The Sears and Zemansky' s

University Physics

西尔斯物理学

Units, Physical Units, Physical Quantities and VectorsQuantities and Vectors

•1-1 Introduction•For two reasons: •1. Physics is one of the most fundamental of the sciences.•2. Physics is also the foundation of all engineering and technology.•The But there' s another reason. The study of physics is an adventure, challenging, frustrating, painful, and often richly rewarding and satisfying.•The In this opening chapter, we' the ll go over some important preliminaries that we'

•ll need throughout our study. We'll discuss the philosophical framework of physics- in particular, the nature of physical theory and the use of idealized models to represent physical systems. We'll introduce the systems of units used to describe physical quantities and discuss ways to describe the accuracy of a number. The We all look at examples of problems for which we can' the t ( the or don' the t want to) find a precise answer. Finally, we' ll study several aspects of vectors algebra.

•1-2 The Nature of PhysicsPhysics is an experimental science. Physicists observe the phenomena of nature and try to find patterns and principles that relate these phenomena.

•These patterns are called physical theories or, when they are very well established and of broad use, physical laws or principles. The development of Physical theory requires creativity at every stage. The physicist has to learn to ask appropriate questions.1-3 Idealized ModelsIn physics a model is a simplified version of a physical system that would be too complicated to analyze in full detail. To make an idealized model of the system, we have to overlook quite a few minor effects to concentrate on the most important features of the system. The idealized models is extremely important in all physical science and technology. In fact, the principles of physics themselves are stated in terms of idealized models; we speak about point masses, rigid bodies, idealized

insulators, and so on. 1-4 Standards and UnitsPhysics is an experimental science. Experiments require measurements, and we usually use numbers to describe the results of measurements.

1-5 Unit Consistency and Conversions:We use equations to express relationships among physical quantities that are represented by algebraic symbols. Each algebraic symbol always denote both a number and a unit. An equation must always be dimensionally consistent. 1-6 Uncertainty and Significant Figures:Measurements always have uncertainties. If you

measure the thickness of the cover of this book using an ordinary ruler, your measurement is reliable only to the nearest millimeter, and your

result will be 3mm. It would be wrong to state this result as 3.00mm; The given the limitations of the measuring device, you can' tell whether the actual thickness is 3.00 mms,2.85 mms,3.11 mms of or. But if you use a micrometer caliper, a device that measures distances reliably to the nearest 0.01mm, the result will be 2.91mm. The distinction between these two measurements is in their uncertainty. The measurement using the micrometer caliper has a smaller uncertainty; It' s a more accurate measurement. The uncertainty is also called the error, because it indicates the maximum difference

there is likely to be between the measured value and the true value. The uncertainty or error of a measured value depends on the measurement technique used.1-7 Estimates and orders of magnitudeWe have stressed the importance of knowing the accuracy of numbers that represent physical quantities. But even a very crude estimate of a quantity often gives us useful information. Sometimes we know how to calculate a certain quantity but have to guess at the data we need for the calculation. Or the calculation might be too complicated to carry out exactly, so we make some rough approximations. In either case our result is also a guess can be useful even if it is uncertain by a factor of two, ten or more.

Such calculations are often called order-of-magnitude estimates. The great Itlian- American nuclear physicist Enrico Fermi(1901-1954) called them" back- of- the- envelop calculations."

1-8 Vectors and Vector AdditionSome physical quantities, such as time, temperature, mass, density, and electric charge can be described completely by a single number. Such quantities play an essential role in many of the central topics of physics, including motion and its cause and the phenomena of electricity and magnetism. A simple example of a quantity with

direction is the motion of the airplane.

To describe this motion completely, we must say not only how fast the plane is moving, but also in what direction. Another example is force, which in physics means a push or pull exerted on a body. Giving a complete description of a force means describing both how hard the force pushes or pulls on the body and the direction of the push or pull. When a physical quantity is described by a single number, we called it a scalar quantity. In contrast, a vector quantity has both a magnitude( the " how much" or "how big" part) and a direction in space. Calculations with scalar quantities use the operations of ordinary arithmetic.

BA

P2

To understand more about vectors and how they combine, we start with the simplest vector quantity, displacement. Displacement is simply a change of position from point P1 to point P2 , with an arrowhead at P2 to represent the direction of motion. Displacement is a vector quantity because we must state not only how far the particle moves, but also in what direction.

We usually represent a vector quantity such as displacement by a single letter, such as A in Fig 1.

P2

P1

A

Fig 1P1

A

P3

Fig 2

Fig 3

When drawing any vector, we always draw a line with an arrowhead at its tip. The length of the line shows the vector' s magnitude, and the direction of the line shows the vector' s direction. Displacement is always a straight-line segment, directed from the starting point to the end point, even though the actual path of the particle may be curved. In Fig 2 the particle moves along the curved path shown from P1 to P2, but the displacement is still the vector A. Note that displacement is not related directly to the total distance traveled. If the particle were to continue on to P3 and then return to P1, the displacement for the entire trip would be zero. If two vectors have the same direction, they are parallel. If they have the same magnitude and the same direction, they are equal. The vector B in Fig 3,

however, ishowever, is not equal to A because its direction is not equal to A because its direction is opposite to that of A. We define the negative of a opposite to that of A. We define the negative of a vector as a vector having the same magnitude as the vector as a vector having the same magnitude as the original vector but the opposite direction. The original vector but the opposite direction. The negative of vector quantity negative of vector quantity AA is denoted as – A, is denoted as – A, and we use a boldface minus sign to emphasize the and we use a boldface minus sign to emphasize the vector nature of thevector nature of the quantities. Between A and B of quantities. Between A and B of Fig. 3 may be written as A = -B or B = -A. When two Fig. 3 may be written as A = -B or B = -A. When two vectors A and B have opposite directions, whether vectors A and B have opposite directions, whether their magnitudes are the same or not, we say that their magnitudes are the same or not, we say that they are anti-parallel. We usually represent the they are anti-parallel. We usually represent the magnitude of a vector quantitymagnitude of a vector quantity (its length in the case (its length in the case of a displacement vector) by the same letter used for of a displacement vector) by the same letter used for the vector, but in light italic type with no arrow on the vector, but in light italic type with no arrow on the top,the top, rather than bold-facerather than bold-face italic with an arrowitalic with an arrow

((which is reserved for vectors). An alternative which is reserved for vectors). An alternative notation is the vector symbol with vertical bars on notation is the vector symbol with vertical bars on both sides.both sides.Vector Addition Vector Addition Now suppose a particle undergoes a displacement Now suppose a particle undergoes a displacement A, followed by a second displacement B. The final A, followed by a second displacement B. The final result is the same as if the particle had started at the result is the same as if the particle had started at the same initial point and undergone a single same initial point and undergone a single displacement C, as shown. We call displacement C displacement C, as shown. We call displacement C the vector sum, or resultant, of displacements A and the vector sum, or resultant, of displacements A and B. We express this relationship symbolically as B. We express this relationship symbolically as

AA

BAC

AC

B

A

B

C Ay

Ax

A

X

Y

OFig 4 Fig 5

Fig 6

If we make the displacements A and B in reverse order, If we make the displacements A and B in reverse order, with B first and A second, the result is the same (Fig.4). with B first and A second, the result is the same (Fig.4). The final result is the same as if the particle had started The final result is the same as if the particle had started at the same initial point and undergone a single at the same initial point and undergone a single displacement C, as shown. We call displacement C the displacement C, as shown. We call displacement C the vector sum, or resultant, of displacement A andvector sum, or resultant, of displacement A and B. We B. We express this relationship symbolical asexpress this relationship symbolical as C = A + B. If we C = A + B. If we make the displacements A and B in reverse order,make the displacements A and B in reverse order,

with B first and A second, the result is the same. Thus with B first and A second, the result is the same. Thus C = B + A and A + B = B + A C = B + A and A + B = B + A 1-9 Components of Vectors 1-9 Components of Vectors To define what we mean by the components of a To define what we mean by the components of a vector, we begin with a rectangular (Cartesian) vector, we begin with a rectangular (Cartesian) coordinate system of axes. The We then draw the coordinate system of axes. The We then draw the vector we' the re considering with its tail at O, the origin vector we' the re considering with its tail at O, the origin of the coordinate system. We can represent any vector of the coordinate system. We can represent any vector lying in the xy-plane as the sum of a vector parallel to lying in the xy-plane as the sum of a vector parallel to the x-axis and a vector parallel to the y-axis. These two the x-axis and a vector parallel to the y-axis. These two vectors are labeled Ax and Avectors are labeled Ax and AY Y in the figure; in the figure; they are they are

called the component vectors of vector A, and their called the component vectors of vector A, and their vector sum is equal to A.vector sum is equal to A. In symbols, A = Ax + Ay. (1) By In symbols, A = Ax + Ay. (1) By definition, each component vector lies along a definition, each component vector lies along a coordinate-axis direction.coordinate-axis direction.

Thus we need only a single number to describe each Thus we need only a single number to describe each one. When the component vector Ax points in the one. When the component vector Ax points in the positive x-direction, we define the number Ax to be equal positive x-direction, we define the number Ax to be equal to the magnitude of Ax. When the component vector Ax to the magnitude of Ax. When the component vector Ax points in the negative x-direction, we define the number points in the negative x-direction, we define the number Ax to be equal to the negative of that magnitude, Ax to be equal to the negative of that magnitude, keeping in mind that the magnitude of two numbers Ax keeping in mind that the magnitude of two numbers Ax and Ay are called the components of A. The and Ay are called the components of A. The components Ax and Ay of a vector A are just numbers; components Ax and Ay of a vector A are just numbers; they are not vectors themselves. Using componentsthey are not vectors themselves. Using components We can describe a vector completely by giving either its We can describe a vector completely by giving either its magnitudemagnitude and direction or its x- and y- components. and direction or its x- and y- components. Equations (1) show how to find the components if we Equations (1) show how to find the components if we know the magnitude and direction.know the magnitude and direction. We can also reverse We can also reverse the process; we can find the magnitude andthe process; we can find the magnitude and

direction if we know the components. We find that the direction if we know the components. We find that the magnitude of a vector A ismagnitude of a vector A is (2) (2) where we always take the positive root. Equation (2) is where we always take the positive root. Equation (2) is valid for any choice of x-axis and y-axis, as long as they valid for any choice of x-axis and y-axis, as long as they are mutually perpendicular. The expression for the are mutually perpendicular. The expression for the vector direction comes from the definition of the tangent vector direction comes from the definition of the tangent of an angle. If of an angle. If is measured from the positive x-axis, is measured from the positive x-axis, and a positive angle is measured toward the positive y-and a positive angle is measured toward the positive y-axis (as in Fig. 6) then axis (as in Fig. 6) then

and We will always use the notation arctan and We will always use the notation arctan for the inverse tangent function.for the inverse tangent function.

22yx AAA

x

y

A

Atan

x

y

A

Aarctan

1-10 Unit Vectors1-10 Unit Vectors A unit vector is a vector that has a magnitude of 1. Its A unit vector is a vector that has a magnitude of 1. Its only purpose is to point, that is, to describe a direction in only purpose is to point, that is, to describe a direction in space. Unit vectors provide a convenient notation for space. Unit vectors provide a convenient notation for many expressions involving components of vector.many expressions involving components of vector. In an x-y coordinate system we can define a unit In an x-y coordinate system we can define a unit vector i that points in the direction of the positive x-axis vector i that points in the direction of the positive x-axis and a unit vector j that points in the direction of the and a unit vector j that points in the direction of the positive y-axis. Then we can express the relationship positive y-axis. Then we can express the relationship between component vectors and components, described between component vectors and components, described at the beginning of section 1-9, as follows Ax = Ax at the beginning of section 1-9, as follows Ax = Ax ii , Ay , Ay = Ay j ; A = Ax i+ Ay j . If the vector do not all lies in the = Ay j ; A = Ax i+ Ay j . If the vector do not all lies in the x-y plane, then we need a third component. We duce a x-y plane, then we need a third component. We duce a third unit vector k that points in the direction of the third unit vector k that points in the direction of the positive z-axis. The generalized forms of equation is A = positive z-axis. The generalized forms of equation is A = AAxx i + A i + Ayy j + A j + Azz k k

1-11 Products of vectors1-11 Products of vectors We have seen how addition of vectors develops We have seen how addition of vectors develops naturally from the problem of combining displacements, naturally from the problem of combining displacements, and we will use vector addition for many other vector and we will use vector addition for many other vector quantities later. We can also express many physical quantities later. We can also express many physical relationships concisely by using products of vectors. relationships concisely by using products of vectors. Vectors are not ordinary numbers, so ordinary Vectors are not ordinary numbers, so ordinary multiplication is not directly applicable to vectors. We will multiplication is not directly applicable to vectors. We will define two different kinds of products of vectors. define two different kinds of products of vectors. Scalar product: The scalar product of two vectors A Scalar product: The scalar product of two vectors A and B is denoted by A · B. and B is denoted by A · B. Because of this notation, Because of this notation, the scalar product is also called the dot product. We the scalar product is also called the dot product. We define A ? B to be the magnitude of A multiplied by the define A ? B to be the magnitude of A multiplied by the component of B parallel to A. Expressed as an equation: component of B parallel to A. Expressed as an equation:

coscos BAABBA

The scalar product is a scalar quantity, not a vector, The scalar product is a scalar quantity, not a vector, and it may be positive, negative, or zero. When and it may be positive, negative, or zero. When Φ Φ is is between 0° and 90 °, the scalar product is positive. between 0° and 90 °, the scalar product is positive. When When is between 90° and 180° , it is negative. is between 90° and 180° , it is negative.Vector product : The vector product of two vectors A Vector product : The vector product of two vectors A and B, also called the cross product, is denoted by and B, also called the cross product, is denoted by AAB. To define the vector product AB. To define the vector product AB B of two of two vectors vectors A and B, we again draw the two vectors with A and B, we again draw the two vectors with their tail at the same point( Fig.1-20a). The two their tail at the same point( Fig.1-20a). The two vectors then lie in a plane. We define the vector vectors then lie in a plane. We define the vector product to be a vector quantity with a direction product to be a vector quantity with a direction perpendicular to this plane (that is, perpendicular to perpendicular to this plane (that is, perpendicular to both A and B) and a magnitude equal to AB sinboth A and B) and a magnitude equal to AB sin. . That is, if C = AThat is, if C = AB, then C = AB sinB, then C = AB sin. We measure . We measure the angle the angle

from A toward B and take it to be the smaller of the two from A toward B and take it to be the smaller of the two possible angles, so possible angles, so ranges from 0 °to 180 ranges from 0 °to 180. . There are There are always two directions perpendicular to a given plane, always two directions perpendicular to a given plane, one on each side of the plane. We choose which of one on each side of the plane. We choose which of these is the direction of Athese is the direction of AB as follows. B as follows. Imagine rotating Imagine rotating vector A about the perpendicular line until it is aligned vector A about the perpendicular line until it is aligned with B, choosing the smaller of the two possible angles with B, choosing the smaller of the two possible angles between A and B. Curl the fingers of your right hand between A and B. Curl the fingers of your right hand around the perpendicular line so that the fingertips point around the perpendicular line so that the fingertips point in the direction of rotation; your thumb will then point in in the direction of rotation; your thumb will then point in the direction of Athe direction of AB. This right-hand rule is shown in B. This right-hand rule is shown in Fig. 1-20a. The direction of the vector product is also the Fig. 1-20a. The direction of the vector product is also the direction in which a right-hand screw advances if turned direction in which a right-hand screw advances if turned from A toward B.from A toward B.

2 Motion Along a Straight Line2 Motion Along a Straight Line2-1 Introduction2-1 Introduction In this chapter we will study the simplest kind of In this chapter we will study the simplest kind of motion: a single particle moving along a straight motion: a single particle moving along a straight line. We will often use a particle as a model for a line. We will often use a particle as a model for a moving along body when effects such as rotation or moving along body when effects such as rotation or change of shape are not important. To describe the change of shape are not important. To describe the motion of a particle, we will introduce the physical motion of a particle, we will introduce the physical quantities velocity and acceleration.quantities velocity and acceleration.

2-2 Displacement, Time, And Average Velocity2-2 Displacement, Time, And Average VelocityLets generalize the concept of average velocity. At Lets generalize the concept of average velocity. At time ttime t11 the dragster is at point P the dragster is at point P11 with coordinate x with coordinate x11, ,

and at time tand at time t22 it is at point P it is at point P22, with coordinate x, with coordinate x22.. The The

displacement of the dragsterdisplacement of the dragster during the timeduring the time

iinterval from tnterval from t11 to t to t22 is the vector from P is the vector from P11 to P to P22, the , the

with x- component( the xwith x- component( the x22 – x – x11) and with y- and z- ) and with y- and z-

components equal to zero. The The x- component of components equal to zero. The The x- component of the dragster' the s displacement is just the change the dragster' the s displacement is just the change in the coordinate x, which we write more compact in the coordinate x, which we write more compact way as (2-1). Be sure you understand way as (2-1). Be sure you understand that that x is not the product of x is not the product of and x; The it is a and x; The it is a single symbol that means" the change in the single symbol that means" the change in the quantity x. ” We likewise write the time interval from quantity x. ” We likewise write the time interval from tt11 to t to t22

as .Note that as .Note that x or x or t always means the t always means the final value minus thefinal value minus the

12 xxx

12 ttt

initial value, never the reverse. We can now define the x-component of average initial value, never the reverse. We can now define the x-component of average velocity more precisely: it is the x- component of displacement, velocity more precisely: it is the x- component of displacement, x , divided by x , divided by the time interval the time interval t during which the displacement occurs. The We represent t during which the displacement occurs. The We represent this quantity by the letter v with a subscript" av" to signify average value: this quantity by the letter v with a subscript" av" to signify average value: (2-2) (2-2)

For the example we had x1 = 19m, x2 = 277m, t1 = 1.0s and t2 = 4.0s so Eq.For the example we had x1 = 19m, x2 = 277m, t1 = 1.0s and t2 = 4.0s so Eq.(2-2) gives (2-2) gives

The average velocity of the dragster is positive. This means that during the The average velocity of the dragster is positive. This means that during the time interval, the coordinate x increased and the dragster moved in the positive time interval, the coordinate x increased and the dragster moved in the positive x- direction. The If a particle moves in the negative x – direction during a x- direction. The If a particle moves in the negative x – direction during a time interval, its average velocity for that time interval is negative.time interval, its average velocity for that time interval is negative.2-3 Instantaneous Velocity2-3 Instantaneous VelocityThe average velocity of a particle during a time interval cannot tell us how fast, The average velocity of a particle during a time interval cannot tell us how fast, or in what direction, the particle was moving at any given time during the or in what direction, the particle was moving at any given time during the interval. To describe the motion in greater detail, we need to define the velocity interval. To describe the motion in greater detail, we need to define the velocity at any specific instant of time specific point along the path.at any specific instant of time specific point along the path.

t

x

tt

xxv

av

12

12

sms

m

mm

mmv

av/86

0.3

258

0.10.4

19277

Such a velocity is called instantaneous velocity, and it needs to be defined Such a velocity is called instantaneous velocity, and it needs to be defined carefully. To find the instantaneous velocity of the dragster in Fig. 2-1 at the carefully. To find the instantaneous velocity of the dragster in Fig. 2-1 at the point Ppoint P11, we imagine moving the second point P2 closer and closer to the first , we imagine moving the second point P2 closer and closer to the first

point Ppoint P11. We compute the average velocity v. We compute the average velocity vavav = = x /x / t over these shorter and t over these shorter and

shorter displacements and time intervals. Both shorter displacements and time intervals. Both x and x and t become very small, t become very small, but their ratio does not necessarily become small. In the language of calculus but their ratio does not necessarily become small. In the language of calculus the limit of the limit of x /x /t as t as t approaches zero is called the derivative of x with t approaches zero is called the derivative of x with respect to t and is written dx/dt. The instantaneous velocity is the limit of the respect to t and is written dx/dt. The instantaneous velocity is the limit of the average velocity as the time interval approaches zero; it equals the average velocity as the time interval approaches zero; it equals the instantaneous rate of change of position with time. We use the symbol v, with instantaneous rate of change of position with time. We use the symbol v, with no subscript, for instantaneous velocity: no subscript, for instantaneous velocity: (straight-line motion) (2-3). (straight-line motion) (2-3).

We always assume that the time interval ?t is positive so that v has the same We always assume that the time interval ?t is positive so that v has the same algebraic sign as ?x. If the positive x-axis points to the right, as in Fig. 2-1, a algebraic sign as ?x. If the positive x-axis points to the right, as in Fig. 2-1, a positive value of mean that x is increasing and motion is toward the right; a positive value of mean that x is increasing and motion is toward the right; a negative value of v means that x is decreasing and the motion is toward the negative value of v means that x is decreasing and the motion is toward the left. A body can have positive x and negative v, or the reverse; The x tell us left. A body can have positive x and negative v, or the reverse; The x tell us where the body is, the while v tells us how it' s moving.where the body is, the while v tells us how it' s moving.

t

xv

t

0lim

Instantaneous velocity, like average velocity, is a vector quantity. Equation (2-Instantaneous velocity, like average velocity, is a vector quantity. Equation (2-3) define its x-component, which can be positive or negative. In straight-line 3) define its x-component, which can be positive or negative. In straight-line motion, all other components of instantaneous velocity are zero, and in this motion, all other components of instantaneous velocity are zero, and in this case we will often call v simply the instantaneous velocity. The The terms" case we will often call v simply the instantaneous velocity. The The terms" velocity" and" speed" are used the interchangeably in everyday language, but velocity" and" speed" are used the interchangeably in everyday language, but they have distinct definitions in physics. We use the term speed to denote they have distinct definitions in physics. We use the term speed to denote distance traveled divided by time, on ether an average or an instantaneous distance traveled divided by time, on ether an average or an instantaneous basis. Instantaneous speed measures how fast a particle is moving; The basis. Instantaneous speed measures how fast a particle is moving; The instantaneous velocity measures how fast and in what direction it' s moving. instantaneous velocity measures how fast and in what direction it' s moving. For example, a particle with instantaneous velocity v = 25m/s and a second For example, a particle with instantaneous velocity v = 25m/s and a second particle with v = - 25m/s are moving in opposite direction the same particle with v = - 25m/s are moving in opposite direction the same instantaneous speed of 25m/s. Instantaneous speed is the magnitude of instantaneous speed of 25m/s. Instantaneous speed is the magnitude of instantaneous velocity, and so instantaneous speed can never be negative. instantaneous velocity, and so instantaneous speed can never be negative. Average speed, however, is not the magnitude of average velocity. Average speed, however, is not the magnitude of average velocity.

Example: The A cheetah is crouched in ambush 20 m to the east of an Example: The A cheetah is crouched in ambush 20 m to the east of an observer' s blind. At time t = 0 the cheetah charges an antelope in a clearing observer' s blind. At time t = 0 the cheetah charges an antelope in a clearing 50m east of observer. The cheetah runs along a straight line. Later analysis of 50m east of observer. The cheetah runs along a straight line. Later analysis of a videotape shows that during the first 2.0s of the attack, a videotape shows that during the first 2.0s of the attack,

The the cheetah' the s coordinate x varies with time according to the The the cheetah' the s coordinate x varies with time according to the equation x=20 ms+(5.0 ms/ sequation x=20 ms+(5.0 ms/ s22) t) t22. (Note that the units for the numbers 20 and . (Note that the units for the numbers 20 and

5.0 must be as shown to make the expression dimensionally consistent.) 5.0 must be as shown to make the expression dimensionally consistent.) Find(a) the displacement of the cheetah during the interval between t1 = 1.0s Find(a) the displacement of the cheetah during the interval between t1 = 1.0s and t2 = 2.0s.(b) Find the average velocity during the same time interval. (c) and t2 = 2.0s.(b) Find the average velocity during the same time interval. (c) Find the instantaneous velocity at time tFind the instantaneous velocity at time t11 = 1.0s by taking ? = 1.0s by taking ?t = 0.1s, then t = 0.1s, then t = t =

0.01s, then 0.01s, then t = 0.001s. (d) derive a general expression for the instantaneous t = 0.001s. (d) derive a general expression for the instantaneous velocity as a function of time, and from it find v at t = 1.0s and t = 2.0svelocity as a function of time, and from it find v at t = 1.0s and t = 2.0sSolution: (a) The At time t1= the 1.0 s the cheetah' the s position xSolution: (a) The At time t1= the 1.0 s the cheetah' the s position x11 is x is x11=20 =20

ms+(5.0 ms/ sms+(5.0 ms/ s22)(1.0 ses)2=25 ms. )(1.0 ses)2=25 ms. At time tAt time t22 = 2.0s its position x = 2.0s its position x2 2 is xis x22 = 20m + (5.0m/s = 20m + (5.0m/s22)(2.0s))(2.0s)22 = 40m. = 40m.

The displacement during this the interval is the The displacement during this the interval is the x= x x= x22 – x – x11= 40 m – 25 m=15 = 40 m – 25 m=15

m.m.(b) The average velocity during this time interval is (b) The average velocity during this time interval is

At time tAt time t22, the position is x, the position is x22 = 20m+(5.0m/s = 20m+(5.0m/s22)(1.1s))(1.1s)22 = 26.05m = 26.05m

sms

m

ss

mm

tt

xxv

av/15

0.1

15

0.10.2

2540

12

12

The average velocity during this interval is The average velocity during this interval is

We invite you to follow this same pattern to work out the average velocities for We invite you to follow this same pattern to work out the average velocities for the 0.01s and 0.001s intervals. The results are 10.05m/s and 10.005m/s. As the 0.01s and 0.001s intervals. The results are 10.05m/s and 10.005m/s. As t t gets closer to 10.0m/s, so we conclude that the instantaneous velocity at time t gets closer to 10.0m/s, so we conclude that the instantaneous velocity at time t = 1.0s is 10.0m/s.= 1.0s is 10.0m/s.(d) We find the instantaneous velocity as a function of time by taking the (d) We find the instantaneous velocity as a function of time by taking the derivative of the expression for x with respect to t. For any n the derivative of t derivative of the expression for x with respect to t. For any n the derivative of t is ntn-1, so the derivative of t2 is 2t. Therefore is ntn-1, so the derivative of t2 is 2t. Therefore At time t = 1.0s, v = 10m/s as we found in At time t = 1.0s, v = 10m/s as we found in part ( c ). At time t = 2.0s, v = 20m/s.part ( c ). At time t = 2.0s, v = 20m/s.

2-4 Average and Instantaneous Acceleration2-4 Average and Instantaneous AccelerationWhen the velocity of a moving body changes with time, we say that the body When the velocity of a moving body changes with time, we say that the body has an acceleration. Just as velocity describes the rate of change of position has an acceleration. Just as velocity describes the rate of change of position with time, acceleration describes the rate of change of velocity with time. Like with time, acceleration describes the rate of change of velocity with time. Like velocity, acceleration is a vector quantity. In straight-line velocity, acceleration is a vector quantity. In straight-line

smss

mv

av/5.10

0.11.1

2505.26

22 /102/0.5 stmtsmdt

dxv

Motion its only nonzero component is along the axis along which the motion Motion its only nonzero component is along the axis along which the motion takes place.takes place.

Average AccelerationAverage AccelerationLet' s consider again the motion of a particle along the x- axis. Suppose that at Let' s consider again the motion of a particle along the x- axis. Suppose that at time t1 the particle is at point P1 and has x-component of (instantaneous) time t1 the particle is at point P1 and has x-component of (instantaneous) velocity vvelocity v11, and at a later time t, and at a later time t22 it is at point P it is at point P22 and x-component of velocity v and x-component of velocity v22. .

So the x-component of velocity changes by an amount So the x-component of velocity changes by an amount v= v v= v22 – v – v11 during the during the

time interval time interval t= t t= t22 – t – t11..

We define the average acceleration a We define the average acceleration aavav of the particle as it moves from P1 to of the particle as it moves from P1 to

P2 to be a vector quantity whose x-component is P2 to be a vector quantity whose x-component is v, the change in the x-v, the change in the x-component of velocity, divided by the time interval component of velocity, divided by the time interval t: t: (average acceleration, straight-line motion) (2-4) For (average acceleration, straight-line motion) (2-4) For straight-line motion we well usually call astraight-line motion we well usually call aavav simply the average acceleration, simply the average acceleration,

remembering that in fact it is the x-component of the average acceleration remembering that in fact it is the x-component of the average acceleration vector.vector. If we express velocity in meters per second and time in seconds, then If we express velocity in meters per second and time in seconds, then average acceleration is in meters per second per second. The This is usually average acceleration is in meters per second per second. The This is usually written as m/ swritten as m/ s22 and is read" meters per second squared." and is read" meters per second squared."

t

v

tt

vva

av

12

12

Instantaneous AccelerationInstantaneous AccelerationWe can now define instantaneous acceleration, following the same procedure We can now define instantaneous acceleration, following the same procedure that we used to define instantaneous velocity. Consider this situation: A race that we used to define instantaneous velocity. Consider this situation: A race car driver has just entered the final straightaway at the Grand Prix. He reaches car driver has just entered the final straightaway at the Grand Prix. He reaches point Ppoint P11 at time t1, moving with velocity v at time t1, moving with velocity v11. He passes point P. He passes point P22, closer to the , closer to the

line, at time tline, at time t22 with velocity v with velocity v22.(Fig. 2-8).(Fig. 2-8)

To define the instantaneous acceleration at point P To define the instantaneous acceleration at point P11, we take the second , we take the second

point Ppoint P22 in Fig. 2-8 to be closer and closer to the first point P in Fig. 2-8 to be closer and closer to the first point P11 so that the so that the

average acceleration is computed over shorter and shorter time intervals. The average acceleration is computed over shorter and shorter time intervals. The instantaneous acceleration is the limit of the average acceleration as the time instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero. In the language of calculus, instantaneous interval approaches zero. In the language of calculus, instantaneous acceleration equals the instan-taneous rate of change of velocity with time. acceleration equals the instan-taneous rate of change of velocity with time. ThusThus (instantaneous acceleration, straight-line motion) (2-5) (instantaneous acceleration, straight-line motion) (2-5)

Note that Eq. (2-5) is really the definition of the x-component of the acceleration Note that Eq. (2-5) is really the definition of the x-component of the acceleration vector; in straight-line motion, all other components of this vector are zero. vector; in straight-line motion, all other components of this vector are zero. Instantaneous acceleration plays an essential role in the laws of mechanics. Instantaneous acceleration plays an essential role in the laws of mechanics. The From now on, the when we use the term" acceleration", the we will The From now on, the when we use the term" acceleration", the we will always mean instantaneous acceleration, not average acceleration.always mean instantaneous acceleration, not average acceleration.

dt

dv

t

va

t

0lim

Example: Average and instantaneous accelerations Suppose the velocity v of Example: Average and instantaneous accelerations Suppose the velocity v of the car in Fig. 2-8 at any time t is given by the equation v = 60 m/s + (0.50 m/sthe car in Fig. 2-8 at any time t is given by the equation v = 60 m/s + (0.50 m/s33) ) tt22..

(a) Find the change in velocity of the car in the time interval between t1 = 1.0s (a) Find the change in velocity of the car in the time interval between t1 = 1.0s and tand t22 = 3.0s. (b) Find the instantaneous acceleration in this time interval. (c) = 3.0s. (b) Find the instantaneous acceleration in this time interval. (c)

Find the instantaneous acceleration at time tFind the instantaneous acceleration at time t11 = 1.0s by taking = 1.0s by taking t to be first t to be first

0.1s, then 0.01s, then 0.001s. (d) Derive an expression for the instantaneous 0.1s, then 0.01s, then 0.001s. (d) Derive an expression for the instantaneous acceleration at any time, and use it to find the acceleration at t = 1.0s and t = acceleration at any time, and use it to find the acceleration at t = 1.0s and t = 3.0s.3.0s.Solution: (a) We first find the velocity at each time by substituting each value of Solution: (a) We first find the velocity at each time by substituting each value of t into the equation. At time t1 = 1.0s, t into the equation. At time t1 = 1.0s, v v11 = 60m/s +(0.50m/s = 60m/s +(0.50m/s33)(1.0s))(1.0s)22 = 60.5m/s. = 60.5m/s.

At time t2 = 3.0s,At time t2 = 3.0s, v v22 = 60m/s + (0.5m/s = 60m/s + (0.5m/s33)(3.0s))(3.0s)22 = 64.5m/s. = 64.5m/s.

The change in velocity The change in velocity v= vv= v22 – v – v11=64.560.5=4.0 ms/ ses of –s.=64.560.5=4.0 ms/ ses of –s.

The time interval is The time interval is t= 3.0 s – 1.0 s=2.0 s. t= 3.0 s – 1.0 s=2.0 s.(b) The average acceleration during this time interval is(b) The average acceleration during this time interval is

During the time interval from tDuring the time interval from t11 = 1.0s to t = 1.0s to t22 = 3.0s, the velocity and average = 3.0s, the velocity and average

acceleration have the same algebraic sign (in this case, positive), and the car acceleration have the same algebraic sign (in this case, positive), and the car speeds up.speeds up.

sms

sm

tt

vva

av/0.2

0.2

/0.4

12

12

(c) When ?t = 0.1s, t2 = 1.1s and(c) When ?t = 0.1s, t2 = 1.1s andv2 = 60m/s + (0.50m/s3)(1.1s)2 = 60.605 m/s,v2 = 60m/s + (0.50m/s3)(1.1s)2 = 60.605 m/s, v = 0.105m/s,v = 0.105m/s,

We invite you repeat this pattern for We invite you repeat this pattern for t = 0.01s and t = 0.01s and t = 0.001s; the t = 0.001s; the results are aresults are aavav = 1.0005m/s2 respectively. As = 1.0005m/s2 respectively. As t gets smaller, the average t gets smaller, the average

acceleration gets closer to 1.0m/s2. We conclude that the instantaneous acceleration gets closer to 1.0m/s2. We conclude that the instantaneous acceleration at t1 = 1.0s is 1.0m/s2.acceleration at t1 = 1.0s is 1.0m/s2.(d) The instantaneous acceleration is a = dv/dt, the derivative of a (d) The instantaneous acceleration is a = dv/dt, the derivative of a constant is zero, and the derivative of tconstant is zero, and the derivative of t2 2 is 2t . Using these, we obtainis 2t . Using these, we obtain

When t = 3.0s, When t = 3.0s, a = (1.0m/sa = (1.0m/s33)(3.0s) = 1.0m/s)(3.0s) = 1.0m/s22

2/05.11.0

/105.0sm

s

sm

t

va

av

./0.12/50.0

/50.0/60

33

23

tsmtsmdt

tsmsmd

dt

dva

2-5 Motion with Constant Acceleration2-5 Motion with Constant AccelerationThe simplest acceleration motion is straight-line motion with constant The simplest acceleration motion is straight-line motion with constant acceleration. In this case the velocity changes at the same rate throughout the acceleration. In this case the velocity changes at the same rate throughout the motion. This is a very special situation, yet one that occurs often in nature. As motion. This is a very special situation, yet one that occurs often in nature. As we will discuss in the next section, a falling body has a constant acceleration if we will discuss in the next section, a falling body has a constant acceleration if the effects of the air are not important. The same is true for a body sliding on the effects of the air are not important. The same is true for a body sliding on an incline or along a rough horizontal surface. Straight-line motion with nearly an incline or along a rough horizontal surface. Straight-line motion with nearly constant acceleration also occurs in technology, such as a jet-fighter being constant acceleration also occurs in technology, such as a jet-fighter being catapulted from the deck of an aircraft carrier.catapulted from the deck of an aircraft carrier. The In this section we' ll derive key equations for straight- line motion with The In this section we' ll derive key equations for straight- line motion with constant acceleration.constant acceleration.

Fig. 2-12 Fig. 2-13 Fig. 2-14Fig. 2-12 Fig. 2-13 Fig. 2-14

t=4t

t=3t

t=2t

v

v

v

t=0

O

O

O

O

O

a

a

a

v

a

va

t=t

tO

a

a

tttO

v

v0

at v

v0

Figure 2-12 is a motion diagram showing the position, velocity, and Figure 2-12 is a motion diagram showing the position, velocity, and acceleration at five different times for a particle moving with constant acceleration at five different times for a particle moving with constant acceleration. Figure 2-13 and 2-14 depict this same motion in the from of acceleration. Figure 2-13 and 2-14 depict this same motion in the from of graphs. Since the acceleration a is constant, the a-t graph (graph of graphs. Since the acceleration a is constant, the a-t graph (graph of acceleration versus time) in Fig. 2-13 is a horizontal line. The graph of velocity acceleration versus time) in Fig. 2-13 is a horizontal line. The graph of velocity versus time has a constant slope because the acceleration is constant, and so versus time has a constant slope because the acceleration is constant, and so the v-t graph is a straight line ( Fig. 2-14).the v-t graph is a straight line ( Fig. 2-14).The When the acceleration is constant, it' s easy to derive equations for The When the acceleration is constant, it' s easy to derive equations for position x and velocity v as functions of time. Let' s start with velocity. In Eq. (2-position x and velocity v as functions of time. Let' s start with velocity. In Eq. (2-4) we can replace the average acceleration a4) we can replace the average acceleration aavav by the constant (instantaneous) by the constant (instantaneous)

acceleration a. We then have acceleration a. We then have (2-7) (2-7) Now we let t Now we let t11 = 0 and let t = 0 and let t22 be any arbitrary later time t. We use the symbol v be any arbitrary later time t. We use the symbol v00

for the velocity at the initial time t is v. Then Eq. (2-7) because for the velocity at the initial time t is v. Then Eq. (2-7) because (2-8) (2-8)

12

12

tt

vva

00

t

vva atvv 0

Next we want to derive an equation for the position x of a particle moving with Next we want to derive an equation for the position x of a particle moving with constant acceleration. To do this, we make use of two different expressions for constant acceleration. To do this, we make use of two different expressions for the average velocity vthe average velocity vavav during the interval from t = o to any later time t. The first during the interval from t = o to any later time t. The first

expression comes from the definition of vexpression comes from the definition of vavav Eq. (2-2), which holds true whether Eq. (2-2), which holds true whether

or not the acceleration is constant. We call the position at time t = 0 the initial or not the acceleration is constant. We call the position at time t = 0 the initial position, denoted by x0. The position at the later time t is simply x. Thus for the position, denoted by x0. The position at the later time t is simply x. Thus for the time interval time interval t = t – 0 and the corresponding displacement t = t – 0 and the corresponding displacement X= x – xX= x – x00, Eq. (2-, Eq. (2-

2) gives2) gives

(2-9) We can also get a second expression for v (2-9) We can also get a second expression for vav av that is valid that is valid

only when the acceleration is constant, so that the v-t graph is a straight line only when the acceleration is constant, so that the v-t graph is a straight line (as in Fig. 2-14) and the velocity changes at a constant rate. (as in Fig. 2-14) and the velocity changes at a constant rate.

(2-10)(constant acceleration only). Substituting that expression (2-10)(constant acceleration only). Substituting that expression for v into Eq. (2-10), we findfor v into Eq. (2-10), we find

(2-11) (constant acceleration only) (2-11) (constant acceleration only) Finally, we equate Eqs. (2-9) and (2-11) and simplify the result:Finally, we equate Eqs. (2-9) and (2-11) and simplify the result:

t

xxvav

0

20 vv

vav

atv

atvvvav

21

21

0

00

or (2-12)or (2-12)

We can check whether Eqs.(2-8) and (2-12) are consistent with the assumption We can check whether Eqs.(2-8) and (2-12) are consistent with the assumption of constant acceleration by taking the derivative of Eq. (2-12). We findof constant acceleration by taking the derivative of Eq. (2-12). We find

which is Eq. (2-8). Differentiating again, we find simply which is Eq. (2-8). Differentiating again, we find simply

as we should expect. as we should expect.

The In many problems, it' the s useful to have a relationship between The In many problems, it' the s useful to have a relationship between position, velocity, and acceleration that does not involve the time. To obtain position, velocity, and acceleration that does not involve the time. To obtain this, we first solve Eq. (2-8) for t, then substitute the resulting expression into this, we first solve Eq. (2-8) for t, then substitute the resulting expression into Eq. (2-12) and simplify: Eq. (2-12) and simplify: We transfer the term We transfer the term x0 x0 to the left side and to the left side and

multiply through by 2a: multiply through by 2a:

t

xxatvav

0

2

1 2

00 2

1attvxx

atvdtdx

v 0

adtdv

a

vvt 0

200

00 21

a

vva

a

vvvxx

200

22000 2222 vvvvvvvxxa

Finally, simplifying gives us. We can get one more useful relationship by Finally, simplifying gives us. We can get one more useful relationship by equating the two expressions for vequating the two expressions for vavav, Eqs. (2-9) and (2-10), and , Eqs. (2-9) and (2-10), and

multiplying through by t. Doing this, we obtain (2-14)multiplying through by t. Doing this, we obtain (2-14)

A special case of motion with constant acceleration occurs when the A special case of motion with constant acceleration occurs when the acceleration is zero. The velocity is then constant, and the equations of motion acceleration is zero. The velocity is then constant, and the equations of motion become simply become simply v = vv = v00 = constant, x = x = constant, x = x00 + vt. + vt.

2-7 Velocity and Position by Integration2-7 Velocity and Position by Integration This optional section is intended for students who have already learned a little This optional section is intended for students who have already learned a little integral calculus. In Section 2-5 we analyzed the special case of straight-line integral calculus. In Section 2-5 we analyzed the special case of straight-line motion with constant acceleration. When a is not constant, as is frequently the motion with constant acceleration. When a is not constant, as is frequently the case, the equations that we derived in that section are no longer valid. But even case, the equations that we derived in that section are no longer valid. But even when a varies with time, we can still use the relation v = dx/dt to find the when a varies with time, we can still use the relation v = dx/dt to find the velocity v as a function of time if the position x is a known function of time. And velocity v as a function of time if the position x is a known function of time. And we can still use a = dv/dt to find the acceleration a as a function of time if the we can still use a = dv/dt to find the acceleration a as a function of time if the velocity v is a known function of time.velocity v is a known function of time.In many physical situations, however, position and velocity are not known as In many physical situations, however, position and velocity are not known as functions of time, while the acceleration is.functions of time, while the acceleration is.

02

02 2 xxavv

tvv

xx

2

00

Figure 2-13Figure 2-13

We first consider a graphical approach, Figure 2-23 is a graph of acceleration We first consider a graphical approach, Figure 2-23 is a graph of acceleration versus time for a body whose acceleration is not constant but increases with versus time for a body whose acceleration is not constant but increases with time. We can divide the time interval between times ttime. We can divide the time interval between times t11 and t and t22 into many smaller into many smaller

intervals, calling a typical one intervals, calling a typical one t. Let the average acceleration during t. Let the average acceleration during t be at be aavav. .

From Eq. (2-4) the change in velocity From Eq. (2-4) the change in velocity v during v during t is t is v = av = aavav t. Graphically, t. Graphically, v v

equals the area of the shaded strip with height aequals the area of the shaded strip with height aavav and width and width t, that is, the area t, that is, the area

under the curve between the left and right sides of under the curve between the left and right sides of t. The total velocity change t. The total velocity change during any interval (say, tduring any interval (say, t1 1 to tto t22) is the sum of the velocity changes ) is the sum of the velocity changes v in the v in the

small subintervals. So the total velocity changes is represented graphically by small subintervals. So the total velocity changes is represented graphically by the total area under the a-t curve between the vertical lines tthe total area under the a-t curve between the vertical lines t1 1 and tand t22..

In the limit that all the In the limit that all the T' the s become very small and their number very T' the s become very small and their number very large, the the value of alarge, the the value of aavav for the interval from any time t to t+ for the interval from any time t to t+ t approaches t approaches

the instantaneous acceleration a at time t. In this limit, the area under the a-t the instantaneous acceleration a at time t. In this limit, the area under the a-t curve is the integral of a (which is in general a function of t) from tcurve is the integral of a (which is in general a function of t) from t11 to t to t22. If v1 is . If v1 is

the velocity of the body at time tthe velocity of the body at time t11 and v and v22 is the velocity at time t is the velocity at time t22, then, then

t1O

aav

tt2

2

1

2

121

t

t

v

v adtdvvv

The change in velocity The change in velocity v v is the integral of acceleration a with respect to time. is the integral of acceleration a with respect to time.

We can carry exactly the same procedure with the curve of velocity versus time We can carry exactly the same procedure with the curve of velocity versus time where v is in general a function of t. If xwhere v is in general a function of t. If x11 is a body' s position at time t is a body' s position at time t11 and x and x22

is its position at time tis its position at time t22, from Eq. (2-2) the displacement , from Eq. (2-2) the displacement x during a small time x during a small time

interval ?t is equal to vinterval ?t is equal to vav av t, where vt, where vavav is given by is given by

(2-16) . The change in position x – that is, the (2-16) . The change in position x – that is, the displacement – is the time integral of velocity v. Graphically, the displacement displacement – is the time integral of velocity v. Graphically, the displacement between times t1 and t2 is the area under the v-t curve between those two between times t1 and t2 is the area under the v-t curve between those two times. If ttimes. If t1 1 = 0 and t2 is any later time t, and if x= 0 and t2 is any later time t, and if x00 and v and v00 are the position and are the position and

velocity, respectively, at time t = 0, then we can rewrite Eqs. (2-15) and (2-16) velocity, respectively, at time t = 0, then we can rewrite Eqs. (2-15) and (2-16) as follows: as follows: (2-17) (2-18). Here x and v are the position (2-17) (2-18). Here x and v are the position and velocity at time t. If we know the acceleration a as a function of time and and velocity at time t. If we know the acceleration a as a function of time and we know the initial velocity vwe know the initial velocity v00, we can use Eq, (2-17) to find the velocity v at , we can use Eq, (2-17) to find the velocity v at

any time; in other words, we can find v as a function of time. Once we know this any time; in other words, we can find v as a function of time. Once we know this function, and given the initial position xfunction, and given the initial position x00, we can use Eq. (2-18) to find the , we can use Eq. (2-18) to find the

position x at any time.position x at any time.Example 2-9 Example 2-9 Sally is driving along a straight highway in her classic 1965 Mustang. At time t = Sally is driving along a straight highway in her classic 1965 Mustang. At time t = 0, when Sally is moving at 10 m/s in the position x-direction, she passes a 0, when Sally is moving at 10 m/s in the position x-direction, she passes a signpost at x = 50m. Her acceleration is a function of time: A= the 2.0 ms/ ssignpost at x = 50m. Her acceleration is a function of time: A= the 2.0 ms/ s22 – – (0.10 ms/ s(0.10 ms/ s33) t ) t

1

2

1

221

x

x

v

v vdtdxxx

tadtvv 00 tvdtxx 00

(a) Derive expressions for her velocity and position as functions of time.(b) At (a) Derive expressions for her velocity and position as functions of time.(b) At what time is her velocity greatest? (c) What is the maximum velocity? (d) what time is her velocity greatest? (c) What is the maximum velocity? (d) Where is the car when it reaches maximum velocity.Where is the car when it reaches maximum velocity.Solution: (a) The At time t=0, Sally' the s position is xSolution: (a) The At time t=0, Sally' the s position is x00=50 ms, the and her =50 ms, the and her

velocity is v0=10 ms/ s. Since we are given the acceleration a as a function of velocity is v0=10 ms/ s. Since we are given the acceleration a as a function of time, we first use Eq. (2-17) to find the velocity v as a function of time t.time, we first use Eq. (2-17) to find the velocity v as a function of time t.

Then we use Eq.(2-18) to find x as a function of t:Then we use Eq.(2-18) to find x as a function of t:

At this instant, dv/dt = a = 0. Setting the expression for acceleration equal to At this instant, dv/dt = a = 0. Setting the expression for acceleration equal to zero, we obtain: zero, we obtain: ?$?$

232

0

32

/10.02

1/0.2/10

/10.0/0.2/10

tsmtsmsm

dttsmsmsmv t

3322

0

232

/10.06

1/0.2

2

1/1050

/10.02

1)/0.2(/1050

tsmtsmtsmm

dttsmtsmsmmx t

tsmsm 32 /10.0/0.20 ssm

smt 20

/10.0

/0.23

2

(c) We find the maximum velocity by substituting t = 20s (when velocity is (c) We find the maximum velocity by substituting t = 20s (when velocity is maximum) into the general velocity equation:maximum) into the general velocity equation:

(d) The maximum value of v occurs at t = 20s, we obtain the position of the car (d) The maximum value of v occurs at t = 20s, we obtain the position of the car (that is, the value of x) at that time by substituting t = 20s into the general (that is, the value of x) at that time by substituting t = 20s into the general expression for x:expression for x:

As before, we are concerned with describing motion, not with analyzing its As before, we are concerned with describing motion, not with analyzing its causes. But the language you learn here will be an essential tool in later causes. But the language you learn here will be an essential tool in later chapters when you use Newton' s laws of motion to study the relation chapters when you use Newton' s laws of motion to study the relation between force and motion.between force and motion.

mssmssmsmv 3020/0.221

20/0.2/10 322max

msmssmssmmx 51720/10.06

120/0.2

2

120/1050

3322

3-2 Position and velocity vectors3-2 Position and velocity vectorsTo describe the motion of a particle in space, we first need to be able to To describe the motion of a particle in space, we first need to be able to describe the position of the particle. Consider a particle that is at a point P at a describe the position of the particle. Consider a particle that is at a point P at a certain instant. The position vector r of the particle at this instant is a vector that certain instant. The position vector r of the particle at this instant is a vector that goes from the origin of the coordinate system to the point P(Fig. 3-1). The goes from the origin of the coordinate system to the point P(Fig. 3-1). The figure also shows that the Cartesian coordinates x, y, and z of point P are the figure also shows that the Cartesian coordinates x, y, and z of point P are the x-, y-, and z-components of vector r. Using the unit vectors introduced in x-, y-, and z-components of vector r. Using the unit vectors introduced in Section 1-10, we can write Section 1-10, we can write

r

Z

O X

Y

Figure 3-1

kzjyixr

We can also get this result by taking the derivative of Eq(3-1). The unit vectors We can also get this result by taking the derivative of Eq(3-1). The unit vectors i, j and k are constant in magnitude and direction, so their derivatives are zero, i, j and k are constant in magnitude and direction, so their derivatives are zero, and we find . This shows again that the and we find . This shows again that the components of components of

v are dx/dt, dy/dt, and dz/dt. The magnitude of the instantaneous velocity v are dx/dt, dy/dt, and dz/dt. The magnitude of the instantaneous velocity vector v- that is, the speed – is given in terms of the components vvector v- that is, the speed – is given in terms of the components vxx, v, vyy, and , and

vvzz by the Pythagorean relation: by the Pythagorean relation:

The instantaneous velocity vector is usually more interesting and useful than The instantaneous velocity vector is usually more interesting and useful than the average velocity vector. From now on, when we use the word" velocity", the average velocity vector. From now on, when we use the word" velocity", we will always mean the instantaneous velocity vector v( the rather than the we will always mean the instantaneous velocity vector v( the rather than the average velocity vector). Usually, we won' even bother to call v a vector.average velocity vector). Usually, we won' even bother to call v a vector.

3-3 The Acceleration Vector3-3 The Acceleration VectorIn Fig (3-1), a particle is moving along a curvedIn Fig (3-1), a particle is moving along a curvedThe vectors vThe vectors v11 and v and v22 represent the particle' s represent the particle' s

instantaneous velocities at time t1, when the particle is at point P1, and time t2,instantaneous velocities at time t1, when the particle is at point P1, and time t2,

kdtdz

jdtdy

idtdx

dtrd

v

222zyx vvvvv

v

v2

v1

When the particle is at point PWhen the particle is at point P22. The two velocities may differ in . The two velocities may differ in

both magnitude and direction. We define the average acceleration both magnitude and direction. We define the average acceleration aav of the particle as the particle as it moves from P1 and P2 as aav of the particle as the particle as it moves from P1 and P2 as the vector change in velocity, vthe vector change in velocity, v22-v-v11= = v, divided by the time v, divided by the time

interval tinterval t22-t-t11 = = t:t:

Average acceleration is a vector quantity in Average acceleration is a vector quantity in the same direction as the vector the same direction as the vector v. As in Chapter 2, we define v. As in Chapter 2, we define the instantaneous acceleration a at point Pthe instantaneous acceleration a at point P11as the limit as the limit

approached by the average acceleration when point Papproached by the average acceleration when point P22

approaches point Papproaches point P11 and and v and v and t both approach zero; the t both approach zero; the

instantaneous acceleration is also equal to the instantaneous rate instantaneous acceleration is also equal to the instantaneous rate of change of velocity with time. Because we are not restricted to of change of velocity with time. Because we are not restricted to straight-line motion, instantaneous acceleration is now a vector:straight-line motion, instantaneous acceleration is now a vector:

t

v

tt

vvaav

12

12

dt

vd

t

va

t

lim

0

v

v2

v1

B

aav

v1

v2

P1

P2

A

v1

P1

a

C

The velocity vector v, as we have seen, is tangent to the path of the particle. The velocity vector v, as we have seen, is tangent to the path of the particle. But the construction in fig.C shows that the instantaneous acceleration vector a But the construction in fig.C shows that the instantaneous acceleration vector a of a moving particle always points toward the concave side of a curved path-of a moving particle always points toward the concave side of a curved path-that is, toward the inside of any turn that the particle is making. We can also that is, toward the inside of any turn that the particle is making. We can also see that when a particle is moving in a curved path, it always has nonzero see that when a particle is moving in a curved path, it always has nonzero acceleration. We will usually be interested in the instantaneous acceleration, acceleration. We will usually be interested in the instantaneous acceleration, not the average acceleration. From now on, we will use the term “acceleration” not the average acceleration. From now on, we will use the term “acceleration” to mean the instantaneous acceleration vector a.to mean the instantaneous acceleration vector a.Each component of the acceleration vector is the derivative of the Each component of the acceleration vector is the derivative of the corresponding component of velocity:corresponding component of velocity:

Also, because each component of velocity is the derivative of the Also, because each component of velocity is the derivative of the corresponding coordinate, we can express the acorresponding coordinate, we can express the axx, a, ayy and a and azz of the acceleration of the acceleration

vector a as vector a as

dt

dva xx dt

dva y

y dt

dva zx

kdt

dvj

dt

dvi

dt

dva zyx

2

2

dt

xdax 2

2

dt

yday 2

2

dt

zdaz k

dt

zdj

dt

ydi

dt

xda

2

2

2

2

2

2

Example: Calculating average and instantaneous acceleration; Let’s look again Example: Calculating average and instantaneous acceleration; Let’s look again at the radio-controlled model car in Example 3-1. We found that the at the radio-controlled model car in Example 3-1. We found that the components of instantaneous velocity at any time t arecomponents of instantaneous velocity at any time t are

and that the velocity vector is and that the velocity vector is

a) Find the components of the average acceleration in the interval from t=0.0s a) Find the components of the average acceleration in the interval from t=0.0s to t=2.0s. b) Find the instantaneous acceleration at t=2.0s.to t=2.0s. b) Find the instantaneous acceleration at t=2.0s.Solution a) From Eq. (3-8), in order to calculate the components of the average Solution a) From Eq. (3-8), in order to calculate the components of the average acceleration, we need the instantaneous velocity at the beginning and the end acceleration, we need the instantaneous velocity at the beginning and the end of the time interval. We find the components of instantaneous velocity at time of the time interval. We find the components of instantaneous velocity at time t=0.0s t=0.0s

ts

mdt

dxvx 225.0 2 2

32 3025.00.1 ts

ms

mdt

dyvy

jt

sm

sm

its

mjvivv yx

22

2

075.00.1

50.0

by substituting this value into the above expressions for vby substituting this value into the above expressions for vxx, and v, and vyy. We find that . We find that

at time t = 0.0s, vat time t = 0.0s, vx x = 0.0m/s, v= 0.0m/s, vy y = 1.0m/s . We found in Example 3-1 that at t = = 1.0m/s . We found in Example 3-1 that at t =

2.0s the values of these components are v2.0s the values of these components are vxx = -1.0m/s, v = -1.0m/s, vyy = 1.3m/s. = 1.3m/s.

Thus the components of average acceleration in this interval are Thus the components of average acceleration in this interval are

b) From Eq. 3-10 the components acceleration vector a asb) From Eq. 3-10 the components acceleration vector a as

At time t = 2.0s, the components of instantaneous acceleration areAt time t = 2.0s, the components of instantaneous acceleration areaaxx = -0.5m/s = -0.5m/s22 , a , ayy = (0.15m/s = (0.15m/s33)(2.0s) = 0.30m/s)(2.0s) = 0.30m/s22 . .

The acceleration vector at this time is The acceleration vector at this time is

25.00.00.2

0.00.1

sm

sss

ms

m

t

va x

xav

215.00.00.2

0.13.1

sm

sss

ms

m

t

va y

yav

js

mis

mjaiaa yx

22 15.050.0

js

mis

ma

22 30.05.0

3-5 Motion in A Circle3-5 Motion in A CircleWhen a particle moves along a curved path, the direction of its velocity When a particle moves along a curved path, the direction of its velocity changes.As we saw in Section 3-3, this means that the particle must have a changes.As we saw in Section 3-3, this means that the particle must have a component of acceleration perpendicular to the path, even if its speed is component of acceleration perpendicular to the path, even if its speed is constant. In this section we’ll calculate the acceleration for the important special constant. In this section we’ll calculate the acceleration for the important special case of motion in a circle.case of motion in a circle.Uniform Circular MotionUniform Circular MotionWhen a particle moves in a circle with constant speed, the motion is called When a particle moves in a circle with constant speed, the motion is called uniform circular motion. There is no component of acceleration parallel uniform circular motion. There is no component of acceleration parallel (tangent) to the path; otherwise, the speed would change. The component of (tangent) to the path; otherwise, the speed would change. The component of acceleration perpendicular (normal) to the path, which cause the direction of acceleration perpendicular (normal) to the path, which cause the direction of the velocity to change, is related in a simple way to the speed of the particle the velocity to change, is related in a simple way to the speed of the particle and the radius of the circle.and the radius of the circle.In uniform circular motion the acceleration is perpendicular to the velocity at In uniform circular motion the acceleration is perpendicular to the velocity at each instant; as the direction of the velocity changes, the direction of the each instant; as the direction of the velocity changes, the direction of the acceleration also changes.acceleration also changes.

s

R

v1

v2

P1

P2

A OB

P1

O

v

v2

v1

P2

Figure A shows a particle moving with constant speed in a circular path Figure A shows a particle moving with constant speed in a circular path radius R with center at O. The particle moves from Pradius R with center at O. The particle moves from P11 to P to P22 in a time in a time t. t.

The vector change in velocity The vector change in velocity v during this time is shown in Fig. B.v during this time is shown in Fig. B. The angles labeled The angles labeled in Fig. A and B are the same because v in Fig. A and B are the same because v11 is is

perpendicular to the line OPperpendicular to the line OP11 and v and v22 is perpendicular to the line OP is perpendicular to the line OP22. .

Hence the triangles OPHence the triangles OP11PP22(Fig. A) and OP(Fig. A) and OP11PP22(Fig. B(Fig. B) ) are similar. Ratios of are similar. Ratios of

corresponding sides are equal, so corresponding sides are equal, so or or

The magnitude aThe magnitude aavav of the average acceleration during of the average acceleration during t is thereforet is therefore

The magnitude of the instantaneous acceleration a at point PThe magnitude of the instantaneous acceleration a at point P11. Also, P. Also, P11

can is the limit of this expression as we take point Pcan is the limit of this expression as we take point P22 closer and closer to closer and closer to

point Ppoint P11::

R

s

t

v

sR

vv 1

t

s

R

v

t

vaav

1

t

s

R

v

t

s

R

va

tt

0

11

0limlim

But the limit of But the limit of s/ s/ t is the speed vt is the speed v11 at point P at point P11. Also, P. Also, P11 can be any point on can be any point on

the path, so we can drop the subscript and let v represent the speed at any the path, so we can drop the subscript and let v represent the speed at any point. Thenpoint. Then

Because the speed is constant, the acceleration is always perpendicular to the Because the speed is constant, the acceleration is always perpendicular to the instantaneous velocity.We conclude: In uniform circular motion, the magnitude instantaneous velocity.We conclude: In uniform circular motion, the magnitude a of the instantaneous acceleration is equal to the square v divided by the a of the instantaneous acceleration is equal to the square v divided by the radius R of the circle. Its direction is perpendicular to v and inward along the radius R of the circle. Its direction is perpendicular to v and inward along the radius. Because the acceleration is always directed toward the center of the radius. Because the acceleration is always directed toward the center of the circle, it is sometimes called centripetal acceleration.circle, it is sometimes called centripetal acceleration.

Non-Uniform Circular MotionNon-Uniform Circular MotionWe have assumed calculate throughout this section that the particle’s speed is We have assumed calculate throughout this section that the particle’s speed is constant. If the speed varies, we call the motion non-uniform circular motion. In constant. If the speed varies, we call the motion non-uniform circular motion. In non-uniform circular motion, still gives the radial component of acceleration non-uniform circular motion, still gives the radial component of acceleration

R

varad

2

R

varad

2

, which is always perpendicular to the instantaneous , which is always perpendicular to the instantaneous velocity and direction toward the center of the circle. But velocity and direction toward the center of the circle. But since the speed v has different values at different points since the speed v has different values at different points in the motion, the value of a rad is not constant. in the motion, the value of a rad is not constant.

In non-uniform circular motion there is also a component of acceleration that is In non-uniform circular motion there is also a component of acceleration that is parallel to the instantaneous velocity. This is the component aparallel to the instantaneous velocity. This is the component a1111 that we that we

discussed in Section 3-3; here we call this component adiscussed in Section 3-3; here we call this component a tantan to emphasize that it to emphasize that it

is tangent to the circle. From the discussion at the end of Section 3-3 we see is tangent to the circle. From the discussion at the end of Section 3-3 we see that the tangential component of acceleration athat the tangential component of acceleration a tantan is equal to the rate of change is equal to the rate of change

of speed. Thus of speed. Thus and and

The vector acceleration of a particle moving in a circle with varying speed is the The vector acceleration of a particle moving in a circle with varying speed is the vector sum of the radial and tangential components of accelerations. The vector sum of the radial and tangential components of accelerations. The tangential component is in the same direction as the velocity if the particle is tangential component is in the same direction as the velocity if the particle is speeding up, and is in the opposite direction if the particle is slowing down. speeding up, and is in the opposite direction if the particle is slowing down. In uniform circular motion there is no tangential component of acceleration, but In uniform circular motion there is no tangential component of acceleration, but the radial component is the magnitude of dv/dt. We have mentioned before that the radial component is the magnitude of dv/dt. We have mentioned before that the two quantities the two quantities |dv/dt| and d|v|/dt are in general not equal. In uniform circular |dv/dt| and d|v|/dt are in general not equal. In uniform circular motion the first is constant and equal to vmotion the first is constant and equal to v22/R; the second is zero./R; the second is zero.

R

varad

2

dt

vda

tan

4 Newton’s Laws of Motion4 Newton’s Laws of Motion4-1 Introduction4-1 IntroductionIn this chapter we will use the kinematic quantities displacement, velocity, and In this chapter we will use the kinematic quantities displacement, velocity, and acceleration along with two new concepts, force and mass, to analyze the acceleration along with two new concepts, force and mass, to analyze the principles of dynamics. These principles can be wrapped up in a neat package principles of dynamics. These principles can be wrapped up in a neat package of three statements called Newton’s laws of motion. The first law states that of three statements called Newton’s laws of motion. The first law states that when the net force on a body is zero, its motion doesn’t change. The second when the net force on a body is zero, its motion doesn’t change. The second law relates force to acceleration when the net forces is not zero. The third law law relates force to acceleration when the net forces is not zero. The third law is a relation between the forces that two interacting bodies exert on each other. is a relation between the forces that two interacting bodies exert on each other. These laws, based on experimental studies of moving bodies, are fundamental These laws, based on experimental studies of moving bodies, are fundamental in two ways. First, they cannot be deduced or proved from other principles. in two ways. First, they cannot be deduced or proved from other principles. Second, they are the foundation of classical mechanics ( also called Newtonian Second, they are the foundation of classical mechanics ( also called Newtonian mechanics). Newton’s laws are not universal, however; they require mechanics). Newton’s laws are not universal, however; they require modification at very high speeds (near the speed of light) and for very small modification at very high speeds (near the speed of light) and for very small size ( such as within the atom).size ( such as within the atom).4-2 Force And Interactions4-2 Force And InteractionsIn everyday language, a force is a push or pull. The concept of force gives us a In everyday language, a force is a push or pull. The concept of force gives us a quantitative description between two bodies or between a body and its quantitative description between two bodies or between a body and its environment.environment.When a force involves direct contact between two bodies, we call it a contact When a force involves direct contact between two bodies, we call it a contact force.force.

Contact forces include the pushes or pulls you exert with your hand, the force Contact forces include the pushes or pulls you exert with your hand, the force of rope pulling on a block to which it is tied, and the friction force that the of rope pulling on a block to which it is tied, and the friction force that the ground exerts on a ball player sliding into home. There are also forces, called ground exerts on a ball player sliding into home. There are also forces, called long-range forces, that act even when the bodies are separated by empty long-range forces, that act even when the bodies are separated by empty space. You’ve experienced long-range forces if you’ve ever played with a pair space. You’ve experienced long-range forces if you’ve ever played with a pair of magnets. Gravity, too, is a long-range force; the sun exerts a gravitational of magnets. Gravity, too, is a long-range force; the sun exerts a gravitational pull on the earth, even over a distance of 150 million kilometers, that keeps the pull on the earth, even over a distance of 150 million kilometers, that keeps the earth in orbit.earth in orbit.Force is a vector quantity; you can push or pull a body in different directions. Force is a vector quantity; you can push or pull a body in different directions. Thus to describe a force, we need to describe the direction in which it acts as Thus to describe a force, we need to describe the direction in which it acts as well as its magnitude, the quantity that describes “how much” or “how hard” the well as its magnitude, the quantity that describes “how much” or “how hard” the force pushes or pulls.force pushes or pulls.When two forces FWhen two forces F11 and F and F22 act at the same time at point A of a body (Fig. 4-2), act at the same time at point A of a body (Fig. 4-2),

experiment shows that the effect on the body’s motion is the same as the effect experiment shows that the effect on the body’s motion is the same as the effect of a single force R equal to the vector sum of the original forces: R = Fof a single force R equal to the vector sum of the original forces: R = F11 + F + F22..

Fig.4-2 Fig. 4-3 Fig.4-2 Fig. 4-3

R

A

F1

F2

O

F

Fx

Fy

More generally, the effect of any number of forces applied at a point on a body More generally, the effect of any number of forces applied at a point on a body is the same as the effect of a single force equal to the vector sum of the forces. is the same as the effect of a single force equal to the vector sum of the forces. This important principle goes by the superposition of forces. This important principle goes by the superposition of forces. The experimental discovery that forces combine according to vector addition is The experimental discovery that forces combine according to vector addition is of the utmost importance. We will use this fact many times throughout our study of the utmost importance. We will use this fact many times throughout our study of physics . It allows us to replace a force by its component vectors, as we did of physics . It allows us to replace a force by its component vectors, as we did with displacements in Section 1-9. For example, in Fig. 4-3, force F acts on a with displacements in Section 1-9. For example, in Fig. 4-3, force F acts on a body at point O. The component vectors of F in the directions OX and OY are body at point O. The component vectors of F in the directions OX and OY are FFxx and F and Fyy. When F. When Fxx and F and Fyy are applied simultaneously, the effect is exactly the are applied simultaneously, the effect is exactly the

same as the effect of the original force F. Any force can be replaced by its same as the effect of the original force F. Any force can be replaced by its component vectors, acting at the same point.component vectors, acting at the same point.4-3 Newton’s First Law4-3 Newton’s First Law A body acted on by no net force moves with constant velocity (which may be A body acted on by no net force moves with constant velocity (which may be zero) and zero acceleration. This is Newton’s first law of motion.zero) and zero acceleration. This is Newton’s first law of motion.It’s important to note that the net force is what matters in Newton’s first law. It’s important to note that the net force is what matters in Newton’s first law. For example, a physics book at rest on a horizontal table top has two forces For example, a physics book at rest on a horizontal table top has two forces acting on it; the downward force of the earth’s gravitational attraction ( a long-acting on it; the downward force of the earth’s gravitational attraction ( a long-range force that acts even if the table top is elevated above the ground) and an range force that acts even if the table top is elevated above the ground) and an upward supporting force exerted by the table top ( a contact force). The upward upward supporting force exerted by the table top ( a contact force). The upward push of the surface is just as great as the downward pull of gravity, so the net push of the surface is just as great as the downward pull of gravity, so the net force acting on the book (that is, the vector sum of the two forces) is zero. force acting on the book (that is, the vector sum of the two forces) is zero.

In agreement with Newton’s first law, if the book is at rest on the table top, it In agreement with Newton’s first law, if the book is at rest on the table top, it remains at rest. We find that if the body is at rest at the start, it remains at rest; remains at rest. We find that if the body is at rest at the start, it remains at rest; if it is initially moving, it continues to move in the same direction with constant if it is initially moving, it continues to move in the same direction with constant speed. These results show that in Newton’s first law, zero net force is speed. These results show that in Newton’s first law, zero net force is equivalent to no force at all. This is just the principle of superposition of forces equivalent to no force at all. This is just the principle of superposition of forces that we saw in Section 4-2.that we saw in Section 4-2.4-4 Newton’s Second Law4-4 Newton’s Second LawMass and ForceMass and ForceMass is a quantitative measure of inertia, which we discussed in Section 4-3. Mass is a quantitative measure of inertia, which we discussed in Section 4-3. The greater its mass, the more a body “resists” being accelerated. The SI unit The greater its mass, the more a body “resists” being accelerated. The SI unit of mass is kilogram. One Newton is the amount of net force that gives an of mass is kilogram. One Newton is the amount of net force that gives an acceleration of one meter per second squared to a body with a mass of one acceleration of one meter per second squared to a body with a mass of one kilogram. We can use this definition to calibrate the spring balances and other kilogram. We can use this definition to calibrate the spring balances and other instruments used to measure forces. Because of the way we have defined the instruments used to measure forces. Because of the way we have defined the newton, it is related to the units of mass, length, and time. For Eq. newton, it is related to the units of mass, length, and time. For Eq. (4-5) to be dimensionally consistent, it must be true that 1newton = (1 kilogram) (4-5) to be dimensionally consistent, it must be true that 1newton = (1 kilogram) (1 meter per second squared).(1 meter per second squared).We can also use Eq. (4-5) to compare a mass with the standard mass and thus We can also use Eq. (4-5) to compare a mass with the standard mass and thus to measure masses. Suppose we apply a constant net force F to a body having to measure masses. Suppose we apply a constant net force F to a body having a known mass ma known mass m11 and we find an acceleration of magnitude a and we find an acceleration of magnitude a11. We then apply . We then apply

the the

maF

same force to another body having an unknown mass msame force to another body having an unknown mass m22, and we find an , and we find an

acceleration of magnitude aacceleration of magnitude a22. Then, according to Eq. (4-5). Then, according to Eq. (4-5)

(4-6) (4-6)

The ratio of the masses is the inverse of the ratio of the accelerations. Figure 4-The ratio of the masses is the inverse of the ratio of the accelerations. Figure 4-12 shows the inverse proportionally between mass and acceleration. In 12 shows the inverse proportionally between mass and acceleration. In principle we could use Eq. (4-6) to measure an unknown mass mprinciple we could use Eq. (4-6) to measure an unknown mass m22, but it is , but it is

usually easier to determine mass indirectly by measuring the body’s weight.usually easier to determine mass indirectly by measuring the body’s weight.When two bodies with masses mWhen two bodies with masses m11 and m and m22 are fastened together, we find that are fastened together, we find that

the mass of the composite body is always mthe mass of the composite body is always m11 + m + m22 (Fig. 4-12). This additive (Fig. 4-12). This additive

property of mass may seem obvious, but it has to be verified experimentally.property of mass may seem obvious, but it has to be verified experimentally.

Fig. 4-6 Fig. 4-6

2

1

1

2

a

a

m

m 2211 amam

m1

F

a1

m2

F

a2

m1+m2

F

a3

Newton’s Second LawNewton’s Second LawWe’ve been careful to state that net force on a body is what causes that We’ve been careful to state that net force on a body is what causes that body to accelerate. Experiment shows that if a combination of forces Fbody to accelerate. Experiment shows that if a combination of forces F11, ,

FF22, F, F33, , ····is applied to as body, the body will have the same acceleration ····is applied to as body, the body will have the same acceleration

(magnitude and direction) as when only a single force is applied, if that (magnitude and direction) as when only a single force is applied, if that single force is equal to the vector sum single force is equal to the vector sum FF11+F+F22+F+F33+ + ···· In other words, the ···· In other words, the

principle of superposition of forces also holds true when the net force is principle of superposition of forces also holds true when the net force is not zero and the body is accelerating.not zero and the body is accelerating.Equation (4-5) relates the magnitude of the net force on a body to the Equation (4-5) relates the magnitude of the net force on a body to the magnitude of the acceleration that it produces. We have also seen that magnitude of the acceleration that it produces. We have also seen that the direction of the net force is the same as the direction of the the direction of the net force is the same as the direction of the acceleration, whether the body’s path is straight or curved. Newton acceleration, whether the body’s path is straight or curved. Newton wrapped up all these relationships and experimental results in a single wrapped up all these relationships and experimental results in a single concise statement that we now call Newton’s second law of motion:concise statement that we now call Newton’s second law of motion:If a net external force acts on a body, the body accelerates. The direction If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. The net force of acceleration is the same as the direction of the net force. The net force vector is equal to the mass of the body times the acceleration of the vector is equal to the mass of the body times the acceleration of the body.body.In symbols, In symbols, (4-7) (4-7)

amF

There are at least four aspects of Newton’s second law that deserve special There are at least four aspects of Newton’s second law that deserve special atten-tion. First, Eq. (4-7) is a vector equation. Usually, we will use it in atten-tion. First, Eq. (4-7) is a vector equation. Usually, we will use it in component from, with a separate equation for each component of force and the component from, with a separate equation for each component of force and the corresponding accelerationcorresponding acceleration

(4-8) (4-8)This set of component equations is equivalent to the single vector equation (4-This set of component equations is equivalent to the single vector equation (4-7). Each component of total force equals the mass times the corresponding 7). Each component of total force equals the mass times the corresponding component of acceleration.component of acceleration.Second, the statement of Newton’s second law refers to external forces. By this Second, the statement of Newton’s second law refers to external forces. By this we mean forces exerted on the body by other bodies in its environment. It’s we mean forces exerted on the body by other bodies in its environment. It’s impossible for a body to affect its own motion by exerting a force on itself; if it impossible for a body to affect its own motion by exerting a force on itself; if it was possible, you could lift yourself to the ceiling by pulling up on your belt! was possible, you could lift yourself to the ceiling by pulling up on your belt! That’s why only external forces are included in the sum That’s why only external forces are included in the sum F in Eqs. (4-7) and (4-F in Eqs. (4-7) and (4-8).8).Third, Eqs. (4-7) and (4-8) are valid only when the mass m is constant. It’s easy Third, Eqs. (4-7) and (4-8) are valid only when the mass m is constant. It’s easy to think of systems whose masses change, such as a leaking tank truck, a to think of systems whose masses change, such as a leaking tank truck, a rocket ship, or a moving railroad car being loaded with coal. But such systems rocket ship, or a moving railroad car being loaded with coal. But such systems are better handled by using the concept of momentum; we’ll get to that in are better handled by using the concept of momentum; we’ll get to that in Chapter 8.Chapter 8.Finally, Newton’s second law is valid only in inertial frames of reference, just Finally, Newton’s second law is valid only in inertial frames of reference, just like the first law. Thus it is not valid in the reference frame of any of the like the first law. Thus it is not valid in the reference frame of any of the acceleratingaccelerating

xx maF yy maF zz maF

Vehicles in Fig. 4-8; relative to any of these frames, the passenger accelerates Vehicles in Fig. 4-8; relative to any of these frames, the passenger accelerates even though the net force on the passenger is zero. We will usually assume even though the net force on the passenger is zero. We will usually assume that the earth is an adequate approximation to an inertial frame, although that the earth is an adequate approximation to an inertial frame, although because of its rotation and orbital motion it is not precisely inertial.because of its rotation and orbital motion it is not precisely inertial.4-5 Mass and Weight4-5 Mass and WeightThe weight of a body is a familiar force. It is the force of the earth’s gravitational The weight of a body is a familiar force. It is the force of the earth’s gravitational attraction for the body. We will study gravitational interactions in detail in attraction for the body. We will study gravitational interactions in detail in chapter 12, but we need some preliminary discussion now. The terms mass chapter 12, but we need some preliminary discussion now. The terms mass and weight are often misused and interchanged in everyday conversation. It is and weight are often misused and interchanged in everyday conversation. It is absolutely essential for you to understand clearly the distinction between these absolutely essential for you to understand clearly the distinction between these two physical quantities.two physical quantities.Mass characterizes the inertial properties of a body. The weight of a body is a Mass characterizes the inertial properties of a body. The weight of a body is a force, a vector quantity.force, a vector quantity.4-6 Newton’s Third Law 4-6 Newton’s Third Law A force acting on a body is always the result of its interaction with another A force acting on a body is always the result of its interaction with another body, so forces always come in pairs. In each of these cases the force that you body, so forces always come in pairs. In each of these cases the force that you exert on the other body is in the opposite direction to the force that body exert exert on the other body is in the opposite direction to the force that body exert on you. Experiments show that whenever two bodies interact, the two forces on you. Experiments show that whenever two bodies interact, the two forces that they exert on the other are always equal in magnitude and opposite in that they exert on the other are always equal in magnitude and opposite in direction. This fact is called Newton’s third law of motion. Expressed in words, If direction. This fact is called Newton’s third law of motion. Expressed in words, If body A exerts a force on body B (an “action”), then body B exerts a force on body A exerts a force on body B (an “action”), then body B exerts a force on body A (a “reaction”).body A (a “reaction”).

These two forces have the same magnitude but are opposite in direction. These two forces have the same magnitude but are opposite in direction. These two forces act on different bodies.These two forces act on different bodies. (4-11) (4-11) In this statement, “action” and “reaction” are the two opposite forces; we In this statement, “action” and “reaction” are the two opposite forces; we some-times refer to them as an action-reaction pair. This is not meant to imply some-times refer to them as an action-reaction pair. This is not meant to imply any cause-and-effect relationship; we can consider either force as the “action” any cause-and-effect relationship; we can consider either force as the “action” and the other as the “reaction”. We often say simply that the forces are “equal and the other as the “reaction”. We often say simply that the forces are “equal and opposite,” meaning that they have equal magnitudes and opposite and opposite,” meaning that they have equal magnitudes and opposite direction.direction. We stress that the two forces described in Newton’s third law act on different We stress that the two forces described in Newton’s third law act on different bodies. This is important in problems involving Newton’s first or second law, bodies. This is important in problems involving Newton’s first or second law, which involve the forces that act on a body. which involve the forces that act on a body. 4-7 Using Newton’s Laws4-7 Using Newton’s LawsNewton’s three laws of motion contain all the basic principles we need to solve Newton’s three laws of motion contain all the basic principles we need to solve a wide variety of problems in mechanics. These laws are very simple in form, a wide variety of problems in mechanics. These laws are very simple in form, but the process of applying them to specific situations can pose real but the process of applying them to specific situations can pose real challenges. This section introduces some useful techniques. challenges. This section introduces some useful techniques. When you use Newton’s first law, , for an equilibrium situation, or When you use Newton’s first law, , for an equilibrium situation, or Newton’s second law, , for a non-equilibrium situation, you must Newton’s second law, , for a non-equilibrium situation, you must apply it to some specific body. It is absolutely essential to decided at the apply it to some specific body. It is absolutely essential to decided at the beginningbeginning

BonAAonB FF

0F

amF

What body you talking about. This may sound trivial, but it isn’t. Once you What body you talking about. This may sound trivial, but it isn’t. Once you have chosen a body, then you have to identify all the forces acting on it. have chosen a body, then you have to identify all the forces acting on it. Don’t get confused between the forces acting on a body and the forces Don’t get confused between the forces acting on a body and the forces exerted by that body on some other body you’ve chosen go into .exerted by that body on some other body you’ve chosen go into . To help identify the relevant forces, draw a free-body diagram. What’s To help identify the relevant forces, draw a free-body diagram. What’s that? It is a diagram showing the chosen body by itself, “free” of its that? It is a diagram showing the chosen body by itself, “free” of its surrounding, with vectors down to show the magnitudes and directions of surrounding, with vectors down to show the magnitudes and directions of all the forces applied to the body by the various other bodies that interact all the forces applied to the body by the various other bodies that interact with it. Be careful to include all the forces acting on the body, but be with it. Be careful to include all the forces acting on the body, but be equally careful not to include any forces that the body exerts on any other equally careful not to include any forces that the body exerts on any other body. In particular, the two forces in an action-reaction pair must never body. In particular, the two forces in an action-reaction pair must never appear in the same free-body diagram because they never act on the appear in the same free-body diagram because they never act on the same body. Furthermore, forces that a body exerts on itself are never same body. Furthermore, forces that a body exerts on itself are never include, since these can’t affect on the body’s motion.include, since these can’t affect on the body’s motion.The cutting-blade assembly on a radial-arm saw has a mass of 5.0 kg. It is The cutting-blade assembly on a radial-arm saw has a mass of 5.0 kg. It is pulled along a pair of frictionless horizontal rails aligned with the x-axis by pulled along a pair of frictionless horizontal rails aligned with the x-axis by a force Fa force Fxx. The position of the blade assembly as a function of time is. The position of the blade assembly as a function of time is

x = (0.18m/sx = (0.18m/s22)t)t22 – (0.030m/s – (0.030m/s33)t)t3 .3 .

Find the net force acting on the blade assembly as a function of time. Find the net force acting on the blade assembly as a function of time. What is the force at time t = 3.0s ? For what times is the force positive? What is the force at time t = 3.0s ? For what times is the force positive? Negative? Zero?Negative? Zero?

F

Solution Figure 4-24b shows the free-body diagram and coordinate axes. The Solution Figure 4-24b shows the free-body diagram and coordinate axes. The forces are the horizontal force Fforces are the horizontal force Fxx, the weight w, and the upward force n that the , the weight w, and the upward force n that the

rail exert to support the cutting head. There is no acceleration in the vertical rail exert to support the cutting head. There is no acceleration in the vertical direction, so the sum of the vertical components of forces must be zero. The direction, so the sum of the vertical components of forces must be zero. The net forces has only a horizontal component, equal to Fnet forces has only a horizontal component, equal to Fxx; from Newton’s second ; from Newton’s second

law, this is equal to malaw, this is equal to maxx. To determine F. To determine Fxx, we first find the acceleration a, we first find the acceleration axx by by

taking the second derivative of x. The second derivative of ttaking the second derivative of x. The second derivative of t22 is 2, and the is 2, and the second derivative of tsecond derivative of t33 is 6t, so is 6t, so

then, from Newton’s second law,then, from Newton’s second law,FFxx = ma = maxx = (5.0kg)[(0.36m/s = (5.0kg)[(0.36m/s33 – (0.18m/s – (0.18m/s33)]t = 1.8N – (0.90N/s)t)]t = 1.8N – (0.90N/s)t

Figure 4-24(b) Figure 4-25(a) Figure 4-25(b) Figure 4-24(b) Figure 4-25(a) Figure 4-25(b)

ts

ms

mdt

xdax 322

2

18.036.0

w

y

Fx

x 0

-1.00

-2.00

1.00

4321t(s)

5

2.00

Fx(N)

0.20

4321t(s)

5

0.40

-0.40

-0.20

0

ax(m/s2)

At time t = 3.0s the force is 1.80N – (0.90 N/s)(3.0s) = -0.90N. The force is At time t = 3.0s the force is 1.80N – (0.90 N/s)(3.0s) = -0.90N. The force is zero when azero when axx = 0, that is, when 0.36m/s = 0, that is, when 0.36m/s22 – (0.18m/s – (0.18m/s33)t = 0. This happens )t = 0. This happens

when t = 2.0s. When t when t = 2.0s. When t 2.0s, F 2.0s, Fxx is positive, and the blade assembly is is positive, and the blade assembly is

being pulled to the right. When t being pulled to the right. When t > 2.0s, F> 2.0s, Fxx is negative, and the horizontal is negative, and the horizontal

force on the assembly is to the left. Figure 4-25 shows graphs of Fforce on the assembly is to the left. Figure 4-25 shows graphs of Fxx and a and axx

as functions of time. You shouldn’t be surprised that the net force and as functions of time. You shouldn’t be surprised that the net force and acceleration are directly proportional; according to Newton’s second law, acceleration are directly proportional; according to Newton’s second law, this is always true.this is always true.

5 Applications of Newton’s Laws5 Applications of Newton’s Laws

5-1 Introduction5-1 IntroductionNewton’ three laws of motion, the foundation of classic mechanics, can be Newton’ three laws of motion, the foundation of classic mechanics, can be stated very simply, as we have seen. stated very simply, as we have seen. 5-2 Using Newton’s First Particles In Equilibrium5-2 Using Newton’s First Particles In EquilibriumWe learned in Chapter 4 that a body is in equilibrium when it is at rest or We learned in Chapter 4 that a body is in equilibrium when it is at rest or moving with constant velocity in an inertial frame of reference. When a moving with constant velocity in an inertial frame of reference. When a particle is at rest or is moving with constant velocity in an inertial frame of particle is at rest or is moving with constant velocity in an inertial frame of reference, the net force acting on it---must be zero.reference, the net force acting on it---must be zero. (5-1) (5-1)

We will usually use this in component form:We will usually use this in component form: (5-2) (5-2)This section is about using Newton’s first law to solve problems dealing This section is about using Newton’s first law to solve problems dealing with bodies in equilibrium. with bodies in equilibrium. 5-3 Using Newton’s Second Law: Dynamics of Particles5-3 Using Newton’s Second Law: Dynamics of ParticlesWe are now ready to discuss dynamics problems, in which we applyWe are now ready to discuss dynamics problems, in which we apply

0F

0 xF

0 yF

0 zF

Newton’s second law to bodies that are acceleration and hence are not in Newton’s second law to bodies that are acceleration and hence are not in equilibrium. In this care force on the body is not zero, but is equal to the mass equilibrium. In this care force on the body is not zero, but is equal to the mass of the body times its acceleration :of the body times its acceleration : (5-3) (5-3)We will use this relation in component form:We will use this relation in component form: (5-4) (5-4)5-4 Frictional Forces5-4 Frictional Forceswe have seen several problems where a body rests or slides on a surface that we have seen several problems where a body rests or slides on a surface that exerts forces on the body, and we have used the terms normal force and exerts forces on the body, and we have used the terms normal force and friction force to describe these forces. Whenever two bodies interaction forces friction force to describe these forces. Whenever two bodies interaction forces constant (touching) of their surfaces, we call the interaction forces contact constant (touching) of their surfaces, we call the interaction forces contact forces. Normal and friction forces are both contact forces.forces. Normal and friction forces are both contact forces.Kinetic and Static FrictionKinetic and Static FrictionFirst, when a body rests or slides on a surface, we can always represent the First, when a body rests or slides on a surface, we can always represent the contact force exerted by the surface on the body in terms of components of contact force exerted by the surface on the body in terms of components of force perpendicular and parallel to the surface. We call the perpendicular force perpendicular and parallel to the surface. We call the perpendicular component vector the normal force, denoted by component vector the normal force, denoted by nn .(Recall that normal is a .(Recall that normal is a synonym for perpendicular) component vector parallel to the surface is the synonym for perpendicular) component vector parallel to the surface is the friction force, denoted by friction force, denoted by f .f . By definition , n and f are always perpendicular to By definition , n and f are always perpendicular to each other. each other.

amF

xmaFyy maF zz maF

The direction of the friction force is always such as to oppose relative motion of The direction of the friction force is always such as to oppose relative motion of the two surfaces.the two surfaces.The kind of friction that acts when a body slides over a surface is called a The kind of friction that acts when a body slides over a surface is called a kinetic fraction force fkinetic fraction force fkk, The adjective “kinetic” and the subscript “k” remind us , The adjective “kinetic” and the subscript “k” remind us

that the two surfaces are moving relative to each other. The magnitude of the that the two surfaces are moving relative to each other. The magnitude of the kinetic friction force usually increases when the normal force increases. In kinetic friction force usually increases when the normal force increases. In many cases the magnitude of the kinetic friction force fmany cases the magnitude of the kinetic friction force fkk is found experimentally is found experimentally

to be approximately proportional to the magnitude n of the normal force. In to be approximately proportional to the magnitude n of the normal force. In such case we can writesuch case we can write (5-5) (5-5)where where µµkk (pronounced “mu-sub-k”) is a constant called the coefficient of kinetic (pronounced “mu-sub-k”) is a constant called the coefficient of kinetic

friction. The more slippery the surface, the smaller the coefficient friction. The more slippery the surface, the smaller the coefficient of friction. of friction. Because it is a quotient of two force magnitudes, Because it is a quotient of two force magnitudes, µµkk is a pure number without is a pure number without

units.units. Friction forces may also act when there is no relative motion. If you try to Friction forces may also act when there is no relative motion. If you try to slide that box of books across the floor, the box may not move at all because slide that box of books across the floor, the box may not move at all because the floor exerts an equal and opposite friction force on the box. This is called a the floor exerts an equal and opposite friction force on the box. This is called a static friction force fstatic friction force fss. For a given pair of surfaces the maximum value, called . For a given pair of surfaces the maximum value, called

(f(fss))maxmax, is approximately proportional to n; we call the proportionality factor µ, is approximately proportional to n; we call the proportionality factor µss

(pronounced “mu-sub-s”) the coefficient of static friction. Some representative (pronounced “mu-sub-s”) the coefficient of static friction. Some representative values of µvalues of µss are shown in Table 5-1. In a particular situation, the actual force of are shown in Table 5-1. In a particular situation, the actual force of

static friction can static friction can

nf kk

have any magnitude between zero (when there is no other force parallel to the have any magnitude between zero (when there is no other force parallel to the surface) and a maximum value given by surface) and a maximum value given by µµssn. In symbolsn. In symbols

(5-6) (5-6)Like Eq. (5-5), this is a relation between magnitudes, not a vector relation.Like Eq. (5-5), this is a relation between magnitudes, not a vector relation.

Fluid Resistance and Terminal SpeedFluid Resistance and Terminal SpeedThe direction of the fluid resistance force acting on a body is always opposite The direction of the fluid resistance force acting on a body is always opposite the direction of the body’s velocity relative to the fluid. The magnitude of the the direction of the body’s velocity relative to the fluid. The magnitude of the fluid resistance force usually increases with the speed of the body through the fluid resistance force usually increases with the speed of the body through the fluid. Contrast this behavior with that of the kinetic friction force between two fluid. Contrast this behavior with that of the kinetic friction force between two surfaces in contact, which we can usually regard as independent of speed. For surfaces in contact, which we can usually regard as independent of speed. For low speeds, the magnitude f of the resisting force of the fluid is approximately low speeds, the magnitude f of the resisting force of the fluid is approximately proportional to the body’s speed v:proportional to the body’s speed v: (5-7) (5-7)Where k is a proportionality constant that depends on the shape and size of the Where k is a proportionality constant that depends on the shape and size of the body and the properties of the fluid. In motion through air at the speed of a body and the properties of the fluid. In motion through air at the speed of a tossed tennis ball or faster, the resisting force is approximately proportional to tossed tennis ball or faster, the resisting force is approximately proportional to vv22 rather than to v. It is then called air drag or simply drag. In this case we rather than to v. It is then called air drag or simply drag. In this case we replace Eq.(5-7)replace Eq.(5-7)

(5-8) (5-8)

nf ss

kvf

2Dvf

Because of the vBecause of the v22 dependence, air drag increases rapidly with increasing dependence, air drag increases rapidly with increasing speed. The air drag on a typical car is negligible at low speeds but speed. The air drag on a typical car is negligible at low speeds but comparable to or greater the force of rolling resistance at highway speeds. comparable to or greater the force of rolling resistance at highway speeds. The value of D depends on the shape and size of the body and on the The value of D depends on the shape and size of the body and on the density of the air. Because of the effects of fluid resistance, an object density of the air. Because of the effects of fluid resistance, an object falling in a fluid will not have a constant acceleration. To describe its falling in a fluid will not have a constant acceleration. To describe its motion, we have to start over, using Newton’s second law.motion, we have to start over, using Newton’s second law. The free-body diagram is shown in Fig.5-22. We take the positive The free-body diagram is shown in Fig.5-22. We take the positive direction to be downward and neglect any force associated with buoyancy direction to be downward and neglect any force associated with buoyancy in the water. There are no x-components, and Newton’s second law givesin the water. There are no x-components, and Newton’s second law gives

Fig.5-22 Fig.5-23 Fig.5-22 Fig.5-23

makvmgFy )(

y

f

W=mgO

g

a

t O

v

vt

tO

y

t

When the rock first starts to move, v = 0, the resisting force is zero, and When the rock first starts to move, v = 0, the resisting force is zero, and the initial acceleration is a = g. As its speed increases, until finally it is the initial acceleration is a = g. As its speed increases, until finally it is equal in magnitude to the weight. At this time, mg – kv = 0, the equal in magnitude to the weight. At this time, mg – kv = 0, the acceleration becomes zero, and there is no further increase in speed. The acceleration becomes zero, and there is no further increase in speed. The final speed vfinal speed vtt, called the terminal speed, is given by mg – kv, called the terminal speed, is given by mg – kv tt = 0, or = 0, or

Figure 5-23 shows how the acceleration, velocity, and position vary with Figure 5-23 shows how the acceleration, velocity, and position vary with time. As time goes by, the acceleration approaches zero, and the velocity time. As time goes by, the acceleration approaches zero, and the velocity approaches vapproaches vtt. The slope of the graph of y versus t becomes constant as . The slope of the graph of y versus t becomes constant as

the velocity becomes constant. To see how the graphs in Fig. 5-23 are the velocity becomes constant. To see how the graphs in Fig. 5-23 are derived, we must find the relation between speed and time during the derived, we must find the relation between speed and time during the interval before the terminal speed is reached. We go back to Newton’s interval before the terminal speed is reached. We go back to Newton’s second law, which we rewrite assecond law, which we rewrite as

After rearranging terms and replacing mg/k by vAfter rearranging terms and replacing mg/k by v tt, we integrate both sides, , we integrate both sides,

noting that v = 0 when t = 0:noting that v = 0 when t = 0:

k

mgvt

kvmgdt

dvm

tv

t

dtm

k

vv

dv00

Which integrate to Which integrate to or or

and finally and finally

Note that v becomes equal to the terminal speed v Note that v becomes equal to the terminal speed v tt only in the limit that only in the limit that

tt; the rock cannot attain terminal speed in any finite length of time.; the rock cannot attain terminal speed in any finite length of time. The derivative of v gives a as a function of time, and the integral of v The derivative of v gives a as a function of time, and the integral of v gives y as a function of time. We leave the derivations for you to gives y as a function of time. We leave the derivations for you to complete, the results are complete, the results are

tm

k

v

vv

t

t

ln tmk

t

ev

v /1

tmk

t evv /1

tmk

gea

tm

k

t ek

mtvy 1

In deriving the terminal speed in Eq. 5-23, we assumed that the fluid In deriving the terminal speed in Eq. 5-23, we assumed that the fluid resistance force was proportional to the speed. For an object falling resistance force was proportional to the speed. For an object falling through the air at high speeds, so that the fluid resistance is proportional to through the air at high speeds, so that the fluid resistance is proportional to vv22 as in Eq. (5-8), we invite you to show that the terminal speed v as in Eq. (5-8), we invite you to show that the terminal speed v tt is given is given

byby

This expression for terminal speed explains the observation that heavy This expression for terminal speed explains the observation that heavy objects in air tend to fall faster than light objects.objects in air tend to fall faster than light objects.

5-5 Dynamics of Circular Motion5-5 Dynamics of Circular MotionWe talked about uniform circular motion in Section 3-5. We showed that We talked about uniform circular motion in Section 3-5. We showed that when a particle moves in a circular path with constant speed, the particle’s when a particle moves in a circular path with constant speed, the particle’s acceleration is always directed toward the center of the circle. The acceleration is always directed toward the center of the circle. The magnitude amagnitude aradrad of the acceleration is constant and is given in terms of the of the acceleration is constant and is given in terms of the

speed v and the radius R of the circle byspeed v and the radius R of the circle by

D

mgvt

R

varad

2

The magnitude of the radial acceleration is given by aThe magnitude of the radial acceleration is given by a radrad = v = v22/R, so the /R, so the

magnitude of the net inward radial force Fmagnitude of the net inward radial force Fnetnet on a particle with mass m must be on a particle with mass m must be

5-6 The Fundamental Forces of Nature 5-6 The Fundamental Forces of Nature Our current understanding is that all forces are expressions of just four distinct Our current understanding is that all forces are expressions of just four distinct classes of fundamental forces, or interactions between particles. Two are classes of fundamental forces, or interactions between particles. Two are familiar in everyday experience. The other two involve interactions between familiar in everyday experience. The other two involve interactions between subatomic particles that we cannot observe with the unaided senses.subatomic particles that we cannot observe with the unaided senses. Of the two familiar classes, gravitational interactions were the first to be Of the two familiar classes, gravitational interactions were the first to be studied in detail. The weight of a body results from the earth’s gravitational studied in detail. The weight of a body results from the earth’s gravitational attraction acting on it.attraction acting on it. The second familiar class of forces, electromagnetic interaction, includes The second familiar class of forces, electromagnetic interaction, includes electric and magnetic forces. All atoms contain positive and negative electric electric and magnetic forces. All atoms contain positive and negative electric charge, so atoms and molecules can exert electric forces on each other. charge, so atoms and molecules can exert electric forces on each other. Contact forces, in-cluding the normal force, friction, and fluid resistance, are the Contact forces, in-cluding the normal force, friction, and fluid resistance, are the combination of all such forces exerted on the atoms of a body by atoms in its combination of all such forces exerted on the atoms of a body by atoms in its surroundings.surroundings.

R

vmmaF radnet

2

These two interactions differ enormously in their These two interactions differ enormously in their strength. The electrical repul-sion between two protons strength. The electrical repul-sion between two protons at a given distance is stronger than their gravitational at a given distance is stronger than their gravitational attraction by a factor of the order of 1035. Gravitational attraction by a factor of the order of 1035. Gravitational forces play no significant role in atomic or molecular forces play no significant role in atomic or molecular structure. But in bodies of astronomical size, positive structure. But in bodies of astronomical size, positive charge and negative charge are usually present in charge and negative charge are usually present in nearly equal amounts, and the resulting electrical nearly equal amounts, and the resulting electrical interactions nearly cancel each other out. Gravitational interactions nearly cancel each other out. Gravitational inter-actions are thus the dominant influence in the inter-actions are thus the dominant influence in the motion of planets and in the internal structure of stars.motion of planets and in the internal structure of stars.The other two interactions are less familiar. One, the The other two interactions are less familiar. One, the strong interaction, is responsible for holding the nucleus strong interaction, is responsible for holding the nucleus of an atom together. Nucleus containof an atom together. Nucleus contain

electrically neutral neutrons and positively charged protons. electrically neutral neutrons and positively charged protons. The charged protons repel each other, and a nucleus could The charged protons repel each other, and a nucleus could not be stable if it were not for the presence of an attrac-tive not be stable if it were not for the presence of an attrac-tive forces of a different kind that counteracts the repulsive forces of a different kind that counteracts the repulsive electrical interactions. In this context the strong interaction is electrical interactions. In this context the strong interaction is also called the nuclear force. It has much shorter range than also called the nuclear force. It has much shorter range than electrical interactions, but within its range it is much stronger.electrical interactions, but within its range it is much stronger. Finally, there is the weak interaction. It plays no direct role Finally, there is the weak interaction. It plays no direct role in the behavior of ordinary matter, but it is of vital important in the behavior of ordinary matter, but it is of vital important interactions among fundamental particles. The weak interactions among fundamental particles. The weak interaction is responsible for a common form of radioactivity interaction is responsible for a common form of radioactivity beta-decay, in which a neutron in a radioactive nucleus is beta-decay, in which a neutron in a radioactive nucleus is transformed in to a proton while ejecting an electron and an transformed in to a proton while ejecting an electron and an essentially massless particle called an antineutrino. essentially massless particle called an antineutrino. During the past several decades a unified theory of the During the past several decades a unified theory of the electromagnetic and weak interactions has been developed.electromagnetic and weak interactions has been developed.

6 Work and Kinetic Energy6 Work and Kinetic Energy6-1 6-1 IntroductionIntroductionThe new method that we’re about to introduce uses the ideas of work and ener-The new method that we’re about to introduce uses the ideas of work and ener-gy. The applications of these ideas go far beyond mechanics, however. The im-gy. The applications of these ideas go far beyond mechanics, however. The im-portance of the energy idea stems from the principle of conservation of energy: portance of the energy idea stems from the principle of conservation of energy: Energy is a quantity that can be converted from one form to another but Energy is a quantity that can be converted from one form to another but cannot be created or destroyedcannot be created or destroyed. We’ll use the energy idea throughout the . We’ll use the energy idea throughout the rest of this book to study a tremendous range of physical phenomena.rest of this book to study a tremendous range of physical phenomena.In this chapter, through, our concentration will be on mechanics. We’ll learn In this chapter, through, our concentration will be on mechanics. We’ll learn about one important form of energy called kinetic energy, or energy of motion, about one important form of energy called kinetic energy, or energy of motion, and how it relates to the concept of work.and how it relates to the concept of work.6-2 6-2 WorkWorkIn physics, work has a much more precise definition. By making use of the defi-In physics, work has a much more precise definition. By making use of the defi-nition we’ll find that in any motion, no matter how complicated, the total work nition we’ll find that in any motion, no matter how complicated, the total work done on a particle by all forces that act on it equals the change in its kinetic done on a particle by all forces that act on it equals the change in its kinetic energy—a quantity that’s related to the particle’s speed.energy—a quantity that’s related to the particle’s speed.The physics’s definition of work is based on these observations. Consider a The physics’s definition of work is based on these observations. Consider a body that undergoes a displacement of magnitude along a straight line. (For body that undergoes a displacement of magnitude along a straight line. (For now, we will assume that any body we discuss can be treated as a particle so now, we will assume that any body we discuss can be treated as a particle so that wethat we

can ignore any rotation or changes in shape of the body) While the body can ignore any rotation or changes in shape of the body) While the body moves, a constant force with magnitude F acts on it in the same direction as moves, a constant force with magnitude F acts on it in the same direction as displacement s. We define the work W done by a constant force F acting on the displacement s. We define the work W done by a constant force F acting on the body under these conditions as body under these conditions as W = Fs (6-1) W = Fs (6-1)The work done on the body is greater if either the force F or the displacement s The work done on the body is greater if either the force F or the displacement s is greater. is greater. When the force F and the displacement When the force F and the displacement ss have different directions, we take have different directions, we take the component of F in the direction of the displacement s, and we define the the component of F in the direction of the displacement s, and we define the work as the product of this component and magnitude of the displacement. The work as the product of this component and magnitude of the displacement. The component of F in the direction of s is F coscomponent of F in the direction of s is F cos, so , so W = Fscos W = Fscos (6-2) (6-2)we are assuming that F and we are assuming that F and are constant during the displacement. If are constant during the displacement. If = 0, = 0, so that F and s are in the same direction, then cos so that F and s are in the same direction, then cos = 1 and we are back to Eq. = 1 and we are back to Eq. 6-1. 6-1. Equation (6-2) has the form of the scalar product of two vectors. We can Equation (6-2) has the form of the scalar product of two vectors. We can write Eq. (6-2) more compactly as write Eq. (6-2) more compactly as W = W = F F • s (• s (6-36-3)) It’sIt’s important to understand that work is a scalar quantity, even through it’s important to understand that work is a scalar quantity, even through it’s calculated by using two vector quantities (force and displacement). A 5-N force calculated by using two vector quantities (force and displacement). A 5-N force toward the east acting on a body that moves 6 m to the east does exactly the toward the east acting on a body that moves 6 m to the east does exactly the same work as a 5-N force toward the north acting on a body that moves 6 m to same work as a 5-N force toward the north acting on a body that moves 6 m to the north. the north.

It’s also important to realize that work can be positive, negative, or zero. This is It’s also important to realize that work can be positive, negative, or zero. This is the essential way in which work as defined in physics differs from the the essential way in which work as defined in physics differs from the “everyday” de-finition of work. When the force has a component in the same “everyday” de-finition of work. When the force has a component in the same direction as the dis-placement (direction as the dis-placement ( between zero and 90 between zero and 90), cos ), cos in Eq. (6-2) is in Eq. (6-2) is positive and the work W is positive. When the force has a component opposite positive and the work W is positive. When the force has a component opposite to the displacement (to the displacement ( bet-ween 90 bet-ween 90° and 180°, cos ° and 180°, cos is negative and the work is negative and the work is negative. When the force is perpendicular to the displacement, is negative. When the force is perpendicular to the displacement, = 90 = 90° and ° and the work done by the force is zero.the work done by the force is zero.

Fig.6-6 (a) (b) (c) (d) Fig.6-6 (a) (b) (c) (d)

F

v

F

n

w

ф

F

v

F

n

w

F

v

F

n

w

F

F

v

n

w

6-3 6-3 Work and Kinetic EnergyWork and Kinetic EnergyThe total work done on a body by external forces is related to the body’s The total work done on a body by external forces is related to the body’s displace-ment, that is, to changes in its position. But the total work is also displace-ment, that is, to changes in its position. But the total work is also related to changes in the speed of the body. To see this, consider Fig. 6-6, related to changes in the speed of the body. To see this, consider Fig. 6-6, which shows several examples of a block on a frictionless table. The forces which shows several examples of a block on a frictionless table. The forces acting on the block are its weight w, the normal force n, and the force F exerted acting on the block are its weight w, the normal force n, and the force F exerted on it by the hand.on it by the hand.In Fig. 6-6a the net force on the block is in the direction of its motion. From In Fig. 6-6a the net force on the block is in the direction of its motion. From Newton’s second law, this means that the block speeds up; from Eq. (6-1), this Newton’s second law, this means that the block speeds up; from Eq. (6-1), this also means that the total work Walso means that the total work Wtottot done on the block is positive. The total work done on the block is positive. The total work

is also positive in Fig.6-6b, but only the component F cosis also positive in Fig.6-6b, but only the component F cos contributes to W contributes to Wtottot. .

The block again speeds up, and this same component F cos The block again speeds up, and this same component F cos is what causes is what causes the acceleration. The total work negative in Fig.6-6c because the net force the acceleration. The total work negative in Fig.6-6c because the net force opposes the displacement; in this case the block slows down. The net force is opposes the displacement; in this case the block slows down. The net force is zero in Fig. 6-6d, so the speed of the block stays the same and the total work zero in Fig. 6-6d, so the speed of the block stays the same and the total work done on the block is zero. We can conclude that when a particle undergoes a done on the block is zero. We can conclude that when a particle undergoes a displacement, it speeds up if Wdisplacement, it speeds up if Wtottot>0, slows down if W>0, slows down if Wtottot<0, and maintains the <0, and maintains the

same speed if Wsame speed if Wtottot = 0. = 0.

Let’s make these observations more quantitative. Consider a particle with mass Let’s make these observations more quantitative. Consider a particle with mass m moving along the x-axis under the action of a constant net force with m moving along the x-axis under the action of a constant net force with magnitude F directed along the positive x-axis (Fig. 6-1). The particle’smagnitude F directed along the positive x-axis (Fig. 6-1). The particle’s

acceleration is constant and given by Newton’s second law, F = ma. Suppose acceleration is constant and given by Newton’s second law, F = ma. Suppose the speed changes the speed changes from vfrom v11 to v to v22 while the particle undergoes a displacement s while the particle undergoes a displacement s

= x= x22 – x – x11 from point x from point x11 to x to x22. Using a constant-acceleration equation, Eq. (2-13), . Using a constant-acceleration equation, Eq. (2-13),

and replacing vand replacing v00 by v by v11, v by v, v by v22, and (x – x, and (x – x00) by s, we have ) by s, we have

When we multiply this equation by m and equate ma to the net force F, we findWhen we multiply this equation by m and equate ma to the net force F, we find

andand

(6-4) (6-4)

The product Fs is the work done by the net force F and thus is equal to the The product Fs is the work done by the net force F and thus is equal to the total work Wtotal work Wtottot done by all the forces acting on the particle. The quantity ½ mv done by all the forces acting on the particle. The quantity ½ mv22

is called the kinetic energy K of the particle.is called the kinetic energy K of the particle.

(6-5) (6-5)

asvv 22

1

2

2

s

vva

2

2

1

2

2

s

vvmmaF

2

2

1

2

2

2

1

2

2 2

1

2

1mvmvFs

2

2

1mvk

Like work, the kinetic energy of a particle is a scalar quantity; it depends only Like work, the kinetic energy of a particle is a scalar quantity; it depends only on the particle’s mass and speed, not its direction of motion. We can interpret on the particle’s mass and speed, not its direction of motion. We can interpret Eq. 6-4 in terms of work and kinetic energy. The first term on the right side of Eq. 6-4 in terms of work and kinetic energy. The first term on the right side of Eq. (6-4) is KEq. (6-4) is K22 = ½ mv = ½ mv22

2 2 , the final kinetic energy of the particle (that is, after the , the final kinetic energy of the particle (that is, after the

displace-ment). The second term is the initial kinetic energy, Kdisplace-ment). The second term is the initial kinetic energy, K11 = ½ mv = ½ mv1122, and , and

difference between these terms is the change in kinetic energy. So Eq. (6-4) difference between these terms is the change in kinetic energy. So Eq. (6-4) says that says that thethe work done by the net force on a particle equals the change in work done by the net force on a particle equals the change in the particle’s kinetic energy:the particle’s kinetic energy:WWtottot = K = K22 – K – K11 = = KK

This result is This result is the work-energy theoremthe work-energy theorem..Because we used Newton’s laws in deriving the work-energy theorem, we can Because we used Newton’s laws in deriving the work-energy theorem, we can use it only in an inertial frame of reference. The speeds that we use to compute use it only in an inertial frame of reference. The speeds that we use to compute the kinetic energies and the distances that we use to compute work must be the kinetic energies and the distances that we use to compute work must be measured in an inertial frame. Note also that the work-energy theorem is valid measured in an inertial frame. Note also that the work-energy theorem is valid in any inertial frame, but the values of Win any inertial frame, but the values of W tottot and K and K22 – K – K11 may differ from one may differ from one

inertial frame to another (because the displacement and speed of a body may inertial frame to another (because the displacement and speed of a body may be diffe-rent in different frame).be diffe-rent in different frame).

6-4 6-4 Work and Energy With Varying ForcesWork and Energy With Varying ForcesSo far in this chapter we’ve considered work done by constant forces only. But So far in this chapter we’ve considered work done by constant forces only. But what happens when you stretch a spring? The more you stretch it, the harder what happens when you stretch a spring? The more you stretch it, the harder you you

to pull, so the force you exert is not constant as the spring is stretched. We’ve to pull, so the force you exert is not constant as the spring is stretched. We’ve also restricted our discussion to straight-line motion. You can think of many also restricted our discussion to straight-line motion. You can think of many situations in which a force that varies in magnitude, direction or both acts on a situations in which a force that varies in magnitude, direction or both acts on a body moving along a curved path. We need to be able to compute the work body moving along a curved path. We need to be able to compute the work done by the force in these more general cases. Fortunately, we’ll find that the done by the force in these more general cases. Fortunately, we’ll find that the work-energy theorem holds true even when varying forces are considered and work-energy theorem holds true even when varying forces are considered and when the body’s path is not straight.when the body’s path is not straight. To add only one complication at a time, let’s consider straight-line motion To add only one complication at a time, let’s consider straight-line motion with a force that is directed along the line but with an x-component F that may with a force that is directed along the line but with an x-component F that may change as the body moves. For example, imagine a train moving on a straight change as the body moves. For example, imagine a train moving on a straight track with the engineer constantly changing the locomotive’s throttle setting or track with the engineer constantly changing the locomotive’s throttle setting or applying the brakes. Suppose a particle moves along the x-axis from point xapplying the brakes. Suppose a particle moves along the x-axis from point x11 to to

xx22. .

Fig. 6-1(a) Fig. 6-11(b) Fig. 6-12 Fig. 6-1(a) Fig. 6-11(b) Fig. 6-12

O

Fx

FaFb

Fc

FdFe

Ff

Δxa Δxf

X

x2 – x1

O

F

Fx

x1 x2

x

x2-x1

O

Fx

x1 x2

x

x2-x1

Figure 6-11(a) is a graph of the x-component of force as a function of the Figure 6-11(a) is a graph of the x-component of force as a function of the partic-le’s coordinate x. To find the work done by this force, we divide the total partic-le’s coordinate x. To find the work done by this force, we divide the total dis-placement into small segments dis-placement into small segments xxaa, , xxbb and so on average force F and so on average force Faa in that in that

seg-ment multiplied by the displacement seg-ment multiplied by the displacement xxa. a. WeWe do this for each segment and do this for each segment and

then add the results for all the segments. The work done by the force in the then add the results for all the segments. The work done by the force in the total displacement from xtotal displacement from x11 to x to x22 is approximately is approximately

W = FW = Fa a xxa a + F + Fb b xxb b + + ….….

As the number of segments becomes very large and the width of each As the number of segments becomes very large and the width of each becomes very small, this sum becomes (in the limit) the integral of F from xbecomes very small, this sum becomes (in the limit) the integral of F from x11 to to

xx22 : :

(6-7) (6-7)Note that Note that FFa a xxa a represents the area of the first vertical strip in Fig. 6-11(b) and represents the area of the first vertical strip in Fig. 6-11(b) and

that the integral in Eq. (6-7) represents the area under the curve of Fig. 6-11(a) that the integral in Eq. (6-7) represents the area under the curve of Fig. 6-11(a) between xbetween x11 and x and x22. On a graph of force as a function of position, the total work . On a graph of force as a function of position, the total work

done by the force is represented by the area under the curve between the initial done by the force is represented by the area under the curve between the initial and final positions. An alternative interpretation of Eq. (6-7) is that the work W and final positions. An alternative interpretation of Eq. (6-7) is that the work W equals the average force that acts over the entire displacement, multiplied by equals the average force that acts over the entire displacement, multiplied by the displacement.the displacement. Equation (6-7) also applies if F, the x-component of the force, is constant. Equation (6-7) also applies if F, the x-component of the force, is constant.

2

1

x

x Fdxw

In that case, F may be taken outside the integral:In that case, F may be taken outside the integral:

but xbut x22 – x – x11 = s, the total displacement of the particle. So in the case of a = s, the total displacement of the particle. So in the case of a

constant force F, Eq. (6-7) says that W = Fs, in agreement with Eq. (6-1). The constant force F, Eq. (6-7) says that W = Fs, in agreement with Eq. (6-1). The interpret-ation of work as the area under the curve of F as a function of x also interpret-ation of work as the area under the curve of F as a function of x also holds for a constant force; W = Fs is the area of a rectangle of height F and holds for a constant force; W = Fs is the area of a rectangle of height F and width s.width s. Now let’s apply what we’ve learned to the stretched spring. If the elongation Now let’s apply what we’ve learned to the stretched spring. If the elongation x is not too great, we find that F is directly proportional to x:x is not too great, we find that F is directly proportional to x:F = kx (6-8)F = kx (6-8)where k is a constant called the force constant (or spring constant) of the where k is a constant called the force constant (or spring constant) of the spring. Equation (6-8) shows that the units of k are force divided by distance, spring. Equation (6-8) shows that the units of k are force divided by distance, N/m in SI units. The observation that elongation is directly proportional to force N/m in SI units. The observation that elongation is directly proportional to force for elongations that are not too great was made by Robert Hooke in 1678 and for elongations that are not too great was made by Robert Hooke in 1678 and is known as Hooke’s law. It really shouldn’t be called a “law”, since it’s a is known as Hooke’s law. It really shouldn’t be called a “law”, since it’s a statement about a specific device and not a fundamental law of nature.statement about a specific device and not a fundamental law of nature. To stretch a spring, we must do work. We apply equal and opposite forces to To stretch a spring, we must do work. We apply equal and opposite forces to ends of the spring and gradually increase the forces. The force at the moving ends of the spring and gradually increase the forces. The force at the moving end does do work. The work done by F when the elongation goes from zero to end does do work. The work done by F when the elongation goes from zero to a maximum value X isa maximum value X is

2

1

2

1 12

x

x

x

x xxFdxFFdxW

We can also obtain this result graphically. The area of the shaded We can also obtain this result graphically. The area of the shaded triangle in Fig.6-14, representing the total work done by the force, is triangle in Fig.6-14, representing the total work done by the force, is equal to half the pro-duct of the base and altitude, orequal to half the pro-duct of the base and altitude, or

This equation also says that the work is the average force kx/2 multiplied This equation also says that the work is the average force kx/2 multiplied by the total displacement X. We see that the total work is proportional to by the total displacement X. We see that the total work is proportional to the square of the final elongation X. To stretch an ideal spring by 2 cm, the square of the final elongation X. To stretch an ideal spring by 2 cm, you must do four times as much work as is needed to stretch it by 1 cm.you must do four times as much work as is needed to stretch it by 1 cm.Equation (6-9) assumes that the spring was originally unstretched. If Equation (6-9) assumes that the spring was originally unstretched. If initially the spring is already stretched a distance xinitially the spring is already stretched a distance x11, the work we must , the work we must

do to stretch it to a greater elongation xdo to stretch it to a greater elongation x22 is is

20 0 2

1kxkxdxFdxw x x

x

F=kx

F

xO

kx

Fig. 6-4

2

2

1

2

1kxkxxW

WORK-ENERGY THEOREM FOR STRAIGHT-LINE MOTION, VARYING WORK-ENERGY THEOREM FOR STRAIGHT-LINE MOTION, VARYING FORCESFORCESIn Section 6-3 we derived the work-energy theorem, WIn Section 6-3 we derived the work-energy theorem, W tottot = K = K22 – K – K11, for the , for the

special case of straight-line motion with a constant net force. We can now special case of straight-line motion with a constant net force. We can now prove that this theorem is true even when the force varies with position. As in prove that this theorem is true even when the force varies with position. As in section 6-3, let’s consider a particle that undergoes a displacement x while section 6-3, let’s consider a particle that undergoes a displacement x while being acted on by a net force with x-component F, which we now allow to vary. being acted on by a net force with x-component F, which we now allow to vary. Just as in Fig. 6-11, we divide the total displacement x into a large number of Just as in Fig. 6-11, we divide the total displacement x into a large number of small segments small segments x. We can apply the work-energy theorem, Eq. (6-6), to each x. We can apply the work-energy theorem, Eq. (6-6), to each segment because the value of F in each small segment is approximately segment because the value of F in each small segment is approximately constant. The change in kinetic energy in segment constant. The change in kinetic energy in segment xxaa is equal to the work F is equal to the work Fa a

xxaa , and so on. The total change of kinetic energy is the sum of the changes in , and so on. The total change of kinetic energy is the sum of the changes in

the individual segments, and thus is equal to the total work done on the particle the individual segments, and thus is equal to the total work done on the particle during the entire displacement. So Wduring the entire displacement. So W tottot = = holds for varying forces as well as holds for varying forces as well as

for constant ones.for constant ones.Here’s an alternative derivation of the work-energy theorem for a force that may Here’s an alternative derivation of the work-energy theorem for a force that may vary with position. It involves making a change of variable from x to v in the vary with position. It involves making a change of variable from x to v in the workwork

2

1

2

2 2

1

2

12

1

2

1

kxkxkxdxFdxw x

x

x

x

6-10

integral. As a preliminary, we note that the acceleration a of the particle can be integral. As a preliminary, we note that the acceleration a of the particle can be expressed in various ways, using a = dv/dt, v = dx/dt, and the chain rule for expressed in various ways, using a = dv/dt, v = dx/dt, and the chain rule for derivatives:derivatives:

(6-11) (6-11)using this result, Eq. (6-7) tells us that the total work done by the net force F isusing this result, Eq. (6-7) tells us that the total work done by the net force F is

(6-12) (6-12)Now (dv/dx)dx is the change in velocity dv during the displacement dx, so in Now (dv/dx)dx is the change in velocity dv during the displacement dx, so in Eq. (6-12) we can substitute dv for (dv/dx)dx. This changes the integration Eq. (6-12) we can substitute dv for (dv/dx)dx. This changes the integration variable from x to v, so we change the limits from xvariable from x to v, so we change the limits from x11 and x and x22 to the to the

corresponding velocities vcorresponding velocities v11 and v and v22 at these points. This gives us at these points. This gives us

The integral of v dv is just vThe integral of v dv is just v22/2. Substituting the upper and lower limits, we /2. Substituting the upper and lower limits, we finally findfinally find (6-13) (6-13)

dx

dvv

dt

dx

dx

dv

dt

dva

2

1

2

1

2

1

x

x

x

x

x

xtot dxdx

dvmvmadxFdxW

2

1

v

vtot mvdvW

2

1

2

2 2

1

2

1mvmvWtot

WORK-ENERGY THEOREM FOR MOTION ALONG A CURVEWORK-ENERGY THEOREM FOR MOTION ALONG A CURVEWe can generalize our definition of work further to include a force that varies in We can generalize our definition of work further to include a force that varies in direction as well magnitude and a displacement that lies along a curved path. direction as well magnitude and a displacement that lies along a curved path. Sup-pose a particle moves from point PSup-pose a particle moves from point P11 to P to P22 along a curve, as shown in Fig. along a curve, as shown in Fig.

6-17a. We divide the portion of the curve between these points into many 6-17a. We divide the portion of the curve between these points into many infinitesimal vector displacements, and we call a typical one of these dl. Each dl infinitesimal vector displacements, and we call a typical one of these dl. Each dl is tangent to the path at its position. Let F be the force at a typical point along is tangent to the path at its position. Let F be the force at a typical point along the path, and let the path, and let be the along between F and dl at this point. Then the small be the along between F and dl at this point. Then the small element of work dw done on the particle during the displacement dl may be element of work dw done on the particle during the displacement dl may be written as written as dW = F cosdW = F cosdl = Fdl = F dl = Fdl = F••dl ,dl ,

where Fwhere F = F cos = F cos is the component of F in the direction parallel to dl (Fig. 6- is the component of F in the direction parallel to dl (Fig. 6-

17b). The total work done by F on the particle as it moves from P17b). The total work done by F on the particle as it moves from P11 to P to P22 is then is then

(6-14) (6-14)

Fig. 6-7a Fig. 6-7b Fig. 6-7a Fig. 6-7b

ldFdlFdlFW P

P

P

P

P

P

2

1

2

1

2

1

cos

dl

P1

P2F

F

F = Fcos

dl

P1

P2

F

We can show that the work-energy theorem, Eq. (6-6), holds true even with We can show that the work-energy theorem, Eq. (6-6), holds true even with varying forces and a displacement along a curved path. The force F is varying forces and a displacement along a curved path. The force F is essentially constant over any given infinitesimal segment dl of the path, so we essentially constant over any given infinitesimal segment dl of the path, so we can apply the work-energy theorem for straight-line motion to that segment can apply the work-energy theorem for straight-line motion to that segment equals the work from dW = Fequals the work from dW = F|| || dl = Fdl done on the particle. Adding up these dl = Fdl done on the particle. Adding up these

infinitesimal quantities of work from all the segments along the whole path infinitesimal quantities of work from all the segments along the whole path gives the total work done, Eq. (6-4), and this equals the total change in kinetic gives the total work done, Eq. (6-4), and this equals the total change in kinetic energy over the whole path. So Wenergy over the whole path. So Wtottot = = K = KK = K22 – K – K11 is true in general, no matter is true in general, no matter

what the path and no matter what the character of the forces. This can be what the path and no matter what the character of the forces. This can be proved more rigorously by using steps like those in Eqs. (6-11) through (6-13).proved more rigorously by using steps like those in Eqs. (6-11) through (6-13).Note that only the component of the force parallel to the path, FNote that only the component of the force parallel to the path, F11, does work on , does work on

the particle, so only this component can change the speed and kinetic energy the particle, so only this component can change the speed and kinetic energy of the particle. The component perpendicular to the path, Fof the particle. The component perpendicular to the path, F = F sin = F sin, has no , has no

effect on the particle’s speed; it only acts to change the particle’s direction.effect on the particle’s speed; it only acts to change the particle’s direction.The integral in Eq. (6-14) is called a line integral. To evaluate this integral in a The integral in Eq. (6-14) is called a line integral. To evaluate this integral in a speci-fic problem, we need some sort of detailed description of the path and of speci-fic problem, we need some sort of detailed description of the path and of how F varies along the path.how F varies along the path.Example 6-9.Example 6-9.At a family picnic you are appointed to push your obnoxious cousin At a family picnic you are appointed to push your obnoxious cousin Throckmorton in a swing. His weight is w, the length of the chains is R, and you Throckmorton in a swing. His weight is w, the length of the chains is R, and you push Throcky until the chains make an angle push Throcky until the chains make an angle 00 with the vertical. To do this, with the vertical. To do this,

you exert a varyingyou exert a varying

horihorizontal force F that starts at zero and gradually increase just enough so that zontal force F that starts at zero and gradually increase just enough so that Throcky and the swing move very slowly and remain very nearly in equilibrium. Throcky and the swing move very slowly and remain very nearly in equilibrium. What is the total work done on Throcky by all forces? What is the work done by What is the total work done on Throcky by all forces? What is the work done by the tension T in the chains? What is the work you do by exerting the force F? the tension T in the chains? What is the work you do by exerting the force F? (Neglect the weight of chains and seat).(Neglect the weight of chains and seat).Solution: The free-body diagram is shown in Fig. 6-18b. We have replaced the Solution: The free-body diagram is shown in Fig. 6-18b. We have replaced the tensions in the two chains with a single tension T. Because Throcky is in equili-tensions in the two chains with a single tension T. Because Throcky is in equili-brium at every point, the net force on him is zero, and the total work done on brium at every point, the net force on him is zero, and the total work done on him by all forces is zero. At any point during the motion the chain force on him by all forces is zero. At any point during the motion the chain force on Throcky is perpendicular to each dl, so the angle between the chain force and Throcky is perpendicular to each dl, so the angle between the chain force and the displace-ment is always 90the displace-ment is always 90°. Therefore the work done by the chain tension °. Therefore the work done by the chain tension is zero.is zero.To compute the work done by F, we have to find out how it varies with the To compute the work done by F, we have to find out how it varies with the angle angle . Throcky is in equilibrium . Throcky is in equilibrium at every point, so from at every point, so from FFxx = 0 we get = 0 we get

F + (-T sin F + (-T sin) = 0,) = 0,and from and from F Fyy = 0 we find = 0 we find

T cos T cos + (-w) = 0 + (-w) = 0By eliminating T from these two equations, we obtainBy eliminating T from these two equations, we obtain F = w tan F = w tan . .The point where F is applied swings through the arc s.The point where F is applied swings through the arc s.The arc length s equals the radius R of the circular path The arc length s equals the radius R of the circular path w

Tsin F

Tcos

y

T

Multiplied by the length Multiplied by the length (in radians), so s = R (in radians), so s = R. Therefore the displacement dl . Therefore the displacement dl corresponding to a small change of angle dcorresponding to a small change of angle d has a magnitude dl = ds = Rd has a magnitude dl = ds = Rd . . The work done by F is The work done by F is

Now we express everything in terms of the varying angle Now we express everything in terms of the varying angle :: . .

If If 00 = 0, there is no displacement; in that case, cos = 0, there is no displacement; in that case, cos 00 = 0 and W = wR. In that = 0 and W = wR. In that

case the work you do is the same as if you had lifted Throcky straight up a case the work you do is the same as if you had lifted Throcky straight up a distance R with a force equal to his weight w. In fact, the quantity R(1-cos distance R with a force equal to his weight w. In fact, the quantity R(1-cos 00) is ) is

the increase in his height above the ground during the displacement, so for any the increase in his height above the ground during the displacement, so for any value of value of 00 the work done by force F is the change in height multiplied by the the work done by force F is the change in height multiplied by the

weight. This is an example of a more general result that we’ll prove in Section weight. This is an example of a more general result that we’ll prove in Section 7-2.7-2.6-5 Power6-5 PowerWhen a quantity of work is done might When a quantity of work is done might W is done during a time interval W is done during a time interval t, the t, the average work done per unit or average power Paverage work done per unit or average power Pav av is defined to be is defined to be

(6-15) (6-15)

dsFldFW cos

00

0 00 cos1sincostan wRdwRRdwW

t

WPav

The rate at which work is done might not be constant. Even when it varies, The rate at which work is done might not be constant. Even when it varies, we can define instantaneous P as the limit of the quotient in Eq. (6-15) at we can define instantaneous P as the limit of the quotient in Eq. (6-15) at t approaches zero:t approaches zero:

(6-16) (6-16)

The SI unit of power is the watt (W), named for the English inventor James The SI unit of power is the watt (W), named for the English inventor James Watt. Watt.

dt

dW

t

WP

t

lim

0

7 Potential Energy and 7 Potential Energy and Energy ConservationEnergy Conservation

7-1 Introduction7-1 Introduction

When a dive jumps off a high board into a swimming pool, she hits the water When a dive jumps off a high board into a swimming pool, she hits the water moving pretty fast, with a lot of kinetic energy. Where does that energy come moving pretty fast, with a lot of kinetic energy. Where does that energy come from? The answer we learned in Chapter 6 was that the gravitational force from? The answer we learned in Chapter 6 was that the gravitational force (her weight) does work on the diver as she falls. The diver’s kinetic energy-(her weight) does work on the diver as she falls. The diver’s kinetic energy-energy associated with her motion-increase by an amount equal to the work energy associated with her motion-increase by an amount equal to the work done.done.

But there is a very useful alternative way to think about work and kinetic But there is a very useful alternative way to think about work and kinetic energy. This new approach is based on the concept of potential energy, energy. This new approach is based on the concept of potential energy, which is energy associated with the position of a system rather than its which is energy associated with the position of a system rather than its motion. In this approach, there is gravitational potential energy even while the motion. In this approach, there is gravitational potential energy even while the diver is standing on the high board. Energy is not added to the earth-diver diver is standing on the high board. Energy is not added to the earth-diver system as the diver falls, but rather a storehouse of energy is transformed system as the diver falls, but rather a storehouse of energy is transformed from one form (potential energy) to another( kinetic energy) as she falls. In from one form (potential energy) to another( kinetic energy) as she falls. In this chapter we’ ll see how this transformation can be understand from the this chapter we’ ll see how this transformation can be understand from the work-energy theorem. work-energy theorem.

7-2 GRAVITATIONAL POTENTIAL ENERGY7-2 GRAVITATIONAL POTENTIAL ENERGYA particle gains or loses kinetic energy because it interacts with other objects A particle gains or loses kinetic energy because it interacts with other objects that exert forces on it. We learned in Chapter 6 that during any interaction the that exert forces on it. We learned in Chapter 6 that during any interaction the change in a particle by the forces that act on it.change in a particle by the forces that act on it.When a body falls without air resistance, gravitational potential energy When a body falls without air resistance, gravitational potential energy decreases and the body’s kinetic energy increases. But in Chapter 6 we used decreases and the body’s kinetic energy increases. But in Chapter 6 we used the work-energy theorem to conclude that the kinetic energy of a falling body the work-energy theorem to conclude that the kinetic energy of a falling body increases because the force of the earth’s gravity (the body’s weight) does increases because the force of the earth’s gravity (the body’s weight) does work on the body. Let’s use the work-energy theorem to show that gravitational work on the body. Let’s use the work-energy theorem to show that gravitational potential energy. potential energy. Let’s consider a body with mass m that moves along the (vertical) y-axis, as in Let’s consider a body with mass m that moves along the (vertical) y-axis, as in Fig 7-1. The forces acting on it are its weight, with magnitude w = mg, and Fig 7-1. The forces acting on it are its weight, with magnitude w = mg, and possibly some other forces; we call the vector sum (resultant) of all the other possibly some other forces; we call the vector sum (resultant) of all the other forces Fforces Fotherother..

Fig 7-1(a) Fig 7-1(b) Fig 7-1(a) Fig 7-1(b)

W = mg

Fother

y1

y2

O

W = mg

Fother

y2

y1

We’ll assume that the body stays close enough to the earth’s surface that the We’ll assume that the body stays close enough to the earth’s surface that the weight is constant. We want to find the work done by the weight when the body weight is constant. We want to find the work done by the weight when the body drops from a height ydrops from a height y11 above the origin to a lower height y above the origin to a lower height y22 (Fig. 7-1a). The (Fig. 7-1a). The

weight and displace-ment are in the same direction, so the work Wweight and displace-ment are in the same direction, so the work Wgravgrav done on done on

the body by its weight is positive; the body by its weight is positive; WWgravgrav = Fs = w(y = Fs = w(y22 – y – y11) = mgy) = mgy11 – mgy – mgy22 (7-1) (7-1)

This expression also gives the correct work when the body moves upward and This expression also gives the correct work when the body moves upward and yy22 is greater than y is greater than y11 (Fig. 7-1b). In that case the quantity (y (Fig. 7-1b). In that case the quantity (y11 – y – y22) is negative, ) is negative,

and Wand Wgravgrav is negative because the weight and displacement are opposite in is negative because the weight and displacement are opposite in

direction.direction.Equation (7-1) shows that we can express WEquation (7-1) shows that we can express Wgravgrav in terms of the values of the in terms of the values of the

quan-tity quan-tity mgy mgy at the beginning and end of the displacement. This quantity, the at the beginning and end of the displacement. This quantity, the product of the weight product of the weight mgmg and the height y above the origin of coordinates, is and the height y above the origin of coordinates, is the gravitational energy, U:the gravitational energy, U:U = U = mgy mgy (7-2)(7-2)Its initial value is UIts initial value is U11 = = mgymgy11 , and its final value is U, and its final value is U22 = = mgymgy22 . . TheThe change in U change in U

is the final value minus the initial value, or is the final value minus the initial value, or U = UU = U22 – U – U11. We can express the . We can express the

work Wwork Wgravgrav done by the gravitational force during the displacement from y done by the gravitational force during the displacement from y11 to y to y22

as as WWgrav grav = U = U11 – U – U22 = -( U = -( U22 – U – U11 ) = - ) = - U (7-3) U (7-3)

The negative sign in front of The negative sign in front of U is essential. When the U is essential. When the body moves up, y increase, the work done by the body moves up, y increase, the work done by the gravitational force is negative, and the gravitational gravitational force is negative, and the gravitational potential energy increase ( potential energy increase ( U >0). When the body U >0). When the body moves down, y decreases, the gravitational force does moves down, y decreases, the gravitational force does positive work, and the gravitational potential energy positive work, and the gravitational potential energy decreases (decreases (U<0).U<0).Conservation of Mechanical Energy ( Gravitational Force Conservation of Mechanical Energy ( Gravitational Force Only )Only )To see what gravitational potential energy is good for, To see what gravitational potential energy is good for, suppose the body’ssuppose the body’s

weight is the only force acting on it, so Fweight is the only force acting on it, so Fotherother = 0. The body is then falling freely = 0. The body is then falling freely

with no air resistance and can be moving either up or down. Let its speed at with no air resistance and can be moving either up or down. Let its speed at point ypoint y11 be v be v11, and let its speed at y, and let its speed at y22 be v be v22. The work-energy theorem, Eq. (6-6), . The work-energy theorem, Eq. (6-6),

says that the total work done on the body equals the change in the body’s in says that the total work done on the body equals the change in the body’s in the body’s kinetic energy; Wthe body’s kinetic energy; Wtot tot = = K = KK = K22 – K – K11. If gravity is the only force that . If gravity is the only force that

acts, then from Eq. (7-3), Wacts, then from Eq. (7-3), Wtottot = W = Wgravgrav = - = -U = UU = U11 – U – U22. Pointing there together, . Pointing there together,

we getwe getU or KU or K22 – K – K11 = U = U11 – U – U2 2 , which we can rewrite as K, which we can rewrite as K11 + U + U11 = K = K22 + +

UU22 , (7-4) or , (7-4) or

(7-5) (7-5)We now define the sum K + U of kinetic and potential energy to be E, the total We now define the sum K + U of kinetic and potential energy to be E, the total mechanical energy of the system. By “system” we mean the body of mass m mechanical energy of the system. By “system” we mean the body of mass m and the earth considered together, because gravitational potential energy U is and the earth considered together, because gravitational potential energy U is a shared pro-perty of both bodies. Then Ea shared pro-perty of both bodies. Then E11 = K = K11 + U + U11 is the total mechanical is the total mechanical

energy at yenergy at y11, and E, and E22 = K = K22 + U + U22 is the total mechanical energy at y is the total mechanical energy at y22. Equation (7-. Equation (7-

4) says that when the body’s weight is the only force doing work on it, E4) says that when the body’s weight is the only force doing work on it, E11 = E = E22. .

Thus is, E is constant; it has the same value at yThus is, E is constant; it has the same value at y11 and y and y22. But since the . But since the

positions ypositions y11 and y and y22 are arbitrary points in the motion of the body, the total are arbitrary points in the motion of the body, the total

mechanical energy E has the same value at all points during the motion, E = K mechanical energy E has the same value at all points during the motion, E = K + U = constant.+ U = constant.

2

2

21

2

1 2

1

2

1mgymvmgymv

A quantity that always has the same value is called a conserved quantity. A quantity that always has the same value is called a conserved quantity. When only the force of gravity does work, the total mechanical energy is When only the force of gravity does work, the total mechanical energy is constant, that is, is conserved. This is our first example of the conservation of constant, that is, is conserved. This is our first example of the conservation of mechanical energy.mechanical energy.Effect of Other ForceEffect of Other ForceWhen other forces act on the body in addition to its weight, then FWhen other forces act on the body in addition to its weight, then Fotherother in Fig. 7-1 in Fig. 7-1

is not zero. The gravitational work Wis not zero. The gravitational work Wgravgrav is still given by Eq. (7-3), but the total is still given by Eq. (7-3), but the total

work Wwork Wtottot is then the sum of W is then the sum of Wgravgrav, and the work done by all forces is W, and the work done by all forces is W tottot = W = Wgravgrav

+ W+ Wotherother. Equating this to the change in kinetic energy, we have. Equating this to the change in kinetic energy, we have

W Wotherother+ W+ Wgravgrav + = K + = K22 – K – K11 (7-6) (7-6)

Also, from Eq. (7-3), WAlso, from Eq. (7-3), Wgravgrav = U = U11 – U – U22, so , so

W Wotherother + U + U11 –U –U22 = K = K22 – K – K11,,

which we can rearrange in the formwhich we can rearrange in the formKK11 + U + U11 = W = Wgravgrav = K = K22 + U + U22 (if forces other than gravity do work) (7-7) (if forces other than gravity do work) (7-7)

Finally, using the appropriate expressions for the various energy terms, we Finally, using the appropriate expressions for the various energy terms, we obtainobtain

(7-8) (7-8) The meaning of Eqs. (7-7) and (7-8) is this: The work done by all forces The meaning of Eqs. (7-7) and (7-8) is this: The work done by all forces other than the gravitational force equals the change in the total mechanical other than the gravitational force equals the change in the total mechanical energy E = K + U of the system, where U is the gravitational potential energy. energy E = K + U of the system, where U is the gravitational potential energy.

2

2

21

2

1 2

1

2

1mgymvWmgymv other

Gravitational Potential Energy For Motion Along A Curved PathGravitational Potential Energy For Motion Along A Curved Path

Figure 7-4 (a) (b) Figure 7-4 (a) (b)the body is acted on by the gravitational force w = mg and possibly by other the body is acted on by the gravitational force w = mg and possibly by other forces whose resultant we call Fforces whose resultant we call Fotherother. To find the work done by the gravitational . To find the work done by the gravitational

force during this displacement, we divide the path up into small segments force during this displacement, we divide the path up into small segments s; a s; a typical segment is shown in Fig. 7-4b. The work done by the gravitational force typical segment is shown in Fig. 7-4b. The work done by the gravitational force over this segment is the scalar product of the force and the displacement. In over this segment is the scalar product of the force and the displacement. In terms of unit vectors, the force is w = mterms of unit vectors, the force is w = mgg = -mg = -mgj j and the displacement is and the displacement is ss = = xxii + + yyjj , so the work done by the gravitational force is , so the work done by the gravitational force is w w ··ss = -mg = -mgjj·(·(xxii + + yyj) = -j) = - mg mg y .y .The work done by gravity is the same as though the body had been displaced The work done by gravity is the same as though the body had been displaced ver-tically a distance ver-tically a distance y, with no horizontal displacement. This is true for every y, with no horizontal displacement. This is true for every seg-ment, so the total work done by the gravitational force is –mg multiplied by seg-ment, so the total work done by the gravitational force is –mg multiplied by the the

y1

y2W = mg

Fother

s

x

y

W = mg

Total vertical displacement (yTotal vertical displacement (y22 – y – y11): ):

WWgravgrav = - mg( y = - mg( y2 2 – y– y11) = mgy) = mgy11 – mgy – mgy2 2 = U= U11 – U – U22 . .

This work done by gravity is the same as though the body had been displaced This work done by gravity is the same as though the body had been displaced ver-tical path. So even if the path a body follows between two points is curved, ver-tical path. So even if the path a body follows between two points is curved, the total work done by the gravitational force depends only on the difference in the total work done by the gravitational force depends only on the difference in height between the two points of the path. This work is unaffected by any height between the two points of the path. This work is unaffected by any horizontal mo-tion that may occur. So we can use the same expression for horizontal mo-tion that may occur. So we can use the same expression for gravitational potential energy whether the body’s path is curved or straight.gravitational potential energy whether the body’s path is curved or straight.

7-3 Elastic Potential Energy7-3 Elastic Potential EnergyWe’ll describe the process of storing energy in a deformable body body such We’ll describe the process of storing energy in a deformable body body such as a spring or rubber band in terms of elastic potential energy. A body is called as a spring or rubber band in terms of elastic potential energy. A body is called elastic if it returns to its original shape and size after being deformed. To be elastic if it returns to its original shape and size after being deformed. To be specific, we’ll consider storing energy in an ideal spring like the ones we specific, we’ll consider storing energy in an ideal spring like the ones we discussed in Section 6-4. To keep such an ideal spring stretched by a distance discussed in Section 6-4. To keep such an ideal spring stretched by a distance x, we must exert a force F = kx, where k is the force constant of the spring. The x, we must exert a force F = kx, where k is the force constant of the spring. The ideal spring is a useful idealization because many other elastic bodies show ideal spring is a useful idealization because many other elastic bodies show this same direct proportionality between force F and displacement x, provided this same direct proportionality between force F and displacement x, provided that x is sufficiently small.that x is sufficiently small.We found in Section 6-4 that the work we must do on the spring to move one We found in Section 6-4 that the work we must do on the spring to move one end form an elongation xend form an elongation x11 to a different elongation x to a different elongation x22 is is

(work done on a spring) (work done on a spring) where k is the force constant of the spring. If we stretch the spring further, we where k is the force constant of the spring. If we stretch the spring further, we do positive work on the spring; if we let the spring relax while holding one end, do positive work on the spring; if we let the spring relax while holding one end, we do negative work on it. We also saw that this expression for work is still we do negative work on it. We also saw that this expression for work is still correct if the spring is compressed, not stretched, so that xcorrect if the spring is compressed, not stretched, so that x11 or x or x22 or both are or both are

negative. Now we need to find the work done by the spring. From Newton’s negative. Now we need to find the work done by the spring. From Newton’s third law the two quan-tities of work are just negatives of each other. Changing third law the two quan-tities of work are just negatives of each other. Changing the signs in this equation, we find that in a displacement from xthe signs in this equation, we find that in a displacement from x11 to x to x22 the spring the spring

does an amount of work Wdoes an amount of work Welel given by given by

(work done by a spring) (work done by a spring)Just as for gravitational work, we can express the work done by the spring in Just as for gravitational work, we can express the work done by the spring in terms of a given quantity at the beginning and end of the displacement. This terms of a given quantity at the beginning and end of the displacement. This quantity is ½kxquantity is ½kx22..We can use Eq. (7-9) to express the work WWe can use Eq. (7-9) to express the work Welel done on the block by the elastic done on the block by the elastic

force in terms of the change in potential energy:force in terms of the change in potential energy:

(7-10). (7-10).The work-energy theorem says that WThe work-energy theorem says that W tottot = K = K22 – K – K11, no matter what kind of , no matter what kind of

forces are on a body. If the elastic force is the only forces that does work on the forces are on a body. If the elastic force is the only forces that does work on the body, thenbody, thenWWtottot = W = Welel = U = U11 – U – U22..

2

1

2

2 2

1

2

1kxkxW

2

2

2

1 2

1

2

1kxkxW

UUUkxkxWel 21

2

1

2

2 2

1

2

1

The work-energy theorem WThe work-energy theorem Wtottot = K = K22 – K – K11 then gives us then gives us

KK11 + U + U11 = K = K22 + U + U22 (if only the elastic force does work) (7-11). (if only the elastic force does work) (7-11).

Here U is given by Eq. (7-9), so Here U is given by Eq. (7-9), so

(if only the elastic force does work)(7- (if only the elastic force does work)(7-12).12).In this case the total mechanical energy E = K + U (the sum of kinetic and In this case the total mechanical energy E = K + U (the sum of kinetic and elastic potential energy) is conserved. For Eq. (7-12) to be strictly correct, the elastic potential energy) is conserved. For Eq. (7-12) to be strictly correct, the ideal spring that we’ve been discussing must also be massless. If the spring ideal spring that we’ve been discussing must also be massless. If the spring has a mass, it will also have kinetic energy as the coils of the spring move has a mass, it will also have kinetic energy as the coils of the spring move back and forth. We can neg-lect the kinetic energy of the spring if its mass is back and forth. We can neg-lect the kinetic energy of the spring if its mass is much less than the mass m of the body attached to the spring. For instance, a much less than the mass m of the body attached to the spring. For instance, a typical automobile has a mass of 1200 kg or more. The springs in its typical automobile has a mass of 1200 kg or more. The springs in its suspension have masses of only a few kilograms, so their mass can be suspension have masses of only a few kilograms, so their mass can be neglected if we want to study how a car bounces on its suspension.neglected if we want to study how a car bounces on its suspension. If forces other than the elastic force also do work on the body, we call their If forces other than the elastic force also do work on the body, we call their work Wwork Wotherother, as before. Then the total work is W, as before. Then the total work is Wtottot = W = Welel + W + Wotherother and the work- and the work-

energy theorem givesenergy theorem givesWWelel + W + Wotherother = K = K22 – K – K1 .1 .

The work done by the spring is still WThe work done by the spring is still Welel = U = U11 – U – U22, so again, so again

KK11 + U + U11 + W + Wotherother = K = K22 + U + U2 2 (if forces other than the elastic force do work) (7-13)(if forces other than the elastic force do work) (7-13)

2

2

2

2

2

1

2

1 2

1

2

1

2

1

2

1kxmvkxmv

andand (if forces (if forces other than the elastic force do work (7-14). other than the elastic force do work (7-14). This equation shows that the work done by all forces This equation shows that the work done by all forces other than the elastic force equals the change in the total other than the elastic force equals the change in the total mechanical energy E = K +U of the system, where U is mechanical energy E = K +U of the system, where U is the elastic potential energy. The “system” is made up of the elastic potential energy. The “system” is made up of the body of mass m and the spring of force constant k. the body of mass m and the spring of force constant k. when Wother is positive, E increases; when Wother is when Wother is positive, E increases; when Wother is negative, E decreases. You should compare Eq. (7-14) negative, E decreases. You should compare Eq. (7-14) to Eq. (7-8), which describes situations in which there is to Eq. (7-8), which describes situations in which there is gravitational potential energy but no elastic potential gravitational potential energy but no elastic potential energy.energy.

2

2

2

2

2

1

2

1 2

1

2

1

2

1

2

1kxmvWkxmv other

Situations With Both Gravitational And Elastic Potential Situations With Both Gravitational And Elastic Potential EnergyEnergyEquations (7-11), (7-12), (7-13), and (7-14) are valid when the Equations (7-11), (7-12), (7-13), and (7-14) are valid when the only potential energy in the system is elastic potential energy. only potential energy in the system is elastic potential energy. What happens when we have both gravitational and elastic forces, What happens when we have both gravitational and elastic forces, such as a block attached to the lower end of a vertically hang-ing such as a block attached to the lower end of a vertically hang-ing spring? We can still use Eq. (7-13spring? We can still use Eq. (7-13), ), but now Ubut now U11 and U and U22 are the are the

initial and final values of the total potential energy, including both initial and final values of the total potential energy, including both gravitational and elastic potential energies. That is, U = Ugravitational and elastic potential energies. That is, U = Ugravgrav + U + Uelel. .

Thus the most general statement of the relationship between Thus the most general statement of the relationship between kinetic energy, potential energy, and work done by other forces is kinetic energy, potential energy, and work done by other forces is

(valid in (valid in general) (7-15).general) (7-15).That is, That is, the work done by all forces other than the the work done by all forces other than the gravitational force or elastic force equals the change in the gravitational force or elastic force equals the change in the total mechanical energy E = K + U of the system, where U istotal mechanical energy E = K + U of the system, where U is

2,2,21,1,1 elgravotherelgrav UUKWUUK

the sum of the gravitational potential energy and the sum of the gravitational potential energy and elastic potential energy.elastic potential energy. 7-4 Conservative And Non-conservative Forces7-4 Conservative And Non-conservative ForcesIn our discussions of potential energy we have talked In our discussions of potential energy we have talked about “storing” kinetic energy by converting it to potential about “storing” kinetic energy by converting it to potential energy. A force that offers this opportunity of two-way energy. A force that offers this opportunity of two-way conversion between kinetic and potential energies is conversion between kinetic and potential energies is called acalled a conservative force. We have seen two We have seen two examples of conservative forces: the gravitational force examples of conservative forces: the gravitational force and the spring force. An essential feature of and the spring force. An essential feature of conservative forces is that their work is always conservative forces is that their work is always reversible.reversible. Anything that we deposit in the energy “bank” Anything that we deposit in the energy “bank” can later be with-drawn without loss. Another important can later be with-drawn without loss. Another important aspect of conservative forces is that a body may move aspect of conservative forces is that a body may move from point 1 to point 2 by variousfrom point 1 to point 2 by various

paths, but the wok done by a conservative force is the same for paths, but the wok done by a conservative force is the same for all of these paths. Thus if a body stays close to the surface of the all of these paths. Thus if a body stays close to the surface of the earth, the gravitational force mg is independent of height, and earth, the gravitational force mg is independent of height, and the work done by this force depends only on the change in the work done by this force depends only on the change in height. If the body moves around a closed path, ending at the height. If the body moves around a closed path, ending at the same point where it started, the total work done by the same point where it started, the total work done by the gravitational force is always zero.gravitational force is always zero. The work done by a conservative force always has these The work done by a conservative force always has these properties:properties:1. It can always be expressed as the difference between the 1. It can always be expressed as the difference between the initial and final values of a potential energy function.initial and final values of a potential energy function.2. It is reversible.2. It is reversible.3. It is independent of the body and depends only on the starting 3. It is independent of the body and depends only on the starting and ending points.and ending points.4. When the starting and ending points are the same, the total 4. When the starting and ending points are the same, the total work is zero. work is zero.

When the only forces that do work are conservative forces, the total mechanical When the only forces that do work are conservative forces, the total mechanical energy E = K + U is constant.energy E = K + U is constant.A force that is not conservative is called a nonconservative force. The work A force that is not conservative is called a nonconservative force. The work done by a nonconservative force cannot be represented by a potential-energy done by a nonconservative force cannot be represented by a potential-energy function. Some nonconservative force, like kinetic friction or fluid resistance, function. Some nonconservative force, like kinetic friction or fluid resistance, cause mechanical energy to be lost or dissipated; a force of this kind is called a cause mechanical energy to be lost or dissipated; a force of this kind is called a dissipative force. There are also nonconservative forces that increase dissipative force. There are also nonconservative forces that increase mechanical energy. The fragments of an exploding firecracker fly off with very mechanical energy. The fragments of an exploding firecracker fly off with very large kinetic energy, thanks to a chemical reaction of gunpowder with oxygen.large kinetic energy, thanks to a chemical reaction of gunpowder with oxygen.

THE LAW OF CONSERVATION OF ENERGYTHE LAW OF CONSERVATION OF ENERGYNonconservative forces cannot be represented in terms of potential energy. But Nonconservative forces cannot be represented in terms of potential energy. But we can describe the effects of these forces in terms of kinds of energy other we can describe the effects of these forces in terms of kinds of energy other than kinetic and potential energy. When a car with locked brakes skids to a than kinetic and potential energy. When a car with locked brakes skids to a stop, the tires and the road surface both become hotter. The energy associated stop, the tires and the road surface both become hotter. The energy associated with this change in the state of the materials is called internal energy. Raising with this change in the state of the materials is called internal energy. Raising the temperature of a body increase its internal energy; lowering the body’s the temperature of a body increase its internal energy; lowering the body’s temperature of a body decreases its internal energy. temperature of a body decreases its internal energy. To see the significance of internal energy, let’s consider a block sliding on a To see the significance of internal energy, let’s consider a block sliding on a rough surface. Friction does negative work on the block as it slides, and the rough surface. Friction does negative work on the block as it slides, and the change in inter-nal energy of the block and surface (both of which get hotter) is change in inter-nal energy of the block and surface (both of which get hotter) is positive.positive.

Careful experiments show that the increase in the internal energy Careful experiments show that the increase in the internal energy is exactly equal to the absolute value of the work done by friction. is exactly equal to the absolute value of the work done by friction. In other words,In other words, UUintint = - W = - Wotherother , ,

where where UUintint is the change in internal energy. If we substitute this is the change in internal energy. If we substitute this

into Eq. (7-7), Eq. (7-13), or (7-15), we findinto Eq. (7-7), Eq. (7-13), or (7-15), we find K K11 + U + U11 - - UUintint = K = K22 + U + U2 .2 .

Writing Writing K = KK = K22 – K – K11 and and U = UU = U22 – U – U11, we can finally express , we can finally express

this asthis asK + K + U + U + UUintint = 0 (law of conservation energy) (7-16). = 0 (law of conservation energy) (7-16).

This remarkable statement is the general form of the law of This remarkable statement is the general form of the law of conservation. In a given process, the kinetic energy, potential conservation. In a given process, the kinetic energy, potential energy, and internal energy of a system may all change. But the energy, and internal energy of a system may all change. But the sum of those changes is always zero. If there is a decrease in one sum of those changes is always zero. If there is a decrease in one form of energy, it is made up for by an increase in the other forms. form of energy, it is made up for by an increase in the other forms. When we ex-pand our definition of energy to include internalWhen we ex-pand our definition of energy to include internal

energy, Eq. (7-16) says that energy is never created or destroyed; it only changes form. No exception to this rule has ever been found. 7-5 Force and Potential EnergyFor the two kinds of conservative forces (gravitational and elastic) we have studied, we started with a description of behavior of the force and derived from that an expression for the potential energy. We can reverse this procedure: If we are given a potential-expression, we can find the corresponding force.

Let’s apply this to a small displacement x. The work done by the force during this displacement is approximately equal to . We have to say “approximately” because may vary a little over the interval x. But it is at least approximately true that and .

UW

xFx xxFx

xFx UxxFx

x

UxFx

)(

dx

xdUxFx

Force and Potential Energy in Three Dimensions We can extend this analysis to three dimensions, where the particle may move in the x-, y-, z-direction, or all at once, under the action of a conservative force that has components Fx, Fy , Fz . Each component of force may be a function of the coordinates x, y and z. The potential-energy function U is also a function of all three space coordinates.

To make these relations exact, we need to take the limits x 0, y 0, and z0. So that these ratios become derivatives. Because U may be a function of all three coordinates, we need to remember that we calculate each of these derivatives, only one coordinate changes at a time.

x

UFX

y

UFy

z

UFz

x

UFx

y

UFy

z

UFz

)( k

z

Uj

y

Ui

x

UF

8 Momentum Impulse and Collisions8 Momentum Impulse and Collisions8-1 Introduction8-1 IntroductionA common theme of all these questions is that they can’t be A common theme of all these questions is that they can’t be answered by directly applying Newton’s second law, answered by directly applying Newton’s second law, F F = m= maa , , because there are forces acting about which we know very litter.because there are forces acting about which we know very litter.Our approach will use two new concepts, momentum and impulse, Our approach will use two new concepts, momentum and impulse, and a new conservation law, conservation of momentum. This and a new conservation law, conservation of momentum. This conservation law is every bit as important as that of conservation conservation law is every bit as important as that of conservation of energy. The law of conservation of momentum is valid even in of energy. The law of conservation of momentum is valid even in situations in which Newton’s laws are inadequate, such as bodies situations in which Newton’s laws are inadequate, such as bodies moving at very high speeds (near as the speed of light) or objects moving at very high speeds (near as the speed of light) or objects on a very small scale (such as the constituents of atoms). Within on a very small scale (such as the constituents of atoms). Within the domain of Newtonian mechanics, conservation of momentum the domain of Newtonian mechanics, conservation of momentum enables us to analyze many situations that would be very difficult enables us to analyze many situations that would be very difficult if we tried to use Newton’s laws directly. Among these are collision if we tried to use Newton’s laws directly. Among these are collision problems, in which two bodies collide and can exert very large problems, in which two bodies collide and can exert very large forces on each other for a short time.forces on each other for a short time.

8-2 Momentum And ImpulseLet’s consider a particle of constant mass m. Because a = dv/dt, we can write Newton’s second law for this particle as

8-1We can take the mass m inside the derivative because it is constant. Thus Newton’s second law says that the net force F acting on a particle equals the time rate of change of the combination mv, the product of the particle’s mass and velocity. We’ll call this combination the momentum, or linear momentum, of the particle. Using the symbol p for momentum, we have

vmdt

d

dt

vdmF

pp = m = mvv (definition of momentum) (8-2) (definition of momentum) (8-2)Momentum is a vector quantity that has a magnitude (mv) and a Momentum is a vector quantity that has a magnitude (mv) and a direction (the same as the velocity vector v). The units of the direction (the same as the velocity vector v). The units of the magnitude of momentum are units of mass times speed; the SI magnitude of momentum are units of mass times speed; the SI units of momentum are kgunits of momentum are kg·m/s. The plural of momentum is ·m/s. The plural of momentum is “momenta”.“momenta”.Substituting Eq.(8-2) into Eq.(8.1), we haveSubstituting Eq.(8-2) into Eq.(8.1), we have

(Newton’s second law in terms of (Newton’s second law in terms of momentum) (8-3).momentum) (8-3).The net force (vector sum of all forces) acting on a particle The net force (vector sum of all forces) acting on a particle equals the time rate of change momentum of the particleequals the time rate of change momentum of the particle. . This, not This, not F F = m= maa , is the form in which Newton originally stated , is the form in which Newton originally stated his second law (though he called momentum the “quantity of his second law (though he called momentum the “quantity of motion”). It is valid only in inertial frames of reference.motion”). It is valid only in inertial frames of reference.We will often express the momentum of a particle in terms of its We will often express the momentum of a particle in terms of its components. If the particle has velocity components vcomponents. If the particle has velocity components vxx, v, vyy, and v, and vzz, ,

its momentum components pits momentum components pxx, p, pyy, ,, ,

dt

pdF

and pz are given by px = mvx , py = mvy , pz = mvz (8-4).These three components equations are equivalent to Eq. (8-2).A particle’s momentum p = mv and its kinetic energy k = ½ mv2 both depend on the mass and velocity of the particle. What is the fundamental difference between these two quantities? A purely mathematical answer is that momentum is a vector whose magnitude is proportional to speed, while kinetic energy is a scalar proportional to the speed squared. But to see the physical difference between momentum and kinetic energy, we must first define a quantity closely related to momentum called impulse.

Let’s first consider a particle acted on by a constant net force Let’s first consider a particle acted on by a constant net force F F during a time interval during a time interval t from tt from t11 to t to t22, (We’ll look at the case of , (We’ll look at the case of

varying forces shortly.) The impulse of the net force, denoted by J, varying forces shortly.) The impulse of the net force, denoted by J, is defined to be product of the net force and the time interval:is defined to be product of the net force and the time interval: J = J = F(tF(t22 – t – t11) = ) = FFt t

(8-5).(8-5).Impulse is a vector quantity, its direction is the same aa the net Impulse is a vector quantity, its direction is the same aa the net force force F .F .To see what impulse is good for, let’s go back Newton’s second To see what impulse is good for, let’s go back Newton’s second law as restated in terms of momentum, Eq. (8-3). If the net force law as restated in terms of momentum, Eq. (8-3). If the net force F is constant, then dp/dt is also constant. In that case, dp/dt is F is constant, then dp/dt is also constant. In that case, dp/dt is equal to total change in momentum pequal to total change in momentum p22 – p – p11 during the time interval during the time interval

tt22 – t – t11, divided by the interval:, divided by the interval:

12

12

tt

PPF

Multiplying this equation by (tMultiplying this equation by (t22 – t – t11), we have), we have

. .Comparing to Eq. (8-5), we end up with a result called the impulse-momentum Comparing to Eq. (8-5), we end up with a result called the impulse-momentum theorem:theorem: (impulse-momentum theorem). (8-6) (impulse-momentum theorem). (8-6)The change in momentum of a particle during a time interval equals the The change in momentum of a particle during a time interval equals the impulse of the net force that acts on the particle during that interval.impulse of the net force that acts on the particle during that interval.The impulse-momentum theorem also holds when forces are not constant. To The impulse-momentum theorem also holds when forces are not constant. To see this, we integrate both sides of Newton’s second law see this, we integrate both sides of Newton’s second law F = dp/dt over time F = dp/dt over time be-tween the limits tbe-tween the limits t11 and t and t22::

The interval on the left is defined to be the impulse J of the net force The interval on the left is defined to be the impulse J of the net force F during F during this interval:this interval: (general definition of impulse) (8-7). (general definition of impulse) (8-7).With this definition the impulse-momentum theorem J = PWith this definition the impulse-momentum theorem J = P22 – P – P1 1 , Eq. (8-6), is , Eq. (8-6), is

va-lid even when the net force va-lid even when the net force F varies with time.F varies with time.We can define an average net force FWe can define an average net force Favav such that even when such that even when F is not F is not

constant, the impulse J is given byconstant, the impulse J is given by J = F J = Favav( t( t2 2 – t– t11) (8-8).) (8-8).

1212 )( ppttF

12 ppJ

122

1

2

1

2

1

pppddtdt

pddtF p

p

t

t

t

t

21

t

t dtF

When When F is constant, F is constant, F = FF = Favav and Eq. (8-8) reduces to Eq. (8-5). and Eq. (8-8) reduces to Eq. (8-5).

Fig.(8-1) Fig.(8-1)Figure (8-1) shows a graph of the x-component of net force Figure (8-1) shows a graph of the x-component of net force FFxx as a function of as a function of

time during a collision. This might represent the force on a soccer ball that is in time during a collision. This might represent the force on a soccer ball that is in contact with a player’s foot from time tcontact with a player’s foot from time t11 to t to t22. The x-component of impulse . The x-component of impulse

during this interval is represented by the blue area under the curve between tduring this interval is represented by the blue area under the curve between t11

and tand t2. 2. This area is equal to the rectangular area bounded by tThis area is equal to the rectangular area bounded by t11, t, t22 , and (F , and (Favav))xx, ,

so (Fso (Favav))xx(t(t2 2 - t- t11) is equal to the impulse of the actual time-varying force during the ) is equal to the impulse of the actual time-varying force during the

same interval.same interval. Impulse and momentum are both vector quantities, and Eqs.(8-5) through (8- Impulse and momentum are both vector quantities, and Eqs.(8-5) through (8-8) are all vector equations. In specific problems it is often easiest to use them in 8) are all vector equations. In specific problems it is often easiest to use them in component form:component form:

(8-9) (8-9)

t1 t2

t2-t1

Fx

(Fav)x

xxxxxav

t

t xx mvmvppttFdtFJ 1212122

1

yyyyyav

t

t yy mvmvppttFdtFJ 1212122

1

MOMENTUM AND KINETIC ENERGY COMPAREDWe can see now see the fundamental difference between momentum and kinetic energy. The impulse-momentum theorem J = p2 – p1says that changes in a particle’s momentum are due to impulse, which depends on the time over which the net force acts. By contrast, the work-energy theorem Wtot = K2 – K1 tells us that kinetic energy changes when work is down on a particle; the total work depends on the distance over which the net force acts. Consider a particle that starts from rest at t1 so that v1 = 0. Its initial momentum is p1 = mv1 = 0, and its initial kinetic energy is K1 = ½ mv2 = 0. Now let a constant net force equal to F act on that particle from time t1 until time t2. During this interval the particle moves a distance s in the direction of the

force. From Eq. (8-6), the particle’s momentum at time tforce. From Eq. (8-6), the particle’s momentum at time t22 is p is p22 = =

pp11 + J = J, where J = F (t + J = J, where J = F (t22 – t – t11) is the impulse that acts on the ) is the impulse that acts on the

particle. So the momentum of a particle equals the impulse that particle. So the momentum of a particle equals the impulse that accelerated it from rest to its present speed; impulse is the accelerated it from rest to its present speed; impulse is the product of the net force that accelerated the particle and the time product of the net force that accelerated the particle and the time required for the acceleration. By comparison the kinetic energy of required for the acceleration. By comparison the kinetic energy of the particle at tthe particle at t22 is K is K22 = W = Wtottot = Fs, the total work done on the = Fs, the total work done on the

particle to accelerate it from rest. The total work is the product of particle to accelerate it from rest. The total work is the product of the net force and the distance required to accelerate the particle.the net force and the distance required to accelerate the particle.Both the impulse-momentum and work-energy theorem are Both the impulse-momentum and work-energy theorem are relationships between force and motion, and both rest on the relationships between force and motion, and both rest on the foundation of Newton’s laws. They are integral particles, relating foundation of Newton’s laws. They are integral particles, relating the motion at two different times separated by a finite interval. By the motion at two different times separated by a finite interval. By contrast, Newton’s second law itself (in either of the forms contrast, Newton’s second law itself (in either of the forms F = F = ma or ma or F = dp/dt) F = dp/dt) is a differential principle, relating the forces to is a differential principle, relating the forces to the rate of change of velocity or momentum at each instant.the rate of change of velocity or momentum at each instant.

Fig. 8-6 (b) Fig. 8-6 (c ) Fig. 8-6 (b) Fig. 8-6 (c ) 8-3 Conservation of momentum8-3 Conservation of momentumThe concept of momentum is particularly important in situations in which we The concept of momentum is particularly important in situations in which we have two or more interacting bodies. Let’s consider first an idealized system have two or more interacting bodies. Let’s consider first an idealized system consisting of two bodies that interact with each other but not with anything else.consisting of two bodies that interact with each other but not with anything else.Let’s go over that again with some new terminology. For any system, the forces Let’s go over that again with some new terminology. For any system, the forces that the particles of the system exert on each other are called internal forces. that the particles of the system exert on each other are called internal forces. Forces exerted on any part of the system by some object outside it are called Forces exerted on any part of the system by some object outside it are called external forces. For the sys-tem we have described, the internal forces are Fexternal forces. For the sys-tem we have described, the internal forces are FB on B on

AA, exerted by particle B on particle A, and F, exerted by particle B on particle A, and FA on BA on B, exerted by particle A on , exerted by particle A on

particle B (Figs. 8-6 b, c). There are no external forces; when this is the case, particle B (Figs. 8-6 b, c). There are no external forces; when this is the case, we have an isolated system.we have an isolated system.The net force on particle A is FThe net force on particle A is FB on AB on A, and the net force on particle B is F, and the net force on particle B is FA on BA on B, so , so

from Eq. (8-3) the rates of change of the momenta of the two particles arefrom Eq. (8-3) the rates of change of the momenta of the two particles are

(8-10) (8-10)

x

y

FB on A

x

y

FA on B

dt

pdF ABonA

dt

pdF BAonB

The momentum of each particle changes, but these changes are not independent; according to Newton’s third law, the two forces FB on A and FA on B are always equal in magnitude and opposite in direction. That is, FB on A = - FA on B so FB on A + FA on B = 0. Adding together the two equations in Eq. (8-10), we have

(8-11).The rates of change of the two momenta are equal and opposite, so that the rate of change of the vector sum PA + PB is zero. We now define the total momentum P of the system of two particles as the vector sum of the momenta of the individual particles.That is,

0

dt

PPd

dt

pd

dt

pdFF BABAAonBBonA

P = PP = PA A + P+ PBB (8-12). (8-12).

Then Eq. (8-11) becomes, finally,Then Eq. (8-11) becomes, finally,

(8-13) (8-13)

The time rate of change of the total momentum P is zero. Hence The time rate of change of the total momentum P is zero. Hence the total momentum of the system is constant, even though the the total momentum of the system is constant, even though the individual momenta of the particles that make up the system can individual momenta of the particles that make up the system can change.change.If external forces are also present, they must be included on the If external forces are also present, they must be included on the left side of Eq. (8-13) along with the internal forces. Then the total left side of Eq. (8-13) along with the internal forces. Then the total momentum is, in general, not constant. momentum is, in general, not constant. But if the vector sum of the external forces is zero, these forces But if the vector sum of the external forces is zero, these forces don’t contribute to the sum, and dp/dt is again zero. Thus we have don’t contribute to the sum, and dp/dt is again zero. Thus we have the following general result:the following general result:If the vector sum of the external forces on a system is zero, the If the vector sum of the external forces on a system is zero, the total momentum of the system is constant.total momentum of the system is constant.

0dt

pdFF AonBBonA

This is the simplest form of the principle of conservation of momentum. This This is the simplest form of the principle of conservation of momentum. This principle is a direct consequence of Newton’s third law. What makes this principle is a direct consequence of Newton’s third law. What makes this principle useful is that it doesn’t depend on the detailed nature of the internal principle useful is that it doesn’t depend on the detailed nature of the internal forces that act between members of the system. This means that we can apply forces that act between members of the system. This means that we can apply conservation of momentum even if (as is often the case) we know very little conservation of momentum even if (as is often the case) we know very little about the internal forces. We have used Newton’s second law to derive this about the internal forces. We have used Newton’s second law to derive this principle, so we have to be careful to use it only in internal frames of reference.principle, so we have to be careful to use it only in internal frames of reference. We can generalize this principle for a system containing any number of We can generalize this principle for a system containing any number of particle A, B, C, particle A, B, C, … interacting only with each other. The total momentum of … interacting only with each other. The total momentum of such a system is such a system is P = PP = PAA + P + PBB + + = m = mAA v vAA + m + mBB v vBB + … (total momentum of a system of + … (total momentum of a system of

particles).particles).We make the same argument as before; the total rate of change of momentum We make the same argument as before; the total rate of change of momentum of the system due to each action-reaction pair of internal forces is zero. Thus of the system due to each action-reaction pair of internal forces is zero. Thus the total rate of change of momentum of the entire system is zero whenever the the total rate of change of momentum of the entire system is zero whenever the vector sum of the exter-nal forces acting on it is zero. The internal forces can vector sum of the exter-nal forces acting on it is zero. The internal forces can change the momentum of individual particle in the system but not the total change the momentum of individual particle in the system but not the total momentum of the system.momentum of the system.8-4 Inelastic Collisions8-4 Inelastic CollisionsIf the forces between the bodies are much larger than any external forces, as is If the forces between the bodies are much larger than any external forces, as is the case in most collisions, we can neglect the external forces entirely and treat the case in most collisions, we can neglect the external forces entirely and treat the bodies as an the bodies as an

isolated system. Then momentum is conserved in the collision, isolated system. Then momentum is conserved in the collision, and the total momentum of the system has the same value before and the total momentum of the system has the same value before and after the collision.and after the collision.If the forces between the bodies are also conservative, so that no If the forces between the bodies are also conservative, so that no mechanical energy is lot lost or gained in the collision, the total mechanical energy is lot lost or gained in the collision, the total kinetic energy of the system is the same after the collision as kinetic energy of the system is the same after the collision as before. Such a collision is called an elastic collision. A collision in before. Such a collision is called an elastic collision. A collision in which the total kinetic energy after the collision is less than that which the total kinetic energy after the collision is less than that before the collision is called an inelastic collision. An inelastic before the collision is called an inelastic collision. An inelastic collision in which the colliding bodies stick together and move as collision in which the colliding bodies stick together and move as one body after the collision is often called a completely inelastic one body after the collision is often called a completely inelastic collision. collision. 8-5 Elastic Collisions8-5 Elastic CollisionsFrom conservation of kinetic energy we haveFrom conservation of kinetic energy we have

22

22

21

21 2

1

2

1

2

1

2

1BBAABBAA vmvmvmvm

And conservation of momentum gives

8-7 Rocket Propulsion

Momentum considerations are particularly useful for analyzing a system is which the mass of parts of the system change with time. In such case we can’t use Newton’s second law directly because m change.

We choose our x-axis to be along the rocket’s direction of motion. The x-component of total momentum at this instant is P1 = mv. In a short time interval da the mass of the rocket changes by an amount dm. This is an inherently negative quantity because the rocket’s mass m decrease with time.

2211 BBAABBAA vmvmvmvm

amF

During dt a positive mass –dm of burned fuel is ejected from the rocket; Let vex be the exhaust speed of this material relative to the rocket; the burned fuel is ejected opposite the direction of motion, so its x-component of velocity relative to the rocket is –vex. The x-component of velocity vfuel of the burned fuel relative to our coordinate system is then

And the x-component of momentum of the ejected mass (-dm) is

And the x-component of momentum of the ejected mass (-dm) is

exexfuel vvvvv )(

))(()( exfuel vvdmvdm

(-dm)vfuel = (-dm)(v – vex)

The rocket’s momentum at this time is

(m + dm)(v+dv).

Thus the total x-component of momentum P2 of rocket plus ejected fuel at time t + dt is

According to our initial assumption, the rocket and fuel are an isolated system.

))(())((2 exvvdmdvvdmmP

))(())(( exvvdmdvvdmmmv

This can be simplified to dmdvdmvmdv ex

dt

dmv

dt

dvm ex

dt

dmvF ex

dt

dm

m

v

dt

dva ex m

dmvdv ex

m

mex

v

v

m

m ex m

mdv

m

mdvvd

00 0

m

mv

m

mvvv exex

0

00 lnln

The ratio m/m0 is the original mass divided by the mass after the fuel has been exhausted.

9 Rotation of Rigid Bodies9 Rotation of Rigid Bodies

9-1 Introduction9-1 IntroductionReal-world bodies can be even more complicated; the forces that act on them Real-world bodies can be even more complicated; the forces that act on them can de-form them---stretching, twisting, and squeezing them. We’ll neglect can de-form them---stretching, twisting, and squeezing them. We’ll neglect these deformations for now and assume that the body has a perfectly definite these deformations for now and assume that the body has a perfectly definite and unchanging shape and size. We call this idealized model a rigid body. This and unchanging shape and size. We call this idealized model a rigid body. This chapter and the next are mostly about rotational motion of a rigid body. We chapter and the next are mostly about rotational motion of a rigid body. We begin with kinetic language for describing rotational motion.begin with kinetic language for describing rotational motion.9-2 Angular Velocity And Acceleration9-2 Angular Velocity And AccelerationIn analyzing rotational motion, let’s think first about a rigid body that rotates In analyzing rotational motion, let’s think first about a rigid body that rotates about a fixed axis. By fixed axis we mean an axis that is at rest in some inertial about a fixed axis. By fixed axis we mean an axis that is at rest in some inertial frame of refer-ence and does not change direction relative to that frame. frame of refer-ence and does not change direction relative to that frame.

Fig 9-1 Fig9-2(a) Fig9- Fig 9-1 Fig9-2(a) Fig9-2(b) 2(b)

O

P

S = r

1rad

S = r

1rad

Figure 9-1 shows a rigid body rotating about a fixed axis that passes through point O and is perpendicular to the plane of the diagram, which we choose to call the xy-plane. One way to describe the rotation of this body would be to choose a particular point P on the body and to keep track of the x- and y-coordinates of this point. This isn’t a terribly convenient method, since it takes two numbers (the two coordinates x and y) to specify the rotational position of the body. Instead, we notice that the line OP is fixed in the body and rotates with it. The angle that this line makes with the +x-axis describes the rotational position of the body; we will use this single quantity as a coordinate for rotation.

The angular coordinate The angular coordinate of a rigid body rotating around a fixed of a rigid body rotating around a fixed axis can be positive or negative. If we choose positive angles to axis can be positive or negative. If we choose positive angles to be measured counterclockwise from the positive x-axis, then the be measured counterclockwise from the positive x-axis, then the angle angle in Fig. 9-1 is positive. If we instead choose the positive in Fig. 9-1 is positive. If we instead choose the positive rotation direction to be clockwise, then rotation direction to be clockwise, then in Fig. 9-1 is negative. in Fig. 9-1 is negative. When we considered the motion of a particle along a straight line, When we considered the motion of a particle along a straight line, it essential to specify the direction of positive displacement along it essential to specify the direction of positive displacement along that line; in discussing rotation around a fixed axis, it’s just as that line; in discussing rotation around a fixed axis, it’s just as essential to specify the direction of positive rotation.essential to specify the direction of positive rotation.In describing rotational motion, the most natural way to measure In describing rotational motion, the most natural way to measure the angle the angle is not in degrees, but in radians. As shown in Fig. 9- is not in degrees, but in radians. As shown in Fig. 9-2a, one radian (1 rad ) is the angle subtended at the center of a 2a, one radian (1 rad ) is the angle subtended at the center of a circle by an arc with a length equal to the radius of the circle. In circle by an arc with a length equal to the radius of the circle. In Fig9-2b an angle subtended by an arc of length s on a circle of Fig9-2b an angle subtended by an arc of length s on a circle of radius r. The value of radius r. The value of (in radians) is equal to s divided by r. (in radians) is equal to s divided by r.

An angle in radians is the ratio of two lengths, so it is a pure number, without dimensions. If s = 3.0 m and r = 2.0 m, then = 1.5, but we will often write this as 1.5 rad to distinguish it from an angle measured in degrees or revolutions.Angular Velocity The coordinate shown in Fig. 9-1 specifies the rotational position of a rigid body at a given instant. We can describe the rotational motion of such a rigid body in terms of the rate of change of . In Fig. 9-3a a reference line OP in a rotating body makes an angle 1 with the +x-axis at time t1. At a later time t2 the angle has changed to 2. We define

r

s or s = r (9-1)

the average angular velocity the average angular velocity avav of the body in the time interval of the body in the time interval t t

= t= t2 2 – t– t11 as the ratio of the angular displacement as the ratio of the angular displacement = = 22 - - 1 1 to to

t:t:

The instantaneous angular velocity The instantaneous angular velocity is limit of is limit of avav as as t t

approaches zero, that is, the derivative of approaches zero, that is, the derivative of with respect to t: with respect to t:

any instant, every part of a rotating rigid body has the same any instant, every part of a rotating rigid body has the same angular velocity. The angular velocity is positive if the body is angular velocity. The angular velocity is positive if the body is rotating in the direction of increasing rotating in the direction of increasing and negative if rotating in and negative if rotating in the direction of decreasing the direction of decreasing . .

tttav

12

12

dt

d

tt

lim

0

Angular AccelerationAngular AccelerationWhen the angular velocity of a rigid body changes, it has an angular When the angular velocity of a rigid body changes, it has an angular acceleration. If acceleration. If 11 and and 22 are the instantaneous angular velocities at times t are the instantaneous angular velocities at times t11

and tand t22, we define the average angular acceleration , we define the average angular acceleration avav over the interval over the interval t = tt = t22 – –

tt11 as the change in angular velocity divided by as the change in angular velocity divided by t:t:

(9-4). (9-4).The instantaneous acceleration The instantaneous acceleration is the limit of is the limit of avav as as t t 0: 0:

(9-5). (9-5). The usual unit of angular acceleration is the radian per second, or rad/sThe usual unit of angular acceleration is the radian per second, or rad/s22. . Henceforth we will use the term “angular acceleration” to mean the Henceforth we will use the term “angular acceleration” to mean the instantaneous angular acceleration rather than the average angular instantaneous angular acceleration rather than the average angular acceleration.acceleration.Because Because = d = d/dt, we can also express angular acceleration as the second /dt, we can also express angular acceleration as the second derivative of the angular coordinate:derivative of the angular coordinate:

(9-6). (9-6).In rotational motion, if the angular acceleration In rotational motion, if the angular acceleration is positive, the angular is positive, the angular velocity velocity is increasing; if is increasing; if is negative, is negative, is decreasing. The rotation is is decreasing. The rotation is speeding up if speeding up if and and have have

tttav

12

12

dt

d

tt

0lim

2

2

dt

d

dt

d

dt

d

the same sign and slowing down if the same sign and slowing down if and and have opposite signs. have opposite signs.

9-3 Rotation With Constant Angular Acceleration9-3 Rotation With Constant Angular AccelerationLet Let 00 be the angular velocity of a rigid body at time t = 0, and let be the angular velocity of a rigid body at time t = 0, and let be its be its

angular velocity at any later time t. The angular acceleration angular velocity at any later time t. The angular acceleration is constant and is constant and equal to the average value for any interval. Using Eq. (9-4) with the interval equal to the average value for any interval. Using Eq. (9-4) with the interval from 0 to t, we find from 0 to t, we find

, or , or = = 00 + + t (9-7). t (9-7).

The product The product t is the total change in t is the total change in between t = 0 and the later time t; the between t = 0 and the later time t; the angular velocity angular velocity at time t is the sum of the initial value at time t is the sum of the initial value 00 and this total and this total

change.change.With constant angular acceleration the angular velocity changes at a uniform With constant angular acceleration the angular velocity changes at a uniform rate, so its average value between 0 and t is the average of the initial and final rate, so its average value between 0 and t is the average of the initial and final values:values: (9-8). (9-8).We also know that We also know that avav is the total angular displacement ( is the total angular displacement ( - - 00) divided by the ) divided by the

time interval (t - 0);time interval (t - 0);

(9-9). (9-9). When we equate Eqs. (9-8) and (9-9) and multiply the result by t, we get When we equate Eqs. (9-8) and (9-9) and multiply the result by t, we get

00

t

00

tav

20

av

(9-10)

To obtain a relation between and t that doesn’t contain , we substitute Eq. (9-7) into Eq. (9-10):

(9-11).Relating Linear And Angular Kinematics

t 00 2

1

tt 000 2

1 2

00 2

1tt

Pr

v

x

y

ω

ω

Pr

atan

x

y

ω

ω

arad

Fig 9-7 Fig 9-8

When a rigid body rotates about a fixed axis, every particle in the When a rigid body rotates about a fixed axis, every particle in the body moves in a cir-cular path. The circle lies in a plane body moves in a cir-cular path. The circle lies in a plane perpendicular to the axis and is centered on the axis. The speed perpendicular to the axis and is centered on the axis. The speed of a particle is directly proportional to the body’s angular velocity; of a particle is directly proportional to the body’s angular velocity; the faster the body rotates, the greater the speed of each particle. the faster the body rotates, the greater the speed of each particle. In Figure 9-7, point P is a constant distance r from the axis of In Figure 9-7, point P is a constant distance r from the axis of rotation, so it moves in a circle of radius r. At any time the angle rotation, so it moves in a circle of radius r. At any time the angle (in radians) and the arc length s are related by (in radians) and the arc length s are related by s = r s = r. . We take the time derivative of this, noting that r is constant for any We take the time derivative of this, noting that r is constant for any specific particle, and take absolute value of both sides:specific particle, and take absolute value of both sides:

Now Now ds/dtds/dtis the absolute value of the rate of change of arc is the absolute value of the rate of change of arc length, which is equal to the instantaneous linear speed v of the length, which is equal to the instantaneous linear speed v of the particle. Analogously, particle. Analogously, dd/dt/dt, the absolute value of the rate of , the absolute value of the rate of change of the angle, is the instantaneous change of the angle, is the instantaneous

dt

dr

dt

ds

angular speed angular speed --- that is, the magnitude of the instantaneous angular velocity --- that is, the magnitude of the instantaneous angular velocity in rad/s. Thus in rad/s. Thus v = r v = r (relation between linear and angular speed) (9-13). (relation between linear and angular speed) (9-13).The farther a point is from the axis, the greater its linear speed. The direction of The farther a point is from the axis, the greater its linear speed. The direction of the linear velocity vector is tangent to its circular path at each point (Fig. (9-7).the linear velocity vector is tangent to its circular path at each point (Fig. (9-7).We can represent the acceleration of a particle moving in a circle in terms of its We can represent the acceleration of a particle moving in a circle in terms of its centri-petal and tangential components, acentri-petal and tangential components, aradrad and a and atantan (Fig. 9-8), as we did in (Fig. 9-8), as we did in

Section 3-5. It would be a good idea to review that section now. We found that Section 3-5. It would be a good idea to review that section now. We found that the tangential component of acceleration athe tangential component of acceleration a tantan, the component parallel to the , the component parallel to the

instantaneous velocity, acts to change the magnitude of the particle’s velocity instantaneous velocity, acts to change the magnitude of the particle’s velocity (i.e., the speed) and is equal to the rate of change of speed. Taking the (i.e., the speed) and is equal to the rate of change of speed. Taking the derivative of Eq. (9-13), we findderivative of Eq. (9-13), we find (9-14). (9-14).

This component of a particle’s acceleration is always tangent to the circular This component of a particle’s acceleration is always tangent to the circular path of the particle.path of the particle.The component of the particle’s acceleration directed toward the rotation axis, The component of the particle’s acceleration directed toward the rotation axis, the centripetal component of acceleration athe centripetal component of acceleration aradrad, is associated with the change of , is associated with the change of

direction of the particle’s velocity. In Section 3-5 we worked out the relation adirection of the particle’s velocity. In Section 3-5 we worked out the relation a radrad

= v= v22/r. We can ex-press this in terms of /r. We can ex-press this in terms of by using Eq. (9-13): by using Eq. (9-13):

r

dt

dr

dt

dva tan

(9-15)(9-15)The vector sum of the centripetal and tangential components of acceleration The vector sum of the centripetal and tangential components of acceleration particle in a rotating body is the linear acceleration a (Fig. 9-8).particle in a rotating body is the linear acceleration a (Fig. 9-8).

9-5 Energy In Rotational Motion9-5 Energy In Rotational MotionA rotating rigid body consists of mass in motion, so it has kinetic energy. We A rotating rigid body consists of mass in motion, so it has kinetic energy. We can ex-press this kinetic energy in terms of the body’s angular velocity and a can ex-press this kinetic energy in terms of the body’s angular velocity and a new quantity called moment of inertia that we will define below. To develop this new quantity called moment of inertia that we will define below. To develop this relationship, we think of the body as being made up of a large number of relationship, we think of the body as being made up of a large number of particles, with masses mparticles, with masses m11, m, m22, , …, at distances r…, at distances r11, r, r22, …, from the axis fo , …, from the axis fo

rotation. We label the particles with the index i: The mass of the ith particle is rotation. We label the particles with the index i: The mass of the ith particle is mmii, and its distance from the axis of rotation is r, and its distance from the axis of rotation is r ii. The particles don’t necessarily . The particles don’t necessarily

all lie in the same plane, so we specify that rall lie in the same plane, so we specify that r ii is the per-pendicular distance is the per-pendicular distance

from the axis to the ith particle.from the axis to the ith particle.When a rigid body rotates about a fixed axis, the speed vWhen a rigid body rotates about a fixed axis, the speed v ii of the ith particle is of the ith particle is

given by Eq. (9-13), vgiven by Eq. (9-13), v ii = r = rii, where , where is the body’s angular speed. Different is the body’s angular speed. Different

particles have different values of r, but particles have different values of r, but is the same for all (otherwise, the is the same for all (otherwise, the body wouldn’t be rigid). The kinetic energy of the ith particle can be expressed body wouldn’t be rigid). The kinetic energy of the ith particle can be expressed asas

rr

varad

22

222

2

1

2

1 iiii rmvm

The total kinetic energy of the body is the sum of the kinetic energies of all its The total kinetic energy of the body is the sum of the kinetic energies of all its particles:particles:

Taking the common factor Taking the common factor 22/2 out of this expression, we get/2 out of this expression, we get

The quantity in parentheses, obtained by multiplying the mass of each particle The quantity in parentheses, obtained by multiplying the mass of each particle by the square of its distance from the axis of rotation and adding these by the square of its distance from the axis of rotation and adding these products, is denoted by I and is called the moment of inertia of the body for this products, is denoted by I and is called the moment of inertia of the body for this rotation axis:rotation axis: (9-16). (9-16).The word “moment” means that I depends on how the body’s mass is The word “moment” means that I depends on how the body’s mass is distributed in space; it has nothing to do with a “moment” of time. For a body distributed in space; it has nothing to do with a “moment” of time. For a body with a given rotation axis and a given total mass, the greater the distance from with a given rotation axis and a given total mass, the greater the distance from the axis to the particles that make up the body, the greater the moment of the axis to the particles that make up the body, the greater the moment of inertia. In a rigid body, the distance rinertia. In a rigid body, the distance r ii are all constant and I is independent of are all constant and I is independent of

how the body is rotating around the given axis.how the body is rotating around the given axis.In terms of moment of inertia I, the rotational kinetic energy K of a rigid body is In terms of moment of inertia I, the rotational kinetic energy K of a rigid body is

(rotational kinetic energy of a rigid body) (9-17). (rotational kinetic energy of a rigid body) (9-17).

222

2

2

22

2

1

2

11 2

1

2

1

2

1iii

irmrmrmk

2222

22

2

11 2

1

2

1 i

iirmrmrmk

i

iirmrmrmI 22

22

2

11

2

2

1 IK

The kinetic energy given by Eq. (9-17) is not a new form of energy; it’s the sum of the kinetic energies of the individual particles that make up the rigid body, written in a compact and convenient form in terms of the moment of inertia. When using Eq. (9-17) must be measured in radians per second, not revolutions or degrees per second, to give K in joules; this is because we used vi = ri in our derivation.Equation (9-17) gives a simple physical interpretation of moment of inertia: the greater the moment of inertia, the greater the kinetic energy of a rigid body rotating with a given angular speed . We learned in Chapter 6 that the kinetic energy of a body equals the amount of work done to accelerate that body from

rest. So the greater a body’s moment of inertia, the harder it is to start the body rest. So the greater a body’s moment of inertia, the harder it is to start the body rotating if it’s at rest and the harder it is to stop its rotation if it’s already rotating rotating if it’s at rest and the harder it is to stop its rotation if it’s already rotating (Fig. 9-12). For this reason, I is also called the rotational inertia. (Fig. 9-12). For this reason, I is also called the rotational inertia.

9-7 Moment of Inertia Calculations9-7 Moment of Inertia CalculationsWhen a rigid body cannot be represented by a few point masses but is a When a rigid body cannot be represented by a few point masses but is a continuous distribution of mass, the sum of masses and distance that defines continuous distribution of mass, the sum of masses and distance that defines moment of inertia Eq. (9-16) becomes an integral. Imagine dividing the body moment of inertia Eq. (9-16) becomes an integral. Imagine dividing the body into small mass elements dm so that all points in a particular element are at into small mass elements dm so that all points in a particular element are at essentially the same perpendicular distance from the axis of rotation. We call essentially the same perpendicular distance from the axis of rotation. We call this distance r, as before. Then the moment of inertia isthis distance r, as before. Then the moment of inertia is

(9-20) (9-20) To evaluate the integral, we have to represent r and dm in terms of the same To evaluate the integral, we have to represent r and dm in terms of the same integration variable. When we have an effectively one-dimensional object, such integration variable. When we have an effectively one-dimensional object, such as the slender rods (a) and (b) in Table 9-2, we can use a coordinate x along as the slender rods (a) and (b) in Table 9-2, we can use a coordinate x along the length and relate dm terms of an increment dx volume dV and the density the length and relate dm terms of an increment dx volume dV and the density of the body. Density is mass per unit of the body. Density is mass per unit

dmrI 2

volume, volume, = dm/dv, so we may also write Eq. (9-20) as = dm/dv, so we may also write Eq. (9-20) as

If the body is uniform in density, then we may take If the body is uniform in density, then we may take outside the outside the integral:integral: (9- (9-21).21).To use this equation, we have to express the volume element dV To use this equation, we have to express the volume element dV in terms of the differentials of the integration variable, such as dV in terms of the differentials of the integration variable, such as dV = dxdydz. The element dV must always be chosen so that all = dxdydz. The element dV must always be chosen so that all points within it are at very nearly the same distance from the axis points within it are at very nearly the same distance from the axis of rotation. The limits on the integral are determined by the shape of rotation. The limits on the integral are determined by the shape and dimensions of the body. For regularly shaped bodies this and dimensions of the body. For regularly shaped bodies this integration can often be carried out quite easily. integration can often be carried out quite easily.

dVrI 2

dVrI 2

10 Dynamics of Rotational 10 Dynamics of Rotational MotionMotion10-1 Introduction10-1 IntroductionIn this chapter we will define a new physical quantity, torque, that In this chapter we will define a new physical quantity, torque, that describes the twisting or turning effort of a force. We’ll find that the describes the twisting or turning effort of a force. We’ll find that the net torque acting on a rigid body determines its angular net torque acting on a rigid body determines its angular acceleration, in the same way that the net force on a body acceleration, in the same way that the net force on a body determines its linear acceleration. We all also look at work and determines its linear acceleration. We all also look at work and power in rotational motion in order to understand such problems power in rotational motion in order to understand such problems as how energy is transmitted by rotating drive shaft in a car. as how energy is transmitted by rotating drive shaft in a car. Finally, we will develop a new conservation principle, conservation Finally, we will develop a new conservation principle, conservation of angular momentum, that is tremendously useful for of angular momentum, that is tremendously useful for understanding the rotational motion of both rigid and nonrigid understanding the rotational motion of both rigid and nonrigid bodies. bodies.

10-2 TorqueWhat is it about a force that determines how effective it is in causing or changing rotational motion? The magnitude and direction of the force are important, but so is the position of the point where the force is applied. The quantitative measure of the tendency of a force to cause or change the rotational motion of a body is called torque. Figure 10-2 shows a body that can rotate about an axis that passes through point O and is perpendicular to the plane of the figure. The body is acted on by three forces, F1, F2, and F3, in the plane of the figure. The tendency of F1 to cause a rotation about point O its magnitude

o

A

Bl2

l1

Lever arm of F2

F2

F1

F3

Lever arm of F1

Line of action of F2

Line of action of F1

depends on its magnitude Fdepends on its magnitude F11. It also depends on the perpendicular . It also depends on the perpendicular

distance ldistance l11 between the between the line of actionline of action of the force (that is, the line of the force (that is, the line

along which the vector lies) and point O. We called the distance lalong which the vector lies) and point O. We called the distance l11

the lever arm (or moment arm) of force Fthe lever arm (or moment arm) of force F11 about O. The twisting about O. The twisting

effort is directly proportional to both Feffort is directly proportional to both F11 and l and l11. We define the . We define the

torque (or moment) of the force Ftorque (or moment) of the force F11 with respect to point O as the with respect to point O as the

product Fproduct F11ll11. We will use the Greek letter . We will use the Greek letter for torque. For a force for torque. For a force

of magnitude F whose line of action is a perpendicular distance l of magnitude F whose line of action is a perpendicular distance l from the point O, the torque is from the point O, the torque is = Fl (10-1). The lever arm of F = Fl (10-1). The lever arm of F11

in Fig. 10-2 is the perpendicular distance OA or lin Fig. 10-2 is the perpendicular distance OA or l11, and the lever , and the lever

arm of Farm of F22 is the perpendicular distance OB or l is the perpendicular distance OB or l22. The line of action . The line of action

of Fof F33 passes through the reference point O, so the lever arm for F passes through the reference point O, so the lever arm for F33

is zero and its torque with respect to point O is zero. is zero and its torque with respect to point O is zero. Force FForce F11 in Fig. 10-2 tends to cause counterclockwise rotation in Fig. 10-2 tends to cause counterclockwise rotation

about O, while Fabout O, while F22 tends to cause clockwise rotation. To distinguish tends to cause clockwise rotation. To distinguish

between these two possibilities, we will choose a positive sense of between these two possibilities, we will choose a positive sense of rotation. With the choice thatrotation. With the choice that

counterclockwise torque are positive and clockwise torques are counterclockwise torque are positive and clockwise torques are negative, the torques of Fnegative, the torques of F11 and F and F22 about O are about O are 11 = +F = +F11ll1 1 , , 22 = - = -

FF22ll22. Figure 10-3 shows a force F applied at a position vector r with . Figure 10-3 shows a force F applied at a position vector r with

respect to the chosen point O. There are several ways to calculate respect to the chosen point O. There are several ways to calculate the torque of this force. One is to find the lever arm l and use the torque of this force. One is to find the lever arm l and use = = Fl. Or we can deter-mine the angle Fl. Or we can deter-mine the angle , between the vectors r and , between the vectors r and F; the lever arm is r sinF; the lever arm is r sin, so , so = rF sin = rF sin. A third method is to . A third method is to represent F in terms of a radial component Frepresent F in terms of a radial component Fradrad along the direction along the direction

of r and a tangential component Fof r and a tangential component Ftantan at right angles, perpendicular at right angles, perpendicular

to r. We call this tangential component because if the body to r. We call this tangential component because if the body rotates, the point where the force acts moves in a circle and this rotates, the point where the force acts moves in a circle and this component is tangent to that circle.) Then Fcomponent is tangent to that circle.) Then Ftantan = Fsin = Fsin, and , and = r = r

(Fsin(Fsin) = F) = Ftantan r. The component F r. The component Fradrad has no torque with respect to has no torque with respect to

O because its lever arm with respect to that point is zero.O because its lever arm with respect to that point is zero.

Fr

10-3 Torque And Angular Acceleration For A Rigid BodyWe are now ready to develop the fundamental relation for the rotational dynamics of a rigid body. We will show that the angular acceleration of a rotating rigid body is directly proportional to the sum of the torque components along the axis of rotation. The proportionality factor is the moment of inertia.

o

r

r1

F1,tan

F1,y

F1,rad

m1

Newton’s second law for the tangential component is F1, tan = m1a1,tan (10-4)We can express the tangential acceleration of the first particle in terms of the angular acceleration , using the Eq.(9-4); a1,tan = r1 . Using this relation and multiplying both sides of Eq.(10-4) by r1, we obtain:F1,tan r1 = m1 r1

2 (10-5)Form Eq.(10-2), F1,tanr1 is just the magnitude of the torque 1 of the net force with respect to the rotation axis. Neither of the components F1,rad or F1y contributes to the torque about the y-axis, since neither tends to change the particle’ rotation about that axis. So 1 = F1,tanr1 is the total torque acting on the particle with respect to the rotation axis. Also, m1r1

2 is I1, the moment of inertia of the particle about the rotation axis. With this in mind, we rewrite Eq. (10-5) as1 = I1 = m1 r1

2.We write an equation like this for every particle in the body and then add all these equations:

233

222

211321 rmrmrm

2iii rm

I

(Rotational analog of Newton’s second law for a rigid body)10-5 Work And Power In Rotational MotionSuppose a tangential force Ftan acts at the of a pivoted disk. The disk rotates through an infinitesimal angle d about a fixed axis during an infinitesimal time interval dt (Fig, 10-19b).

o

RR

d

dsFtan

Ftan

The work done by the force Ftan while a point on the rim moves a distance ds is dW = Ftands. If d is measured in radians, then ds = Rd and dW = Ftan Rd. Now FtanR is the torque due to the force Ftan, so dW = d. (10-22)The total work W done by the torque during an angular displacement from 1 to 2 is

2

1

dW (Work done by a torque).

If the torque is constant while the angle changes by a finite amount = 2- 1, then

12W (Work done by a constant torque).

When a torque does work on a rotating rigid body, the kinetic energy changes by an amount equal to the work done.

dIddt

dId

dt

dIdId )(

21

22 2

1

2

12

1

IIdIWtot

dt

d

dt

dW

dW/dt is the rate of doing work, or power P, and d/dt is angular velocity , so P = .

10-6 Angular MomentumEvery rotational quantity that we have encountered in Chapter 9 and 10 is the analog of some quantity in the translational motion of a particle. The analog of momentum of a particle is angular momentum, a vector quantity denoted as L. Its relation to momentum P is exactly the same as the

relation of torque to force, = r F. For a particle with constant mass m, velocity v, momentum P = mv, and position vector r relative to the origin O of an inertial frame, we define angular momentum L as

vmrprL

(Angular momentum of a particle).

o

P=mv

mvsin

L=rsin

L

z

y

x

z

ox

y

ri

mi

vi=ri

Direction of Li

In Fig. 10-20 a particle moves in the xy-plane; its position vector r and momentum p = mv are shown.

mvlmvrL sin (10-28)

Where l is the perpendicular distance from the line of v to o. when a net force F acts on a particle, its velocity and momentum change, so its angular momentum may also change. We can show that the rate of change of angular momentum is equal to the torque of the net force.

amrvmvdt

vdmrvm

dt

rd

dt

Ld

Fr

dt

Ld(for a particle acted on by net force)

The rate of change of angular momentum of a particle equals the net force acting on it.We can use to find the total angular momentum of a rigid

body rotating about the z-axis with angular speed .

2)( iiiiii rmrrmL (10-30)

The total angular momentum of the slice of the body lying in the xy-plane is the sum Li of the angular momenta Li of the particle. Summing Eq. (10-30), we have

IrmLL iii )( 2

Where I is the moment of inertia of the slice about the z-axis.The angular velocity vector also lies along the rotation axis, as we discussed at the end of Section 9-2.hence for a rigid body rotating around an axis of symmetry, L and are in the same direction. So we have the vector relationship

IL (for a rigid body rotating around a symmetry axis)

dt

Ld

(for any system of particle). (10-32)

10-7 Conservation Of Angular MomentumWhen a system has several parts, the internal forces that the parts exert on each other cause changes in the angular momenta of the parts, but the total angular momentum doesn’t change. Here’s an example. Consider two bodies A and B that interact with each other but not with anything else. Suppose body A exerts a force FAonB on body B; the corresponding torque is AonB . According to Eq. (10-32), this torque is equal to the rate of change of angular momentum of B;

dt

Ld BAonB

At the same time, body B exerts a force FBonA on body A, with a corresponding torque

dt

Ld ABonA

Form Newton’s third law, FBonA = - FAonB. Furthermore, if the forces act along the same line, their lever arms with respect to the chosen axis are equal. Thus the torques of these two forces are equal opposite, and BonA= -AonB. So if we add the two previous equations, we find

0dt

Ld

dt

Ld BA

0

dt

Ld

(zero net external torque) 10-34

It also forms the basis for the principle of conservation of angular momentum. When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved)10-8 Gyroscopes And PrecessionIn all of the situations we’ve looked at so far in this chapter, the axis of rotation either has stayed fixed or has moved and kept the same direction. But a variety of new physical phenomena, some quite unexpected, can occur when the axis of rotation can change direction.

Periodic MotionPeriodic Motion 13-1 Introduction13-1 Introduction The motion repeats itself over and over. We call The motion repeats itself over and over. We call

this periodic motion or oscillation. this periodic motion or oscillation. Understanding periodic motion will be essential Understanding periodic motion will be essential for our later study of waves, sound, alternating for our later study of waves, sound, alternating electric currents, and light.electric currents, and light.

A body that undergoes periodic motion always A body that undergoes periodic motion always has a stable equilibrium position. When it is has a stable equilibrium position. When it is moved away from this position and released, a moved away from this position and released, a force or torque comes into play to pull it back force or torque comes into play to pull it back toward equilibrium.toward equilibrium.

13-2 The Causes of Oscillation13-2 The Causes of OscillationIt’s simplest to define our coordinate system so that the origin O is It’s simplest to define our coordinate system so that the origin O is at the equilibrium position, where the spring is neither stretched at the equilibrium position, where the spring is neither stretched nor compressed. Then x is the x-component of the displacement nor compressed. Then x is the x-component of the displacement of the body from equilibrium and is also the change in length of of the body from equilibrium and is also the change in length of the spring. The x-component of acceleration a is given by a = F/m. the spring. The x-component of acceleration a is given by a = F/m. Whenever the body is displaced from its equilibrium position, Whenever the body is displaced from its equilibrium position, the spring force tends to restore it to the equilibrium position. We the spring force tends to restore it to the equilibrium position. We call a force with this character a restoring force. Oscillation can call a force with this character a restoring force. Oscillation can occur only when there is a restoring force tending to return the occur only when there is a restoring force tending to return the system to equilibrium. system to equilibrium. The amplitude of the motion, denoted by A, is the maximum The amplitude of the motion, denoted by A, is the maximum magnitude of displacement from equilibrium; that is, the maximum magnitude of displacement from equilibrium; that is, the maximum value of x. it is always positive.value of x. it is always positive. The period, T, is the time for one cycle. It is always positive. The period, T, is the time for one cycle. It is always positive. The frequency, f, is the number of cycles in a unit of time. It is The frequency, f, is the number of cycles in a unit of time. It is always positive.always positive. The angular frequency, The angular frequency, ωω, is 2, is 2ππtimes the frequency: times the frequency: ωω=2=2ππf.f.

13-3 Simple Harmonic Motion13-3 Simple Harmonic Motion The very simplest kind of oscillation occurs when the restoring The very simplest kind of oscillation occurs when the restoring force F is directly proportional to the displacement from force F is directly proportional to the displacement from equilibrium x. The x-component of force the spring exerts on the equilibrium x. The x-component of force the spring exerts on the body is the negative of this, so the x-component of force F on the body is the negative of this, so the x-component of force F on the body isbody is

kxF (restoring force exerted by an ideal spring) 13-3

This equation gives the correct magnitude and sign of the force, This equation gives the correct magnitude and sign of the force, whether x is positive, negative, or zero.whether x is positive, negative, or zero. When the restoring force is directly proportional to the displace-When the restoring force is directly proportional to the displace-ment from equilibrium, as given by 13-3, the oscillation is called ment from equilibrium, as given by 13-3, the oscillation is called simple harmonic motions abbreviated SHM. The acceleration simple harmonic motions abbreviated SHM. The acceleration

xm

k

dt

xda

2

2

(simple harmonic motion) 13- 4

The minus sign means the acceleration and displacement always have opposite signs. A body that undergoes simple harmonic motion is called a harmonic oscillator.Equation of Simple Harmonic Motion

x

A

A t+

o x

t

t = 0

x = A cos( t + )

·

参考圆(circle of reference)

To explore the properties of simple harmonic motion, we must express the displacement x of the oscillating body as a function of time, x(t).

The circle in which the ball moves so that its projection matches the motion of the oscillating body is called the circle of reference; we will call the point Q the reference point. We take the circle of reference to lie in the xy-plane, with the origin O at the center of the circle. The x-component of the phasor at time t is just the x-coordinate of the point Q:

cosAx 13-5

This is also the x-coordinate of the shadow P, which is the projection Q onto the x-axis. Hence the acceleration of the shadow P along the x-axis is equal to the x-component of the acceleration vector of the reference point Q.

AaQ2 13-6

coscos 2Aaa Q 13-7

m

k2 or

m

k

(Simple harmonic motion)

m

kf

2

1

2

k

m

fT

221

Displacement, Velocity, And Acceleration In SHMWe still need to find the displacement x as a function of time for a harmonic oscillator. Equation (13-4) for a body in simple harmonic motion along the x-axis is identical to Eq. (13-8) for the x-coordinate of the reference point in uniform circular motion with constant angular speed . mk /

)cos( tAx (displacement in SHM) 13-13

In simple harmonic motion the position is a periodic sinusoidal function of time.The period T is the time for one complete cycle of oscillation. Thus if we start at time t= 0, the time T to complete one cycle is given by

2 Tm

kT or k

mT 2

The constant in Eq. (13-13) is called the phase angle. It tells us at what point in the cycle the motion was at t = 0.

We denote the position at t = 0 by x0. putting t = 0 and x = x0 in equation 13-13, we get

cos0 Ax 13-14

sin( )dx

v A tdt

13-15

)cos(22

2

tAdt

dx

dt

dva 13-16

13-4 Energy In Simple Harmonic MotionWe can learn even more about simple harmonic motion by using energy consideration. The kinetic energy of the body is K = 1/2mv2, and the potential energy of the spring is U = 1/2kx2. There are no nonconservation forces that do work, so the total mechanical energy E = K + U is conserved;

tconskAkxmvE tan2

1

2

1

2

1 222

tconskxmvE tan2

1

2

1 22

Total mechanical energy in SHM

13-5 Applications Of Simple Harmonic MotionVertical SHMSuppose we hang a spring with force constant k and suspend from it a body with mass m.

X=0l

mg

lkF

mglk

kxmgxlkFnet

Angular SHM

I or Idt

d

2

2

I

andI

f

21

13-6 The Simple PendulumA simple pendulum is an idealized model consisting of a point mass suspended by a massless, unstretchable string. When the point mass is pulled to one side of its straight down equilibrium position and released, it oscillates about

The equilibrium position.

x

L

m

mg

mgcosmgsin

T

sinmgF L

xmgmgF x

L

mgF

L

g

m

Lmg

m

k

/L

gf

2

1

2

g

L

fT

212

13-7 The Physical PendulumA physical pendulum is any real pendulum, using a body of finite size, as contrasted to the idealized model of the simple pendulum with all the mass concentrated at a single point.

mg

mgcosθmgsinθ

dθ

O

dsinθ

)sin)(( dmg

)(mgd

2

2

)(dt

dIImgd

I

mgd

dt

d

2

2

13-8 Damped OscillationsRed-world systems always have some dissipative forces, however, and oscillations do die out with time unless we provide some means for replacing the dissipated mechanical energy. The decrease in amplitude caused by dissipative forces is called damping, and the corresponding motion is called damped oscillation.

bvkxF mabvkx

2

2

dt

xdm

dt

dxbkx )cos()2/( tAex tmb

2

2

4m

b

m

k 0

4 2

2

m

b

m

k

kmb 2 tata eCeCx 2121

13-9 Forced Oscillations, Resonance, And Chaos

The Tacoma Narrows Bridge collapsed four months and six days after it was opened for traffic.

19 Mechanical Waves19-1 Introduction Waves can occur whenever a system is disturbed from its equilibrium position and when the disturbance can travel or propagate from one region of the system to another. This chapter and the next two are about mechanical waves, waves that travel within some material called a medium. We’ll begin by deriving the basic equations for describing waves, including the important special case of periodic waves in which the pattern of the wave repeats itself as the wave propagates. Not all waves are mechanical in nature. Another broad class is electromagnetic waves, including light, radio waves, infrared and ultraviolet radiation, x-rays, and gamma rays. No medium is needed for electromagnitude waves; they can travel through empty space.

19-2 Type Of Mechanical WavesA mechanical wave is a disturbance that travels through some material or substance called the medium for the wave.

Because the displacements of the medium are perpendicular or transverse to the direction of travel of the wave along the medium, this is called transverse wave.If the motions of the particles of the medium are back and forth along the same direction that the wave travels, we call this a longitudinal wave.

Transverse wave

0t

4T

t

2T

t

Tt43

Tt Tt

45

4T

t

2T

t

Tt43

Tt

Tt45

Tt23

0t

longitudinal wave.

球面波

波面

波线

柱面波

波线x y

Ray

These examples have three things in common. First, in each case the disturbance travels propagates with a definite speed through the medium. This speed is called the speed of propagation, or simply the wave speed. It is determined in each case by the mechanical properties of the medium. We will use the symbol v for wave speed. Second, the medium itself does not travel through space; its individual particles undergo back-and-forth or up-and-down motions around their equilibrium positions. The overall pattern of the wave disturbance is what travels. Third, to set any of these systems into motion, we have put in energy by doing mechanical work on the system. The wave motion transports this energy from one region of the medium to another. Waves transport energy, but not matter, from one region to another.

19-3 Periodic WavesIn particular, suppose we move the string up and down in simple harmonic motion with amplitude A, frequency f,

angular frequency = 2f, and period T=1/f = 2 / . When a sinusoidal wave passes through a medium, every particle in the medium undergoes simple harmonic motion with the same frequency. For a periodic wave, the shape of the string at any instant is a repeating pattern. The length of one complete wave pattern is the distance from one crest to the next, or from one trough to the next, or from any point to the corresponding point on the next repetition of the wave shape. We call this distance the wavelength of the wave, denoted by . The wave pattern travels with constant speed v and advances a distance of one wavelength in a time interval of one period T. so the wave speed v is given by v = /T, or , because f = 1/T, v = f (periodic wave).

19-4 Mathematical Description Of A Wave For this description we need the concept of a wave function, a function that describes the position of any particle

in the medium at any time. We will concentrate on sinusoidal waves, in which each particle undergoes simple harmonic motion about its equilibrium position.Wave Function For A Sinusoidal WaveSuppose that the displacement of a particle at the left end of string (x = 0), where the wave originates, is given by

ftAtAtxy 2sinsin),0( 19-2

The wave disturbance travels from x = 0 to some point x to the right of the origin in an amount of time given by x/v, where v is the wave speed. So the motion of point x at time t is the same as the motion of point x = 0 at the earlier time t – x/v. Hence we can find the displacement of point x at time t by simply replacing t in Equation (19-2) by (t – x/v).

v

xtfA

v

xtAtxy 2sinsin),( 19-3

x

T

tAtxy 2sin),( 19-4

We get another convenient form of the wave function if we define a quantity k, called the wave number:

2

k (wave number) (19-5)

vk (periodic wave) (19-6)

)sin(),( kxtAtxy

(sinusoidal wave moving in +x-direction)

(19-7)

Wavelength

A

x

y

Period

A

t

y

A

t1 时刻的波形

O

x

u

xx 1

x1

t1+Δt 时刻的波形

Particle Velocity And Acceleration In A sinusoidal WaveFrom the wave function we can get an expression for the transverse velocity of any particle in a transverse wave. We call this vy to distinguish it from the wave propagation speed v. To find the transverse velocity vy at a particular point x, we take the derivative of the wave function y(x,t) with respect to t, keeping x constant. If the wave function is

)sin(),( kxtAtxy then

kxtAt

txytxvy

cos(),(

),( (19-9)

The acceleration of any particle is the second partial derivative of y(x,t) with respect to t:

),()sin(),(

),( 222

2

txykxtAt

txytxay

19-10

We can also compute partial derivatives of y(x,t) with respect to x, holding t constant.

),()sin(),( 22

2

2

txykkxtAkx

txy

(19-11)

Form Eqs.(19-10) and (19-11) and the relation = vk we see that

22

2

22

22

/),(

/),(v

kxtxy

ttxy

2

2

22

2 ),(1),(

t

txy

vx

txy

(wave equation) (19-12)

Equation (19-12), called the wave equation, is one of the most important equations in all of physics.

20 Wave Interference and Normal Modes

20-1 Introduction When a wave strikes the boundaries of its medium, all or part of the wave is reflected. When you yell at a building wall or a cliff face some distance away, the sound wave is reflected from the rigid surface, and an echo comes back. When you flip the end of a rope whose far end is tied to a rigid support, a pulse travels the length of the rope and is reflected overlap in the same region of the medium. This overlapping of waves is called interference. 20-2 Boundary Conditions For A String And The Principle Of SuperpositionAs a simple example of wave reflections and the role of the boundary of a wave medium, let’s look again at transverse waves on a stretched string. What happens when a wave pulse or a sinusoidal wave arrives at the end of the string?

2L

L

23L

If the end is fastened to a rigid support, it is a fixed end that cannot move. The arriving wave exerts a force on the support; the reaction to this force, exerted by the support on the string, “kicks back” on the string and sets up a reflected pulse or wave traveling in the reverse direction. The conditions at the end of the string, such as a rigid support or the complete absence of transverse force, are called boundary conditions.The formation of the reflected pulse is similar to the overlap of two pulses traveling in opposite directions. The principle of superposition Combining the displacements of the separate pulse at each point to obtain the actual displacement is an example of the principle of superposition; when two waves overlap, the actual displacement of any time is obtained by adding the displacement the point would have if only the first wave were present and the displacement it would have if only the second wave were present.

describes the resulting motion in this situation is obtained by adding the two wave functions for the two separate waves. 20-3 Standing Waves On A String We have talked about the reflection of a wave pulse on a string when it arrives at a boundary point. We will again approach the problem by considering the superposition of two waves propagating through the string, one representing the original or incident wave and the other representing the wave reflected at the fixed end. The general term interference is used to describe the result of two or more waves passing through the same region at the same time. Here, instead, the wave pattern in the same position along the string, the amplitude fluctuates. There are particular points called nodes that never move at all. Midway between the nodes is points called antinodes, where the amplitude of motion is greatest. Because the wave pattern doesn’t appear to be moving in either direction along the string, it is called a standing wave.

(To emphasize the difference, a wave that does move along the string is called a traveling wave). The principle of superposition explains how the incident and reflected wave combine to form a standing wave. At a node the displacements of the two waves in red and blue are always equal and opposite and cancel each other out. This cancellation is called destructive interference. Midway between the nodes are the points of the greatest amplitude or antinodes, marked A. At the antinodes the displacement of the two waves in red and blue are always identical, giving a large resultant displacement; this phenomenon is called constructive interference. We can see from the figure that the distance between successive nodes or between successive antinodes is one half-wavelength, or 2. We can derive a wave function for the standing wave of Fig. 20-6 by adding the wave functions y1(x, t) and y2(x, t) for two waves with equal amplitude, period, and wavelength traveling in opposite directions. Here y1(x, t) represents an incident wave traveling

to the left along the +x-axis, arriving at the point x = 0 and being reflected; y2(x, t) represents the reflected wave traveling to the right from x = 0. We noted in Section 20-2 that the wave reflected from a fixed end of a string is inverted, so we give a negative sign to one of the waves;

)sin(,1 kxtAtxy kxtAtxy sin,2

kxtkxtAtxytxytxy sinsin,,, 21

tkxAtxytxytxy cossin2,,, 21

(traveling to the left),

(traveling to the right).

(20-1)

The standing wave amplitude Asw is twice the amplitude A of the original traveling waves:

Asw = 2A

Equation (20-1) has two factors; a function of x and a function of t. The factor 2Asinkx shows that at each instant the shape of the string is a sine curve.

But unlike a wave traveling along a string, the wave shape stays in the same position, oscillating up and down as described by cos t factor. This is in contrast to the phase differences between oscillations of adjacent points that we see with a wave traveling in one direction. We can use Eq. (20-1) to find the positions of the nodes; these are the points for which sin kx = 0, so the displacement is always zero. This occurs when kx =0, , 2, 3, , or, using k = 2,

2

3,

2

2,

2,0

3,

2,,0

kkk

x

In particular, there is a node at x = 0, as there should be, since this point is a fixed end of the string.

A standing wave, unlike a traveling wave, does not transfer energy from one end to the other. The two waves that form it would individually carry equal amounts of power in opposite directions. There is a local flow of energy from each node to the adjacent antinodes and back, but the average rate of energy transfer is zero at every point.