrower regions is fast and its speed in the widerregions is slow. By Newton’s second law, forces (or pressures) must cause thechanges in speed (the accelerations). When the element nears a narrow region,the higher pressure behind it accelerates it so that it then has a greater speed inthe narrow region. When it nears a wide region, the higher pressure ahead of itdecelerates it so that it then has a lesser speed in the wide region.

Bernoulli’s equation is strictly valid only to the extent that the fluid is ideal. Ifviscous forces are present, thermal energy will be involved. We take no accountof this in the derivation that follows.

Fig. 14-19 Fluid flows at a steady ratethrough a length L of a tube, from the inputend at the left to the output end at the right.From time t in (a) to time t # !t in (b), theamount of fluid shown in purple enters theinput end and the equal amount shown ingreen emerges from the output end.

p1

L

Input

v1

y1

(a)

(b)

y

v2

p2

y2

y

x

t

t + !t

x

Output

halliday_c14_359-385hr.qxd 26-10-2009 21:40 Page 374

• Bernoulli’s Equation tells us how tolook at Conservation of Energy foran ideal fluid.

• Let y1, v1 and p1 be the elevation,speed, and pressure of the fluid atone point and let y2, v2 and p2 bethe elevation, speed, and pressure ofthe fluid at another point along astream line.

• p1+ 12ρv

21 +ρgy1 = p2+ 1

2ρv22 +ρgy2

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Special Cases of Bernoulli’s Equation

Let’s look at when v1 = v2 = 0. Bernoulli’s Equation becomes,

p2 = p1 + ρg(y1 − y2)

Let’s also look at when y1 = y2 Bernoulli’s Equation becomes

p1 + 12ρv

21 = p2 + 1

2ρv22

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Special Cases of Bernoulli’s Equation

Let’s look at when v1 = v2 = 0. Bernoulli’s Equation becomes,

p2 = p1 + ρg(y1 − y2)

Let’s also look at when y1 = y2 Bernoulli’s Equation becomes

p1 + 12ρv

21 = p2 + 1

2ρv22

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Proof of Bernoulli’s Equation

• First, we apply Energy Conservation: W = ∆K

• ∆K = 12∆mv22 − 1

2∆mv21 = 12ρ∆V (v22 − v21)

• Next, we look at the work done by gravity

• Wg = −∆mg(y2 − y1) = −ρg∆V (y2 − y1)• And the work done on and by the fluid.

• F∆x = (pA)(∆x) = p(A∆x) = p∆V

• So work done on system is p1∆V and work done by thesystem is −p2∆V . Adding them together,

• Wp = −(p2 − p1)∆V

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Proof of Bernoulli’s Equation

• First, we apply Energy Conservation: W = ∆K

• ∆K = 12∆mv22 − 1

2∆mv21 = 12ρ∆V (v22 − v21)

• Next, we look at the work done by gravity

• Wg = −∆mg(y2 − y1) = −ρg∆V (y2 − y1)

• And the work done on and by the fluid.

• F∆x = (pA)(∆x) = p(A∆x) = p∆V

• So work done on system is p1∆V and work done by thesystem is −p2∆V . Adding them together,

• Wp = −(p2 − p1)∆V

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Proof of Bernoulli’s Equation

• First, we apply Energy Conservation: W = ∆K

• ∆K = 12∆mv22 − 1

2∆mv21 = 12ρ∆V (v22 − v21)

• Next, we look at the work done by gravity

• Wg = −∆mg(y2 − y1) = −ρg∆V (y2 − y1)• And the work done on and by the fluid.

• F∆x = (pA)(∆x) = p(A∆x) = p∆V

• So work done on system is p1∆V and work done by thesystem is −p2∆V . Adding them together,

• Wp = −(p2 − p1)∆V

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Proof of Bernoulli’s Equation

• First, we apply Energy Conservation: W = ∆K

• ∆K = 12∆mv22 − 1

2∆mv21 = 12ρ∆V (v22 − v21)

• Next, we look at the work done by gravity

• Wg = −∆mg(y2 − y1) = −ρg∆V (y2 − y1)• And the work done on and by the fluid.

• F∆x = (pA)(∆x) = p(A∆x) = p∆V

• So work done on system is p1∆V and work done by thesystem is −p2∆V . Adding them together,

• Wp = −(p2 − p1)∆V

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Proof of Bernoulli’s Equation (cont.)

• Combining these two work contributions and our change inkinetic energy,

• −ρg∆V (y2 − y1)−∆V (p2 − p1) = 12ρ∆V (v22 − v21)

• We can cancel the ∆V and separate the terms accordingto subscript to get

• p1 + 12ρv

21 + ρgy1 = p2 + 1

2ρv22 + ρgy2

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Proof of Bernoulli’s Equation (cont.)

• Combining these two work contributions and our change inkinetic energy,

• −ρg∆V (y2 − y1)−∆V (p2 − p1) = 12ρ∆V (v22 − v21)

• We can cancel the ∆V and separate the terms accordingto subscript to get

• p1 + 12ρv

21 + ρgy1 = p2 + 1

2ρv22 + ρgy2

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Bernoulli’s Equation: Example Problem

383PROB LE M SPART 2

sec. 14-10 Bernoulli’s Equation•55 How much work is done by pressure in forcing 1.4 m3 of water through a pipe having an internal diameter of 13 mm if thedifference in pressure at the two ends of the pipe is 1.0 atm?

•56 Suppose that two tanks, 1 and 2, each with a large opening atthe top, contain different liquids.A small hole is made in the side ofeach tank at the same depth h below the liquid surface, but thehole in tank 1 has half the cross-sectional area of the hole in tank 2.(a) What is the ratio r1/r2 of the densities of the liquids if the massflow rate is the same for the two holes? (b) What is the ratioRV1/RV2 of the volume flow rates from the two tanks? (c) At one in-stant, the liquid in tank 1 is 12.0 cm above the hole. If the tanks areto have equal volume flow rates, what height above the hole mustthe liquid in tank 2 be just then?

•57 A cylindrical tank with a large diameter is filledwith water to a depth D ! 0.30 m. A hole of cross-sectional areaA ! 6.5 cm2 in the bottom of the tank allows water to drain out.(a) What is the rate at which water flows out, in cubic meters persecond? (b) At what distance below the bottom of the tank is thecross-sectional area of the stream equal to one-half the area ofthe hole?

•58 The intake in Fig. 14-47 hascross-sectional area of 0.74 m2 andwater flow at 0.40 m/s. At the outlet,distance D ! 180 m below the in-take, the cross-sectional area issmaller than at the intake and thewater flows out at 9.5 m/s intoequipment.What is the pressure dif-ference between inlet and outlet?

•59 Water is moving with a speed of 5.0 m/s through a pipewith a cross-sectional area of 4.0 cm2. The water graduall descends10 m as the pipe cross-sectional area increases to 8.0 cm2. (a) Whatis the speed at the lower level? (b) If the pressure at the upperlevel is 1.5 " 105 Pa, what is the pressure at the lower level?

•60 Models of torpedoes are sometimes tested in a horizontalpipe of flowing water, much as a wind tunnel is used to test modelairplanes. Consider a circular pipe of internal diameter 25.0 cm anda torpedo model aligned along the long axis of the pipe.The modelhas a 5.00 cm diameter and is to be tested with water flowing past itat 2.50 m/s. (a) With what speed must the water flow in the part ofthe pipe that is unconstricted by the model? (b) What will the pres-sure difference be between the constricted and unconstricted partsof the pipe?

•61 A water pipe having a 2.5 cm inside diameter carrieswater into the basement of a house at a speed of 0.90 m/s and apressure of 170 kPa. If the pipe tapers to 1.2 cm and rises to thesecond floor 7.6 m above the input point, what are the (a) speedand (b) water pressure at the second floor?

••62 A pitot tube (Fig. 14-48) is used to determine the air-speed of an airplane. It consists of an outer tube with a number ofsmall holes B (four are shown) that allow air into the tube; thattube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole A at the front end of the device, whichpoints in the direction the plane is headed. At A the air becomesstagnant so that vA ! 0. At B, however, the speed of the air pre-sumably equals the airspeed v of the plane. (a) Use Bernoulli’s

ILW

SSM

SSM

equation to show that

,

where r is the density of the liquid in the U-tube and h is the differ-ence in the liquid levels in that tube. (b) Suppose that the tube con-tains alcohol and the level difference h is 26.0 cm. What is theplane’s speed relative to the air? The density of the air is 1.03 kg/m3

and that of alcohol is 810 kg/m3.

v ! A 2#gh#air

Fig. 14-47 Problem 58.

Reservoir

Generatorbuilding

Outlet

IntakeD

air!

Hole A

Liquid

v

B

!h

B

Air

Fig. 14-48 Problems 62 and 63.

••63 A pitot tube (see Problem 62) on a high-altitude aircraftmeasures a differential pressure of 180 Pa. What is the aircraft’sairspeed if the density of the air is 0.031 kg/m3?

••64 In Fig. 14-49, water flowsthrough a horizontal pipe and thenout into the atmosphere at a speedv1 ! 15 m/s. The diameters of theleft and right sections of the pipe are5.0 cm and 3.0 cm. (a) What volumeof water flows into the atmosphereduring a 10 min period? In the left section of the pipe, what are (b)the speed v2 and (c) the gauge pressure?

••65 A venturi meter is used to measure the flowspeed of a fluid in a pipe. The meter is connected between twosections of the pipe (Fig. 14-50); the cross-sectional area A of theentrance and exit of the meter matches the pipe’s cross-sectionalarea. Between the entrance and exit, the fluid flows from thepipe with speed V and then through a narrow “throat” of cross-

WWWSSM

d 2v1v2 d 1

Fig. 14-49 Problem 64.

Aa

12

h

Pipe Pipe

A

Meterentrance

MeterexitVenturi meter

Manometer

v

V

Fig. 14-50 Problems 65 and 66.

halliday_c14_359-385hr.qxd 26-10-2009 21:41 Page 383

The intake has cross-sectional area of0.74 m2 and water flows at 0.40 m/s. Atthe outlet, distance D = 180 m belowthe intake, the cross-sectional area issmaller than at the intake and waterflows out at 9.5 m/s into the generator.What is the pressure difference betweeninlet and outlet?

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Bernoulli’s Equation: Example Problem (cont)

383PROB LE M SPART 2

sec. 14-10 Bernoulli’s Equation•55 How much work is done by pressure in forcing 1.4 m3 of water through a pipe having an internal diameter of 13 mm if thedifference in pressure at the two ends of the pipe is 1.0 atm?

•56 Suppose that two tanks, 1 and 2, each with a large opening atthe top, contain different liquids.A small hole is made in the side ofeach tank at the same depth h below the liquid surface, but thehole in tank 1 has half the cross-sectional area of the hole in tank 2.(a) What is the ratio r1/r2 of the densities of the liquids if the massflow rate is the same for the two holes? (b) What is the ratioRV1/RV2 of the volume flow rates from the two tanks? (c) At one in-stant, the liquid in tank 1 is 12.0 cm above the hole. If the tanks areto have equal volume flow rates, what height above the hole mustthe liquid in tank 2 be just then?

•57 A cylindrical tank with a large diameter is filledwith water to a depth D ! 0.30 m. A hole of cross-sectional areaA ! 6.5 cm2 in the bottom of the tank allows water to drain out.(a) What is the rate at which water flows out, in cubic meters persecond? (b) At what distance below the bottom of the tank is thecross-sectional area of the stream equal to one-half the area ofthe hole?

•58 The intake in Fig. 14-47 hascross-sectional area of 0.74 m2 andwater flow at 0.40 m/s. At the outlet,distance D ! 180 m below the in-take, the cross-sectional area issmaller than at the intake and thewater flows out at 9.5 m/s intoequipment.What is the pressure dif-ference between inlet and outlet?

•59 Water is moving with a speed of 5.0 m/s through a pipewith a cross-sectional area of 4.0 cm2. The water graduall descends10 m as the pipe cross-sectional area increases to 8.0 cm2. (a) Whatis the speed at the lower level? (b) If the pressure at the upperlevel is 1.5 " 105 Pa, what is the pressure at the lower level?

•60 Models of torpedoes are sometimes tested in a horizontalpipe of flowing water, much as a wind tunnel is used to test modelairplanes. Consider a circular pipe of internal diameter 25.0 cm anda torpedo model aligned along the long axis of the pipe.The modelhas a 5.00 cm diameter and is to be tested with water flowing past itat 2.50 m/s. (a) With what speed must the water flow in the part ofthe pipe that is unconstricted by the model? (b) What will the pres-sure difference be between the constricted and unconstricted partsof the pipe?

•61 A water pipe having a 2.5 cm inside diameter carrieswater into the basement of a house at a speed of 0.90 m/s and apressure of 170 kPa. If the pipe tapers to 1.2 cm and rises to thesecond floor 7.6 m above the input point, what are the (a) speedand (b) water pressure at the second floor?

••62 A pitot tube (Fig. 14-48) is used to determine the air-speed of an airplane. It consists of an outer tube with a number ofsmall holes B (four are shown) that allow air into the tube; thattube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole A at the front end of the device, whichpoints in the direction the plane is headed. At A the air becomesstagnant so that vA ! 0. At B, however, the speed of the air pre-sumably equals the airspeed v of the plane. (a) Use Bernoulli’s

ILW

SSM

SSM

equation to show that

,

where r is the density of the liquid in the U-tube and h is the differ-ence in the liquid levels in that tube. (b) Suppose that the tube con-tains alcohol and the level difference h is 26.0 cm. What is theplane’s speed relative to the air? The density of the air is 1.03 kg/m3

and that of alcohol is 810 kg/m3.

v ! A 2#gh#air

Fig. 14-47 Problem 58.

Reservoir

Generatorbuilding

Outlet

IntakeD

air!

Hole A

Liquid

v

B

!h

B

Air

Fig. 14-48 Problems 62 and 63.

••63 A pitot tube (see Problem 62) on a high-altitude aircraftmeasures a differential pressure of 180 Pa. What is the aircraft’sairspeed if the density of the air is 0.031 kg/m3?

••64 In Fig. 14-49, water flowsthrough a horizontal pipe and thenout into the atmosphere at a speedv1 ! 15 m/s. The diameters of theleft and right sections of the pipe are5.0 cm and 3.0 cm. (a) What volumeof water flows into the atmosphereduring a 10 min period? In the left section of the pipe, what are (b)the speed v2 and (c) the gauge pressure?

••65 A venturi meter is used to measure the flowspeed of a fluid in a pipe. The meter is connected between twosections of the pipe (Fig. 14-50); the cross-sectional area A of theentrance and exit of the meter matches the pipe’s cross-sectionalarea. Between the entrance and exit, the fluid flows from thepipe with speed V and then through a narrow “throat” of cross-

WWWSSM

d 2v1v2 d 1

Fig. 14-49 Problem 64.

Aa

12

h

Pipe Pipe

A

Meterentrance

MeterexitVenturi meter

Manometer

v

V

Fig. 14-50 Problems 65 and 66.

halliday_c14_359-385hr.qxd 26-10-2009 21:41 Page 383

• We use Bernoulli’s Equation with point1 at the inlet and point 2 at the outlet.

p1 + 12ρv

21 + ρgy1 = p2 + 1

2ρv22 + ρgy2

• We’ll solve it for the pressure difference

p2 − p1 = ρgD + 12ρ(v21 − v22)

p2 − p1 = 1.7× 106 Pa

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Bernoulli’s Equation: Example Problem (cont)

383PROB LE M SPART 2

sec. 14-10 Bernoulli’s Equation•55 How much work is done by pressure in forcing 1.4 m3 of water through a pipe having an internal diameter of 13 mm if thedifference in pressure at the two ends of the pipe is 1.0 atm?

•56 Suppose that two tanks, 1 and 2, each with a large opening atthe top, contain different liquids.A small hole is made in the side ofeach tank at the same depth h below the liquid surface, but thehole in tank 1 has half the cross-sectional area of the hole in tank 2.(a) What is the ratio r1/r2 of the densities of the liquids if the massflow rate is the same for the two holes? (b) What is the ratioRV1/RV2 of the volume flow rates from the two tanks? (c) At one in-stant, the liquid in tank 1 is 12.0 cm above the hole. If the tanks areto have equal volume flow rates, what height above the hole mustthe liquid in tank 2 be just then?

•57 A cylindrical tank with a large diameter is filledwith water to a depth D ! 0.30 m. A hole of cross-sectional areaA ! 6.5 cm2 in the bottom of the tank allows water to drain out.(a) What is the rate at which water flows out, in cubic meters persecond? (b) At what distance below the bottom of the tank is thecross-sectional area of the stream equal to one-half the area ofthe hole?

•58 The intake in Fig. 14-47 hascross-sectional area of 0.74 m2 andwater flow at 0.40 m/s. At the outlet,distance D ! 180 m below the in-take, the cross-sectional area issmaller than at the intake and thewater flows out at 9.5 m/s intoequipment.What is the pressure dif-ference between inlet and outlet?

•59 Water is moving with a speed of 5.0 m/s through a pipewith a cross-sectional area of 4.0 cm2. The water graduall descends10 m as the pipe cross-sectional area increases to 8.0 cm2. (a) Whatis the speed at the lower level? (b) If the pressure at the upperlevel is 1.5 " 105 Pa, what is the pressure at the lower level?

•60 Models of torpedoes are sometimes tested in a horizontalpipe of flowing water, much as a wind tunnel is used to test modelairplanes. Consider a circular pipe of internal diameter 25.0 cm anda torpedo model aligned along the long axis of the pipe.The modelhas a 5.00 cm diameter and is to be tested with water flowing past itat 2.50 m/s. (a) With what speed must the water flow in the part ofthe pipe that is unconstricted by the model? (b) What will the pres-sure difference be between the constricted and unconstricted partsof the pipe?

•61 A water pipe having a 2.5 cm inside diameter carrieswater into the basement of a house at a speed of 0.90 m/s and apressure of 170 kPa. If the pipe tapers to 1.2 cm and rises to thesecond floor 7.6 m above the input point, what are the (a) speedand (b) water pressure at the second floor?

••62 A pitot tube (Fig. 14-48) is used to determine the air-speed of an airplane. It consists of an outer tube with a number ofsmall holes B (four are shown) that allow air into the tube; thattube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole A at the front end of the device, whichpoints in the direction the plane is headed. At A the air becomesstagnant so that vA ! 0. At B, however, the speed of the air pre-sumably equals the airspeed v of the plane. (a) Use Bernoulli’s

ILW

SSM

SSM

equation to show that

,

where r is the density of the liquid in the U-tube and h is the differ-ence in the liquid levels in that tube. (b) Suppose that the tube con-tains alcohol and the level difference h is 26.0 cm. What is theplane’s speed relative to the air? The density of the air is 1.03 kg/m3

and that of alcohol is 810 kg/m3.

v ! A 2#gh#air

Fig. 14-47 Problem 58.

Reservoir

Generatorbuilding

Outlet

IntakeD

air!

Hole A

Liquid

v

B

!h

B

Air

Fig. 14-48 Problems 62 and 63.

••63 A pitot tube (see Problem 62) on a high-altitude aircraftmeasures a differential pressure of 180 Pa. What is the aircraft’sairspeed if the density of the air is 0.031 kg/m3?

••64 In Fig. 14-49, water flowsthrough a horizontal pipe and thenout into the atmosphere at a speedv1 ! 15 m/s. The diameters of theleft and right sections of the pipe are5.0 cm and 3.0 cm. (a) What volumeof water flows into the atmosphereduring a 10 min period? In the left section of the pipe, what are (b)the speed v2 and (c) the gauge pressure?

••65 A venturi meter is used to measure the flowspeed of a fluid in a pipe. The meter is connected between twosections of the pipe (Fig. 14-50); the cross-sectional area A of theentrance and exit of the meter matches the pipe’s cross-sectionalarea. Between the entrance and exit, the fluid flows from thepipe with speed V and then through a narrow “throat” of cross-

WWWSSM

d 2v1v2 d 1

Fig. 14-49 Problem 64.

Aa

12

h

Pipe Pipe

A

Meterentrance

MeterexitVenturi meter

Manometer

v

V

Fig. 14-50 Problems 65 and 66.

halliday_c14_359-385hr.qxd 26-10-2009 21:41 Page 383

• We use Bernoulli’s Equation with point1 at the inlet and point 2 at the outlet.

p1 + 12ρv

21 + ρgy1 = p2 + 1

2ρv22 + ρgy2

• We’ll solve it for the pressure difference

p2 − p1 = ρgD + 12ρ(v21 − v22)

p2 − p1 = 1.7× 106 Pa

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Bernoulli’s Equation: Example Problem (cont)

383PROB LE M SPART 2

sec. 14-10 Bernoulli’s Equation•55 How much work is done by pressure in forcing 1.4 m3 of water through a pipe having an internal diameter of 13 mm if thedifference in pressure at the two ends of the pipe is 1.0 atm?

•56 Suppose that two tanks, 1 and 2, each with a large opening atthe top, contain different liquids.A small hole is made in the side ofeach tank at the same depth h below the liquid surface, but thehole in tank 1 has half the cross-sectional area of the hole in tank 2.(a) What is the ratio r1/r2 of the densities of the liquids if the massflow rate is the same for the two holes? (b) What is the ratioRV1/RV2 of the volume flow rates from the two tanks? (c) At one in-stant, the liquid in tank 1 is 12.0 cm above the hole. If the tanks areto have equal volume flow rates, what height above the hole mustthe liquid in tank 2 be just then?

•57 A cylindrical tank with a large diameter is filledwith water to a depth D ! 0.30 m. A hole of cross-sectional areaA ! 6.5 cm2 in the bottom of the tank allows water to drain out.(a) What is the rate at which water flows out, in cubic meters persecond? (b) At what distance below the bottom of the tank is thecross-sectional area of the stream equal to one-half the area ofthe hole?

•58 The intake in Fig. 14-47 hascross-sectional area of 0.74 m2 andwater flow at 0.40 m/s. At the outlet,distance D ! 180 m below the in-take, the cross-sectional area issmaller than at the intake and thewater flows out at 9.5 m/s intoequipment.What is the pressure dif-ference between inlet and outlet?

•59 Water is moving with a speed of 5.0 m/s through a pipewith a cross-sectional area of 4.0 cm2. The water graduall descends10 m as the pipe cross-sectional area increases to 8.0 cm2. (a) Whatis the speed at the lower level? (b) If the pressure at the upperlevel is 1.5 " 105 Pa, what is the pressure at the lower level?

•60 Models of torpedoes are sometimes tested in a horizontalpipe of flowing water, much as a wind tunnel is used to test modelairplanes. Consider a circular pipe of internal diameter 25.0 cm anda torpedo model aligned along the long axis of the pipe.The modelhas a 5.00 cm diameter and is to be tested with water flowing past itat 2.50 m/s. (a) With what speed must the water flow in the part ofthe pipe that is unconstricted by the model? (b) What will the pres-sure difference be between the constricted and unconstricted partsof the pipe?

•61 A water pipe having a 2.5 cm inside diameter carrieswater into the basement of a house at a speed of 0.90 m/s and apressure of 170 kPa. If the pipe tapers to 1.2 cm and rises to thesecond floor 7.6 m above the input point, what are the (a) speedand (b) water pressure at the second floor?

••62 A pitot tube (Fig. 14-48) is used to determine the air-speed of an airplane. It consists of an outer tube with a number ofsmall holes B (four are shown) that allow air into the tube; thattube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole A at the front end of the device, whichpoints in the direction the plane is headed. At A the air becomesstagnant so that vA ! 0. At B, however, the speed of the air pre-sumably equals the airspeed v of the plane. (a) Use Bernoulli’s

ILW

SSM

SSM

equation to show that

,

where r is the density of the liquid in the U-tube and h is the differ-ence in the liquid levels in that tube. (b) Suppose that the tube con-tains alcohol and the level difference h is 26.0 cm. What is theplane’s speed relative to the air? The density of the air is 1.03 kg/m3

and that of alcohol is 810 kg/m3.

v ! A 2#gh#air

Fig. 14-47 Problem 58.

Reservoir

Generatorbuilding

Outlet

IntakeD

air!

Hole A

Liquid

v

B

!h

B

Air

Fig. 14-48 Problems 62 and 63.

••63 A pitot tube (see Problem 62) on a high-altitude aircraftmeasures a differential pressure of 180 Pa. What is the aircraft’sairspeed if the density of the air is 0.031 kg/m3?

••64 In Fig. 14-49, water flowsthrough a horizontal pipe and thenout into the atmosphere at a speedv1 ! 15 m/s. The diameters of theleft and right sections of the pipe are5.0 cm and 3.0 cm. (a) What volumeof water flows into the atmosphereduring a 10 min period? In the left section of the pipe, what are (b)the speed v2 and (c) the gauge pressure?

••65 A venturi meter is used to measure the flowspeed of a fluid in a pipe. The meter is connected between twosections of the pipe (Fig. 14-50); the cross-sectional area A of theentrance and exit of the meter matches the pipe’s cross-sectionalarea. Between the entrance and exit, the fluid flows from thepipe with speed V and then through a narrow “throat” of cross-

WWWSSM

d 2v1v2 d 1

Fig. 14-49 Problem 64.

Aa

12

h

Pipe Pipe

A

Meterentrance

MeterexitVenturi meter

Manometer

v

V

Fig. 14-50 Problems 65 and 66.

halliday_c14_359-385hr.qxd 26-10-2009 21:41 Page 383

• We use Bernoulli’s Equation with point1 at the inlet and point 2 at the outlet.

p1 + 12ρv

21 + ρgy1 = p2 + 1

2ρv22 + ρgy2

• We’ll solve it for the pressure difference

p2 − p1 = ρgD + 12ρ(v21 − v22)

p2 − p1 = 1.7× 106 Pa

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Bernoulli’s Equation: Example Problem (cont)

383PROB LE M SPART 2

sec. 14-10 Bernoulli’s Equation•55 How much work is done by pressure in forcing 1.4 m3 of water through a pipe having an internal diameter of 13 mm if thedifference in pressure at the two ends of the pipe is 1.0 atm?

•56 Suppose that two tanks, 1 and 2, each with a large opening atthe top, contain different liquids.A small hole is made in the side ofeach tank at the same depth h below the liquid surface, but thehole in tank 1 has half the cross-sectional area of the hole in tank 2.(a) What is the ratio r1/r2 of the densities of the liquids if the massflow rate is the same for the two holes? (b) What is the ratioRV1/RV2 of the volume flow rates from the two tanks? (c) At one in-stant, the liquid in tank 1 is 12.0 cm above the hole. If the tanks areto have equal volume flow rates, what height above the hole mustthe liquid in tank 2 be just then?

•57 A cylindrical tank with a large diameter is filledwith water to a depth D ! 0.30 m. A hole of cross-sectional areaA ! 6.5 cm2 in the bottom of the tank allows water to drain out.(a) What is the rate at which water flows out, in cubic meters persecond? (b) At what distance below the bottom of the tank is thecross-sectional area of the stream equal to one-half the area ofthe hole?

•58 The intake in Fig. 14-47 hascross-sectional area of 0.74 m2 andwater flow at 0.40 m/s. At the outlet,distance D ! 180 m below the in-take, the cross-sectional area issmaller than at the intake and thewater flows out at 9.5 m/s intoequipment.What is the pressure dif-ference between inlet and outlet?

•59 Water is moving with a speed of 5.0 m/s through a pipewith a cross-sectional area of 4.0 cm2. The water graduall descends10 m as the pipe cross-sectional area increases to 8.0 cm2. (a) Whatis the speed at the lower level? (b) If the pressure at the upperlevel is 1.5 " 105 Pa, what is the pressure at the lower level?

•60 Models of torpedoes are sometimes tested in a horizontalpipe of flowing water, much as a wind tunnel is used to test modelairplanes. Consider a circular pipe of internal diameter 25.0 cm anda torpedo model aligned along the long axis of the pipe.The modelhas a 5.00 cm diameter and is to be tested with water flowing past itat 2.50 m/s. (a) With what speed must the water flow in the part ofthe pipe that is unconstricted by the model? (b) What will the pres-sure difference be between the constricted and unconstricted partsof the pipe?

•61 A water pipe having a 2.5 cm inside diameter carrieswater into the basement of a house at a speed of 0.90 m/s and apressure of 170 kPa. If the pipe tapers to 1.2 cm and rises to thesecond floor 7.6 m above the input point, what are the (a) speedand (b) water pressure at the second floor?

••62 A pitot tube (Fig. 14-48) is used to determine the air-speed of an airplane. It consists of an outer tube with a number ofsmall holes B (four are shown) that allow air into the tube; thattube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole A at the front end of the device, whichpoints in the direction the plane is headed. At A the air becomesstagnant so that vA ! 0. At B, however, the speed of the air pre-sumably equals the airspeed v of the plane. (a) Use Bernoulli’s

ILW

SSM

SSM

equation to show that

,

where r is the density of the liquid in the U-tube and h is the differ-ence in the liquid levels in that tube. (b) Suppose that the tube con-tains alcohol and the level difference h is 26.0 cm. What is theplane’s speed relative to the air? The density of the air is 1.03 kg/m3

and that of alcohol is 810 kg/m3.

v ! A 2#gh#air

Fig. 14-47 Problem 58.

Reservoir

Generatorbuilding

Outlet

IntakeD

air!

Hole A

Liquid

v

B

!h

B

Air

Fig. 14-48 Problems 62 and 63.

••63 A pitot tube (see Problem 62) on a high-altitude aircraftmeasures a differential pressure of 180 Pa. What is the aircraft’sairspeed if the density of the air is 0.031 kg/m3?

••64 In Fig. 14-49, water flowsthrough a horizontal pipe and thenout into the atmosphere at a speedv1 ! 15 m/s. The diameters of theleft and right sections of the pipe are5.0 cm and 3.0 cm. (a) What volumeof water flows into the atmosphereduring a 10 min period? In the left section of the pipe, what are (b)the speed v2 and (c) the gauge pressure?

••65 A venturi meter is used to measure the flowspeed of a fluid in a pipe. The meter is connected between twosections of the pipe (Fig. 14-50); the cross-sectional area A of theentrance and exit of the meter matches the pipe’s cross-sectionalarea. Between the entrance and exit, the fluid flows from thepipe with speed V and then through a narrow “throat” of cross-

WWWSSM

d 2v1v2 d 1

Fig. 14-49 Problem 64.

Aa

12

h

Pipe Pipe

A

Meterentrance

MeterexitVenturi meter

Manometer

v

V

Fig. 14-50 Problems 65 and 66.

halliday_c14_359-385hr.qxd 26-10-2009 21:41 Page 383

• We use Bernoulli’s Equation with point1 at the inlet and point 2 at the outlet.

p1 + 12ρv

21 + ρgy1 = p2 + 1

2ρv22 + ρgy2

• We’ll solve it for the pressure difference

p2 − p1 = ρgD + 12ρ(v21 − v22)

p2 − p1 = 1.7× 106 Pa

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Test 3 - Review

Test 3 - Information

• Test is Wednesday from 3:30 pm to 4:30 pm in LH-014

• Remember a calculator and formula sheet.

• Test covers Chapters 10, 11, 12, 13, 15 (Homeworks 8 to11)

• Three questions of multiple parts (Similar style anddifficulty as previous tests)

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Chapter 10 - Rotation

• Kinematics of Angular Variables - Replace linear variablewith corresponding angular variable. x→ θ, v → ω, anda→ α

• Rotational Kinetic Energy K = 12Iω

2

• Parallel Axis Theorem I = Icm +MR2

• Newton’s 2nd Law τnet = Iα

• Work-Kinetic Energy Thm. W = τ∆θ and ∆K = W

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Moments of Inertia

Some Common Moments of Inertia

25510-7 CALCU LATI NG TH E ROTATIONAL I N E RTIAPART 1

Table 10-2

Some Rotational Inertias

Axis

Hoop about central axis

Axis

Annular cylinder (or ring) about

central axis R

I = MR 2 (b) (a) I = M(R 12 + R 2

2)

R 2

R 1

Thin rod about axis through center

perpendicular to length

(e) I = ML 2

L

Axis

Axis Axis

Hoop about any diameter

Slab about perpendicular axis through

center

(i) (h) I = MR 2 I = M(a 2 + b 2)

R

b a

Axis

Solid cylinder (or disk) about

central axis

(c) I = MR 2

R L

Axis

Solid cylinder (or disk) about

central diameter

(d) I = MR 2 + ML 2

R L

Axis

Thin spherical shell

about any diameter

(g) I = MR 2

2R

Solid sphere about any diameter

(f) I = MR 2

2R

Axis

1 __ 2 1 __

2

2 __ 5

1 __ 4

2 __ 3

1 __ 2

1 __ 12

1 __ 12

1 __ 12

perpendicular to the plane of the figure, and another axis through point P parallelto the first axis. Let the x and y coordinates of P be a and b.

Let dm be a mass element with the general coordinates x and y. The rota-tional inertia of the body about the axis through P is then, from Eq. 10-35,

which we can rearrange as

(10-37)

From the definition of the center of mass (Eq. 9-9), the middle two integrals ofEq. 10-37 give the coordinates of the center of mass (multiplied by a constant)and thus must each be zero. Because x2 ! y2 is equal to R2, where R is the dis-tance from O to dm, the first integral is simply Icom, the rotational inertia of thebody about an axis through its center of mass. Inspection of Fig. 10-12 shows thatthe last term in Eq. 10-37 is Mh2, where M is the body’s total mass. Thus,Eq. 10-37 reduces to Eq. 10-36, which is the relation that we set out to prove.

I " ! (x2 ! y2) dm # 2a ! x dm # 2b ! y dm ! ! (a2 ! b2) dm.

I " ! r2 dm " ! [(x # a)2 ! ( y # b)2] dm,

Fig. 10-12 A rigid body in cross section,with its center of mass at O.The parallel-axistheorem (Eq. 10-36) relates the rotational in-ertia of the body about an axis through O tothat about a parallel axis through a pointsuch as P, a distance h from the body’s centerof mass. Both axes are perpendicular to theplane of the figure.

dm

r

P

h

a b

x – a

y – b

com O

Rotation axis through

center of mass

Rotation axis through P

y

x

We need to relate therotational inertia aroundthe axis at P to that aroundthe axis at the com.

CHECKPOINT 5

The figure shows a book-like object (one side islonger than the other) and four choices of rotationaxes, all perpendicular to the face of theobject. Rank the choices according to the rotationalinertia of the object about the axis, greatest first.

(1) (2) (3) (4)

halliday_c10_241-274hr2.qxd 29-09-2009 13:12 Page 255

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Chapter 11 - Rolling, Torque, and AngularMomentum

• Rolling Condition: vcm = ωR

• Torque: |τ | = r⊥F• Direction from right hand rule

• Angular momentum: ~̀= ~r × ~p = m(~r × ~v)

• Newton’s 2nd law:

~τnet =d~̀

dt

• Angular momentum of Rigid Body: L = Iω

• Conservation of Angular Momentum: ~Li = ~Lf

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Chapter 12 - Equilibrium and Elasticity

• Static Equilibrium occurs when∑~F = 0 and

∑~τ = 0

• When we look at forces in x and y direction, we get threeindependent equations from these two statements, sum ofthe forces in the x and y direction and sum of the ztorques.

• The key to solving equilibrium problems is a drawing veryclear free body diagrams. Don’t forget Newton’s ThirdLaw when drawing forces on free body diagrams.

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Chapter 13 - Gravitation

• Newton’s Law of Gravitation

F = Gm1m2

r2

where G = 6.67× 10−11 N ·m2/kg2. Gravity is always anattractive force.

• Gravitational Potential Energy

U = −GMm

r

• Escape speed: Speed at which an object will escape thegravitational pull of an astronomical body of mass M andradius R

vescape =

√2GM

R

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Chapter 13 - Kepler’s Laws

The law of orbits All planets move in elliptical orbits with theSun at one focus.

The law of areas A line joining any planet to the sun sweepsout equal areas in equal times. (This statement isequivalent to Conservation of AngularMomentum)

The law of periods

T 2 =

(4π2

GM

)r3

Phys. 131 -11/26/2012

DerekStampone

Outline

Review

Pascal’s andArchimedes’Principle

Fluids - Cont.

Ideal Fluids inMotion

The Equation ofContinuity

Bernoulli’sEquation

Test 3 -Review

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Chapter 15 - Oscillations

• The period is the time required for one complete oscillationor cycle. It is related to the frequency by T = 1/f

• Simple harmonic motion: When restoring force isproportional to the displacement.

• The Linear oscillator: Restoring force from Hooke’s LawF = −kx

ω =

√k

mT = 2π

√m

k

• Pendulums

T = 2π√I/κ torsion pendulum

T = 2π√L/g simple pendulum

T = 2π√I/mgh physical pendulum